View Full Version : Relative v > c

abcdefg

2009-Oct-29, 12:11 PM

We have two identical rockets.

One rocket is located on one side of the earth and one is located directly on the other side of the earth.

At a predetermined time, each takes off with burns such that they each reach 0.99c, one rocket is therefore going along the position x-axis and the other is heading along the negative x-axis.

Question 1: What is the relative speed of the rockets? 1.98c

Question 2: If each shot a laser at each other as a signal, will they ever see the light? Yes.

Kwalish Kid

2009-Oct-29, 12:50 PM

What is your point?

Perikles

2009-Oct-29, 12:57 PM

Question 1: What is the relative speed of the rockets? 1.98cOf course not, because you can't just add the velocities like that. There is a factor in there involving a v^2/c^2 (I forget what exactly) which always limits the addition of velocities to a maximum of c. So if you are in one rocket, you see the Earth moving way at 0.99c, and the other rocket moving away in the same direction at something like 0.995c. This answers the second question as well.

zerocold

2009-Oct-29, 01:31 PM

He's putting this issue under a third observer, here we go again :D

abcdefg

2009-Oct-29, 01:31 PM

What is your point?

Well, first off, there is a point as they are accelerating at which from rocket to rocket, the instantaneous v of each is .5 c. Taking the position of either rocket, the other rocket disappears since length contraction becomes 0. After that point, length contraction crosses into the complex number system.

Next, if I shot a laser from the front of the rocket to the back and out some hole toward the other rocket and I had clocks measuring the time such that I calculate the speed if the light, a speed of c to the frame would never reach the other rocket moving at relative speed v = 1.98c. It is claimed the speed of light to an inertial frame is always c.

abcdefg

2009-Oct-29, 01:31 PM

He's putting this issue under a third observer, here we go again :D

No I am not. I am taking the position of either of the rockets.

abcdefg

2009-Oct-29, 01:35 PM

Of course not, because you can't just add the velocities like that. There is a factor in there involving a v^2/c^2 (I forget what exactly) which always limits the addition of velocities to a maximum of c. So if you are in one rocket, you see the Earth moving way at 0.99c, and the other rocket moving away in the same direction at something like 0.995c. This answers the second question as well.

This calculation does not control the speed of the rockets. These rockets are at .99c.

For example, here (http://math.ucr.edu/home/baez/physics/Relativity/SR/experiments.html#moving-source_tests) lists experiments with particlaes moving faster than .99c. Assume one is shot one way and one the other.

This is legitimate.

macaw

2009-Oct-29, 01:38 PM

Taking the position of either rocket, the other rocket disappears since length contraction becomes 0.

Bad word salad, as usual.

Q1: Prove that "the other rocket disappears"

Q2: prove that "length contraction becomes zero"

After that point, length contraction crosses into the complex number system.

Wrong again.

Q3: Prove that "length contraction crosses into complex numbers"

Next, if I shot a laser from the front of the rocket to the back and out some hole toward the other rocket and I had clocks measuring the time such that I calculate the speed if the light, a speed of c to the frame would never reach the other rocket moving at relative speed v = 1.98c.

Bzzt, wrong, it reaches the other rocket. I know that this is a shock to you but physics doesn't work the way you confabulate.

It is claimed the speed of light to an inertial frame is always c.

...and it is. The fact that you don't know this, is just your perenial problem.

macaw

2009-Oct-29, 01:40 PM

This calculation does not control the speed of the rockets. These rockets are at .99c.

For example, here (http://math.ucr.edu/home/baez/physics/Relativity/SR/experiments.html#moving-source_tests) lists experiments with particlaes moving faster than .99c. Assume one is shot one way and one the other.

So what? the light speed is still c in the respective experiments. That is, unless you are abcdefg and you don't understand what the text is saying.

zerocold

2009-Oct-29, 01:40 PM

Let me understand where is the "contradiction" ,not saying if is right or wrong, just trying to understand where is your ultimate goal.

The relative speed between the 2 spaceships can't be higher than c

The relative speed between a 3th guy, can be

The light will achieve the spaceships in the frame between them

The light will not reach the spaceships in a observer's frame

That is?, not saying is right or wrong, just waiting for the debate...or macaw :D

Edit, lol he came faster than i expected

macaw

2009-Oct-29, 01:43 PM

No I am not. I am taking the position of either of the rockets.

No, you are not. You are taking the position of a third entity, the first rocket is moving with +.99c, the seconf rocket is moving with -.99c producing what is known as a closing speed of 1.98c wrt the said third entity. But you wouldn't happen to know that and you are inadvertently think otherwise as reflected by your new "discoveries".

Perikles

2009-Oct-29, 01:44 PM

This calculation does not control the speed of the rockets. These rockets are at .99c.

.That is a particularly meaningless statement if you do not define your frame of reference. If you are standing on the surface of the Earth, the statement is true. The calculation does 'control' the speed of the rocket, in the sense that in the frame of reference of one rocket, it 'controls' the speed of the second rocket as measured in that reference frame. If you do not grasp this, there is not much hope.

macaw

2009-Oct-29, 01:46 PM

That is?, not saying is right or wrong, just waiting for the debate...or macaw :D

Edit, lol he came faster than i expected

Faster than the speed of light :lol:

zerocold

2009-Oct-29, 01:47 PM

lol, i hope the debate will be civilized :)

These are cool debates, i must say

abcdefg

2009-Oct-29, 01:58 PM

Quote:

Originally Posted by abcdefg

Taking the position of either rocket, the other rocket disappears since length contraction becomes 0.

Bad word salad, as usual.

Q1: Prove that "the other rocket disappears"

Q2: prove that "length contraction becomes zero"

Answer Q1, Q2:

Let's look at length contraction from either rocket.

Obviously, as each burns to .99c and exactly in the same way in opposite directions, there exists an instant where the relative v is c, ie one is moving in the direction of the negative x-axis and one in the positive direction.

L' = L(sqrt( 1 - (v/c)2) = L(sqrt( 1 - (c/c)2) = L*0 = 0.

That implies the length of the rocket becomes 0. Now, a length of zewro means the rocket disappears.

macaw

2009-Oct-29, 02:03 PM

Answer Q1, Q2:

Let's look at length contraction from either rocket.

Obviously, as each burns to .99c and exactly in the same way in opposite directions, there exists an instant where the relative v is c, ie one is moving in the direction of the negative x-axis and one in the positive direction.

L' = L(sqrt( 1 - (v/c)2) = L(sqrt( 1 - (c/c)2) = L*0 = 0.

Bad answer:

1. Massive objects cannot attain c.

2. Length contraction is a relativistic effect, it does not happen in the proper frame of the rocket, so contrary to your new "discovery", the length of the rocket is still...L :-)

That implies the length of the rocket becomes 0.

Nope, bad juju. See above.

Now, a length of zewro means the rocket disappears.

Still bad juju. You failed Q1,2. Let's see what you do with Q3.

abcdefg

2009-Oct-29, 02:06 PM

Wrong again.

Q3: Prove that "length contraction crosses into complex numbers"

[Edit] Answer to Q3

Oh, once each rocket reaches .99c and rocket A decides to perform a length contraction calculation, it is

L' = L sqrt( 1 - (v/c)2 ) = L sqrt( 1 - (1.98c/c)2 ) )

= L sqrt( -2.9204 )

abcdefg

2009-Oct-29, 02:07 PM

So what? the light speed is still c in the respective experiments. That is, unless you are abcdefg and you don't understand what the text is saying.

I am betting I know exactly what I am saying.

abcdefg

2009-Oct-29, 02:09 PM

No, you are not. You are taking the position of a third entity, the first rocket is moving with +.99c, the seconf rocket is moving with -.99c producing what is known as a closing speed of 1.98c wrt the said third entity. But you wouldn't happen to know that and you are inadvertently think otherwise as reflected by your new "discoveries".

Sorry, I am not trying do "discoveries" in this ATM.

I am not taking the position of a 3rd entity. I am taking the position of a rocket. The relative spedd is 1.98c from one rocket to another.

macaw

2009-Oct-29, 02:12 PM

Oh, once each rocket reaches .99c and rocket A decides to perform a length contraction calculation, it is

L' = L sqrt( 1 - (v/c)2 ) = L sqrt( 1 - (1.98c/c)2 ) )

= L sqrt( -2.9204 )

Bad juju again. The v that goes in the length contraction formula is the relative speed. You are plugging in the closing[b] speed and you wonder why you are getting rubbish . Take a deep breath, read the difference between the two speeds and try again.

Q4: What is the relative speed between the rockets?

Q5: What is the [b]correct length contraction?

Please do not rush trying to answer with another set of mistakes. Your elementary mistakes are not "discoveries".

macaw

2009-Oct-29, 02:13 PM

Sorry, I am not trying do "discoveries" in this ATM.

I am not taking the position of a 3rd entity.

Then you don't really know what you are doing <shrug>

I am taking the position of a rocket. The relative spedd is 1.98c from one rocket to another.

Nope, it isn't. Try again. Correctly this time. pay attention, Perikles already clued you in with the correct formula.

abcdefg

2009-Oct-29, 02:13 PM

Let me understand where is the "contradiction" ,not saying if is right or wrong, just trying to understand where is your ultimate goal.

The relative speed between the 2 spaceships can't be higher than c

The relative speed between a 3th guy, can be

The light will achieve the spaceships in the frame between them

The light will not reach the spaceships in a observer's frame

That is?, not saying is right or wrong, just waiting for the debate...or macaw :D

Edit, lol he came faster than i expected

Edit, lol he came faster than i expected

LOL he was quick was he not?

My ultimate goal is to show relativity cannot handle this motion and to show it must be stated that show SR applies only to relative v < c.

Next, I want to show the speed of light is an absolute constant, otherwise causality is broken.

Light will reach the other spaceship.

Perikles

2009-Oct-29, 02:14 PM

Obviously, as each burns to .99c and exactly in the same way in opposite directions, there exists an instant where the relative v is c, ie one is moving in the direction of the negative x-axis and one in the positive direction.This does not happen - read my post above.

I am betting I know exactly what I am saying.Possibly, but it is complete nonsense. I can see now why others are not posting in this thread.

abcdefg

2009-Oct-29, 02:15 PM

Bad answer:

1. Massive objects cannot attain c.

2. Length contraction is a relativistic effect, it does not happen in the proper frame of the rocket, so contrary to your new "discovery", the length of the rocket is still...L :-)

Good answer I had.

Neither rocket attained c. I thought I stated that clearly.

zerocold

2009-Oct-29, 02:17 PM

You are wrong on this abc the speed between the 2 ships should be 0.99999XXXc, in their frame

macaw

2009-Oct-29, 02:19 PM

Of course not, because you can't just add the velocities like that. There is a factor in there involving a v^2/c^2 (I forget what exactly) which always limits the addition of velocities to a maximum of c. So if you are in one rocket, you see the Earth moving way at 0.99c, and the other rocket moving away in the same direction at something like 0.995c. This answers the second question as well.

Yep. Hint for abcdefg:

Rocket 1 moves with v1=+v wrt the observer

Rocket 2, moves with v2=-v wrt the observer. This is equivalent with saying that the observer is moving with +v wrt Rocket 2.

So, we have the situaltion:

Observer moving with +v wrt Rocket 2

Rocket 1 moving with +v wrt observer.

Q6 for abcdefg: What is the speed of Rocket 1 wrt Rocket 2. Please do not answer 2v. :lol:

macaw

2009-Oct-29, 02:20 PM

Good answer I had.

Neither rocket attained c. I thought I stated that clearly.

Then they definitely cannot attain 1.98 c, right? :lol:

macaw

2009-Oct-29, 02:21 PM

[My ultimate goal is to show relativity cannot handle this motion and to show it must be stated that show SR applies only to relative v < c.

You failed. Now you have to answer Q1-Q6.

macaw

2009-Oct-29, 02:23 PM

You are wrong on this abc the speed between the 2 ships should be 0.99999XXXc, in their frame

True but he doesn't want to learn. He'd much rather post mistakes and try to pass them as "discoveries".

abcdefg

2009-Oct-29, 02:23 PM

Bad juju again. The v that goes in the length contraction formula is the relative speed. You are plugging in the [b]closing[b] speed and you wonder why you are getting rubbish . Take a deep breath, read the difference between the two speeds and try again.

Q4: What is the relative speed between the rockets?

Q5: What is the correct length contraction?

Please do not rush trying to answer with another set of mistakes. Your elementary mistakes are not "discoveries".

Answer Q4. I answered 1.98c.

Here, if you have a car driving one way on a highway and one the other each at .4c, the relative speed between them is .8c. You add them in this case.

Now, if I had a car driving at .4c and tried to accelerate to c I could not do this. That would product v < c because nother travels faster than the spedd of light.

You see, I am looking past this though. Neither are breaking this speed limit rule, so the answer is 1.98c.

Answer to Q5: What is the correct length contraction?

I already this answered and I am not going to change the answer.

L' = L sqrt( -2.9204 ).

abcdefg

2009-Oct-29, 02:24 PM

Then they definitely cannot attain 1.98 c, right? :lol:

No, neither can reach c, but the two combined to reach v < 2c.

dgavin

2009-Oct-29, 02:25 PM

We have two identical rockets.

One rocket is located on one side of the earth and one is located directly on the other side of the earth.

At a predetermined time, each takes off with burns such that they each reach 0.99c, one rocket is therefore going along the position x-axis and the other is heading along the negative x-axis.

Question 1: What is the relative speed of the rockets? 1.98c

Question 2: If each shot a laser at each other as a signal, will they ever see the light? Yes.

The Relativistic velosity of the rockets is .99c, the direction of thier travel is irrelevent except for red/blue shift determinations.

The apperant velosity of the rockets is 1.98c. While it looks like they are veling at FTL velosities compared to each other, they are not. Them important velosity is how fast the ships are moving reletive to a rest frame.

The yes, they would see the light. The light would eventualy catch up to Ship 2, fired from ship 1.

Here is how to think of it. The light leaves ship 1 that is in motion. Ship one sees light moving at c according to his time dilated clock in m/s. Light as it moves through space, moves at c according to a non time dialted clock. Ship 2 also sees the light moving at c according to his time dilated clock in m/s.

Because Ship 1 and Ship 2 have the same relativistic frame, thier clocks are the same.

Because they measure light in a time dialated manner, the actual distance light covers in the time dilated state, is greater then in a non time dilated state. This is because at realtivist velosities, space is also length contacted.

This is the basis of the Lorentz formula's, and of Relativity itself.

abcdefg

2009-Oct-29, 02:26 PM

This does not happen - read my post above.

Possibly, but it is complete nonsense. I can see now why others are not posting in this thread.

First, why is it nonsense to produce this thought experiment?

This is supposed to be science with the goal to explore.

The thought experiment is legal.

Second, the two can combine to speeds v < 2c but neither can reach c on its own.

macaw

2009-Oct-29, 02:27 PM

Answer Q4. I answered 1.98c.

Bad juju.Persisting in a bad error is not a mark of intelligence. You were given all the tools to produce the correct answer, try again.

Answer to Q5: What is the correct length contraction?

I already this answered and I am not going to change the answer.

L' = L sqrt( -2.9204 ).

Worse juju since it is based on the bad juju from above :lol:

Try again.

macaw

2009-Oct-29, 02:28 PM

The Relativistic velosity of the rockets is .99c, the direction of thier travel is irrelevent except for red/blue shift determinations.

The apperant velosity of the rockets is 1.98c. While it looks like they are veling at FTL velosities compared to each other, they are not. Them important velosity is how fast the ships are moving reletive to a rest frame.

The yes, they would see the light. The light would eventualy catch up to Ship 2, fired from ship 1.

Here is how to think of it. The light leaves ship 1 that is in motion. Ship one sees light moving at c according to his time dilated clock in m/s. Light as it moves through space, moves at c according to a non time dialted clock. Ship 2 also sees the light moving at c according to his time dilated clock in m/s.

Because Ship 1 and Ship 2 have the same relativistic frame, thier clocks are the same.

Because they measure light in a time dialated manner, the actual distance light covers in the time dilated state, is greater then in a non time dilated state. This is because at realtivist velosities, space is also length contacted.

This is the basis of the Lorentz formula's, and of Relativity itself.

Nice word salad. You are confusing the issues (and abcdefg) even more.

abcdefg

2009-Oct-29, 02:30 PM

You are wrong on this abc the speed between the 2 ships should be 0.99999XXXc, in their frame

How does this limit get set?

Once cannot exceed c but their opposite travels can as a total.

These experiments (math.ucr.edu/home/baez/physics/Relativity/SR/experiments.html#Top) accelerate particles faster than .99c. Now, shoot one particle in one direction and one in the other. What is their relative speed? It is greater than c.

macaw

2009-Oct-29, 02:30 PM

No, neither can reach c, but the two combined to reach v < 2c.

Nope, they don't. You really, really need to learn what relative speed is. You were given all the tools, now it is time you take a long pause from posting and you go learn the subject. Hint: the resultant relative speed is <c. Always.

macaw

2009-Oct-29, 02:31 PM

How does this limit get set?

Once cannot exceed c but their opposite travels can as a total.

These experiments (math.ucr.edu/home/baez/physics/Relativity/SR/experiments.html#Top) accelerate particles faster than .99c. Now, shoot one particle in one direction and one in the other. What is their relative speed? It is greater than c.

Nope, it is <c. Magic, eh?

abcdefg

2009-Oct-29, 02:33 PM

Yep. Hint for abcdefg:

Rocket 1 moves with v1=+v wrt the observer

Rocket 2, moves with v2=-v wrt the observer. This is equivalent with saying that the observer is moving with +v wrt Rocket 2.

So, we have the situaltion:

Observer moving with +v wrt Rocket 2

Rocket 1 moving with +v wrt observer.

Q6 for abcdefg: What is the speed of Rocket 1 wrt Rocket 2. Please do not answer 2v. :lol:

Answer to Q6

First, you are introducing a 3rd observer to resolve this. Yes, you can do this and it will work for this 3rd observer. But, from rocket to rocket, the relative speed is 1.98c.

macaw

2009-Oct-29, 02:34 PM

Answer to Q6

First, you are introducing a 3rd observer to resolve this.

You did. But you don't realize it yet. It will come to you, give it another 1500 posts:lol:

But, from rocket to rocket, the relative speed is 1.98c.

Nope, it isn't. The correct answer is 1.98/(1+.99^2) c < c

Q7: prove that it isn't :lol:

Perikles

2009-Oct-29, 02:34 PM

Answer to Q6But, from rocket to rocket, the relative speed is 1.98c.OK - you win. Your Nobel Prize is in the post.

abcdefg

2009-Oct-29, 02:39 PM

Bad juju.Persisting in a bad error is not a mark of intelligence. You were given all the tools to produce the correct answer, try again.

Worse juju since it is based on the bad juju from above :lol:

Try again.

Hey, what is juju BTW?

Persisting in a bad error is not a mark of intelligence. You were given all the tools to produce the correct answer, try again.

Let's see, you have -v and +v and add them together and get 0? Nope, that is not correct.

One is traveling in one direction at .99c and the other is going the other direction at .99c and they reach light speed entanglement and their engines stall such that the total relative speed cannot exceed c. Now, if the other rocket did not exist, then the other rocket could attain .99c. Nope.

The answer is 1.98c.

Nevertheless, it remains a seminal tenet of anti-relativityism (for lack of a better term) that the trivial Sagnac effect somehow "disproves relativity". Those who espouse this view sometimes claim that the expressions "c+v" and "c-v" appearing in the derivation of the phase shift are prima facie proof that the speed of light is not c with respect to some inertial coordinate system. When it is pointed out that those quantities do not refer to the speed of light, but rather to the sum and difference of the speed of light and the speed of some other object, both with respect to a single inertial coordinate system, which can be as great as 2c according to special relativity, the anti-relativityists are undaunted, and merely proceed to construct progressively more convoluted and specious "objections"

http://www.mathpages.com/rr/s2-07/2-07.htm

Here is some help with understanding this.

abcdefg

2009-Oct-29, 02:41 PM

OK - you win. Your Nobel Prize is in the post.

No, this is not what I am doing.

abcdefg

2009-Oct-29, 02:42 PM

Nope, they don't. You really, really need to learn what relative speed is. You were given all the tools, now it is time you take a long pause from posting and you go learn the subject. Hint: the resultant relative speed is <c. Always.

Wrong. That is what you were taught.

In reality, LT does not work for v >=c, that is all.

This does not force all relative motion v to be less than c as we are seeing.

abcdefg

2009-Oct-29, 02:43 PM

Nope, it is <c. Magic, eh?

Correct, each is travelling v < c.

But, what is their relative speed? It is greater than c.

MDT-1

2009-Oct-29, 02:46 PM

The distance between the rockets will increase at a rate of 1.98c. A laser light will eventually impact the other rocket, regardless of the distance between them at the time the laser was turned on. The laser light will be closing the distance on the rocket it is heading towards. Once the laser is fired from the rocket, that rocket's position is meaningless. The light is then chasing the .99c rocket at a speed of c. It will eventually catch it.

dgavin

2009-Oct-29, 02:46 PM

Nice word salad. You are confusing the issues (and abcdefg) even more.

Explaining the priciple behind the math in plain english is NOT word salad.

abcdefg

2009-Oct-29, 02:50 PM

You did. But you don't realize it yet. It will come to you, give it another 1500 posts:lol:

Funny. The two rockets could have simply been in a common frame in space and each burned in opposite directions at .99c. Where exasctly is this 3rd observer? There is not one.

Nope, it isn't. The correct answer is 1.98/(1+.99^2) c < c

Q7: prove that it isn't :lol:

Answer to Q7

I do not know where you got the above and so I would have a hard time proving it to be false.

Further, I do not know how you would control the speed of the rockets with each going .99c in opposite directions by imposing an equation on them.

macaw

2009-Oct-29, 02:50 PM

Explaining the priciple behind the math in plain english is NOT word salad.

That is, if you explain it correctly rather than comounding the confusion.

abcdefg

2009-Oct-29, 02:53 PM

The distance between the rockets will increase at a rate of 1.98c. A laser light will eventually impact the other rocket, regardless of the distance between them at the time the laser was turned on. The laser light will be closing the distance on the rocket it is heading towards. Once the laser is fired from the rocket, that rocket's position is meaningless. The light is then chasing the .99c rocket at a speed of c. It will eventually catch it.

Wow, someone understands the speed of light is an absolute constant.

Thus, it does not matter what a rocket is doing. Light will proceed at c and catch the other rocket which is traveling slower at .99c, that is correct.

Let's cement this logic.

Einstein: "I certainly knew that the principle of the constancy of the velocity of light is something quite independent of the relativity postulate and I considered what would be more probable, the principle of the constancy of c, as was demanded by Maxwell’s equations, or the constancy of c, exclusively for an observer sitting at the light source. I decided in favor of the first, since I was convinced that each light [ray] should be defined by frequency and intensity alone, quite independently of whether it comes from a moving or a resting light source."

http://philsci-archive.pitt.edu/archive/00001743/01/Norton.doc

Einstein wrote (Einstein Archive, 20-040; translation based on Stachel, 2002, p. 189):Einstein: Your attempt to replace special relativity with the assumption that the velocity of light is constant relative to the source of light was first advocated by Ritz. This assumption is compatible with Michelson’s experiment and with aberration.

http://philsci-archive.pitt.edu/archive/00001743/01/Norton.doc

macaw

2009-Oct-29, 02:54 PM

Answer to Q7

I do not know where you got the above and so I would have a hard time proving it to be false.

So, we are done, you realize that all you have been posting is mistakes. You can learn how to correct them by reading (and learning) a chapter of relativistic kinematics. Start with the correct derivation of relative speed

Further, I do not know how you would control the speed of the rockets with each going .99c in opposite directions by imposing an equation on them.

You don't. The laws of physics control the speed. You only need to make the effort to learn them. Much better use of your time than posting elementary mistakes that don't pass the mustard.

abcdefg

2009-Oct-29, 02:54 PM

I will be back later.

macaw

2009-Oct-29, 02:56 PM

Einstein wrote (Einstein Archive, 20-040; translation based on Stachel, 2002, p. 189):Einstein: Your attempt to replace special relativity with the assumption that the velocity of light is constant relative to the source of light was first advocated by Ritz. This assumption is compatible with Michelson’s experiment and with aberration.

http://philsci-archive.pitt.edu/archive/00001743/01/Norton.doc

...but incompatible with Sagnac, Fizeau and Ives-Stilwell and a bunch of other experiments. Nice try but you fail again :lol:

macaw

2009-Oct-29, 02:58 PM

I will be back later.

try much later, after you studied relative speed vs. closing speed. Once you clear your confusions, you will come up with another angle, I am sure. This one got dulled quickly.

iquestor

2009-Oct-29, 02:59 PM

MDT-1 said:

Once the laser is fired from the rocket, that rocket's position is meaningless. The light is then chasing the .99c rocket at a speed of c. It will eventually catch it.

To me, this is the important point. Once the laser is fired from rocket 1, it will approach rocket 2 at the speed of light. The relative velocity of the two rockets are entirely moot, since the laserlight is not part of either system. This is a basic tenant of TOR. Therefore, Since the laster travels at c, and c > .99c, it will overtake the rocket.

macaw

2009-Oct-29, 03:03 PM

MDT-1 said:

To me, this is the important point. Once the laser is fired from rocket 1, it will approach rocket 2 at the speed of light. The relative velocity of the two rockets are entirely moot, since the laserlight is not part of either system. This is a basic tenant of TOR. Therefore, Since the laster travels at c, and c > .99c, it will overtake the rocket.

Yes, of course but it will take abcdefg another 1500 posts to get this concept since he already "knows" that relativity is wrong :-)

NorthernBoy

2009-Oct-29, 03:03 PM

Well, first off, there is a point as they are accelerating at which from rocket to rocket, the instantaneous v of each is .5 c. Taking the position of either rocket, the other rocket disappears since length contraction becomes 0. After that point, length contraction crosses into the complex number system.

Oh god.

OK, here's how this one works. I see one rocket travelling at 0.99c to the left, and one travelling at 0.99c to the right. In my frame, the distance between them does indeed increase at pretty much 2c. No problem there, nothing in my frame is moving at greater than c, no laws being broken.

If we instead ask what speed the observers in one of the rockets will see the other rocket moving at, then we work this out to be just a little under c. So, in each rocket's frame, nothing is moving at greater than c either.

No paradox, no contradiction, no problem.

NorthernBoy

2009-Oct-29, 03:07 PM

Sorry, I am not trying do "discoveries" in this ATM.

I am not taking the position of a 3rd entity. I am taking the position of a rocket. The relative spedd is 1.98c from one rocket to another.

No, it is not, no matter how many times you assert otherwise.

If the speed of one rocket viewed from earth is v, and the speed of the other is w (in opposite directions), then the speed of one rocket as seen from the other is

(v+w)/(1+vw/c^2)

You are incorrectly adding their speeds, so, of course, get meaningless answers.

NorthernBoy

2009-Oct-29, 03:11 PM

First, why is it nonsense to produce this thought experiment?

Because, when doing so, you are applying an incorrect expression for addition of velocities. Several people have pointed it out to you.

Instead of ignoring the point again and again, why not actually discuss the central disagreement for once.

First, do you understand why people are criticising this step of your thought experiment?

Secondly, can you please explain why you do not think that you should use the equation I give above for the relative velocities, but instead should use simple addition?

This is as bad as someone claiming that adding a litre of water to a litre of ethanol gives you two litres of mixture.

Swift

2009-Oct-29, 03:13 PM

<snip>

i hope the debate will be civilized

What zerocold said. Keep this civil. Lose the personal attacks and the :lol: (which I take as laughing at the person, not sharing a joke). If anyone can't do that, then they should stay out of this thread, or additional actions will be taken. If you think someone is not following these rules, report the post.

NorthernBoy

2009-Oct-29, 03:18 PM

Because Ship 1 and Ship 2 have the same relativistic frame, thier clocks are the same.

So two ships moving apart at close to c both have the same relativistic frame?

This is a very ATM claim, and you should start your own thread to discuss it.

The rest was soo poorly written as to be close to meaningless, I'm afraid.

abcdefg

2009-Oct-29, 04:30 PM

So, we are done, you realize that all you have been posting is mistakes. You can learn how to correct them by reading (and learning) a chapter of relativistic kinematics. Start with the correct derivation of relative speed

You don't. The laws of physics control the speed. You only need to make the effort to learn them. Much better use of your time than posting elementary mistakes that don't pass the mustard.

In the design of LT and the kinematics of SR

Einstein:

Let there be given a stationary rigid rod; and let its length be l as measured by a measuring-rod which is also stationary. We now imagine the axis of the rod lying along the axis of x of the stationary system of co-ordinates, and that a uniform motion of parallel translation with velocity v along the axis of x in the direction of increasing x is then imparted to the rod. We now inquire as to the length of the moving rod, and imagine its length to be ascertained by the following two operations:--

Furthermore,

Einstein stated tB - tA = rAB/(c-v)

As such, it is hypothesized that v < c or you come up with a negative time.

Thus, all kinematical calculations are under the postulate that v < c.

So, your equation is simply built under that condition and that is proof by vacuous implication. In oither words, you have built into your equations that v < c and thus, it is forced to be true by assumption.

These v's are calculated according to the constant acceleration of a and BT. They sum to 1.98c.

In addition in your kinematics, light leaves rocket one at c regardless of the motion of the rocket. It arrives at the other rocket at a distance t(c + .99c) in an effort to determine dx/dtau. Well, you lost the distance the distance the light source rocket traveled during that time with this analysis. So, you only have half the total distance.

abcdefg

2009-Oct-29, 04:38 PM

Quote:

Originally Posted by abcdefg

Einstein wrote (Einstein Archive, 20-040; translation based on Stachel, 2002, p. 189):Einstein: Your attempt to replace special relativity with the assumption that the velocity of light is constant relative to the source of light was first advocated by Ritz. This assumption is compatible with Michelson’s experiment and with aberration.

http://philsci-archive.pitt.edu/arch.../01/Norton.doc

...but incompatible with Sagnac, Fizeau and Ives-Stilwell and a bunch of other experiments. Nice try but you fail again :lol:

I am not a Ritz person.

I am pointing out Einstein's position. So your comment does not apply to me.

In particular, I posted my agreemet with this statement.

Einstein: "I certainly knew that the principle of the constancy of the velocity of light is something quite independent of the relativity postulate and I considered what would be more probable, the principle of the constancy of c, as was demanded by Maxwell’s equations, or the constancy of c, exclusively for an observer sitting at the light source. I decided in favor of the first, since I was convinced that each light [ray] should be defined by frequency and intensity alone, quite independently of whether it comes from a moving or a resting light source."

http://philsci-archive.pitt.edu/archive/00001743/01/Norton.doc

Key Concept: Special Relativity in One Sentence.

All speeds are relative, except for the speed of light, which is absolute.

http://www.astronomy.ohio-state.edu/~ryden/ast162_6/notes23.html

In particular, I do not run with the argument that the speed of light is c in the frame, that is light emission theory.

abcdefg

2009-Oct-29, 04:40 PM

Quote:

Originally Posted by iquestor

MDT-1 said:

To me, this is the important point. Once the laser is fired from rocket 1, it will approach rocket 2 at the speed of light. The relative velocity of the two rockets are entirely moot, since the laserlight is not part of either system. This is a basic tenant of TOR. Therefore, Since the laster travels at c, and c > .99c, it will overtake the rocket.

Yes, of course but it will take abcdefg another 1500 posts to get this concept since he already "knows" that relativity is wrong :-)

Perhaps you should have read my opening statement.

We have two identical rockets.

One rocket is located on one side of the earth and one is located directly on the other side of the earth.

At a predetermined time, each takes off with burns such that they each reach 0.99c, one rocket is therefore going along the position x-axis and the other is heading along the negative x-axis.

Question 1: What is the relative speed of the rockets? 1.98c

Question 2: If each shot a laser at each other as a signal, will they ever see the light? Yes.

abcdefg

2009-Oct-29, 04:41 PM

Oh god.

OK, here's how this one works. I see one rocket travelling at 0.99c to the left, and one travelling at 0.99c to the right. In my frame, the distance between them does indeed increase at pretty much 2c. No problem there, nothing in my frame is moving at greater than c, no laws being broken.

If we instead ask what speed the observers in one of the rockets will see the other rocket moving at, then we work this out to be just a little under c. So, in each rocket's frame, nothing is moving at greater than c either.

No paradox, no contradiction, no problem.

Macaw already tried this.

This is a 3rd observer.

We are taking the position of either rocket and so this does not apply.

abcdefg

2009-Oct-29, 04:44 PM

No, it is not, no matter how many times you assert otherwise.

If the speed of one rocket viewed from earth is v, and the speed of the other is w (in opposite directions), then the speed of one rocket as seen from the other is

(v+w)/(1-vw/c^2)

You are incorrectly adding their speeds, so, of course, get meaningless answers.

Again, you are introducing a 3rd observer in an attempt to understand the problem.

There are only two observers and we can lose the earth and just have them start in a frame by themselves.

There is no 3rd observer.

NorthernBoy

2009-Oct-29, 04:47 PM

In oither words, you have built into your equations that v < c and thus, it is forced to be true by assumption.

It is also observed to be true, which rather contradicts what you wrote up there.

Take a particle that decays int two daughter particles, which will move at 0.99999c relative to it, and accelerate it to 0.99c, and we get the addition of velocities just as others have been telling you.

I know that you don't like this fact, but we've been doing your rocket experiment millions of times per day in labs for years now, and always the numbers come out in line with standard SR and not in line with your bizarre distortion of it.

NorthernBoy

2009-Oct-29, 04:50 PM

Again, you are introducing a 3rd observer in an attempt to understand the problem.

There are only two observers and we can lose the earth and just have them start in a frame by themselves.

There is no 3rd observer.

No, I don't need a third observer. Set these rockets off as you say, and we know that each will view the speed of the other as being less than c. You are apparently claiming that each will see the other doing a speed greater than c, but you have not justified this.

As I say above, we've done the experiment, and observed it.

abcdefg

2009-Oct-29, 04:51 PM

Because, when doing so, you are applying an incorrect expression for addition of velocities. Several people have pointed it out to you.

Instead of ignoring the point again and again, why not actually discuss the central disagreement for once.

First, do you understand why people are criticising this step of your thought experiment?

Secondly, can you please explain why you do not think that you should use the equation I give above for the relative velocities, but instead should use simple addition?

This is as bad as someone claiming that adding a litre of water to a litre of ethanol gives you two litres of mixture.

This is not a valid comparison above.

Further, the addition of velocities does not apply because they are moving in opposite directions.

I have already been through this with macaw.

If you are going .4c and you shoot a particle at .8c in the same direction, the sum velocity will be < c.

These two are moving in opposite directions at .99c. Now folks will assert this is a kind of sum velocity argument. But it is not simply because there is no addition to velocity in the same direction. Further, the addition of velocity lesson is simply to proven one object cannot exceed c which is fine and not done here.

abcdefg

2009-Oct-29, 04:54 PM

No, I don't need a third observer. Set these rockets off as you say, and we know that each will view the speed of the other as being less than c. You are apparently claiming that each will see the other doing a speed greater than c, but you have not justified this.

As I say above, we've done the experiment, and observed it.

Where exactly do you have a reference frame moving to the left at .99c and it compares another moving to the right at .99c?

You do not.

I will think about doing the kinematics to this mathematically .

I will be back.

NorthernBoy

2009-Oct-29, 04:56 PM

Question 1: What is the relative speed of the rockets? 1.98c

You see, this is your normal sneaky trick, that you arogantly assume no-one will pick you up on (despite everyone catching you in it every time).

You, as always, leave out the question of which observer you are claiming the measurement for.

The only observer who would measure a speed of separation of 1.98c is an observer on earth. As you yourself have repeatedly told us, your situation has no such observer, so no, the relative speed is not 1.98c according to either of your two observers.

Why do you insist on always making this same dishonest switch? Why do you come up with a number that applies to one observer, and then either try to remove that observer, or to pretend that another one made it?

If you are so convinced that you are right why do you have to use such a dishonest argument to try to prove your point? Do you see anyone else trying to do this? Do you see anyone else needing to underspecify the situation to leave wriggle room to avoid being pinned down?

This is a direct question, which you are required to answer. Why do you refuse to state to which observer an observation is attributed?

NorthernBoy

2009-Oct-29, 04:57 PM

Where exactly do you have a reference frame moving to the left at .99c and it compares another moving to the right at .99c?

CERN. Except that we get much, much closer to c than one part in one hundred there.

Did you seriously not know that this is what we did there?:confused:

Perikles

2009-Oct-29, 05:02 PM

(v+w)/(1-vw/c^2).Do you by chance have a sign wrong here? If v=w=0.99c you get a very large result. :confused:

NorthernBoy

2009-Oct-29, 05:06 PM

This is not a valid comparison above.

Further, the addition of velocities does not apply because they are moving in opposite directions.

Now that is just silly. The formula works for particles moving in opposite directions, or in the same direction.

NorthernBoy

2009-Oct-29, 05:08 PM

Do you by chance have a sign wrong here? If v=w=0.99c you get a very large result. :confused:

Yes, well caught, it needs to be a plus sign on the bottom. Original post edited to correct this.

abcdefg

2009-Oct-29, 05:56 PM

It is also observed to be true, which rather contradicts what you wrote up there.

Take a particle that decays int two daughter particles, which will move at 0.99999c relative to it, and accelerate it to 0.99c, and we get the addition of velocities just as others have been telling you.

I know that you don't like this fact, but we've been doing your rocket experiment millions of times per day in labs for years now, and always the numbers come out in line with standard SR and not in line with your bizarre distortion of it.

OK, let's take that particle at 0.99999c amd accelerate it to 0.99999c just as everyone has been telling me.

If this acceleration is in the direction of the positive x-axis, then you make sense if the first particle is moving in that direction. But, if the particle is shot in the direction of the negative x-axis, then this shot particle will return to the earth frame.

So, velocity additions are directionally based.

abcdefg

2009-Oct-29, 05:58 PM

Now that is just silly. The formula works for particles moving in opposite directions, or in the same direction.

This means when a particle is shot down the positive x-axis at .99c, and then accelerated back the other direction, it never goes back. That is odd.

abcdefg

2009-Oct-29, 05:59 PM

Rocket A is the one moving left and rocket B is the one moving right.

Rocket A is a rocket in a rocket in a rocket.

So, Rocket A opens up the back and the rocket in a rocket, C, takes off with the same a and burn time. Rocket C is now back in the same frame as when A and B started.

So, that is currently a total speed of v between A and B, but C is still not in the frame of B. So, that is still not enough.

Now, rocket C opens up the back and rocket D takes off at a with the same burn time. It enters the frame of B after the burn. D is now v compared to C.

The piecewise addition is 2v as the total speed between the two or D could have never entered the frame of B.

slang

2009-Oct-29, 06:06 PM

abcdefg, without looking them up or copying them from somewhere, give definitions for the following terms, and for each, describe what that definition means with respect to the situation described in the OP.

Be as precise as possible. Don't look it up, just repeating some quote won't give anyone insight into what the definition means to you. Don't give a lame answer like "I use them just as everyone else."

frame of reference

inertial

coordinate system

When you're done, you can go read this thread (http://www.bautforum.com/forum-introductions-feedback/93226-how-make-discussions-atm-ideas-less-frustrating-suggestion.html) and understand why I'm asking this.

NorthernBoy

2009-Oct-29, 06:08 PM

OK, let's take that particle at 0.99999c amd accelerate it to 0.99999c just as everyone has been telling me.

Still you do not say according to whom these numbers are.

You need to say relative to what, or to whom, these things are meant to be.

That being the case, the above makes no sense at all. If the particle is already going at 0.99999c, then you do not need to accelerate it again to reach this speed, it is already doing it.

I suspect that you were trying to describe some other situation, but in your normal way, you have not explained the situation fully, and so have left ambiguity or outright error in it.

Can you possibly try again?

And can you please do so without changing the situation which you first proposed? Again, this is a very bad habit of yours, to start off in one situation, then to switch as soon as someone engages you in discussion about it.

NorthernBoy

2009-Oct-29, 06:12 PM

This means when a particle is shot down the positive x-axis at .99c, and then accelerated back the other direction, it never goes back. That is odd.

Nope, I can't make any sense of this sentence at all.

abcdefg, your written English is getting worse and worse. If you are claiming that you cannot move a particle in the direction of positive x at this speed, and then slow it and move it iin the direction of egative x, then you are mistaken.

That is such an absurd claim, though, that I cannot believe that this is what you really meant.

I can promise you that if you are willing to stick to one thought experiment, and to stop your ridiculous habit of refusing to say which obervations are made by whom, that I will be able to help you with this.

I can also promise you that if you insist on posting things in such mangled English, that this will never go anywhere.

Please, read your posts back, and understand that you need to make them readable to other people, not just to you. Your writing is really really weak, and it is painful to have to extract your meaning before I can comment on it.

abcdefg

2009-Oct-29, 06:16 PM

abcdefg, without looking them up or copying them from somewhere, give definitions for the following terms, and for each, describe what that definition means with respect to the situation described in the OP.

Be as precise as possible. Don't look it up, just repeating some quote won't give anyone insight into what the definition means to you. Don't give a lame answer like "I use them just as everyone else."

frame of reference

inertial

coordinate system

When you're done, you can go read this thread (http://www.bautforum.com/forum-introductions-feedback/93226-how-make-discussions-atm-ideas-less-frustrating-suggestion.html) and understand why I'm asking this.

I will answer off the top of my head.

1) frame of reference - Taking the position of a particular inertial frame and performing the LT calcs of other frames relative to the chosen one.

2) inertial - I use this to describe the members of one frame being "at rest" with one another.

3) coordinate system - Well this could be Cartesian, Minkowsky and more general topological spaces.

But, in SR, it would be a Euclidian Cartesian space within the inertial frame and it would be Minkowsky looking at other frames.

How does this affect the fact one rocket is heading down the positive x-axis at .99c and the other is heading down the negative x-axis such that .99c -.99c -.99c = .99c - 2*.99c = -.99c?

2v or not 2v, that is the question.

macaw

2009-Oct-29, 06:21 PM

I am not a Ritz person.

Based on what you have posted, you are. So where is the answer to Q1-Q7?

NorthernBoy

2009-Oct-29, 06:21 PM

I will answer off the top of my head.

1) frame of reference - Taking the position of a particular inertial frame and performing the LT calcs of other frames relative to the chosen one.

2) inertial - I use this to describe the members of one frame being "at rest" with one another.

3) coordinate system - Well this could be Cartesian, Minkowsky and more general topological spaces.

But, in SR, it would be a Euclidian Cartesian space within the inertial frame and it would be Minkowsky looking at other frames.

Oh, that's just horrible. If you are going to redefine these terms in that way, then of course no-one is going to understand what you are talking about.

An inertial frame is simply one in which newton's laws apply, i.e. one which is not accelerating or rotating. Within that, objects can be moving in any way that they like, but an object not acted upon by a force will continue moving in a straight line at a constant speed.

Point 1) just makes no sense at all.

slang

2009-Oct-29, 06:24 PM

I will answer off the top of my head.

Thanks, that's the idea. I won't be the one judging the correctness, I'm not qualified for that. Some around here will be able to do so, I'm sure.

1) frame of reference - Taking the position of a particular inertial frame and performing the LT calcs of other frames relative to the chosen one.

Thanks. You're defining what frame is, by using the word frame. That's not very helpful. So, the next word to define, in the context of SR discussions is:

frame

3) coordinate system - Well this could be Cartesian, Minkowsky and more general topological spaces.

That's not defining what it is and what it means to you, that's just listing some kinds.

zerocold

2009-Oct-29, 06:31 PM

I don't know how to use the LT in this case, of course i'm not an expert on this matters, wonder if someone could do the calculation, to get the speed between both ships

Is interesting, because IMO, there should not be communication between both rockets for the 3th observer, while there could be communication in-between the rockets for the rockets themselve, of course , is just my opinion.

abcdefg

2009-Oct-29, 06:34 PM

Nope, I can't make any sense of this sentence at all.

abcdefg, your written English is getting worse and worse. If you are claiming that you cannot move a particle in the direction of positive x at this speed, and then slow it and move it iin the direction of egative x, then you are mistaken.

That is such an absurd claim, though, that I cannot believe that this is what you really meant.

I can promise you that if you are willing to stick to one thought experiment, and to stop your ridiculous habit of refusing to say which obervations are made by whom, that I will be able to help you with this.

I can also promise you that if you insist on posting things in such mangled English, that this will never go anywhere.

Please, read your posts back, and understand that you need to make them readable to other people, not just to you. Your writing is really really weak, and it is painful to have to extract your meaning before I can comment on it.

OK, sorry and thanks. If something is unclear, please say so and I will keep writing it differently until it makes sense to you. I do not mind.

I am trying to say if a particle is moving in the positive x-direction at .99c and there is spherical acceleration in all directions off that particle, then those accelerated in the direction of the positive x-axis are limited to less than .01c. Those in the negative x-axis direction are limited to -1.99c.

In general, the following equation holds for any alteration to the speed of a relativistic particle.

-c < v0 + v cos(Θ) < c.

where v0 is the speed of the relativistic particle

and Θ is the angle to the direction of the particle/object's travel.

abcdefg

2009-Oct-29, 06:37 PM

Thanks, that's the idea. I won't be the one judging the correctness, I'm not qualified for that. Some around here will be able to do so, I'm sure.

Thanks. You're defining what frame is, by using the word frame. That's not very helpful. So, the next word to define, in the context of SR discussions is:

frame

That's not defining what it is and what it means to you, that's just listing some kinds.

Frame is a collection of observers at rest with one another.

A coordinate system is some general set of points operating under a distance formula.

abcdefg

2009-Oct-29, 06:39 PM

Based on what you have posted, you are. So where is the answer to Q1-Q7?

I keep a log of answers to your questions.

Answer log macaw

Q1, Q2 http://www.bautforum.com/against-mainstream/95693-relative-v-c.html#post1610203

Q3 http://www.bautforum.com/against-mainstream/95693-relative-v-c.html#post1610214

Q4, Q5 http://www.bautforum.com/against-mainstream/95693-relative-v-c-2.html#post1610255

Q6 http://www.bautforum.com/against-mainstream/95693-relative-v-c-2.html#post1610255

Q7 http://www.bautforum.com/against-mainstream/95693-relative-v-c-2.html#post1610276

WayneFrancis

2009-Oct-29, 06:40 PM

We have two identical rockets.

One rocket is located on one side of the earth and one is located directly on the other side of the earth.

At a predetermined time, each takes off with burns such that they each reach 0.99c, one rocket is therefore going along the position x-axis and the other is heading along the negative x-axis.

Question 1: What is the relative speed of the rockets? 1.98c

Wrong, their relative speed to each other is not 1.98c

Their relative speed is computed by the formula

V1 + V2 / ( 1 + (V1V2/c2))

so it is .99 + .99 / (1 + .992/c2)

1.98/ (1 + 0.9801)

or 1.98/1.9801

or 0.99994949750012625624968435937579c

0.99994949750012625624968435937579c != 1.98c

Question 2: If each shot a laser at each other as a signal, will they ever see the light? Yes.

So you've shown you don't know what formulas to use...congratulations.

abcdefg

2009-Oct-29, 06:41 PM

CERN. Except that we get much, much closer to c than one part in one hundred there.

Did you seriously not know that this is what we did there?:confused:

I do not have a problem with that.

That has nothing to do with what we are taking about.

We are not talking about the absolute standard of the velocity limiting c.

We are talking about total relative velocity.

These are two different concepts.

macaw

2009-Oct-29, 06:41 PM

I am trying to say if a particle is moving in the positive x-direction at .99c and there is spherical acceleration in all directions off that particle, then those accelerated in the direction of the positive x-axis are limited to less than .01c.

Yes, if you believe in Ritz theory. No, if you know relativity. Speeds do not just add up in the real world, the composition for speed is more complicated.

So, with v1=.99c and with:

w=(v1+v2)/(1+v1*v2/c^2)

you have to answer the new question:

Q8: what is the correct upper limit for v2? (Hint, it is not .01c as you claimed)

Those in the negative x-axis direction are limited to -1.99c.

Nope.

Q9: what is the correct upper limit for v2 in the negative x direction. Hint, it is not -1.99c

So, you now have Q1-Q9 to answer

abcdefg

2009-Oct-29, 06:42 PM

Based on what you have posted, you are. So where is the answer to Q1-Q7?

Think this though.

I posted the light would catch the other rocket.

Clearly, this implies I support an absolute constant c and not one riding with the frame.

WayneFrancis

2009-Oct-29, 06:43 PM

Well, first off, there is a point as they are accelerating at which from rocket to rocket, the instantaneous v of each is .5 c. Taking the position of either rocket, the other rocket disappears since length contraction becomes 0. After that point, length contraction crosses into the complex number system.

Next, if I shot a laser from the front of the rocket to the back and out some hole toward the other rocket and I had clocks measuring the time such that I calculate the speed if the light, a speed of c to the frame would never reach the other rocket moving at relative speed v = 1.98c. It is claimed the speed of light to an inertial frame is always c.

wow your more confused in the post then your last big post.

macaw

2009-Oct-29, 06:44 PM

Think this though.

I posted the light would catch the other rocket.

Clearly, this implies I support an absolute constant c and not one riding with the frame.

Just answer Q1-Q9. Ok?

WayneFrancis

2009-Oct-29, 06:45 PM

This calculation does not control the speed of the rockets. These rockets are at .99c.

Only to a third observer on the earth

To them self their v = 0 and the other is at 0.99994949750012625624968435937579c

For example, here (http://math.ucr.edu/home/baez/physics/Relativity/SR/experiments.html#moving-source_tests) lists experiments with particlaes moving faster than .99c. Assume one is shot one way and one the other.

This is legitimate.

No this is you confused and not understanding SR

macaw

2009-Oct-29, 06:52 PM

To them self their v = 0 and the other is at 0.99994949750012625624968435937579c

Nice.

No this is you confused and not understanding SR

It is precisely Ritz' theory.

abcdefg

2009-Oct-29, 06:55 PM

Wrong, their relative speed to each other is not 1.98c

Their relative speed is computed by the formula

V1 + V2 / ( 1 + (V1V2/c2))

so it is .99 + .99 / (1 + .992/c2)

1.98/ (1 + 0.9801)

or 1.98/1.9801

or 0.99994949750012625624968435937579c

0.99994949750012625624968435937579c != 1.98c

So you've shown you don't know what formulas to use...congratulations.

Thanks you.

Now, I will see if I can get your answer to work.

I start with ship A which is the one moving left.

I burn to attain 0.99994949750012625624968435937579c. Now, I should be in the same frame as ship B.

Nope, I am not, I am moving at 0.99994949750012625624968435937579c -.99c which is close to the original launch frame or close to the earth frame. I did not make it into the frame of ship B because it is traveling at .99c relative to the original launch frame.

So, this is not the total relative velocity.

Your equation did not work.

abcdefg

2009-Oct-29, 06:56 PM

wow your more confused in the post then your last big post.

OK, I will redo it.

WayneFrancis

2009-Oct-29, 06:58 PM

Answer to Q7

I do not know where you got the above and so I would have a hard time proving it to be false.

Exactly, you don't know what formula to use thus you can't prove it false. But you've been given the formula. The fact you don't understand it is another fact. You can lead a horse to water...

Further, I do not know how you would control the speed of the rockets with each going .99c in opposite directions by imposing an equation on them.

.99c according to who? The observer on Earth. The rest of your complaint seems to be leaving something out between your brain and what your fingers typed. You might as well complain about ice cream tasting blue.

abcdefg

2009-Oct-29, 06:58 PM

Nice.

It is precisely Ritz' theory.

Ritz' theory of light emission has to do with light spedd adding to the source speed. We are talking about frames.

I will let Einstein tell you about the independence between the two concepts of light and frames.

Einstein: "I certainly knew that the principle of the constancy of the velocity of light is something quite independent of the relativity postulate."

http://philsci-archive.pitt.edu/archive/00001743/01/Norton.doc

NorthernBoy

2009-Oct-29, 07:00 PM

I do not have a problem with that.

That has nothing to do with what we are taking about.

Of course it does. We have one particle travelling at nearly c in one direction, and one travelling at c in the opposite direction, and you are asking questions about their relative speed.

How is this not analagous to your two rockets problem?

macaw

2009-Oct-29, 07:00 PM

Thanks you.

Nope, I am not, I am moving at 0.99994949750012625624968435937579c -.

You can't, you are subtracting speeds again.

Q10: What is the correct computation?

Your equation did not work.

The questions are accumulating, you now have Q1-Q10 to answer.

macaw

2009-Oct-29, 07:01 PM

I will let Einstein tell you about the independence between the two concepts of light and frames.

That would not be necessary, please answer Q1-Q10.

NorthernBoy

2009-Oct-29, 07:03 PM

.99c according to who? The observer on Earth.

He's still doing it, isn't he, refusing to add the critically important fact about which observer he's talking about.

I do hope he stops soon.

WayneFrancis

2009-Oct-29, 07:06 PM

OK, let's take that particle at 0.99999c amd accelerate it to 0.99999c just as everyone has been telling me.

If this acceleration is in the direction of the positive x-axis, then you make sense if the first particle is moving in that direction. But, if the particle is shot in the direction of the negative x-axis, then this shot particle will return to the earth frame.

So, velocity additions are directionally based.

pleas read http://en.wikipedia.org/wiki/Velocity

abcdefg

2009-Oct-29, 07:19 PM

Yes, if you believe in Ritz theory. No, if you know relativity. Speeds do not just add up in the real world, the composition for speed is more complicated.

So, with v1=.99c and with:

w=(v1+v2)/(1+v1*v2/c^2)

you have to answer the new question:

Q8: what is the correct upper limit for v2? (Hint, it is not .01c as you claimed)

Answer Q8

Now, let's say what this means.

First there exists a rest frame and two other observers one at v1 and one at v2.

So, this is a 3 observer model. So, right off it is wrong.

But, under this false model as compared to the one offered on this thread is.

w=(v1+v2)/(1+v1*v2/c^2) < c

(v1+v2) < c( (1+v1*v2/c^2) )

c(v1+v2) < c^2 + v1*v1

cv1 + cv2 < c^2 + v1 * v2

cv2 - v1 * v2 < c^2 - cv1

v2( c - v1 ) < c( c - v1 )

v2 < c.

WayneFrancis

2009-Oct-29, 07:23 PM

Thanks you.

Now, I will see if I can get your answer to work.

I start with ship A which is the one moving left.

I burn to attain 0.99994949750012625624968435937579c. Now, I should be in the same frame as ship B.

You are not getting it. I'll explain this fully, unlike you.

Ships A and B launch in opposite directions.

Ships C launches with B in the same direction of B

All ships appear to accelerate to .99c from an observer on Earth.

After they all accelerated up to .99c, from the point of view of the Earth, The following will be true

A observes B & C speeding away from them at 0.99994949750012625624968435937579c

A observes Earth speeding away from them at .99c

B & C observe A speeding away from them at 0.99994949750012625624968435937579c

B & C observe Earth speeding away from them at .99c

If at this point in time C starts accelerating toward A then B would observe C to have the same time and space contraction as A when it reached a relative velocity, from B, of 0.99994949750012625624968435937579c.

This is not the same as being in the same frame becuase A & C are still seperated by a large distance and it would require more changes of velocity from one or both of the ships, A & C, before they are actually in the same frame.

This all brings us to your last big post where we kept explaining to you that you where not taking into account all the pertinent details.

Nope, I am not, I am moving at 0.99994949750012625624968435937579c -.99c which is close to the original launch frame or close to the earth frame. I did not make it into the frame of ship B because it is traveling at .99c relative to the original launch frame.

So, this is not the total relative velocity.

Your equation did not work.

No you have mixed frames. IE if you don't know how to apply the formulas then yes you will get garbage answers. So once again you have only shown that you are not educated in SR. No big revelation to the rest of us.

abcdefg

2009-Oct-29, 07:23 PM

Yes, if you believe in Ritz theory. No, if you know relativity. Speeds do not just add up in the real world, the composition for speed is more complicated.

So, with v1=.99c and with:

w=(v1+v2)/(1+v1*v2/c^2)

you have to answer the new question:

Q8: what is the correct upper limit for v2? (Hint, it is not .01c as you claimed)

Nope.

Q9: what is the correct upper limit for v2 in the negative x direction. Hint, it is not -1.99c

So, you now have Q1-Q9 to answer

Answer Q9

Correct limit in the negative direction -c.

Now, what is the total relative velocity differential?

|-c| + |c| = 2c.

This is the most general equation to describe the total relative velocity change to a moving particle.

In general, the following equation holds for any alteration to the speed of a relativistic particle.

-c < v0 + v cos(Θ) < c.

where v0 is the speed of the relativistic particle

and Θ is the angle to the direction of the particle/object's travel.

abcdefg

2009-Oct-29, 07:25 PM

Just answer Q1-Q9. Ok?

Answer log macaw

Q1, Q2 http://www.bautforum.com/against-mainstream/95693-relative-v-c.html#post1610203

Q3 http://www.bautforum.com/against-mainstream/95693-relative-v-c.html#post1610214

Q4, Q5 http://www.bautforum.com/against-mainstream/95693-relative-v-c-2.html#post1610255

Q6 http://www.bautforum.com/against-mainstream/95693-relative-v-c-2.html#post1610255

Q7 http://www.bautforum.com/against-mainstream/95693-relative-v-c-2.html#post1610276

Q8 http://www.bautforum.com/against-mainstream/95693-relative-v-c-4.html#post1610539

Q9 http://www.bautforum.com/against-mainstream/95693-relative-v-c-4.html#post1610541

macaw

2009-Oct-29, 07:30 PM

Answer Q8

Now, let's say what this means.

First there exists a rest frame and two other observers one at v1 and one at v2.

So, this is a 3 observer model. So, right off it is wrong.

But, under this false model as compared to the one offered on this thread is.

w=(v1+v2)/(1+v1*v2/c^2) < c

(v1+v2) < c( (1+v1*v2/c^2) )

c(v1+v2) < c^2 + v1*v1

cv1 + cv2 < c^2 + v1 * v2

cv2 - v1 * v2 < c^2 - cv1

v2( c - v1 ) < c( c - v1 )

v2 < c.

Good, you got the right answer. Now , let's see the answers to the other questions, Q1-Q7,Q9-10. Looks like you are contradicting yourself, you are back to the stance v>c. Why?

macaw

2009-Oct-29, 07:33 PM

Answer Q9

Correct limit in the negative direction -c.

Q11: How did you get the answer? Show the steps, like in Q8.

Now, what is the total relative velocity differential?

|-c| + |c| = 2c.

The above is closing speed, not relative speed.

Q12: Explain the difference between closin and relative speed so we can see that you are starting to learn.

macaw

2009-Oct-29, 07:34 PM

Answer log macaw

Q1, Q2 http://www.bautforum.com/against-mainstream/95693-relative-v-c.html#post1610203

Q3 http://www.bautforum.com/against-mainstream/95693-relative-v-c.html#post1610214

Q4, Q5 http://www.bautforum.com/against-mainstream/95693-relative-v-c-2.html#post1610255

Q6 http://www.bautforum.com/against-mainstream/95693-relative-v-c-2.html#post1610255

Q7 http://www.bautforum.com/against-mainstream/95693-relative-v-c-2.html#post1610276

Q8 http://www.bautforum.com/against-mainstream/95693-relative-v-c-4.html#post1610539

Q9 http://www.bautforum.com/against-mainstream/95693-relative-v-c-4.html#post1610541

Bad answers to all of them. Especially since you are contradicting the correct answers for Q8,9 you just managed to find. You realize that you are just contradicting your own solutions, right? Try again.

dgavin

2009-Oct-29, 07:46 PM

Thanks you.

Now, I will see if I can get your answer to work.

I start with ship A which is the one moving left.

I burn to attain 0.99994949750012625624968435937579c. Now, I should be in the same frame as ship B.

Nope, I am not, I am moving at 0.99994949750012625624968435937579c -.99c which is close to the original launch frame or close to the earth frame. I did not make it into the frame of ship B because it is traveling at .99c relative to the original launch frame.

So, this is not the total relative velocity.

Your equation did not work.

I think you are missing that Velocities are relative and not additive. Ship 1 and 2 are moving each at .99c in oppostie (or some) direction relative to a rest frame. Relative to each other, they are moving at 0.99994949750012625624968435937579c. However when Ship 1 measures the distance covered between itself and Ship 2, relative to thier originating point and from its point of view, then it can calculate that to an apparent seperation velosity of 1.98c.

The reaosn this happens is that both Ship 1 and Ship 2 when they measure from thier point of view, are in both a time dilated state and length contracted state. In this state space and time are compressed to the point that the other ship is moving at 0.99994949750012625624968435937579c relative to them and the originating point is moving at .99c relative to them.

Another way to look at it is like this. Motion=energy=mass (simplified). As something moves faster it increases in energy. According to e=mc2 it means it is increasing in mass. As it's mass increases, it's gravitional effects on space/time are larger also. The faster something moves, the more it -warps- space time. Time moves slower for them, while at the same time, they are moving though compressed (less) space.

If somehow you could have any object with mass at a relative velosity of >=c you have a problem. At light speed a object with mass has infinate energry, which also implies, infinate mass and infinate gravity. The moment any object with mass reached lightspeed, it would collapse into a black hole like object from it's own infinate gravity.

You can have appearent seperation velocities that seem like they are >c, but the actual velocities of any object with mass must always be < c. Only massless particles or waves (photons) can have a v=c, and only objects with negative masses (hypothetical tachyons) can have a v>c .

abcdefg

2009-Oct-29, 07:52 PM

Exactly, you don't know what formula to use thus you can't prove it false. But you've been given the formula. The fact you don't understand it is another fact. You can lead a horse to water...

.99c according to who? The observer on Earth. The rest of your complaint seems to be leaving something out between your brain and what your fingers typed. You might as well complain about ice cream tasting blue.

I said this already, .99c according to the acceleration equations.

v(BT) = c tanh(a BT/c)

macaw

2009-Oct-29, 07:55 PM

I think you are missing that Velocities are relative and not additive. Ship 1 and 2 are moving each at .99c in oppostie (or some) direction relative to a rest frame. Relative to each other, they are moving at 0.99994949750012625624968435937579c. However when Ship 1 measures the distance covered between itself and Ship 2, relative to thier originating point and from its point of view, then it can calculate that to an apparent seperation velosity of 1.98c.

Correct.

The reaosn this happens is that both Ship 1 and Ship 2 when they measure from thier point of view, are in both a time dilated state and length contracted state.

Incorrect.

Another way to look at it is like this. Motion=energy=mass (simplified). As something moves faster it increases in energy. According to e=mc2 it means it is increasing in mass.

Incorrect.

As it's mass increases, it's gravitional effects on space/time are larger also.

A classical fallacy.

The faster something moves, the more it -warps- space time. Time moves slower for them, while at the same time, they are moving though compressed (less) space.

Another fallacy.

At light speed a object with mass has infinate energry, which also implies, infinate mass and infinate gravity.

Another fallacy.

The moment any object with mass reached lightspeed, it would collapse into a black hole like object from it's own infinate gravity.

Yet another incorrect statement.

You can have appearent seperation velocities that seem like they are >c,

Actually, they are.

abcdefg

2009-Oct-29, 07:58 PM

Of course it does. We have one particle travelling at nearly c in one direction, and one travelling at c in the opposite direction, and you are asking questions about their relative speed.

How is this not analagous to your two rockets problem?

Well, I an in one rocket and I know my speed is v(BT) = c tanh(a BT/c) according to the acceleration equations.

This rocket is B moving to the right.

Now, I apply a reverse burn of a and BT and then the speed is -v(BT)=-.99c from its original value. This only made it to the earth frame. This is not enough.

Now, I have to do it again, a burn of a at BT. At this point, the rocket B is in the frame of A and the total speed change from the original B frame is -v(BT) - v(BT) = -2*v(BT) = -2*.99c = -1.98c

The total relative speed between the two frames is 1.98c.

macaw

2009-Oct-29, 08:03 PM

total speed change from the original B frame is -v(BT) - v(BT) = -2*v(BT) .

Nope, it isn't, you are making the same mistake again. There is no surprise you are getting wrong answers. So, why don't you go back and answer Q1-Q12. Please do not point to your previously already refuted posts. Answer in line.

I know my speed is v(BT) = c tanh(a BT/c) according to the acceleration equations.

What you don't know is that the above has been derived using the correct speed composition formula, i.e. w=(u+v)/(1+uv/c^2) and not the incorrect one that you insist on using. I gave you the site (http://users.telenet.be/vdmoortel/dirk/Physics/Acceleration.html) where you learned it. The irony is that you sometimes even link in the site. See the first formula in the derivation?

pzkpfw

2009-Oct-29, 08:35 PM

Out of interest I used the (correct) formula w=(v1+v2)/(1+v1*v2/c^2) to calculate the relative closing speed of two cars heading for a head-on crash at 100 km/h each.

Turned out they were closing at what seemed to be 200 km/h as we'd "expect". So I then multiplied their speed by 10 and re-calculated...

All speeds in km/h

c

1,079,252,848.800

(correct) ("expected") (difference)

v1 v2 w v1+v2 (v1+v2)-w

100.000 100.000 200.000 200.000 0.000

1,000.000 1,000.000 2,000.000 2,000.000 0.000

10,000.000 10,000.000 20,000.000 20,000.000 0.000

100,000.000 100,000.000 199,999.998 200,000.000 0.002

1,000,000.000 1,000,000.000 1,999,998.283 2,000,000.000 1.717

10,000,000.000 10,000,000.000 19,998,283.095 20,000,000.000 1,716.905

100,000,000.000 100,000,000.000 198,297,563.311 200,000,000.000 1,702,436.689

1,000,000,000.000 1,000,000,000.000 1,076,121,453.793 2,000,000,000.000 923,878,546.207

(scroll right)

Turns out they need to be going pretty darn fast before the relativistic effects become noticable. (Note how w tends to c).

It's hardly surprising that people need to cling to the idea that relative speed is v1 + v2 (or v1 - v2, whatever). It works fine for the speeds we deal with in normal life. I remember myself in high school absolutely refusing to "believe" my physics teacher when he pointed this out to me. I'd have learned more, and faster, if I were more open to change of my preconceptions.

macaw

2009-Oct-29, 08:49 PM

Out of interest I used the (correct) formula w=(v1+v2)/(1+v1*v2/c^2) to calculate the relative closing speed of two cars heading for a head-on crash at 100 km/h each.

I'd have learned more, and faster, if I were more open to change of my preconceptions.

I find it ironic that abcdefg is attempting to use the formula for accelerated motion that was derived based on the relativistic speed composition in order to .... prove his incorrect use of the galilean speed composition.

NorthernBoy

2009-Oct-29, 08:56 PM

The total relative speed between the two frames is 1.98c.

Again, you have left out the critical part. You have not said to whom this measurement applies.

It is not a difficult point, and I've politely pointed out to you that it is dishonest of you to omit it, yet you still omit it.

Why?

nauthiz

2009-Oct-29, 09:15 PM

I'm a bit late to the thread, but it seems to me that there's a need to specify the reference frame from which the measurement of speed is being made. I don't think relativity says that an observer at Z can't see the relative speed between two objects at X and Y as greater than c. It's saying that from X you can't see Y moving faster than c relative to yourself, and vice versa. (This being fine because there's no such thing as an objective reference frame in relativity.)

So to take the numbers from the 2nd-to-last row of pzkpfw's table, let's say two rockets are traveling away from Earth in opposite directions at 0.927c. An observer from Earth would see them both traveling away at that speed, and calculate their velocity relative to each other as 1.853c. However, that says nothing about what either of the rockets would see: each of them sees the other rocket as receding at only 0.997c.

NorthernBoy

2009-Oct-29, 09:22 PM

I'm a bit late to the thread, but it seems to me that there's a need to specify the reference frame from which the measurement of speed is being made.

Yes, there is. abcdefg refuses to do this, though, which is the cause of all of his problems.

macaw

2009-Oct-29, 09:23 PM

I'm a bit late to the thread, but it seems to me that there's a need to specify the reference frame from which the measurement of speed is being made. I don't think relativity says that an observer at Z can't see the relative speed between two objects at X and Y as greater than c. It's saying that from X you can't see Y moving faster than c relative to yourself, and vice versa. (This being fine because there's no such thing as an objective reference frame in relativity.)

So to take the numbers from the 2nd-to-last row of pzkpfw's table, let's say two rockets are traveling away from Earth in opposite directions at 0.927c. An observer from Earth would see them both traveling away at that speed, and calculate their velocity relative to each other as 1.853c. However, that says nothing about what either of the rockets would see: each of them sees the other rocket as receding at only 0.997c.

Yes, we have expained this same exact pair of concepts several different times. Maybe you have more success.

abcdefg

2009-Oct-29, 09:25 PM

I will be back shortly or so with the math.

abcdefg

2009-Oct-29, 09:27 PM

Yes, there is. abcdefg refuses to do this, though, which is the cause of all of his problems.

Uhhhh, yes I am.

But remember, the acceleration equations are for one launch frame and one accelerating observer.

Now, I applied two different components causing this relative v > c.

So, it is the sum of two values and that is OK.

I'll back in a little while.

David Holland

2009-Oct-29, 09:30 PM

You keep saying there is no third observer. Who is measuring their speed at .99c. There is a third observer implicit in your thought experiment.

nauthiz

2009-Oct-29, 10:36 PM

Yes, we have expained this same exact pair of concepts several different times. Maybe you have more success.

I figured that was likely the case, but thought I'd give it another shot anyway. I know I didn't really grasp the concept until someone explained it to me the right way.

Then again, I was approaching it from a standpoint of, "What am I not understanding?". . .

(ETA: Turns out what I wasn't really understanding is that there is no such thing as global frame of reference. I was so used to implicitly assuming the existence of such a thing that I didn't even realize I was doing it, so it'd keep sneaking in and confusing me.)

abcdefg

2009-Oct-29, 10:40 PM

You keep saying there is no third observer. Who is measuring their speed at .99c. There is a third observer implicit in your thought experiment.

This would be relative to the launch frame based on the acceleration equations.

NorthernBoy

2009-Oct-29, 10:52 PM

Uhhhh, yes I am.

Yes you are what?

I mentioned the fact that you keep leaving out critical details, and you reply 'Yes I am".

That makes no sense, it is not a response to the point made, and adds nothing to the discussion. is this your new technique, to move from proof by assertion and obfuscation to actually writing "responses" which are not?

You then add

Now, I applied two different components

You cannot "apply" components. You can apply transformations, boosts, or accelerations, but not components.

Can you please stick to standard English?

NorthernBoy

2009-Oct-29, 10:53 PM

This would be relative to the launch frame based on the acceleration equations.

How bizarre. When I tried to talk about things relative to that frame, you objected, and said that I could not, as there was no observer there.

As soon as it suits you, though, you refer to it yourself.

Is this how you think adult discussion works; that you get to scream "no fair" when someone else uses a frame, but can drop it in yourself when you think that it bolsters your argument?

macaw

2009-Oct-29, 10:57 PM

I figured that was likely the case, but thought I'd give it another shot anyway. I know I didn't really grasp the concept until someone explained it to me the right way.

Then again, I was approaching it from a standpoint of, "What am I not understanding?". . .

(ETA: Turns out what I wasn't really understanding is that there is no such thing as global frame of reference. I was so used to implicitly assuming the existence of such a thing that I didn't even realize I was doing it, so it'd keep sneaking in and confusing me.)

Actually, the difference between separation (closing) speed and relative speed makes for a very interesting subject. Maybe when all the dust settles I will post a short essay. Until then, let's watch abcdefg trying to disprove relativity (again).

abcdefg

2009-Oct-29, 11:02 PM

OK, shoot light from A to B right after the burn time BT for each.

The total distance between the two at the instant of the end of the burn is based on the acceleration equations.

2* c^2/a [ cosh(a BT/c) - 1 ] = K and is a known value.

Now, in tradition, the light will move such that the time to move from A to B = the time to move from B to A.

The time to move from A to B is t1 = K/(c-v).

Thus, the roundtrip time is 2*t1 = 2K/(c-v).

Now, at this point u < c and everyone is happy. But is the not the end.

While light is moving to strike B, A is not sitting still in space and is moving left at v(BT) = c tanh(a BT/c). Thus, A moves in an opposite direction to the light.

So, when the light strikes B, A moved left a distance vt1.

So, for light to get back to the point of emission in space from A, it will take 2*t1 as shown above.

But, in that time A moved left a distance v*(2*t1).

Thus, light must travel double the distance expected and thus the total relative velocity is double v.

Also, assume A wants to get to the frame B.

First A burns a for BT and acquires v from the previous frame. That is v.

But, A is now back to the earth frame and not to B.

Thus, an additional burn is required at a for BT to enter the frame of B.

The total speed is v + v to make that happen.

abcdefg

2009-Oct-29, 11:04 PM

Actually, the difference between separation (closing) speed and relative speed makes for a very interesting subject. Maybe when all the dust settles I will post a short essay. Until then, let's watch abcdefg trying to disprove relativity (again).

You have not considered the absolute constant speed of light in your analysis.

You will have a tough time making everything behave the way you want with an absolute constant speed of light.

abcdefg

2009-Oct-29, 11:06 PM

Actually, the difference between separation (closing) speed and relative speed makes for a very interesting subject. Maybe when all the dust settles I will post a short essay. Until then, let's watch abcdefg trying to disprove relativity (again).

Oh, one more thing.

I have not tried to disprove relativity yet.

That is the next ATM.

This ATM is to establish light speed is an absolute constant and not a constant to the frame.

abcdefg

2009-Oct-29, 11:09 PM

He's still doing it, isn't he, refusing to add the critically important fact about which observer he's talking about.

I do hope he stops soon.

No, I am using two rocket frames, a common launch frame and the acceleration equations.

I am looking at what happens when relativistic speeds are approached.

Why?

These are pseudo uniform motion frames when compare to the original launch frame.

abcdefg

2009-Oct-29, 11:11 PM

You can't, you are subtracting speeds again.

Q10: What is the correct computation?

The questions are accumulating, you now have Q1-Q10 to answer.

Answer Q10. This does not work when light is emitted and bounced back frame to frame when one is .99c in the positive direction and one is .99c in the negative direction.

That is why it is not correct.

abcdefg

2009-Oct-29, 11:16 PM

Of course it does. We have one particle travelling at nearly c in one direction, and one travelling at c in the opposite direction, and you are asking questions about their relative speed.

How is this not analagous to your two rockets problem?

OK, in this sense it is the same but the total relative v > c though.

I bounce light between the two A and B and light does not behave as though total v < c. It behaves as though the total relative velocity is 1.98c.

abcdefg

2009-Oct-29, 11:17 PM

That would not be necessary, please answer Q1-Q10.

Q1, Q2 http://www.bautforum.com/against-mainstream/95693-relative-v-c.html#post1610203

Q3 http://www.bautforum.com/against-mainstream/95693-relative-v-c.html#post1610214

Q4, Q5 http://www.bautforum.com/against-mainstream/95693-relative-v-c-2.html#post1610255

Q6 http://www.bautforum.com/against-mainstream/95693-relative-v-c-2.html#post1610255

Q7 http://www.bautforum.com/against-mainstream/95693-relative-v-c-2.html#post1610276

Q8 http://www.bautforum.com/against-mainstream/95693-relative-v-c-4.html#post1610539

Q9 http://www.bautforum.com/against-mainstream/95693-relative-v-c-4.html#post1610541

Q10. http://www.bautforum.com/against-mainstream/95693-relative-v-c-5.html#post1610772

slang

2009-Oct-29, 11:20 PM

No, I am using two rocket frames, [one] common launch frame and the acceleration equations.

Frame is a collection of observers at rest with one another. (http://www.bautforum.com/against-mainstream/95693-relative-v-c-3.html#post1610478)

two + one = three (collections of) observers.

abcdefg

2009-Oct-29, 11:29 PM

You are not getting it. I'll explain this fully, unlike you.

Ships A and B launch in opposite directions.

Ships C launches with B in the same direction of B

All ships appear to accelerate to .99c from an observer on Earth.

After they all accelerated up to .99c, from the point of view of the Earth, The following will be true

A observes B & C speeding away from them at 0.99994949750012625624968435937579c

A observes Earth speeding away from them at .99c

B & C observe A speeding away from them at 0.99994949750012625624968435937579c

B & C observe Earth speeding away from them at .99c

If at this point in time C starts accelerating toward A then B would observe C to have the same time and space contraction as A when it reached a relative velocity, from B, of 0.99994949750012625624968435937579c.

This is not the same as being in the same frame becuase A & C are still seperated by a large distance and it would require more changes of velocity from one or both of the ships, A & C, before they are actually in the same frame.

This all brings us to your last big post where we kept explaining to you that you where not taking into account all the pertinent details.

No you have mixed frames. IE if you don't know how to apply the formulas then yes you will get garbage answers. So once again you have only shown that you are not educated in SR. No big revelation to the rest of us.

You are applying the equations correctly.

Somebody gave you equations you you just applied them.

You did not consider sending light signals back and forth to see the result.

These equations do not work when light is emitted from A to B and back again.

This post shows a total relative v > c.

http://www.bautforum.com/against-mainstream/95693-relative-v-c-5.html#post1610759

abcdefg

2009-Oct-29, 11:34 PM

Frame is a collection of observers at rest with one another. (http://www.bautforum.com/against-mainstream/95693-relative-v-c-3.html#post1610478)

two + one = three (collections of) observers.

Yes, but the launch frame is incidental and for bookeeping only.

The real action is these rockets going in opposite directions at .99c with light sent to each.

I used the launch frame only to establish the individual acceleration equations to each A and B.

This way I can establish a known distance at the instant the burn stops and establish the v's of A and B in a predictable fashion.

Once this is achieved, light signals can be sent back and forth to prove the total relative v is > c.

slang

2009-Oct-29, 11:50 PM

Oh, so now you're going from "I don't need a third observer" to "I need a third observer but he doesn't really count because he doesn't actually do anything"? (paraphrased, not actual quotes, caveat something)

Yes, but the launch frame is incidental and for bookeeping only.

(resisting halloween jokes on boo-keeping) What bookkeeping?

abcdefg

2009-Oct-29, 11:56 PM

I think you are missing that Velocities are relative and not additive. Ship 1 and 2 are moving each at .99c in oppostie (or some) direction relative to a rest frame. Relative to each other, they are moving at 0.99994949750012625624968435937579c. However when Ship 1 measures the distance covered between itself and Ship 2, relative to thier originating point and from its point of view, then it can calculate that to an apparent seperation velosity of 1.98c.

The reaosn this happens is that both Ship 1 and Ship 2 when they measure from thier point of view, are in both a time dilated state and length contracted state. In this state space and time are compressed to the point that the other ship is moving at 0.99994949750012625624968435937579c relative to them and the originating point is moving at .99c relative to them.

Another way to look at it is like this. Motion=energy=mass (simplified). As something moves faster it increases in energy. According to e=mc2 it means it is increasing in mass. As it's mass increases, it's gravitional effects on space/time are larger also. The faster something moves, the more it -warps- space time. Time moves slower for them, while at the same time, they are moving though compressed (less) space.

If somehow you could have any object with mass at a relative velosity of >=c you have a problem. At light speed a object with mass has infinate energry, which also implies, infinate mass and infinate gravity. The moment any object with mass reached lightspeed, it would collapse into a black hole like object from it's own infinate gravity.

You can have appearent seperation velocities that seem like they are >c, but the actual velocities of any object with mass must always be < c. Only massless particles or waves (photons) can have a v=c, and only objects with negative masses (hypothetical tachyons) can have a v>c .

Let me be clear, I did not say v > c as an individual object.

I said the total or maybe group relative velocity > c.

That is different.

No single object with mass may acquire c and I support this logic.

abcdefg

2009-Oct-29, 11:58 PM

Nope, it isn't, you are making the same mistake again. There is no surprise you are getting wrong answers. So, why don't you go back and answer Q1-Q12. Please do not point to your previously already refuted posts. Answer in line.

What you don't know is that the above has been derived using the correct speed composition formula, i.e. w=(u+v)/(1+uv/c^2) and not the incorrect one that you insist on using. I gave you the site (http://users.telenet.be/vdmoortel/dirk/Physics/Acceleration.html) where you learned it. The irony is that you sometimes even link in the site. See the first formula in the derivation?

Your equation does not hold up to the round trip calculation using light to decide total v < c.

This post will show how that is.

http://www.bautforum.com/against-mainstream/95693-relative-v-c-5.html#post1610759

abcdefg

2009-Oct-29, 11:59 PM

Out of interest I used the (correct) formula w=(v1+v2)/(1+v1*v2/c^2) to calculate the relative closing speed of two cars heading for a head-on crash at 100 km/h each.

Turned out they were closing at what seemed to be 200 km/h as we'd "expect". So I then multiplied their speed by 10 and re-calculated...

All speeds in km/h

c

1,079,252,848.800

(correct) ("expected") (difference)

v1 v2 w v1+v2 (v1+v2)-w

100.000 100.000 200.000 200.000 0.000

1,000.000 1,000.000 2,000.000 2,000.000 0.000

10,000.000 10,000.000 20,000.000 20,000.000 0.000

100,000.000 100,000.000 199,999.998 200,000.000 0.002

1,000,000.000 1,000,000.000 1,999,998.283 2,000,000.000 1.717

10,000,000.000 10,000,000.000 19,998,283.095 20,000,000.000 1,716.905

100,000,000.000 100,000,000.000 198,297,563.311 200,000,000.000 1,702,436.689

1,000,000,000.000 1,000,000,000.000 1,076,121,453.793 2,000,000,000.000 923,878,546.207

(scroll right)

Turns out they need to be going pretty darn fast before the relativistic effects become noticable. (Note how w tends to c).

It's hardly surprising that people need to cling to the idea that relative speed is v1 + v2 (or v1 - v2, whatever). It works fine for the speeds we deal with in normal life. I remember myself in high school absolutely refusing to "believe" my physics teacher when he pointed this out to me. I'd have learned more, and faster, if I were more open to change of my preconceptions.

This is good and accurate using the mainstream off the shelf tools.

But, this post shows the correct interpretation using light signals.

http://www.bautforum.com/against-mainstream/95693-relative-v-c-5.html#post1610759

abcdefg

2009-Oct-30, 12:05 AM

Again, you have left out the critical part. You have not said to whom this measurement applies.

It is not a difficult point, and I've politely pointed out to you that it is dishonest of you to omit it, yet you still omit it.

Why?

I am not trying to omit it.

You require that and I do not.

I am quite sure you are aware of the v calculated when acceleration is applied compared to the launch frame.

But, in this case the speeds are so high that we are operating in a land of pseudo uniform motion.

Once these relativistic speeds are attained, the absolue speed of light becomes a valuable tool in deciding v.

In particular, when light is emitted from A in the direction of B, there is no way a scientific person would argue A sits still at these speeds. Thus A continues on to the left.

This post

http://www.bautforum.com/against-mainstream/95693-relative-v-c-5.html#post1610759

shows once the logic that A continues on to the left is accepted, it is impossible that total relative v < c.

pzkpfw

2009-Oct-30, 12:05 AM

But, this post shows the correct interpretation using light signals.

http://www.bautforum.com/against-mainstream/95693-relative-v-c-5.html#post1610759

Actually I don't at all see how you get to your conclusion "Thus, light must travel double the distance expected and thus the total relative velocity is double v." in that post.

You are both:

* Using light to measure the distance between two moving objects, then complaining that it gives different results than if those objects were not moving.

* Refusing to be clear about whose point of view is used to measure the speeds and distances; thus you throw relativity out while still trying to use it to disprove itself.

macaw

2009-Oct-30, 12:07 AM

Answer Q10. This does not work when light is emitted and bounced back frame to frame when one is .99c in the positive direction and one is .99c in the negative direction.

That is why it is not correct.

There is no light bouncing back in Q10. So, please answer the question you have been asked.

macaw

2009-Oct-30, 12:09 AM

Q1, Q2 http://www.bautforum.com/against-mainstream/95693-relative-v-c.html#post1610203

Q3 http://www.bautforum.com/against-mainstream/95693-relative-v-c.html#post1610214

Q4, Q5 http://www.bautforum.com/against-mainstream/95693-relative-v-c-2.html#post1610255

Q6 http://www.bautforum.com/against-mainstream/95693-relative-v-c-2.html#post1610255

Q7 http://www.bautforum.com/against-mainstream/95693-relative-v-c-2.html#post1610276

Q8 http://www.bautforum.com/against-mainstream/95693-relative-v-c-4.html#post1610539

Q9 http://www.bautforum.com/against-mainstream/95693-relative-v-c-4.html#post1610541

Q10. http://www.bautforum.com/against-mainstream/95693-relative-v-c-5.html#post1610772

Pointing to posts that have been shown wrong will not cut it, I have asked you to answer in the thread, please stop dodging and start answering in earnest.

macaw

2009-Oct-30, 12:11 AM

Your equation does not hold up to the round trip calculation using light to decide total v < c.

There is no round trip calculation. You are attempting to resurrect the threads that were already shut down and proven wrong. Stick with the subject.

abcdefg

2009-Oct-30, 12:13 AM

I'm a bit late to the thread, but it seems to me that there's a need to specify the reference frame from which the measurement of speed is being made. I don't think relativity says that an observer at Z can't see the relative speed between two objects at X and Y as greater than c. It's saying that from X you can't see Y moving faster than c relative to yourself, and vice versa. (This being fine because there's no such thing as an objective reference frame in relativity.)

So to take the numbers from the 2nd-to-last row of pzkpfw's table, let's say two rockets are traveling away from Earth in opposite directions at 0.927c. An observer from Earth would see them both traveling away at that speed, and calculate their velocity relative to each other as 1.853c. However, that says nothing about what either of the rockets would see: each of them sees the other rocket as receding at only 0.997c.

there's a need to specify the reference frame

Nope, there is no need.

And, the off the shelf equation does not work.

We are able to decide motion meaning v based on the acceleration equations.

Now, normally, what happens in deciding v is that SR takes the position that c is constant in some frame and not others.

This is important what I just said.

But, that is absolutely false.

Even Einstein agreed, c is an absolute constant.

Einstein: "I certainly knew that the principle of the constancy of the velocity of light is something quite independent of the relativity postulate and I considered what would be more probable, the principle of the constancy of c, as was demanded by Maxwell’s equations, or the constancy of c, exclusively for an observer sitting at the light source. I decided in favor of the first, since I was convinced that each light [ray] should be defined by frequency and intensity alone, quite independently of whether it comes from a moving or a resting light source."

http://philsci-archive.pitt.edu/archive/00001743/01/Norton.doc

So, what I am trying to say is the frame or geocentric model of thought imposed by SR fails in this condition.

On one hand SR, claims c is constant in the rest frame, and on the other using R of S, SR claims it is not constant in other frames.

If someone knows what they are doing, they can force SR into all kinds of paradoxes for this violation of logic.

So, in this case, we apply the acceleration equations for each rocket and we arrive at a total relative v > c based on light signal transmission.

This post spells that out.

http://www.bautforum.com/against-mainstream/95693-relative-v-c-5.html#post1610759

macaw

2009-Oct-30, 12:18 AM

OK, shoot light from A to B right after the burn time BT for each.

The total distance between the two at the instant of the end of the burn is based on the acceleration equations.

2* c^2/a [ cosh(a BT/c) - 1 ] = K and is a known value.

Now, in tradition, the light will move such that the time to move from A to B = the time to move from B to A.

No, it is not. This was explained to you in a previous thread.

The time to move from A to B is t1 = K/(c-v).

Thus, the roundtrip time is 2*t1 = 2K/(c-v).

No, it is not. You are just introducing new errors , the return path is

t2=K/(c+v).

So , contrary to your claims t1 and t2 are not equal.

abcdefg

2009-Oct-30, 12:25 AM

There is no round trip calculation. You are attempting to resurrect the threads that were already shut down and proven wrong. Stick with the subject.

LOL, you are good.

No, they were not proven wrong.

Nope, this is round trip light using relative motion.

Bet you have not done that before.

Normally, you cannot do this but given collinear relative motion, it is allowed.

This round trip relative motion calculation reveals a total relative v > c.

I am surprised you did not follow it.

abcdefg

2009-Oct-30, 12:29 AM

Yes you are what?

I mentioned the fact that you keep leaving out critical details, and you reply 'Yes I am".

That makes no sense, it is not a response to the point made, and adds nothing to the discussion. is this your new technique, to move from proof by assertion and obfuscation to actually writing "responses" which are not?

You then add

You cannot "apply" components. You can apply transformations, boosts, or accelerations, but not components.

Can you please stick to standard English?

When I said components, I meant that normally SR takes everthing as a unit.

That means, v < c period.

How, I let A move left at -vt and B moves right at vt and therefore, I am operating with them as components and not a group.

Then, I use a light transmission and discover a total relative v > c.

I think I need to post some pictures of the events as they unfold.

It appears to be unclear.

macaw

2009-Oct-30, 12:32 AM

LOL, you are good.

No, they were not proven wrong.

Nope, this is round trip light using relative motion.

But you need to calculate correctly. Like I showed you in the previous threads. So, I showed it to you one more time.

abcdefg

2009-Oct-30, 12:32 AM

Actually I don't at all see how you get to your conclusion "Thus, light must travel double the distance expected and thus the total relative velocity is double v." in that post.

You are both:

* Using light to measure the distance between two moving objects, then complaining that it gives different results than if those objects were not moving.

* Refusing to be clear about whose point of view is used to measure the speeds and distances; thus you throw relativity out while still trying to use it to disprove itself.

No, first off, I am not trying to throw relativity out. I am only claiming it is limited in its operation because of the constant speed of light.

But, let me first match some pictures to the equations in my post (http://www.bautforum.com/against-mainstream/95693-relative-v-c-5.html#post1610759)as they unfold and hopefully it will become clear.

I will be back later.

macaw

2009-Oct-30, 12:48 AM

No, first off, I am not trying to throw relativity out. I am only claiming it is limited in its operation because of the constant speed of light.

But, let me first match some pictures to the equations in my post (http://www.bautforum.com/against-mainstream/95693-relative-v-c-5.html#post1610759)as they unfold and hopefully it will become clear.

I will be back later.

You will need to answer Q1-Q12 correctly before you get anywhere. You haven't done that, despite posting repeatedly the same errors as the ones in the previously closed threads.

abcdefg

2009-Oct-30, 12:54 AM

OK, I want to think about this some more.

We are moving way too fast and are assuming the difficult acceleration equations for v and distance for example that are not part of normal physics.

I am also doing a round trip light calculation with relative motion which is also difficult and not part of normal physics, and in fact not in the mainstream.

Those coming by will not see the underlying argument being taken.

I need to slow down and do more pictures and more explanation with links.

Thanks for your patience.

I will be back.

macaw

2009-Oct-30, 12:57 AM

OK, I want to think about this some more.

We are moving way too fast and are assuming the difficult acceleration equations for v and distance for example that are not part of normal physics.

They are part of normal physics, you only need to learn how to apply them.

I am also doing a round trip light calculation with relative motion which is also difficult and not part of normal physics, and in fact not in the mainstream.

This is also not true, the calculation is very easy.

I need to slow down and do more pictures and more explanation with links.

We don't need more pictures, you need to learn what you are dealing with. You are on your third thread, with the same story and with the same exact misconceptions.

dgavin

2009-Oct-30, 01:20 AM

The reason this happens is that both Ship 1 and Ship 2 when they measure from their point of view, are in both a time dilated state and length contracted state.

Incorrect.

It is correct within the context of the subject of this topic that has ship 1 and ship 2 both moving at .99 c.

Show us one legitimate source that indicates something traveling at relativistic speeds is not time dilated nor length contracted.

Another way to look at it is like this. Motion=energy=mass (simplified). As something moves faster it increases in energy. According to e=mc2 it means it is increasing in mass.

Incorrect.

As it's mass increases, it's gravitational effects on space/time are larger also.

A classical fallacy.

Are you saying there is no such thing as relativistic mass? Source please, as that is rather ATM.

Relativistic mass is from motion energy. http://en.wikipedia.org/wiki/Mass_in_general_relativity

All mass has gravity, relativistic or not. http://www.newton.dep.anl.gov/askasci/phy00/phy00806.htm

Colliders routinely, make use of both these principles to produce the effects they get.

The faster something moves, the more it -warps- space time. Time moves slower for them, while at the same time, they are moving though compressed (less) space.

Another fallacy.

Really then explain why a scientist has figured out a way to potentially use light itself to alter the flow of time but twisting space, from lights on relativistic gravitational warping of space? http://www.physorg.com/news63371210.html

What do you think causes both the time dilation and length contraction anyway?

Magic isn't an acceptable answer. Einstein already answered it by his two equivalency principles. You can't pull them out and still be talking about relativity. Energy and mass are equivalent, mass has gravity. You increase the Energy, you increase the gravity. This is true even for simple thermal heating of objects, granted that it is not easily measurable at those small quantities.

At light speed a object with mass has infinite energy, which also implies, infinite mass and infinite gravity.

Another fallacy.

I think I already showed that I was right here above. To summarize again Energy and mass are equivalent, mass has gravity. To move any mass at light-speed exactly requires it to have accumulated infinite energy, if it has infinite energy it has infinite relativistic mass and infinite relativistic gravity. So to say that is incorrect is to contradict relativity and the equivalency principles it was built on.

The moment any object with mass reached light-speed, it would collapse into a black hole like object from it's own infinite gravity.

Yet another incorrect statement.

In this case you are correct! Did that catch you off guard?

Black Holes have limited mass and limited gravity.

Something with mass that was accelerated to exactly light speed, would have infinite energy, infinite mass, and infinite gravity, and likely collapse out of any current frame of existence entirely. I think this condition is best equated by the same conditions that existed just before the big bang. Such a happening if possible (which it is not) would end up forming it's own universe.

You can have apparent separation velocities that seem like they are >c,

Actually, they are.

No, that is their apparent velocity taken from an outside observer, that can see them as moving apart at a >c velocity, relative to each other relative to the observer.

Scientific semantics here, Apparent Velocities are what appear to be happening from an outside observers point of view. The relativistic velocities of each object are still <c to the same the outside observer.

abcdefg

2009-Oct-30, 01:21 AM

Objective of this thread.

First, this thread seeks to establish that light is an absolute constant and not a constant to the frame. If light speed is constant to a frame as normally expressed in the mainstream, then given an object A moving v in the negative x-axis direction and one moving v in the positive x-axis direction, B, both with v > .5c, light will never reach from A to B. Thus, it must be the case that light is an absolute constant in space and not to the frame and then light will always reach from A to B and back again. Otherwise, it cannot.

Further, these are the words of Einstein.

Einstein: "I certainly knew that the principle of the constancy of the velocity of light is something quite independent of the relativity postulate and I considered what would be more probable, the principle of the constancy of c, as was demanded by Maxwell’s equations, or the constancy of c, exclusively for an observer sitting at the light source. I decided in favor of the first, since I was convinced that each light [ray] should be defined by frequency and intensity alone, quite independently of whether it comes from a moving or a resting light source."http://philsci-archive.pitt.edu/archive/00001743/01/Norton.doc

Second, it is not the object to bring down SR. SR will be forced though in a position to admit that total v < c.

Now, SR provides an equation for two objects at relativistic speeds meaning close to c, showing that total v < c.

However, this thread will show that a round trip light calculation will prove total v > c.

Once this is shown, this will only mean ST is reliable at speeds < .5c. If v > = .5c, then paradoxes form.

So, to do this the following fact is needed, no object with mass can equal the speed of light in speed.

abcdefg

2009-Oct-30, 01:27 AM

They are part of normal physics, you only need to learn how to apply them.

This is also not true, the calculation is very easy.

We don't need more pictures, you need to learn what you are dealing with. You are on your third thread, with the same story and with the same exact misconceptions.

We will slow down some.

I am quite aware you see these calcs as every day and so do many here.

Yet, some do not.

We will not be so selfish as to only operate at these levels.

Next, if you are so certain about things, you have yet to mathematically handle this post.

http://www.bautforum.com/against-mainstream/95693-relative-v-c-5.html#post1610759

If you are indeed following all this logic, then you will be able to operate under the context of this post. It is quite straightforward.

macaw

2009-Oct-30, 01:29 AM

It is correct within the context of the subject of this topic that has ship 1 and ship 2 both moving at .99 c.

Time dilation, length contraction play no role in this simple exercise.

Are you saying there is no such thing as relativistic mass? Source please, as that is rather ATM.

What I told you is that the variation of math with speed does not affect the way bodies gravitate.

.

if it has infinite energy it has infinite relativistic mass and infinite relativistic gravity.

Repeating the same mistake doesn't make it right.

Something with mass that was accelerated to exactly light speed, would have infinite energy, infinite mass, and infinite gravity,

Nope, same mistake again. BTW, you cannot create black holes by accelerating mass.

macaw

2009-Oct-30, 01:30 AM

Next, if you are so certain about things, you have yet to mathematically handle this post.

http://www.bautforum.com/against-mainstream/95693-relative-v-c-5.html#post1610759

.

I already pointed out the errors in the above post.

dgavin

2009-Oct-30, 01:35 AM

Time dilation, length contraction play no role in this simple exercise.

What I told you is that the variation of math with speed does not affect the way bodies gravitate.

.

Repeating the same mistake doesn't make it right.

Nope, same mistake again. BTW, you cannot create black holes by accelerating mass.

Then you are contridicting relativity, according to the sources I provided to support my position.

However further depate on this topic would need it's own thread at this point.

Back to the topic at hand.

abcdefg

2009-Oct-30, 01:40 AM

I already pointed out the errors in the above post.

What post?

I would like to see the math.

macaw

2009-Oct-30, 01:41 AM

Then you are contridicting relativity, according to the sources I provided to support my position.

I contradicted your misconceptions. I am sorry I couldn't convince you.

macaw

2009-Oct-30, 01:43 AM

What post?

I would like to see the math.

Post 153. You are making the same exact mistakes as in the previously closed threads.

abcdefg

2009-Oct-30, 02:14 AM

Post 153. You are making the same exact mistakes as in the previously closed threads.

You said,

No, it is not. You are just introducing new errors , the return path is

t2=K/(c+v).

So , contrary to your claims t1 and t2 are not equal.

Prove this and you are way off.

macaw

2009-Oct-30, 02:18 AM

You said,

No, it is not. You are just introducing new errors , the return path is

t2=K/(c+v).

So , contrary to your claims t1 and t2 are not equal.

Prove this and you are way off.

I have taught you this in the previous two threads. The ones that are closed. So, do not drag back your errors from tose threads.

abcdefg

2009-Oct-30, 02:26 AM

I have taught you this in the previous two threads. The ones that are closed. So, do not drag back your errors from tose threads.

You are smarter than this.

I wonder why you will not follow through with the light signals.

You will see total v > c.

abcdefg

2009-Oct-30, 02:41 AM

As I see it, I owe pictures for the events of this weird round trip light event between A and B where v(a) = .99c and v(B) = .99c.

Later.

pzkpfw

2009-Oct-30, 02:57 AM

Objective of this thread.

First, this thread seeks to establish that light is an absolute constant and not a constant to the frame. If light speed is constant to a frame as normally expressed in the mainstream, then given an object A moving v in the negative x-axis direction and one moving v in the positive x-axis direction, B, both with v > .5c, light will never reach from A to B. Thus, it must be the case that light is an absolute constant in space and not to the frame and then light will always reach from A to B and back again. Otherwise, it cannot.

You keep jumping ahead with your ideas, based on your own interpretations - which are very ATM.

Stop. Take a breath. Start with showing why the bit bolded in the quote above is correct.

i.e. the bit where you think v one way plus v the other way makes a relative speed of 2v (to the things moving at v). (i.e. The bit where you think if v is > 0.5 c that means their relative speed is > c and therefore light can't go from one to another).

Why do you think you can just add velocities together like this?

And no, nothing you've posted yet backs this up. Don't bother linking back to that same post again. Doing that a 21st time won't help.

Edit to add:

i.e.

The mainstream view is that two rockets each going 0.99c in opposite directions (from the point of view of an observer "stationary" between them) will see each other receding at 0.999949498c, not at 1.98c as you would claim.

Thus light shone by one will reach the other.

Before you use your claim to "prove" other things, you need to "prove" this claim.

abcdefg

2009-Oct-30, 03:07 AM

You keep jumping ahead with your ideas, based on your own interpretations - which are very ATM.

Stop. Take a breath. Start with showing why the bit bolded in the quote above is correct.

i.e. the bit where you think v one way plus v the other way makes a relative speed of 2v. (The bit where you think if v is > 0.5 c that means their relative speed is > c and therefore light can't go from one to another).

Why do you think you can just add velocities together like this?

And no, nothing you've posted yet backs this up. Don't bother linking back to that same post again. Doing that a 21st time won't help.

Well, that post does establish the procedure of the light protocol. It is not good for SR.

However, your post forces me to listen as to why you will not apply it.

I will think about this.

I am already making pictures and that may not be enough.

I will think some more.

However, why not think yourself and consider the round trip light travel.

Why is that a problem?

macaw

2009-Oct-30, 03:12 AM

.

Why is that a problem?

Because you are severely confused about the basics: you can't add the velocities as you were taught in high school. At relativistic speeds other formulas take over.

You can't claim that light needs t1=L/(c-v) in one direction and that it needs the same time in the opposite direction.

These are basic things that you don't seem to grasp, no matter how many different explanations you have received.

pzkpfw

2009-Oct-30, 03:17 AM

Well, that post does establish the procedure of the light protocol. It is not good for SR.

However, your post forces me to listen as to why you will not apply it.

I will think about this.

I am already making pictures and that may not be enough.

I will think some more.

However, why not think yourself and consider the round trip light travel.

Why is that a problem?

You say "it is not good for SR", but SR provides base details that you ignore.

Take these rockets going at 0.99c. One shines a light at the other. Since their relative speed is less than c, the light can clearly reach one from the other. The second one, in reply, shines a light back to the first. The light does reach back to the first.

I don't see any reason why it "matters" that both rockets are moving while the light travels from one to the other. Your derivation and conclusion made no sense to me.

edit to add: this mainstream reference shows the formula for composition of velocities, from which you can derive the formula used in these last few pages http://en.wikipedia.org/wiki/Special_relativity#Composition_of_velocities .

nauthiz

2009-Oct-30, 03:32 AM

Even if we do assume some sort of objective reference frame such that A and B see the distance between each other growing at 1.98c, the light would still reach. It's moving at 1c, and both A and B are moving at 0.99c, so the light is moving 0.01c faster. It'd take a while, but it would catch up.

(Not that that's actually the way the world works, mind you. . . )

dgavin

2009-Oct-30, 03:40 AM

I contradicted your misconceptions. I am sorry I couldn't convince you.

What misconceptions? I provided sources supporting all the specifics of what I was saying.

I hate to ask this Macaw, but are you simply gainsaying what anyone posts on this thread just to see the reactions?

The habit I see being displayed here is simple contridiction without much supporting material...

Even when others post material supprting thier positions, you provide none supporting yours. Simply saying "No, you are wrong" is not debate, nor is it helping the topics out. If anything it might be hindering them some.

macaw

2009-Oct-30, 03:46 AM

What misconceptions? I provided sources supporting all the specifics of what I was saying.

I hate to ask this Macaw, but are you simply gainsaying what anyone posts on this thread just to see the reactions?

You claimed that objects increase their gravitational characteristics by moving at higher speeds. This is wrong.

You claimed that by moving at near light speeds you can create black holes. This is also wrong.

Need more?

dgavin

2009-Oct-30, 04:16 AM

You keep jumping ahead with your ideas, based on your own interpretations - which are very ATM.

Stop. Take a breath. Start with showing why the bit bolded in the quote above is correct.

i.e. the bit where you think v one way plus v the other way makes a relative speed of 2v (to the things moving at v). (i.e. The bit where you think if v is > 0.5 c that means their relative speed is > c and therefore light can't go from one to another).

Why do you think you can just add velocities together like this?

And no, nothing you've posted yet backs this up. Don't bother linking back to that same post again. Doing that a 21st time won't help.

Edit to add:

i.e.

The mainstream view is that two rockets each going 0.99c in opposite directions (from the point of view of an observer "stationary" between them) will see each other receding at 0.999949498c, not at 1.98c as you would claim.

Thus light shone by one will reach the other.

Before you use your claim to "prove" other things, you need to "prove" this claim.

I'd like to add to this for abcdefg's benefit concering his round trip light senerio.

Given:

A) A lightsource has a specific known wavelength when the source is at rest.

b) .99c Ship 1 fires a beam of lightfrom this lightsource back at Ship 2 that is also receeding at .99c that has a reflector that will bounce the light beam back.

What Happens:

The light being emitted in opposite direction of the lightsources travel is red shifted by the lambda factor that relates to it's reltavistic Velosity of each observer. In the case of the stationary thrid observer he see's it redshifted by the lambda of .99c, in the case of ship 2 they see it red shifted by a lambda of a 0.999949498c velocity.

The light hits the mirror on ship 2 and rebounds back to ship 1. The stationary thrid observer he see's it redshifted by the lambda of 1.98c, and can use this to verify the already known appearant speed of the two ships from each other. Ship 1 eventualy receives the light back but it is now red shifted by lambda of a 1.999898996c velosity.

Change the senerios so that Ship 2 is Chasing Ship 1 at .99c and the.

Ship 1 emits light which is redshifted by the lambda of a 0.999949498c velocity (lambda of .99c for the third stationary observer) , Ship 2 meets the beam head on and reflects it back, blue shifting it by the beam by the lambda of a 0.999949498c velocity (blue shift lambda of .99c for the third stationary observer). When ship 1 receives it back he sees the original non red/blue shifted frequency of the light source. The Thrid observer also see's the same original frequency.

So there really isn't a problem with the round trip of a beam of light. It's just been redshifted (or red and blue shift in second senerio) by the velocity of the emitter, and of the rebounder by thier realtive velosities each other.

Does this help out with your thought experimant on the round trip light any?

macaw

2009-Oct-30, 04:27 AM

The light hits the mirror on ship 2 and rebounds back to ship 1. The stationary thrid observer he see's it redshifted by the lambda of 1.98c,

Absolutely not. This is not how the relativistic Doppler effect works. Plug in the correct speeds (hint: NOT 1.98c) into the correct formulas and see what you'll get.

Ship 1 eventualy receives the light back but it is now red shifted by lambda of a 1.999898996c velosity.

Incorrect. You need to use the correct speeds and the correct formulas. You aren't doing either.

Does this help out with your thought experimant on the round trip light any?

It increases the confusion.

dgavin

2009-Oct-30, 04:29 AM

You claimed that objects increase their gravitational characteristics by moving at higher speeds. This is wrong.

This is another source thats totaly contradics what you are saying See Item 5 here : http://helios.gsfc.nasa.gov/qa_gp_gr.html#particle

Mass, even relatavistic mass has gravity according to the equivlency priciple. I think I proved my point adqaquately that I'm not the one that is incorrect here.

macaw

2009-Oct-30, 04:32 AM

This is another source thats totaly contradics what you are saying See Item 5 here : http://helios.gsfc.nasa.gov/qa_gp_gr.html#particle

Mass, even relatavistic mass has gravity according to the equivlency priciple. I think I proved my point adqaquately that I'm not the one that is incorrect here.

The "mass" that intervenes in gravitation is the proper mass,not the relativistic mass. There is danger in trying to do physcs by web quote mining. You are making a well known rookie mistake (http://math.ucr.edu/home/baez/physics/Relativity/BlackHoles/black_fast.html) in terms of thinking that speed increases the gravitation and that you can form black holes by accelerating mass at relativistic speeds. You are also hijacking the thread by introducing new errors.

dgavin

2009-Oct-30, 04:38 AM

Absolutely not. This is not how the relativistic Doppler effect works. Plug in the correct speeds (hint: NOT 1.98c) into the correct formulas and see what you'll get.

Incorrect. You need to use the correct speeds and the correct formulas. You aren't doing either.

It increases the confusion.

You evidently did not read at all what I had written.

And I'm putting it into words, and not math, to try and explain what would happen to the person. Not to leave him floundering over the math without any explanations or a foundation.

And I think abcdefg is quite capable of answering a question to put to him on his own. There was no need for you to answer it for him. You are not him, so you have no idea what he might think of my explanation.

pzkpfw

2009-Oct-30, 04:41 AM

All of the GPS stuff from post #170 onwards is off topic for this thread, and treads on issues raised in previous threads by abcdefg that are now closed and not to be returned to unless abcdefg comes up with substantial new things to add.

I'm off home soon, so here I simply ask that everybody drop it.

Later, from home, I will split those posts off to a new thread - which will be locked.

macaw

2009-Oct-30, 04:43 AM

You evidently did not read at all what I had written.

And I'm putting it into words, and not math, to try and explain what would happen to the person. Not to leave him floundering over the math without any explanations or a foundation.

Unfortunately, what you put into words is very wrong. Do the calculations for the relativistic Doppler effect and you will manage to prove it to yourself.

NorthernBoy

2009-Oct-30, 07:32 AM

This ATM is to establish light speed is an absolute constant and not a constant to the frame.

YOu need to explain what you mean by absolute.

The "relative" claim in SR is an easy one to define, it says that any observer will see light travelling at c relative to themselves.

What do you mean by absolute for light? It is not at all clear what the statement means, so please define it before you proceed to try to prove it.

NorthernBoy

2009-Oct-30, 07:36 AM

No, I am using two rocket frames, a common launch frame and the acceleration equations.

I am looking at what happens when relativistic speeds are approached.

So now you admit that there are three frames, when you tried to deny it earlier?

Anyway, to answer the above point, that does not change the fact that you are not defining which frame your measurements are valid in. You are refusing to do it because you know that of you were accurate, you would be proven to be wrong.

Which leaves us with the inescapable conclusion that you are deliberately trying to obfuscate.

Really, if you had any point, you would not run scared from defining for EVERY observation, which frame it applied to. Again and again, you are failing to do it.

NorthernBoy

2009-Oct-30, 07:37 AM

OK, in this sense it is the same but the total relative v > c though.

So why did you deny it at first? Why do you post things that you know not to be true?

NorthernBoy

2009-Oct-30, 07:39 AM

I am not trying to omit it.

You require that and I do not.

So why not include it? You know that omiting it allows you to conflate frames, and allows you to confuse who is seeing what.

If you were able to stop obfuscating like this, and still prove something, then you would do so.

Please, grow up, do what you need to do to prove your point, or retract your stupid claims. If you can't do it in the required way, you have nothing.

slang

2009-Oct-30, 08:02 AM

abcdefg, I've shown (http://www.bautforum.com/1610781-post140.html) you, using your own definitions, that you are using a third [collection of] observer[s]. As you've been told by others on page 1. Or are we now going to require your special definition of the verb "use"? And "observer"?

dgavin

2009-Oct-30, 08:02 AM

Unfortunately, what you put into words is very wrong. Do the calculations for the relativistic Doppler effect and you will manage to prove it to yourself.

Very well.

Ship to Ship round trip:

Given:

previous parameters of OP

and

Frequency of source (if at rest) = 540THz

Red shifted frequency detected at ship 2 = 3.395Thz

Red shifted frequency detected after reflected back to ship 1 = 21.346Ghz

If you then take the difference of the starting and ending frequencies, and reverse the calculation, you get a calculated velocity of 1.999898985c.

Not a perfect match but with my calculators rounding it's explainable, and close enough to prove my point.

However I will admit I make a mistake in the second scenario I posted. there would be no red/blue shift for Ship to ship observations at all, as the are moving the same direction at same velocity. My bad there.

HenrikOlsen

2009-Oct-30, 08:27 AM

OThis ATM is to establish light speed is an absolute constant and not a constant to the frame.

So your fundamental idea is that light speed varies with the velocity of the observer's frame, is that it?

And you're trying to establish this with a thought experiment?

And you're doing this despite the constancy of light speed regardless of observer's frame is what every experiment done yet to measure it has shown?

If that is so we could have saved 200 wasted posts that were addressing the consequences of your mistake rather than the actual mistake.

NorthernBoy

2009-Oct-30, 08:42 AM

However, why not think yourself and consider the round trip light travel.

Why is that a problem?

It is not a problem at all. For two objects each moving at 0.99c relative to me sitting here observing, the light can bounce back and forwards quite happily between them. You need to tell us why you believe that it is not so.

When the light is travelling left to right, it is catching the rightward particle at 0.01c. After it catches it up, and bounces off, it is travelling to the left at c, and so is catching the leftward particle at 0.01c, and so obviously will eventually catch it.

In order for it not to be able to make the round trip, I'd need to see the light travelling more slowly than c, but this is not what happens.

And yes, we've observed the light emitted by particles which are travelling at very close to c relative to the lab. The light travels at c relative to the lab, just as we'd expect.

NorthernBoy

2009-Oct-30, 08:48 AM

Mass, even relatavistic mass has gravity according to the equivlency priciple. I think I proved my point adqaquately that I'm not the one that is incorrect here.

No-one working in physics would nowadays use relativistic mass as a concept. Mass means rest mass, always.

I know that in the 50s, a different formulation was used, but since what you call mass is identical to what everyone else calls energy, and since we already have a word for that, relativistic mass was done away with.

You are therefore using terminology in a nonstandard way, which is going to cause confusion.

macaw

2009-Oct-30, 09:49 AM

Very well.

Ship to Ship round trip:

Given:

previous parameters of OP

and

Frequency of source (if at rest) = 540THz

Red shifted frequency detected at ship 2 = 3.395Thz

How did you calculate this value? Please show the symbolic formula. A cursory look says it's wrong.

Red shifted frequency detected after reflected back to ship 1 = 21.346Ghz

If you then take the difference of the starting and ending frequencies, and reverse the calculation, you get a calculated velocity of 1.999898985c.

Definitely wrong.

NorthernBoy

2009-Oct-30, 10:01 AM

So your fundamental idea is that light speed varies with the velocity of the observer's frame, is that it?

Unfortunately, he's now retreated again to a position that no-one can understand. I suspect that he'll launch a new and more convoluted idea from there soon.

His latest assertion, that light speed is absolute, needs to be defined, as so far as I can see, it means nothing. Relative speed is easy to talk about, it means how fast something moves past something else, or how fast an observer sees something.

What does absolute speed mean, though? Does he mean that there exists a preferred frame, in which light is always c, and that frames moving relative t that see something different? Does he mean that it is always c relative to the source, or something different?

If something different, what?

Of course he will not answer this in any clear way. The best we'll get is some trite response such as "not relative to anything, absolute", as he knows full well that if he were to define his terms and his observers, then disproving his foolishness is just the work of a few lines of text.

I've never seen another proponent argue in such bad faith, which is really saying something.

HenrikOlsen

2009-Oct-30, 10:10 AM

However, why not think yourself and consider the round trip light travel.

Ok. Done.

These are two images of the same situation, I changed their velocities relative to the Earth to be .5c in order to fit things on the page when I use the standard scale for length that makes c=1.

Note that neither rocket is accelerating at any point after t=0 as that only serves to utterly confuse things, they did their acceleration some time in the past and they coordinated it so they pass the earth simultaneously.

Note that simultaneity actually applies in this special case. As the distance is 0, observers on both rockets and on the Earth all agree that the rockets passed each other at the same time they both passed the earth, this defines t=0 for each of the three observers.

The people in the rockets are shining flashes at precise previously agreed on intervals counted by clocks they each calibrated to an atomic clock they themselves built to the same specifications, and they have agreed that the flashes will shift frequency after receiving the first flash from the other rocket.

The images show displacement along the x axis, time moving down the y axis

Red is the timeline of Rocket 1, Green of Earth and Blue of Rocket 2.

The magenta lines are light emitted by the rockets before they see anything from the other, Orange is light emitted after.

As mentioned before, scale of the x axis is set so light moves at x=y.

Here's the first image, it shows things seen from the Earth, you'll see that the observer on the Earth can confirm that the flashes are sent by clocks that are in agreement.

http://www.bautforum.com/attachment.php?attachmentid=11132&stc=1&d=1256897307

Here's the same situation seen from Rocket 1.

http://www.bautforum.com/attachment.php?attachmentid=11134&stc=1&d=1256897611

Note that the world lines for the flashes from R2 are what the observer in R1 can work back to given when they were observed.

Note that the world line for Rocket 2 was derived from the common position of R1, R2 and the Earth at t=0 and the intersection of the world line of the first sent flash and the world line back from the first received orange flash, it was not calculated from any formula.

When I measure the relative velocity of R2 seen from R1 in the image I see it's 0.8c.

If we use the formula previous mentioned for the addition of relative velocities, (v1+v2)/(1+v1v2/c^2)=1/(1+0.5c*0.5c/(1c*1c))=1c/1.25=0.8c

You'll note that the formula actually agrees with what R1 will observe.

Incidentally, this is one way of deriving that formula as being the natural consequence of c being constant in all inertial frames.

NB, the two images are not to the same scale except that the ratio between time and distance is the same on both, do not try to transfer measurements from one to the other.

dgavin, please note that neither time dilation, length contraction nor relativistic mass were used in this demonstration, though it does demonstrate time dilation too, since R1 sees the flashes from R2 as being sent by a clock that is going slower even though Earth can inform both rockets that their clocks agree.

captain swoop

2009-Oct-30, 10:27 AM

Dgavin and Macaw, Address the OP. If you want to continue your argument then start your own thread, don't hijack this one

abcdefg

2009-Oct-30, 02:51 PM

Light will be used to map out space such that it can be shown total v > c.

The initial condition this thought experiment is two rockets A and B are in the same frame.

Each will burn BT at a one along the positive x-axis and one along the negative x-axis such that each acquires a speed of .99c.

The distance between them after the burn is 2* c^2/a [ cosh(a BT/c) - 1 ]. Please see the equations for uniform acceleration at this link.

Further set K = 2* c^2/a [ cosh(a BT/c) - 1 ] and K is a known constant.

http://www.bautforum.com/against-mainstream/95749-uniform-accelerated-motion-under-sr.html

Here is a picture at this point.

Picture1 (http://www.proofofabsolutemotion.com/images2/B1.gif)

Some discussion is necessary. Please note. One cannot blindly claim you may approach 0.99c. In particular, if the original launch frame had been proceeding along the positive x-axis at .5c already, then rocket B could not attain .99c relative to the launch frame. It would have been limited to c - 0.5c. But, how can you know this limit in advance? You cannot because you would then know the absolute motion of the launch frame and that is not possible at this time. So, this thought experiment will allow this jump, but this condition needed to be brought to the forefront.

Next, both of these 0.99c speeds are reliable because of the acceleration equations.

The goal is to shoot a laser in order to map the distance of the two rockets while in motion to determine where v > c. Rocket A will shoot a laser to rocket B and B will reflect that laser back to A. A will time the event in its proper time.

Below is the picture thus far.

Picture2 (http://www.proofofabsolutemotion.com/images2/B2.gif)

Now, for light to catch up with B, light must travel the current distance between A and B which is K plus the distance B travels through space.

Thus, while light moves ct, B moves 0,99ct. So, light must travel

ct = K + 0.99ct.

ct - 0.99c = K

t = K/0.01.

Below is the picture thus far.

Picture4 (http://www.proofofabsolutemotion.com/images2/B4.gif)

For light to bounce back to the original emission point in space, when A emitted the laser, it will take another t.

Thus, a laser emitted from A to B and back to the emission point will take a time t.

Now, is A still at the point of light emission? No, it is not.

In time 2t, A moved left 2*0.99c. Thus, light has still not made it back to A, it has farther to go.

Below is the picture thus far.

Picture5 (http://www.proofofabsolutemotion.com/images2/B5.gif)

To calculate the distance light traveled to get to the emission point, it is

2(K + vt) = 2( K + 0.99cK/0.01c) = 200K

But, light has not yet reached back to rocket A. However, 200K is the distance we would expect light to travel for 2 frames with relative motion exactly 0.99c.

Picture6 (http://www.proofofabsolutemotion.com/images2/B6.gif)

Light still has further to go to catch A.

During the travel of light to strike B and back to the emission point, A traveled a distance 2*vt or 1.98t = 1.98* K/0.01 = 198K.

So, light must now travel a distance 198K plus the distance rocket A moves away from the light in time t.

Thus,

ct2 = 198K + vt2 = 198K + .99c

t2 = 198K/0.01c

The additional distance light traveled is 198K + vt2 = 198K + 0.99c(198K/0.01c) = 19800K.

Picture7 (http://www.proofofabsolutemotion.com/images2/B7.gif)

But, this is exactly what we would expect light to travel when catching an object moving at 0.99c.

Thus, to sum up the complete round trip of light, light travels a total distance such that the total relative velocity between rocket A and B is 1.98c.

MDT-1

2009-Oct-30, 03:03 PM

Very nice post, abcdefg. :clap:

abcdefg

2009-Oct-30, 03:36 PM

You say "it is not good for SR", but SR provides base details that you ignore.

Take these rockets going at 0.99c. One shines a light at the other. Since their relative speed is less than c, the light can clearly reach one from the other. The second one, in reply, shines a light back to the first. The light does reach back to the first.

I don't see any reason why it "matters" that both rockets are moving while the light travels from one to the other. Your derivation and conclusion made no sense to me.

edit to add: this mainstream reference shows the formula for composition of velocities, from which you can derive the formula used in these last few pages http://en.wikipedia.org/wiki/Special_relativity#Composition_of_velocities .

The composition of velocity applies a composition of R of S which necessarily injects a 3rd party observer. This derivation does not apply a metric from the A observer to the B observer and back. Light is a fixed metric that can be used in timing as well as distance measurements because it is a contant c regardless or any source motion.

Yes, it is correct that light will strike either A or B and any moving body in the universe because any individual cannot exceed c.

However, when measuring the distance light travels between A and B, light travels too far to describe a total relative v < c. The below post is a more detailed analysis.

http://www.bautforum.com/1611268-post201.html

abcdefg

2009-Oct-30, 03:37 PM

Even if we do assume some sort of objective reference frame such that A and B see the distance between each other growing at 1.98c, the light would still reach. It's moving at 1c, and both A and B are moving at 0.99c, so the light is moving 0.01c faster. It'd take a while, but it would catch up.

(Not that that's actually the way the world works, mind you. . . )

Agreed.

abcdefg

2009-Oct-30, 03:45 PM

You keep jumping ahead with your ideas, based on your own interpretations - which are very ATM.

Stop. Take a breath. Start with showing why the bit bolded in the quote above is correct.

i.e. the bit where you think v one way plus v the other way makes a relative speed of 2v (to the things moving at v). (i.e. The bit where you think if v is > 0.5 c that means their relative speed is > c and therefore light can't go from one to another).

Why do you think you can just add velocities together like this?

And no, nothing you've posted yet backs this up. Don't bother linking back to that same post again. Doing that a 21st time won't help.

Edit to add:

i.e.

The mainstream view is that two rockets each going 0.99c in opposite directions (from the point of view of an observer "stationary" between them) will see each other receding at 0.999949498c, not at 1.98c as you would claim.

Thus light shone by one will reach the other.

Before you use your claim to "prove" other things, you need to "prove" this claim.

Oh, the bold was discounting Ritz' theory or emission theory.

I will be specific.

In emission theory, light moves with the frame and is a constant to the frame.

Thus, in the direction of travel light move c + va and opposite the direction it moves as a speed c - va, where va is the absolute speed of the frame.

So, let's look at rocket A. Light would emit from A in the opposite direction at c - .99c = .01c.

Therefore, light can never reach rocket B.

This is an additional proof that emission theory cannot implement causality and fails.

Maybe I was not clear, but I am absolutely against Ritz' theory because it fails to implement causality after 0.5c which is what I was trying to say.

There is only one possible behavior to light that implements causality and that is light is an absolute constant i.e. the light cone is absolute.

Since that is the case and that is what I believe, light emits from A at a constant c and thus will eventually catch up to rocket B moving at only 0.99c.

abcdefg

2009-Oct-30, 03:51 PM

I'd like to add to this for abcdefg's benefit concering his round trip light senerio.

Given:

A) A lightsource has a specific known wavelength when the source is at rest.

b) .99c Ship 1 fires a beam of lightfrom this lightsource back at Ship 2 that is also receeding at .99c that has a reflector that will bounce the light beam back.

What Happens:

The light being emitted in opposite direction of the lightsources travel is red shifted by the lambda factor that relates to it's reltavistic Velosity of each observer. In the case of the stationary thrid observer he see's it redshifted by the lambda of .99c, in the case of ship 2 they see it red shifted by a lambda of a 0.999949498c velocity.

The light hits the mirror on ship 2 and rebounds back to ship 1. The stationary thrid observer he see's it redshifted by the lambda of 1.98c, and can use this to verify the already known appearant speed of the two ships from each other. Ship 1 eventualy receives the light back but it is now red shifted by lambda of a 1.999898996c velosity.

Change the senerios so that Ship 2 is Chasing Ship 1 at .99c and the.

Ship 1 emits light which is redshifted by the lambda of a 0.999949498c velocity (lambda of .99c for the third stationary observer) , Ship 2 meets the beam head on and reflects it back, blue shifting it by the beam by the lambda of a 0.999949498c velocity (blue shift lambda of .99c for the third stationary observer). When ship 1 receives it back he sees the original non red/blue shifted frequency of the light source. The Thrid observer also see's the same original frequency.

So there really isn't a problem with the round trip of a beam of light. It's just been redshifted (or red and blue shift in second senerio) by the velocity of the emitter, and of the rebounder by thier realtive velosities each other.

Does this help out with your thought experimant on the round trip light any?

This is interesting. I never thought of applying the relativistic doppler effect such that the total shift in frequency exceeds would be expected for total v < c. It would be in fact be double.

I did it as a distance metric.

abcdefg

2009-Oct-30, 03:55 PM

YOu need to explain what you mean by absolute.

The "relative" claim in SR is an easy one to define, it says that any observer will see light travelling at c relative to themselves.

What do you mean by absolute for light? It is not at all clear what the statement means, so please define it before you proceed to try to prove it.

This means it is impossible to add or subtract to its speed in the vacuum of space. Therefore, if light is emitted by a relativistic particle and at the same time from the earth frame at the same time and position and direction, the two light beams would be coincident for all time t.

abcdefg

2009-Oct-30, 03:59 PM

So now you admit that there are three frames, when you tried to deny it earlier?

Anyway, to answer the above point, that does not change the fact that you are not defining which frame your measurements are valid in. You are refusing to do it because you know that of you were accurate, you would be proven to be wrong.

Which leaves us with the inescapable conclusion that you are deliberately trying to obfuscate.

Really, if you had any point, you would not run scared from defining for EVERY observation, which frame it applied to. Again and again, you are failing to do it.

I am not using a third observer. There is a launch frame but I do not use it for calculations.

This post clearly shows only two observers are used.

http://www.bautforum.com/1611268-post201.html

abcdefg

2009-Oct-30, 04:06 PM

So your fundamental idea is that light speed varies with the velocity of the observer's frame, is that it?

And you're trying to establish this with a thought experiment?

And you're doing this despite the constancy of light speed regardless of observer's frame is what every experiment done yet to measure it has shown?

If that is so we could have saved 200 wasted posts that were addressing the consequences of your mistake rather than the actual mistake.

I am not doing any of this above at all. If light is not an absolute constant, then light emission theory applies and light is constant to the frame. This is false and I can run this view into a constradiction vs causality.

Here are some quotes from Einstein. I take the same position he does.

Einstein: "I certainly knew that the principle of the constancy of the velocity of light is something quite independent of the relativity postulate and I considered what would be more probable, the principle of the constancy of c, as was demanded by Maxwell’s equations, or the constancy of c, exclusively for an observer sitting at the light source. I decided in favor of the first, since I was convinced that each light [ray] should be defined by frequency and intensity alone, quite independently of whether it comes from a moving or a resting light source.

http://philsci-archive.pitt.edu/archive/00001743/01/Norton.doc

Einstein wrote (Einstein Archive, 20-040; translation based on Stachel, 2002, p. 189):

Einstein: Your attempt to replace special relativity with the assumption that the velocity of light is constant relative to the source of light was first advocated by Ritz. This assumption is compatible with Michelson’s experiment and with aberration.

http://philsci-archive.pitt.edu/archive/00001743/01/Norton.doc

Additional Link (http://books.google.com/books?id=OAsQ_hFjhrAC&pg=PA189&lpg=PA189&dq=%22Your+attempt+to+replace+special+relativity+w ith+the+assumption+that+the+velocity+of+light+is+c onstant+relative+to+the+source+of+light+was+first+ advocated+by+Ritz%22&source=bl&ots=dXbWHoAa1H&sig=5op_mk28U243wzVzhDxaWTrI8Jk&hl=en&ei=HkjoSs_GKNCJtgeV7vmGBw&sa=X&oi=book_result&ct=result&resnum=1&ved=0CAgQ6AEwAA#v=onepage&q=%22Your%20attempt%20to%20replace%20special%20rel ativity%20with%20the%20assumption%20that%20the%20v elocity%20of%20light%20is%20constant%20relative%20 to%20the%20source%20of%20light%20was%20first%20adv ocated%20by%20Ritz%22&f=false)

Therefore, light is a constant c period, regardless and independent of any frame.

NorthernBoy

2009-Oct-30, 04:07 PM

Light will be used to map out space such that it can be shown total v > c.

You have again just posted a more complicated situation when pressed, rather than answering the questions, as you are required to do so. You have a very bad habit of doing this.

Rather than just trying to avoid them, please answer the questions, as they apply to the simpler examples which were being discussed earlier. The board requires that you do so, please stop avoiding them.

abcdefg

2009-Oct-30, 04:07 PM

So why did you deny it at first? Why do you post things that you know not to be true?

This is false and a comparison of apples and oranges.

NorthernBoy

2009-Oct-30, 04:10 PM

I am not using a third observer. There is a launch frame but I do not use it for calculations.

But I did, and when I did so, you objected.

Now, please answer the questions which you have been asked. Why not start with explaining why you believe that light cannot bounce back and forwards between to particles who an observer will see as having a speed of separation of over 1c.

Your refusal to deal with questions is in breach of the rules, your answering them is not optional if you wish to post here.

NorthernBoy

2009-Oct-30, 04:12 PM

This is false and a comparison of apples and oranges.

Please stop avoiding the questions which you are required to answer.

Your infantile games are not amusing, and you'd do yourself a favour if you dropped them, and fulfilled your obligations.

NorthernBoy

2009-Oct-30, 04:17 PM

This means it is impossible to add or subtract to its speed in the vacuum of space. Therefore, if light is emitted by a relativistic particle and at the same time from the earth frame at the same time and position and direction, the two light beams would be coincident for all time t.

That says nothing different to the standard SR claim that the speed of light is independent of the source, so I'll ask again, what do you mean by saying that it is absolute?

You are implying that you mean something different from the standard view that all observers will measure the speed of light to be c relative to their own frame.

So, please, what is it that differs in your "absolute speed" assertion?

abcdefg

2009-Oct-30, 04:18 PM

So why not include it? You know that omiting it allows you to conflate frames, and allows you to confuse who is seeing what.

If you were able to stop obfuscating like this, and still prove something, then you would do so.

Please, grow up, do what you need to do to prove your point, or retract your stupid claims. If you can't do it in the required way, you have nothing.

This is not true.

I depend on setting up these frame characteristics with the acceleration equations of SR.

This way I can depend on the 0.99c for each A and B and know these velocities are true.

So, I circumvent a third observer which you claim I need. I do not.

If I were sitting in rocket A for example, my eyes would tell me I am not moving, but the acceleration equations which are a reliable calculation of v tell me my actual speed is 0.99c though I cannot feel it.

Now, it is true the "earth frame" has some underlying motion that is unknown. It is estimated to be around 18.55 miles per second in the direction of the orbit around the sun. Obviously, the milky way is doing something, but this number is close.

Either way, the earth launch frame is moving very little when compared to 0.99c such that I can think of it as uniform motion with a very small error condition built into it ie around 9.96X10-5c

For this reason, I do not need to attach a third frame to operate on this problem.

abcdefg

2009-Oct-30, 04:20 PM

But I did, and when I did so, you objected.

Now, please answer the questions which you have been asked. Why not start with explaining why you believe that light cannot bounce back and forwards between to particles who an observer will see as having a speed of separation of over 1c.

Your refusal to deal with questions is in breach of the rules, your answering them is not optional if you wish to post here.

I do not know how many times I need to say I am not using a third frame.

Find it in this post please.

http://www.bautforum.com/1611268-post201.html

I am being extremely specific. I am only using two frames.

abcdefg

2009-Oct-30, 04:21 PM

abcdefg, I've shown (http://www.bautforum.com/1610781-post140.html) you, using your own definitions, that you are using a third [collection of] observer[s]. As you've been told by others on page 1. Or are we now going to require your special definition of the verb "use"? And "observer"?

I am not using a third frame for calculations period. Here are the calculations using light as a metric for distance. There are only two frames in use.

http://www.bautforum.com/1611268-post201.html

WayneFrancis

2009-Oct-30, 04:30 PM

You are applying the equations correctly.

Yes I know I'm applying the equations correctly AND I know you are not.

Somebody gave you equations you you just applied them.

Yes, I'd like to thank my high school physics teacher from 1986, Mr Perry.

You did not consider sending light signals back and forth to see the result.

What does "sending" light signals back and forth matter?

It doesn't matter how "long" the test takes to do. It will come up with the same measurement for velocity. IE I would send 2 light pulses from one rocket to the other. I would measure the time difference between the first pulses round trip and the 2 pulses round trip. I can then compare that to the time difference between when I released the first and second pulse and work out the relative velocity of the object the 2 light pulses bounced off of.

These equations do not work when light is emitted from A to B and back again.

Maybe not in your bizzaro universe but the universe the rest of us live in doing this measurement is pretty easy. In fact I don't even need to use 2 light pulses. I can use 1 pulse and measure the dopler shift of the light returning back to determine the speed of the object.

This post shows a total relative v > c.

no your posts only show a lack of understanding special relativity, a tendancy to switch frames of reference without regards to the implications of those frames and a clear ability to mix values from different frames up to produce a bogus result.

http://www.bautforum.com/against-mainstream/95693-relative-v-c-5.html#post1610759

abcdefg

2009-Oct-30, 04:30 PM

Unfortunately, he's now retreated again to a position that no-one can understand. I suspect that he'll launch a new and more convoluted idea from there soon.

His latest assertion, that light speed is absolute, needs to be defined, as so far as I can see, it means nothing. Relative speed is easy to talk about, it means how fast something moves past something else, or how fast an observer sees something.

What does absolute speed mean, though? Does he mean that there exists a preferred frame, in which light is always c, and that frames moving relative t that see something different? Does he mean that it is always c relative to the source, or something different?

If something different, what?

Of course he will not answer this in any clear way. The best we'll get is some trite response such as "not relative to anything, absolute", as he knows full well that if he were to define his terms and his observers, then disproving his foolishness is just the work of a few lines of text.

I've never seen another proponent argue in such bad faith, which is really saying something.

You can type whatever you want.

Since I came here I said the speed of light is a constant in space regardless of the opinion of the frame.

Perhaps we are not communicating and your interpretation is such that you believe my position on matters is changing, but I can assure you it is not.

Furthermore, out of the ATM's I've seen since I got here, I have supplied more math proof and math than all combined.

Since logic and math is the "real" language of physics, I suggest you communicate with me on those terms and that will clear up any misconceptions.

In math, it is right or it is wrong.

For example, you keep accusing me of using a 3rd observer and being dishonest. Yet, you will not find any evidence of a 3rd observer in the math calculations except to supply a launch frame and that is basically it.

The launch frame has almost nothing to do with the entire problem and is not used on calculating the distance metric for light travel between AS and B which is the point of this thread.

WayneFrancis

2009-Oct-30, 04:34 PM

Yes, but the launch frame is incidental and for bookeeping only.

But this is the ONLY frame in the scenario that will see the ships seperating at a speed of 1.98c which doesn't mean that light from one of those ships would have to travel at >c to reach the other. Once the light is released from one of the ships toward the other it will catch up to the other ship because the speed of the other ship is <c from any other point in space/time

You honestly don't think that if I'm traveling at .99c to the right and shoot a laser to the left that it will only travel at .1c to the left do you?

The real action is these rockets going in opposite directions at .99c with light sent to each.

I used the launch frame only to establish the individual acceleration equations to each A and B.

This way I can establish a known distance at the instant the burn stops and establish the v's of A and B in a predictable fashion.

Once this is achieved, light signals can be sent back and forth to prove the total relative v is > c.

The only way it v > c is in your bizzaro world where you don't understand the equations to use.

slang

2009-Oct-30, 04:37 PM

Some discussion is necessary. Please note. One cannot blindly claim you may approach 0.99c. In particular, if the original launch frame had been proceeding along the positive x-axis at .5c already, then rocket B could not attain .99c relative to the launch frame.

.5c with respect to what? Hey look! A fourth frame!

Geo Kaplan

2009-Oct-30, 04:43 PM

Since logic and math is the "real" language of physics, I suggest you communicate with me on those terms and that will clear up any misconceptions.

In math, it is right or it is wrong.

Taken at face value, those statements are correct. However, equations devoid of context can mislead. Equations presented in an inappropriate context will mislead. And if the reader insists on changing the context, then no communication will have occurred. So your prescription is not guaranteed to succeed. Indeed from the other exchanges in which you've participated, it's clear that it's largely futile. That's why my working hypothesis continues to be that you are the Samuel Beckett of physics.

WayneFrancis

2009-Oct-30, 04:54 PM

Ship A and Ship B are traveling towards point C from opposite direction.

Ship A and B measure point C approaching at .99c

Ship A and B measure each other approaching at 0.99994949750012625624968435937579c

Ship A and B pass each other at point c

After 10 seconds, in their local frame, Ship A shoots a 450nm laser Ship B.

abcdefg, given the above scenario please answer the following

1. How long does ship A think the round trip for the light takes, ie Ship A's frame of reference.

2. What is the frequency of the light when it returns to ship A?

3. What is the frequency of the light when it hits ship B, according to Ship B.

4. If there is a stationary observer at Point C what frequency would they see the laser light as?

5. How does any of these answers contradict with what SR says? IE How do any of these answers show a v > c?

czeslaw

2009-Oct-30, 04:58 PM

Here is a velocity addition formula:

v(1,2)=[v(1)+v(2)] / [1+v(1)v(2)/c^2]

http://en.wikipedia.org/wiki/Velocity-addition_formula

A third observer sees that distance between the two objects increases with a velocity 2c but an observer on the object sees it increases with 1c.

abcdefg

2009-Oct-30, 05:00 PM

Ok. Done.

These are two images of the same situation, I changed their velocities relative to the Earth to be .5c in order to fit things on the page when I use the standard scale for length that makes c=1.

Note that neither rocket is accelerating at any point after t=0 as that only serves to utterly confuse things, they did their acceleration some time in the past and they coordinated it so they pass the earth simultaneously.

Note that simultaneity actually applies in this special case. As the distance is 0, observers on both rockets and on the Earth all agree that the rockets passed each other at the same time they both passed the earth, this defines t=0 for each of the three observers.

The people in the rockets are shining flashes at precise previously agreed on intervals counted by clocks they each calibrated to an atomic clock they themselves built to the same specifications, and they have agreed that the flashes will shift frequency after receiving the first flash from the other rocket.

The images show displacement along the x axis, time moving down the y axis

Red is the timeline of Rocket 1, Green of Earth and Blue of Rocket 2.

The magenta lines are light emitted by the rockets before they see anything from the other, Orange is light emitted after.

As mentioned before, scale of the x axis is set so light moves at x=y.

Here's the first image, it shows things seen from the Earth, you'll see that the observer on the Earth can confirm that the flashes are sent by clocks that are in agreement.

http://www.bautforum.com/attachment.php?attachmentid=11132&stc=1&d=1256897307

Here's the same situation seen from Rocket 1.

http://www.bautforum.com/attachment.php?attachmentid=11134&stc=1&d=1256897611

Note that the world lines for the flashes from R2 are what the observer in R1 can work back to given when they were observed.

Note that the world line for Rocket 2 was derived from the common position of R1, R2 and the Earth at t=0 and the intersection of the world line of the first sent flash and the world line back from the first received orange flash, it was not calculated from any formula.

When I measure the relative velocity of R2 seen from R1 in the image I see it's 0.8c.

If we use the formula previous mentioned for the addition of relative velocities, (v1+v2)/(1+v1v2/c^2)=1/(1+0.5c*0.5c/(1c*1c))=1c/1.25=0.8c

You'll note that the formula actually agrees with what R1 will observe.

Incidentally, this is one way of deriving that formula as being the natural consequence of c being constant in all inertial frames.

NB, the two images are not to the same scale except that the ratio between time and distance is the same on both, do not try to transfer measurements from one to the other.

dgavin, please note that neither time dilation, length contraction nor relativistic mass were used in this demonstration, though it does demonstrate time dilation too, since R1 sees the flashes from R2 as being sent by a clock that is going slower even though Earth can inform both rockets that their clocks agree.

I agree these are correct the correct applications of handling 0,5c given a three observer model.

However, let's assume Rocket A accelerated to 0.99999999c and the person inside was sleeping during the whole thing and did not even realize he/she left the earth.

Earth never liked A very much so earth is not watching either A and earth was responsible for preprogramming the ship to accelerate to 0,.99999999c without A knowing. Actually, earth was mad at A for presenting ATM ideas.

Now, A wakes up after the burn and wonders what happened.

A applies relativity and concludes he/she can accelerate in any direction to 0.99999999c.

Is this true? Nope. Yet, SR absolutely allows it in these circumstances. SR is wrong here and will give A incorrect answers.

Now, one may argue if the earth frame is added, then everything will be OK.

Well, at the instant the earth and the ship were moving in the same direction as earth's orbit around the sun. This is estimated to be 18.55 miles per second. Now, again, A is limited without knowing it in the forward direction.

More generally, the application of SR is limited by the actual underlying unknown motion of the frame but SR does not know what this is.

So, SR is based on an error condition it has no way of determining and further cannot provide any guidance for a solution whatsoever.

Furthermore, this light distance travel in this post (http://www.bautforum.com/1611268-post201.html) refutes the above analysis.

Light travels to far for a velocity composition argument under SR.

abcdefg

2009-Oct-30, 05:26 PM

But I did, and when I did so, you objected.

Now, please answer the questions which you have been asked. Why not start with explaining why you believe that light cannot bounce back and forwards between to particles who an observer will see as having a speed of separation of over 1c.

Your refusal to deal with questions is in breach of the rules, your answering them is not optional if you wish to post here.

Light can bounce back and forth between any two observers in the universe regardless of speed.

Light emits at a constant c from the frame regardless of its motion and will find any observer because that observer is traveling v < c.

NorthernBoy

2009-Oct-30, 05:30 PM

You can type whatever you want.

Since I came here I said the speed of light is a constant in space regardless of the opinion of the frame..

A frame does not have an opinion.

Can you please stick to physics, rather than metaphysics or personification of inanimate objects?

Perhaps we are not communicating and your interpretation is such that you believe my position on matters is changing, but I can assure you it is not.

No, as I've mentioned previously, it is at times very ambiguous, and at these times, you will generally refuse to clarify. When I tell you, for example, that you have not said which frame a measurement is in, you tell me that this does not matter, rather than just clarifying which one you were speaking about.

You are doing it here now, too, by not explaining what you mean by the speed of light being absolute.

I'll ask again, can you please give a clear and concise explanation of what you mean by absolute speed, and how it differs from the relativistic tenet that whatever the speed of an observer, they will always measure that the speed is c relative to them?

captain swoop

2009-Oct-30, 05:32 PM

abcdefg.

OK so you are back to arguing the exact same points that were the OP of your previous ATM threads.

As you have had your go at that ATM I am closing this thread.

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