# Thread: Relative v > c

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## Relative v > c

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.

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3. Originally Posted by abcdefg
Question 1: What is the relative speed of the rockets? 1.98c
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.

4. He's putting this issue under a third observer, here we go again

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Originally Posted by Kwalish Kid
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.

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Originally Posted by zerocold
He's putting this issue under a third observer, here we go again
No I am not. I am taking the position of either of the rockets.

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Originally Posted by Perikles
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 lists experiments with particlaes moving faster than .99c. Assume one is shot one way and one the other.

This is legitimate.

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Originally Posted by abcdefg
Taking the position of either rocket, the other rocket disappears since length contraction becomes 0.

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.

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Originally Posted by abcdefg
This calculation does not control the speed of the rockets. These rockets are at .99c.

For example, here 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.

10. 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

Edit, lol he came faster than i expected

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Originally Posted by abcdefg
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".

12. Originally Posted by abcdefg
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.

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Originally Posted by zerocold

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

Edit, lol he came faster than i expected
Faster than the speed of light

14. lol, i hope the debate will be civilized

These are cool debates, i must say

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Quote:
Originally Posted by abcdefg
Taking the position of either rocket, the other rocket disappears since length contraction becomes 0.
Originally Posted by macaw

Q1: Prove that "the other rocket disappears"
Q2: prove that "length contraction becomes zero"
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.

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Originally Posted by abcdefg
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.

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.

Now, a length of zewro means the rocket disappears.
Still bad juju. You failed Q1,2. Let's see what you do with Q3.

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Originally Posted by macaw

Wrong again.
Q3: Prove that "length contraction crosses into complex numbers"

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 )

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Originally Posted by macaw
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.

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Originally Posted by macaw
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.

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Originally Posted by abcdefg
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 [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".

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Originally Posted by abcdefg
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.

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Originally Posted by zerocold
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

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.

23. Originally Posted by abcdefg
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.
Originally Posted by abcdefg
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.

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Originally Posted by macaw

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 :-)
Neither rocket attained c. I thought I stated that clearly.

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

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Originally Posted by Perikles
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.

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Originally Posted by abcdefg
Neither rocket attained c. I thought I stated that clearly.
Then they definitely cannot attain 1.98 c, right?

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Originally Posted by abcdefg
[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.

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Originally Posted by zerocold
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".

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Originally Posted by macaw
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".

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?