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.Those in the negative x-axis direction are limited to -1.99c.
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
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.
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.
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/arch.../01/Norton.doc
pleas read http://en.wikipedia.org/wiki/Velocity
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.
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.
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 < v_{0} + v cos(Θ) < c.
where v_{0} is the speed of the relativistic particle
and Θ is the angle to the direction of the particle/object's travel.
Answer log macaw
Q1, Q2 http://www.bautforum.com/against-mai...ml#post1610203
Q3 http://www.bautforum.com/against-mai...ml#post1610214
Q4, Q5 http://www.bautforum.com/against-mai...ml#post1610255
Q6 http://www.bautforum.com/against-mai...ml#post1610255
Q7 http://www.bautforum.com/against-mai...ml#post1610276
Q8 http://www.bautforum.com/against-mai...ml#post1610539
Q9 http://www.bautforum.com/against-mai...ml#post1610541
Q11: How did you get the answer? Show the steps, like in Q8.
The above is closing speed, not relative speed.Now, what is the total relative velocity differential?
|-c| + |c| = 2c.
Q12: Explain the difference between closin and relative speed so we can see that you are starting to learn.
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 .
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.
Another 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.
At light speed a object with mass has infinate energry, which also implies, infinate mass and infinate gravity.
Another fallacy.
Yet another incorrect statement.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,
Actually, they are.
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.
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 where you learned it. The irony is that you sometimes even link in the site. See the first formula in the derivation?I know my speed is v(BT) = c tanh(a BT/c) according to the acceleration equations.
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...
(scroll right)Code: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
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.
I don't see any Ice Giants.