Relative Relativistic Velocities

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Discussion Overview

The discussion revolves around the concept of relative relativistic velocities, specifically how to calculate the speed of one spaceship as observed from another spaceship's frame of reference when both are moving at 0.51c in opposite directions. The scope includes theoretical considerations and mathematical reasoning related to relativistic velocity addition.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • One participant expresses confusion about calculating the speed of spaceship B in A's frame of reference.
  • Another participant suggests that the concept of relativity implies that speeds cannot simply be added, as this would lead to superluminal speeds.
  • A participant provides the relativistic velocity addition formula and calculates the speed of B in A's frame, arriving at a specific value.
  • Another participant reiterates the use of the velocity addition formula and presents a similar calculation, questioning an earlier solution while arriving at the same result.
  • A participant discusses the labeling of the spaceships and frames of reference, suggesting a different approach to the problem while maintaining that both labeling methods are valid.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the best approach to the problem, and multiple viewpoints on the calculation and labeling of frames remain present throughout the discussion.

Contextual Notes

The discussion includes various assumptions about the definitions of velocity in different frames and the implications of relativistic effects, which are not fully resolved.

Chinkylee
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Relative Relativistic Velocities!

Suppose 2 space ships are moving at 0.51c, one moving left, the other one moving right.
A<------- -------->B
0.51c 0.51c

the speeds measured from a stationary observer. What is the speed of B in A's frame of reference?

Been thinking about the answer but can't find anything so confused now.
 
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Oh, i think you will find your answer by saying everything remains relative. I know that some people will state that you need to add the speeds up from both frames, but if this was to hold, both of them would find each other moving away from each other faster than light.
 


The speed of B in A's frame of reference is given by
v=(.51+.51)c/(1+.51X.51).
 


Google the phrase "relativistic velocity addition". This is well-known and there is a lot of material available.
 


The velocity addition formula with c=1 is w=(u+v)/(1+uv). In this case, u=-0.51, w=0.51, and v is what we're trying to find. So solve for v: w+wuv=u+v, w-u=(1-wu)v, v=(w-u)/(1-wu)=0.81.

Clem's solution "w unknown, u=v=0.51" looked wrong to me, but I'm getting the same result using "v unknown, w=-u=0.51".
 


Fredrik said:
The velocity addition formula with c=1 is w=(u+v)/(1+uv). In this case, u=-0.51, w=0.51, and v is what we're trying to find. So solve for v: w+wuv=u+v, w-u=(1-wu)v, v=(w-u)/(1-wu)=0.81.

Clem's solution "w unknown, u=v=0.51" looked wrong to me, but I'm getting the same result using "v unknown, w=-u=0.51".
Look at the way u, v and w are defined in terms of the diagram here, with A moving at v to the right relative to B and B moving at u to the right relative to C, and A moving at w to the right relative to C. Now relabel the spaceship on the left in the OP's diagram as "C", relabel the spaceship on the right as "A", and have the observer in the middle be "B", like so:

C<------- B -------->A

Then shift to C's rest frame, where both B and A are moving to the right, with A having a higher speed than B:

C ------->B ------------->A

...and this will then match the diagram on the page I linked to above, with B moving at u=0.51c to the right in C's frame, and A moving at v=0.51c to the right in B's frame.
 


I see. I used the labeling BCA instead of CBA. This is of course just as correct, but more complicated then necessary.
 

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