Does the presence of B change the velocity of C with respect to A?

[O Great One]

Let's say there are two observers A and C. C is moving .5c with respect to A. If we introduce a third observer which we will call B and B is moving at .2c with respect to A and is moving in the same direction as C and is inbetween A and C, and we now apply the velocity addition formula we get that C is now moving at (.5/1.06) or .47c relative to A.
So, does this mean that without B, C is moving at .5c relative to A and that with B, C is moving at .47 relative to A?
Does just the mere presence of B change the velocity of C with respect to A?

What am I missing?

 

[Meir Achuz]

You may be using a classical result that v_CA=v_CB+v_BA that is not true in relativity.
B has no effect on v_CA.

 

[Doc Al]

Let's say there are two observers A and C. C is moving .5c with respect to A. If we introduce a third observer which we will call B and B is moving at .2c with respect to A and is moving in the same direction as C and is inbetween A and C, and we now apply the velocity addition formula we get that C is now moving at (.5/1.06) or .47c relative to A.

How did you apply the velocity addition formula? It seems that you have assumed that if the speed of B with respect to A is 0.2c, then the speed of C with respect to B must be 0.3c. Not so.

So, does this mean that without B, C is moving at .5c relative to A and that with B, C is moving at .47 relative to A?
Does just the mere presence of B change the velocity of C with respect to A?

Of course not.

[oops... clem beat me too it]

 

[O Great One]

How did you apply the velocity addition formula? It seems that you have assumed that if the speed of B with respect to A is 0.2c, then the speed of C with respect to B must be 0.3c. Not so.

Yes. That is what I was assuming. So, does that mean that the velocity of C relative to B is such that when you apply the velocity addition formula you get a velocity of C relative to A of .5c?

 

[neopolitan]

I'm a visual sort of guy, and there may be others out there, so I have attached a jpg of velocity addition for a slightly different scenario.

The image is taken from the perspective of A, for whom B is moving at 0.25c and C is moving at 0.666c.

Look at the red dashed lines, these are B's equivalents to A's blue dashed lines.

Using those you can see that for B, C is moving at 0.5c and A is moving at (minus) 0.25c.

$$v_{cB} = \frac{v_{cA} - v_{bA}}{1 + \frac{v_{bA}}{c}.\frac{v_{cA}}{c}}$$

$$v_{cB} = \frac{0.66c - 0.25c}{1 + 0.25 * 0.666}$$

$$v_{cB} = 0.5c$$

Alternatively,

$$v_{cA} = \frac{v_{cB} - v_{bB}}{1 + \frac{v_{bB}}{c}.\frac{v_{cB}}{c}}$$

$$v_{cA} = \frac{0.66c - 0.5c}{1 + 0.5 * 0.666}$$

$$v_{cA} = 0.25c$$

cheers,

neopolitan

PS just to make it clear, v_{a[\sub] on the diagram means the velocity of A. However, that velocity has to be relative to something (in the diagram which is taken from the perspective of A, it is 0m/s, that's why it is a vertical line). In the equations I have added a capital letter indicating who the velocity is relative to, so vaA[\sub] would be the velocity of A relative to A, and vcA[\sub] would be the velocity of C relative to A.}

 

[Doc Al]

Yes. That is what I was assuming. So, does that mean that the velocity of C relative to B is such that when you apply the velocity addition formula you get a velocity of C relative to A of .5c?

Of course!

You can use the velocity addition formula to figure out the speed of C relative to B. (It turns out to be c/3, not .3c.)

 

[MeJennifer]

To stretch it a bit further, the generalized velocity addition is neither commutative nor associative (see for instance Ungar).