Adding Velocities with Special Relativity

[Severian596]

I'm independently studying SR right now. A practice problem in Hogg's text here, Chapter 4, problem 4-7 says (slightly modified):

In an interplanetary rocket race there are three rockets: X, Y, and Z. Slow team X is traveling in their old rocket at speed 0.9c relative to the finish line. They are passed by faster team Y, and X observes Y to pass at 0.9c relative to themselves. But team Y observes fastest team Z to pass Y's own rocket at 0.9c. What are the speeds of teams X, Y, and Z relative to the rest frame of the finish line?

Hint: The answer is not 0.9c, 1.8c, and 2.7c!

Just prior, Hogg gives the equation to find the velocity of a cantalope thrown by B who's moving with respect to rest frame A. The final measured velocity w of the cantalope (thrown by B at velocity v who is himself moving at velocity u) as measured by A is:

$$w = \frac{u+v}{1+\frac{uv}{c^{2}}}$$

I will state below how I solved the problem and would appreciate it if someone offers consent or denial, thanks!

Severian's Solution
Let $$X_R$$, $$Y_R$$, and $$Z_R$$ be the velocities of rockets X, Y, and Z with respect to the Rest frame finish line.

Rocket X's velocity with respect to the finish line
We are told at the start that $$X_R = 0.9c$$.

Rocket Y's velocity with respect to the finish line
We must calculate Y's velocity with respect to the finish line.
$$Y_R=\frac{X_{R} + 0.9c}{1+\frac{(X_{R})(0.9c)}{c^{2}}}=0.99447514c$$

Rocket Z's velocity with respect to the finish line
We must first calculate Z's velocity with respect to Y, then finally use Y's perceived velocity of Z to find Z's velocity with respect to X.
$$Z_Y=\frac{Y_{R} + 0.9c}{1+\frac{(Y_{R})(0.9c)}{c^{2}}}=0.99970846c$$
therefore
$$Z_R=\frac{Z_{Y} + 0.9c}{1+\frac{(Z_{Y})(0.9c)}{c^{2}}}=0.99998466c$$

Final Results
$$X_R = 0.9c$$ (this required no calculation)
$$Y_R = 0.99447514c$$ (this required one calculation)
$$Z_R = 0.99998466c$$ (this required two calculations)

 

[Severian596]

I have a specific question about calculating Z's velocity with respect to Y. Should I instead use 0.9c as Y's velocity instead of $$Y_R$$?

$$Z_Y=\frac{Y_{R} + 0.9c}{1+\frac{(Y_{R})(0.9c)}{c^{2}}}=0.99970846c$$

As written I used the same value for Y's velocity with respect to rest frame R as I did for Y's velocity with respect to Z. I have a feeling that's not correct, and that I should instead use

$$Z_Y=\frac{0.9c + 0.9c}{1+\frac{(0.9c)(0.9c)}{c^{2}}}=0.99447514c$$

Same thing with the first equation, except the number will happen to work out the same because $$X_{R}=0.9c$$

 

[Severian596]

I'll just bump my own thread once, to make sure if someone could please comment on it they get the chance. I'm not asking for much, just an evaluation of the equations to see if I set up the problem right. Please note my second post where I discuss possible corrections to the velocity of Z with respect to Y...but again, I'm not sure if they're correct.

thanks and I won't bump this post again.

 

[Doc Al]

Severian's Solution
Let $$X_R$$, $$Y_R$$, and $$Z_R$$ be the velocities of rockets X, Y, and Z with respect to the Rest frame finish line.

Rocket X's velocity with respect to the finish line
We are told at the start that $$X_R = 0.9c$$.

Right!

Rocket Y's velocity with respect to the finish line
We must calculate Y's velocity with respect to the finish line.
$$Y_R=\frac{X_{R} + 0.9c}{1+\frac{(X_{R})(0.9c)}{c^{2}}}=0.99447514c$$

Right!

Rocket Z's velocity with respect to the finish line
We must first calculate Z's velocity with respect to Y, then finally use Y's perceived velocity of Z to find Z's velocity with respect to X.

No, Z's velocity with respect to Y is given as 0.9c.

$$Z_Y=\frac{Y_{R} + 0.9c}{1+\frac{(Y_{R})(0.9c)}{c^{2}}}=0.99970846c$$

This is actually the answer for $Z_R$. Note that the reasoning should exactly parallel what you did to find out $Y_R$

therefore
$$Z_R=\frac{Z_{Y} + 0.9c}{1+\frac{(Z_{Y})(0.9c)}{c^{2}}}=0.99998466c$$

No. An unneeded (and incorrect) step.

 

[Severian596]

Note that the reasoning should exactly parallel what you did to find out $Y_R$

Excellent! Thanks for the helpful feedback. With your help I now see that there is no need to resolve $$Z_R$$ twice.

In summary I believe the answers are found through the following:

$$X_R = 0.9c$$

$$Y_R=\frac{X_{R} + 0.9c}{1+\frac{(X_{R})(0.9c)}{c^{2}}}=0.99447514c$$

$$Z_R=\frac{Y_{R} + 0.9c}{1+\frac{(Y_{R})(0.9c)}{c^{2}}}=0.99970846c$$

Again thank you very much Doc_Al

 

[Doc Al]

looks good

Again thank you very much Doc_Al

My pleasure. (Your answers look good to me.)