Adding Velocities with Special Relativity

In summary, the conversation discusses a practice problem in Hogg's text about interplanetary rocket race involving three rockets: X, Y, and Z. The problem requires finding the speeds of all three rockets relative to the rest frame of the finish line. The conversation also includes a hint given by Hogg for finding the velocity of a cantalope thrown by a moving object. The solution to the problem involves calculating each rocket's velocity with respect to the finish line, using the given equation, and following the same reasoning for each rocket. The final results are X_R = 0.9c, Y_R = 0.99447514c, and Z_R = 0.99970846c. The conversation also includes a
  • #1
Severian596
286
0
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:

[tex]w = \frac{u+v}{1+\frac{uv}{c^{2}}}[/tex]

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

Severian's Solution
Let [tex]X_R[/tex], [tex]Y_R[/tex], and [tex]Z_R[/tex] 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 [tex]X_R = 0.9c[/tex].

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

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.
[tex]Z_Y=\frac{Y_{R} + 0.9c}{1+\frac{(Y_{R})(0.9c)}{c^{2}}}=0.99970846c[/tex]
therefore
[tex]Z_R=\frac{Z_{Y} + 0.9c}{1+\frac{(Z_{Y})(0.9c)}{c^{2}}}=0.99998466c[/tex]

Final Results
[tex]X_R = 0.9c[/tex] (this required no calculation)
[tex]Y_R = 0.99447514c[/tex] (this required one calculation)
[tex]Z_R = 0.99998466c[/tex] (this required two calculations)
 
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  • #2
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 [tex]Y_R[/tex]?

[tex]Z_Y=\frac{Y_{R} + 0.9c}{1+\frac{(Y_{R})(0.9c)}{c^{2}}}=0.99970846c[/tex]

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

[tex]Z_Y=\frac{0.9c + 0.9c}{1+\frac{(0.9c)(0.9c)}{c^{2}}}=0.99447514c[/tex]

Same thing with the first equation, except the number will happen to work out the same because [tex]X_{R}=0.9c[/tex]
 
  • #3
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.
 
  • #4
Severian596 said:
Severian's Solution
Let [tex]X_R[/tex], [tex]Y_R[/tex], and [tex]Z_R[/tex] 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 [tex]X_R = 0.9c[/tex].
Right!
Rocket Y's velocity with respect to the finish line
We must calculate Y's velocity with respect to the finish line.
[tex]Y_R=\frac{X_{R} + 0.9c}{1+\frac{(X_{R})(0.9c)}{c^{2}}}=0.99447514c[/tex]
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.
[tex]Z_Y=\frac{Y_{R} + 0.9c}{1+\frac{(Y_{R})(0.9c)}{c^{2}}}=0.99970846c[/tex]
This is actually the answer for [itex]Z_R[/itex]. Note that the reasoning should exactly parallel what you did to find out [itex]Y_R[/itex]
therefore
[tex]Z_R=\frac{Z_{Y} + 0.9c}{1+\frac{(Z_{Y})(0.9c)}{c^{2}}}=0.99998466c[/tex]
No. An unneeded (and incorrect) step.
 
  • #5
Doc Al said:
Note that the reasoning should exactly parallel what you did to find out [itex]Y_R[/itex]
Excellent! Thanks for the helpful feedback. With your help I now see that there is no need to resolve [tex]Z_R[/tex] twice.

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

[tex]X_R = 0.9c[/tex]

[tex]Y_R=\frac{X_{R} + 0.9c}{1+\frac{(X_{R})(0.9c)}{c^{2}}}=0.99447514c[/tex]

[tex]Z_R=\frac{Y_{R} + 0.9c}{1+\frac{(Y_{R})(0.9c)}{c^{2}}}=0.99970846c[/tex]

Again thank you very much Doc_Al
 
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  • #6
looks good

Severian596 said:
Again thank you very much Doc_Al
My pleasure. (Your answers look good to me.)
 

1. How do you add velocities with Special Relativity?

In Special Relativity, velocities are added using the relativistic velocity addition formula: v = (u + w)/(1 + uw/c^2), where v is the resulting velocity, u is the velocity of one object, w is the velocity of another object, and c is the speed of light in a vacuum. This formula takes into account the time dilation and length contraction effects of Special Relativity.

2. Is the addition of velocities in Special Relativity the same as in classical physics?

No, the addition of velocities in Special Relativity is different from classical physics. In classical physics, velocities are simply added together. However, in Special Relativity, the relativistic velocity addition formula must be used to account for the effects of time dilation and length contraction.

3. What happens when the velocities being added are close to the speed of light?

If the velocities being added are close to the speed of light, the resulting velocity will also be close to the speed of light. This is because according to Special Relativity, the speed of light is the maximum speed at which anything can travel. Therefore, velocities approaching the speed of light cannot be added together in the same way as in classical physics.

4. Can velocities ever be added to exceed the speed of light in Special Relativity?

No, according to Special Relativity, the speed of light is the maximum speed at which anything can travel. Therefore, it is not possible to add velocities together and exceed the speed of light. This is known as the cosmic speed limit.

5. Why is it important to use Special Relativity when adding velocities?

It is important to use Special Relativity when adding velocities because it provides a more accurate understanding of how objects move at high speeds. In classical physics, the addition of velocities does not take into account the effects of time dilation and length contraction, which are essential to understanding the behavior of objects moving at or near the speed of light.

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