# Three Reference Frame Relativity Problem - Velocity Addition or Proper Lengths

Tags:
1. Oct 1, 2014

### mailman123

1. The problem statement, all variables and given/known data
Anna and Bob have identical meter sticks. Anna an observer on Earth, sees Bob traveling in a spaceship at 0.5c away from her. Bob leaves his spaceship in an escape pod, moving away from Earth and the spaceship at 0.1c with respect to the spaceship. How long is Bob's meter stick according to Anna?

2. Relevant equations
x' = gamma*x
x' = gamma*(x-vt)
u = (u' + v)/(1+u'v/c2)

3. The attempt at a solution
So first I used the velocity addition equation to find the velocity of the escape pod with respect to Anna (treating the velocity of the escape pod as u' and the velocity of the spaceship with respect to Anna as v). With this, I found that the velocity of the pod relative to Earth was u=0.57c. I then used the length contraction equation (since the meter stick is proper length in the frame of Bob's shuttle), using u as my new v. This got me one answer, 0.821m.

Then, to check my answer I decided to use length contraction and time dilation to find the answer. First I converted the length of the meter stick from the pods frame to the shuttles and found it to be 0.995m. I then used this to find the time. I was a little iffy on whether I could still do the simplified length contraction equation for my second transformation, so I actually tried using both and neither "worked" with the above answer. I then tried using the time I got (cant find the numbers in my mess of work but it was ~3*10^-10 I think) and the 0.995m in the full lorentz transformation. This got me a different answer, a good .02-3 off. I also tried using just the x'=gamma*x again with similar issues.

I kind of lost my work for the end of the second method, but which ever way I chose for the second method, my answer never seemed to match my first. Did I make a math error in the first approach? Or is one of the approaches invalid?

2. Oct 1, 2014

### Orodruin

Staff Emeritus
If I understand your second approach correctly, this is what you tried:
1. Compute the length of the rod in the spaceship frame by length contraction of the rest length in the pod frame.
2. Compute the length of the rod in the Earth frame by length contraction of the contracted length in the spaceship frame.
This is generally not allowed because the length contraction formula is only valid when going from the rest frame of an object to a frame that is moving relative to it (i.e., only valid for step 1). This is based on the way that the length contraction formula is derived. Your first method is correct.

Note: The $\gamma$ factor for the combined velocity of the pod in the Earth frame should be $\gamma_1\gamma_2(1+v_1 v_2)$. Your second method (if it is what I described above) would give you $\gamma_1 \gamma_2$ so it would be off by a factor $1+v_1 v_2 = 1+0.1\times 0.5 = 1.05$. In your answer that would make a difference of 0.04 which seems a bit bigger than what you are quoting...

3. Oct 1, 2014

### mailman123

Yeah, upon rereading my post the second part was a little confusing. I actually tried doing the general Lorentz transformation for the second step, but got lazy and decided I'd just do length contraction twice because "length is length! Right?" . So all this confusion because I was too lazy :D. Thanks for your help :D.

PS I managed to find my old answer and it actually was 0.86 so it was .04 off like you suggested it should be.

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted