# Solving Lorentz Transform problem using only length contraction

1. Sep 16, 2014

### Tubefox

1. The problem statement, all variables and given/known data

A traveler in a rocket of length 2d sets up a coordinate system S' with origin O' anchored at the exact middle of the rocket and the x' axis along the rocket’s length. At t' = 0 she ignites a flashbulb at O'. (a) Write down the coordinates t'_F, x'_F, and t'_B, x'_B for the arrival of the light at the front and back of the rocket. (b) Now consider the same experiment as observed in a frame S relative to which the rocket is traveling at speed v (with S and S' arranged in the standard configuration). Use the Lorentz transformation to find the coordinates xF , t_F and xB , tB of the arrival of the two signals.

(This is the part my question is on) Repeat but do not use Lorentz transformation, just use length contraction and the fact that the speed of light is the same in every reference frame. Follow these steps:

1) Sketch the contracted rocket in the S frame at t=0. It moves with speed v.

(I did this.)

2) Write a formula for the time tF when the front of the rocket gets to xF

3) Write a formula for the time tF when the light gets to xF

4) Equate (2) and (3) to solve for xF and tF

5) Repeat for xB and tB

2. Relevant equations
$L = \frac{L_0}{\gamma}$
$x=\gamma(x' + vt')$
$t=\gamma(t' + \frac{vx'}{c^2})$

3. The attempt at a solution

Here's the prime frame measurements:

$x'_F=d \\ t'_F = \frac{d}{c}\\ x'_B = -d\\ t'_B= \frac{d}{c}$

And according to the Lorentz transform equations, here's what we have for S:

$x_F=\gamma d (1+\frac{v}{c})\\ t_F=\frac{\gamma d}{c}(1+\frac{v}{c})\\ x_B=\gamma d(\frac{v}{c}-1)\\ t_B=\frac{\gamma d}{c}(1-\frac{v}{c})$

That's all fine. The issue arises when I try to follow the procedure outlined in the problem statement. It appears to be nonsense. Why would it take the same amount of time for the rocket to get to $x_F$ as it would for the light to get to $x_F$? Even so, how can I solve for two unknowns based on one equation? Could someone, at the very least, clarify the wording a little bit?

Thanks.

2. Sep 16, 2014

### Staff: Mentor

I assume that's a typo. I think you meant the front and back of the ship.

In any case, why do you think it would take the same amount of time for the light to reach the front and back of the ship as seen from the S frame? Where's your analysis using length contraction only?

3. Sep 17, 2014

### Tubefox

No, that is not a typo. At least not on my part.

4. Sep 17, 2014

### Orodruin

Staff Emeritus
Where is the front of the rocket after a time t?
Where is the light sent towards the front after time t?
What relation must be satisfied for the time when the light hits the front of the rocket?

5. Sep 17, 2014

### Tubefox

$x_F$

Says right there in the procedure that it's at $x_F$

It's equal to the time it takes the front of the rocket to reach $x_F$

Last edited by a moderator: Sep 17, 2014
6. Sep 17, 2014

### Orodruin

Staff Emeritus
You misunderstand me. Those are questions I want you to answer by giving expressions in terms of d, c, t and v.

Edit: Without using Lorentz transformations, only referring to length contraction and the speed of light being constant.

Edit 2: And the reason the light takes the same time to reach $x_F$ as the front takes is because it is defined as the point where the light catches up ...

7. Sep 17, 2014

### Staff: Mentor

By definition, $x_F$ is the location of the event "light reaches front of rocket". So of course the front of the rocket and the light must be at the same location when the light reaches that point.

8. Sep 17, 2014

### nrqed

Where does it say that ?

9. Sep 17, 2014

### Staff: Mentor

Ah, so you're done? The answer is $x_F$ and $t_F$?

Seriously, you must solve for those coordinates in terms of the given parameters (d, v, c).