# Homework Help: Relativity (Spaceship Problem)

1. Sep 5, 2013

### Gemini_Cricket

1. The problem statement, all variables and given/known data
Spaceship 1 passes spaceship 2 with a relative speed v. An observer in spaceship 1 measures a time interval ∆t for spaceship 2 to pass by. Find the length of spaceship 2 as measured in its own rest frame, i.e., find the proper length of spaceship 2 in terms of ∆t.

2. Relevant equations
Time dilation and length contraction formulas:
L=vΔt
L'=vΔt'
Δt'=γΔt

3. The attempt at a solution
Since I'm looking for L' in terms of Δt, I just plugged the equation for Δt' into the L' equation. It doesn't seem right to me though. I'm unsure if I need to do a Lorentz transformation for this problem, and I don't really know how to do one. Though looking at the velocity addition rule in my book there is also a Δx involved... I'm lost.

2. Sep 5, 2013

### Staff: Mentor

If spaceship 2 is traveling with speed v relative to observer 1, and it takes time interval Δt for spaceship 2 to pass this observer, as reckoned by observer 1 from his frame of reference, how long is spaceship 2? This is the contracted length of spaceship 2. How long is spaceship 2 in its own rest frame of reference?

3. Sep 5, 2013

### Simon Bridge

Welcome to PF;
The regular physics you are used to applies so long as you don't change reference frames.
When you change reference frames, you have to do a Lorentz transformation.

4. Sep 5, 2013

### Gemini_Cricket

Ship 2 in it's own frame should be a length of L' = vΔt'.

Here's what I've come up with. Using instead Δt=γΔt' and the above equation. I get L'=vΔt/γ.

5. Sep 5, 2013

### Simon Bridge

Check:
Are moving lengths bigger or smaller than stationary lengths?
L is the length of ship2 in ship1's frame - should it be bigger or smaller than L'?
Is L' bigger or smaller than L?

6. Sep 5, 2013

### Gemini_Cricket

Moving lengths are smaller. In ship 1's frame, ship 2 is moving, so L should be smaller than L'.

7. Sep 5, 2013

### Gemini_Cricket

I am very confused here. You say that L is the length of ship 2 in ship 1's frame. Wouldn't that just make the answer to the problem L = vΔt?

8. Sep 5, 2013

### Simon Bridge

You defined the variables when you said:
$\small v\Delta t$ is, indeed the length of ship2 in ship1's frame.
Is that the length you are asked to find?

9. Sep 6, 2013

### Gemini_Cricket

No, I am asked to find Ship 2 in Ship 2's frame.

Well I worked it out equating the velocities. It's just L' = L/γ though. I need it in terms of Δt, and I don't believe the previous step, L' = LΔt'/Δt, counts does it? That's in terms of both times...

10. Sep 6, 2013

### Simon Bridge

From your earlier definitions, since γ>1:
This says that the the length of ship2 in ship2's frame is shorter than the length of ship2 in ship1's frame. Is that what you meant to say?

Usually the primed variable is as observed from the unprimed frame.

11. Sep 6, 2013

### Gemini_Cricket

Why would ship 2 be shorter in ship 2's frame? I thought moving objects appeared smaller. Wouldn't 2 be stationary to 2, but moving relative to 1? And therefore, 2 would be larger to 2 than to 1. Also, 1 would appear smaller to 2? I am probably looking at this way wrong... I appreciate your patience.

My book is throwing me way off. It has these equations written backwards (it tells you on a side note)... Okay, now I'm pretty sure it has to be L' = vΔt/γ now that I look at it.

12. Sep 6, 2013

### Staff: Mentor

Yes. Good reasoning.

Try again. L' is the length of the spaceship (which is at rest in 2's frame of reference) as measured by 2. Look at your quote above. What does the equation for L' have to be if, as we know, γ is greater than 1?

13. Sep 6, 2013

### Gemini_Cricket

okay so it should be gamma used as a multiplier instead of a divisor.

I get L'={gamma}v{delta}t. I hope that's right because it makes sense to me after rereading this post.

Last edited: Sep 6, 2013
14. Sep 6, 2013

### Staff: Mentor

Yes. I know it seems confusing now, but it takes some getting used to. You'll get there. We all struggled with this stuff.

15. Sep 6, 2013

### Gemini_Cricket

Yeah I just learned a hard lesson to not pay attention to the equations in the book, because they need to be somewhat "derived" for each problem. Thanks for all the help. Its kind of funny, I had this as my first answer. Yet I erased it in lieu of my friend having a seemingly better explanation haha.

16. Sep 6, 2013

### Simon Bridge

That's right - the equations given to you only apply for a particular approach. You need to understand the approach to use the equations.

It's actually easier to derive the equations to the physics of what is going on that to try to apply definitions made by someone else.

If the equation you have is telling you something different to what the physics is telling you - then it is the wrong equation. It doesn't matter where you got the equation or what kind of reputation or experience they have or what material it's chiseled into. It's part of the power of science, the source of the core arrogance of science, and also of the core humility. It also means that you have a way of checking your work to see if you are right, without having to ask some authority.

For this problem: if you'd said something like L2 is the length of ship2 in ship2's frame, then L2' is the same length in the frame of someone who is moving wrt ship2. Then you get $\small L_2^\prime = L_2/\gamma$ and solve for $\small L_2$ consistent with the stock equations.

But since you defined your variables differently from how the text book uses them, you got mixed up.