# Relativity (Spaceship Problem)

Gemini_Cricket

## Homework Statement

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.

## Homework Equations

Time dilation and length contraction formulas:
L=vΔt
L'=vΔt'
Δt'=γΔt

## 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.

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?

Homework Helper
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.

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.

Homework Helper
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?

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

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?

Homework Helper
You defined the variables when you said:
Ship 2 in it's own frame should be a length of L' = vΔt'.

##\small v\Delta t## is, indeed the length of ship2 in ship1's frame.
Is that the length you are asked to find?

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...

Homework Helper
It's just L' = L/γ though
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.

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.

Mentor
And therefore, 2 would be larger to 2 than to 1.

Yes. Good reasoning.

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.

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?

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.

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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.

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.