Special relativity problem

If a spaceship accelerates at a set amount (say 1g) for certain amount of time (say 1 year) from Earths point of view what length of time do the crew measure?

Right, earths frame of reference the velocity of the spaceship:

v(t) = at

Am I right in thinking the time period experienced by the crew would be this:



I figured I should substitute v to get a term that is just a function of t and integrate it? I keep getting nonsense answers...

Like so:



Negative time??

The problem is that you are trying to lorentz transform a euclidean acceleration which isn't a 4 vector and is thus not well behaved under lorentz transforms. You need to consider proper acceleration i.e. acceleration wrt proper time. This yields Rindler coordinates.

Rindler coordinates? Never heard of them. Are they related to the maths I wrote in my above post?

(It's late so apologies if I'm being dense)

yes, what you said is very much the same . I suppose my answer was more of a mathematician answer
They're actually quite interesting. They are the special relativity equivalent of Schwarzchild coordinates and they even have an event horizon! In fact, the actual event horizon on black holes looks so much like Rindler coordinates we can start doing quantum mechanics on the horizon (well, QFT) and Hawking radiation just plops out.

Your problem is right at the beginning with the equation v(t) = at


Think about what this means. This means that the observer on earth sees the spaceship moving with a constant acceleration. However, this cannot be possible as if you were to wait long times you would eventually get v > c! So really the acceleration cannot be constant, it is a function of time. Furthermore, the value of the acceleration is dependent on the frame you are observing. You can see this by taking:

$x' = \gamma(x-vt)$

And differentiating it twice with respect to t' (messy), in the x direction, you get:

$a' = \frac{1}{\gamma^3(1-uv/c^2)^3}a$


Now the trick to solving questions such as the one you asked is to define an instantaneous rest frame moving at v(t) along with the spaceship. This makes u = v and t' = tau (the proper time) in the above equation giving:


$a' = \frac{1}{(1-v^2/c^2)^{3/2}}a = \frac{dv'}{d\tau}$

Rearranging we get:

$\frac{dv}{dt} = \frac{dv'}{d\tau}(1-v^2/c^2)^{3/2}$

so

$\frac{dv}{d\tau} = \frac{dt}{d\tau}\frac{dv'}{d\tau}(1-v^2/c^2)^{3/2} = (1-v^2/c^2)\frac{dv'}{d\tau}$



Integration gives:

$v(\tau) = ctanh(v'(\tau)/c)$

Using this value of v we can now write:

$\frac{dt}{d\tau} = (1-v^2/c^2)^{-1/2} = cosh(\frac{v'(\tau)}{c})$

You can integrate this depending on the physical situation. In the case where

$v' = f\tau$

with f a constant, you get:

$t = t_{0} + \frac{c}{f}sinh({f\tau}{c})$



So in short, it's nowhere near as straightforward as you thought it would be!

However, this cannot be possible as if you were to wait long times you would eventually get v > c! So really the acceleration cannot be constant, it is a function of time.

Ok... That is a lot more complicated than I thought!

Using an instantaneous rest frame moving along with the spaceship is the bit I need to get my head round.

When you say:



Is this assuming the spaceship produces constant thrust? In order to make my problem simpler I'd assumed the thrust of the engines increased as the (relative?) mass increase - and I'd make sure I put values in that wouldn't take the velocity above c.

I can see that this wouldn't make a whole lot of sense as a tactic though, because you could still accelerate the ship reasonably (just less than the constant) when you reached maximum thrust, and it’s like the problem above.

Would forcing the acceleration to be constant (as long as you only did it while v < c) not be possible?

Thanks a lot for your answer. I'm going to have to think about it for a while...

Aha! I just realised what you did wrong.

Because I too was thinking that you should be able to assume a = constant and still get a positive answer. Furthermore, this problem should tend towards the real answer when you're dealing with very small times (and hence velocities).

Your problem is you got the equation the wrong way round. It should be:

$d\tau = (1-v^2/c^2)^{1/2}dt$

Whereas you essentially put a -1/2 as the exponent. This is why it is extremely important to define your symbols carefully and understand what they mean.

Look at the equation you wrote at the top. When t' = tau, i.e. when you're moving in a frame with the spaceship such that the spaceship experiences proper time tau, you find that the proper time is greater than the coordinate time. But in reality the person travelling really fast ages less compared to the relative, stationary observer. So that equation doesn't make sense.


With the equation in the correct form, integration gives:

http://www.wolframalpha.com/input/?i=integrate+%281-%28a%5E2t%5E2%29%2Fc%5E2%29%5E1%2F2+dt

Which makes more sense. It's still wrong though, for the reasons described earlier. The real answer requires the use of the instantaneous rest frame. What is so useful about this frame is that the spaceship is not moving in it. So then you can define a constant acceleration without running into any problems from lorentz transformations.

Ah! I can't belive I didn't check that...

Ok, that makes sense. But yes, I was oversimplifying how acceleration works in relativity.