# Simple question about time dilation in accelerated reference frames

1. Nov 26, 2011

### enfield

If one wants to calculate the elapsed time from the perspective of an object A moving at velocity, v, for time, t, relative to a stationary object B, all you have to do is calculate:

$$\int_{t_o}^{t_f}\frac{t}{\gamma}$$ Of course, $$\gamma$$ has no dependence on t because v is constant, so we get: $$\int_{t_o}^{t_f}\frac{t}{\gamma} = \frac{t_f-t_o}{\gamma}$$

My question is if this can be plainly extended to accelerated reference frames where the velocity is changing. In other words is this equation valid for calculating elapsed time when velocity varies : $$\int_{t_o}^{t_f}\frac{t}{\gamma(t)}$$ where $$\gamma(t)$$ is the lorentz factor at time t?

I will try to make this more clear later. Any insight is appreciated.

2. Nov 26, 2011

### atyy

It doesn't work for an accelerated frame.

It does, however, work as the formula for the elapsed proper time along an accelerated worldline, where the time and velocity of the accelerated object are relative to an inertial frame. The quantity calculated is a geometrical invariant (ie. an accelerated frame would calculate the same value, except the formula is more complicated in an accelerated frame).

The elapsed proper time along any worldline defines what we mean by an ideal clock travelling on the worldline.

Last edited: Nov 26, 2011
3. Nov 27, 2011

### Passionflower

If we have the case of one dimensional constant proper acceleration and identify:

t = proper time for home stayer
tau = proper time of the traveler
alpha = constant proper acceleration

Then we get:
$$\Large \tau=c \;\;{\it arcsinh} \left( {\frac {t\alpha}{c}} \right) {\alpha}^{-1}$$
and
$$\Large t=c \;\; \sinh \left( {\frac {\tau\,\alpha}{c}} \right) {\alpha}^{-1}$$
I am trying to think what you think is wrong about

Last edited: Nov 27, 2011
4. Nov 27, 2011

### atyy

The formula works as the elapsed proper time for an accelerated observer in an inertial frame (not for an accelerated frame).

5. Nov 27, 2011

### Passionflower

Why do you think it does not work for an accelerated frame?

6. Nov 27, 2011

### atyy

The proper time for any worldline, inertial or accelerated, in any frame, inertial or accelerated, comes from integrating the root of ds2. Only in an inertial frame is ds=√(dt2-dx2)=dt√(1-v2).

7. Nov 27, 2011

### Passionflower

What do you think is the instantaneous velocity of the co-moving frame if not v(t)?
I think in case of one dimensional constant proper acceleration we have:
$$\Large v_{{t}}=\sqrt {{\frac {{t}^{2}}{{t}^{2}+{\alpha}^{-2}}}}$$
If we plug this into the Lorentz factor we get the following integral:
$$\Large \int _{0}^{{\it time}}\!\sqrt {1-{\frac {{t}^{2}}{{t}^{2}+{\alpha}^{-2 }}}}{dt}$$
Which becomes (if we assume c is not equal to 1):
$$\Large \tau=c \;\;{\it arcsinh} \left( {\frac {t\alpha}{c}} \right) {\alpha}^{-1}$$

Last edited: Nov 27, 2011
8. Nov 27, 2011

### atyy

It's the instantaneous velocity of the accelerated worldline relative to an inertial frame.

9. Nov 27, 2011

### Passionflower

Atty, if you think I am wrong perhaps you could construct the integral the way you think is right without using gamma?

10. Nov 27, 2011

### atyy

I think the OP's and your equations are right for an accelerated worldline described in an inertial coordinates.

For an accelerated worldline in an accelerated frame, we'd integrate the root of ds2=guvdxudxv, exactly the same, except that ds2≠dt2-dx2, since guv≠diag(1,-1,-1,-1) in an accelerated frame.

11. Nov 27, 2011

### enfield

I think I misused the term accelerated reference frame. An "accelerated worldline described in an inertial coordinates" was the situation I was wondering about. Sorry about that!

Anyway, I'm happy intuition is right in this case. Wikipedia wasn't making that connection very explicit to me.

Passionflower, thanks or those equations :) and the explanations of them. Yeah, the case of constant proper acceleration is all I was really interested in.