Simple question about time dilation in accelerated reference frames

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Main Question or Discussion Point

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:


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

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 : [tex] \int_{t_o}^{t_f}\frac{t}{\gamma(t)} [/tex] where [tex] \gamma(t) [/tex] is the lorentz factor at time t?

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

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  • #2
atyy
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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.
 
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  • #3
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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:
[tex]\Large \tau=c \;\;{\it arcsinh} \left( {\frac {t\alpha}{c}} \right) {\alpha}^{-1}[/tex]
and
[tex]\Large t=c \;\; \sinh \left( {\frac {\tau\,\alpha}{c}} \right) {\alpha}^{-1}
[/tex]
It doesn't work for an accelerated frame.
I am trying to think what you think is wrong about
enfield said:
[tex] \int_{t_o}^{t_f}\frac{t}{\gamma(t)} [/tex] where [tex] \gamma(t) [/tex] is the lorentz factor at time t?
 
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  • #4
atyy
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I am trying to think what you think is wrong about ...
The formula works as the elapsed proper time for an accelerated observer in an inertial frame (not for an accelerated frame).
 
  • #5
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The formula works as the elapsed proper time for an accelerated observer in an inertial frame (not for an accelerated frame).
Why do you think it does not work for an accelerated frame?
 
  • #6
atyy
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Why do you think it does not work for an accelerated frame?
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
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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).
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:
[tex]\Large v_{{t}}=\sqrt {{\frac {{t}^{2}}{{t}^{2}+{\alpha}^{-2}}}}[/tex]
If we plug this into the Lorentz factor we get the following integral:
[tex]\Large \int _{0}^{{\it time}}\!\sqrt {1-{\frac {{t}^{2}}{{t}^{2}+{\alpha}^{-2
}}}}{dt}[/tex]
Which becomes (if we assume c is not equal to 1):
[tex]\Large \tau=c \;\;{\it arcsinh} \left( {\frac {t\alpha}{c}} \right) {\alpha}^{-1}[/tex]
 
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  • #8
atyy
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What do you think is the instantaneous velocity of the co-moving frame if not v(t)?
It's the instantaneous velocity of the accelerated worldline relative to an inertial frame.
 
  • #9
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Atty, if you think I am wrong perhaps you could construct the integral the way you think is right without using gamma?
 
  • #10
atyy
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Atty, if you think I am wrong perhaps you could construct the integral the way you think is right without using gamma?
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
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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.
 

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