Proper time in an acceleration frame

In summary, the conversation discusses the proper time, Δτ, in relation to coordinate time, t, in an inertial frame. The equation dτ = √(-gμνdxμ/dλdxv/dλ)dλ is used, with the substitution dλ = dt, to find the proper time. The equation Δτ = ∫√(-(-1+v^2/c^2))dt is then discussed, with the approximation Δτ ≈ Δt - (g^2*Δt^2)/(6c^2), where Δt = 2√(L/g). The conversation also mentions the use of hyperbolic trigonometric functions, and points out that the given answer
  • #1
LCSphysicist
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Homework Statement
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Relevant Equations
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$$d\tau = \sqrt{-g_{\mu v}\frac{dx^{\mu}}{d\lambda} \frac{dx^{v}}{d \lambda}} d \lambda $$
Calling ##d \lambda = d t##, t is coordinate time in an inertial frame.
$$\Delta \tau = \int \sqrt{-(-1+v^2/c^2)} dt$$
$$\Delta \tau \approx \Delta t - \frac{g^2*\Delta t^2}{6c^2}$$
where $$\Delta t = 2 \sqrt{L/g}, L = 30,000ly$$

So we just need to substitute it. Is this right?
 
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  • #2
Your second equation of time integration gives ##\tau(t)##. To integrate it we need to know ##v(t)##. You can get it from the relation of v(t) and constant acceleration below
[tex]\frac{d}{dt}\frac{v}{\sqrt{1-v^2/c^2}}=g[/tex]
By further integrating v(t) with time you get x(t) so you can get ##\tau(x)## to know proper time for one way the amount of which should be doubled for a round trip.
 
  • #3
Herculi said:
Homework Statement:: .
Relevant Equations:: .

View attachment 278319
$$d\tau = \sqrt{-g_{\mu v}\frac{dx^{\mu}}{d\lambda} \frac{dx^{v}}{d \lambda}} d \lambda $$
Calling ##d \lambda = d t##, t is coordinate time in an inertial frame.
$$\Delta \tau = \int \sqrt{-(-1+v^2/c^2)} dt$$
$$\Delta \tau \approx \Delta t - \frac{g^2*\Delta t^2}{6c^2}$$
where $$\Delta t = 2 \sqrt{L/g}, L = 30,000ly$$

So we just need to substitute it. Is this right?
No, it's not right. I don't think you've understood the problem at all. The proper acceleration must be ##g##. You've given no explanation on how you calculated that integral.

Have you heard of the hyberbolic trig functions?

The answer you give can't be right. You're going to get ##\Delta \tau## to be negative.
 

FAQ: Proper time in an acceleration frame

1. What is proper time in an acceleration frame?

Proper time in an acceleration frame is the time measured by a clock that is at rest in that frame. It is the time experienced by an observer who is moving along with the clock.

2. How is proper time related to time dilation?

Proper time is related to time dilation in that it is the time measured by a clock in a frame that is not undergoing acceleration. Time dilation occurs when an object is moving at a high velocity, causing time to appear to pass slower for that object compared to a stationary observer.

3. Does proper time change in different frames of reference?

Yes, proper time can change in different frames of reference. This is due to the phenomenon of time dilation, which causes time to appear to pass at different rates for observers in different frames of reference.

4. How is proper time calculated in an acceleration frame?

Proper time in an acceleration frame can be calculated using the equation t' = t/√(1-v²/c²), where t is the time measured by an observer at rest in the frame, v is the velocity of the moving object, and c is the speed of light.

5. What is the significance of proper time in relativity?

Proper time is significant in relativity because it is the time experienced by an observer who is moving along with a clock. It is a fundamental concept in understanding the effects of time dilation and the relativity of time in different frames of reference.

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