Derivation of formula for time dilation

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SUMMARY

This discussion focuses on deriving the gravitational time dilation formula, specifically T=To(1/(sqrt(1-(2GM)/(Rc2))), which is applicable in the context of the Schwarzschild solution. Participants emphasize the necessity of the Schwarzschild metric and the significance of the radial coordinate R, clarifying that it must be greater than the Schwarzschild radius for the formula to hold true. The conversation also touches on the relationship between gravitational potential and time dilation, highlighting that the derivation can be approached without delving deeply into General Relativity (GR) concepts.

PREREQUISITES
  • Understanding of the Schwarzschild metric
  • Familiarity with gravitational time dilation concepts
  • Knowledge of basic principles of General Relativity
  • Ability to interpret mathematical expressions related to spacetime metrics
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  • Study the derivation of the Schwarzschild metric in detail
  • Learn about gravitational potential in the context of General Relativity
  • Explore the relationship between time dilation and gravitational fields
  • Investigate the implications of time dilation in practical astrophysical scenarios
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Students and professionals in physics, particularly those specializing in General Relativity, astrophysics, or theoretical physics, will benefit from this discussion. It is also valuable for educators seeking to explain complex concepts of time dilation and spacetime metrics.

wpan
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I know how to derive the lorentz time dilation equation. I am wondering how to derive the equation for gravitational time dilation: T=To(1/(sqrt(1-(2GM)/(Rc2)))
 
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wpan said:
I know how to derive the lorentz time dilation equation. I am wondering how to derive the equation for gravitational time dilation: T=To(1/(sqrt(1-(2GM)/(Rc2)))
The way you have it written, the formula applies only to the Schwarzschild solution, where R is the Schwarzschild radius. In that case, the way to derive it is to write down the Schwarzschild metric, and look at g00. For a motionless test particle (dr = dθ = dφ = 0),

ds2 = (1 - 2GM/Rc2) dt2,

and from this, ds/dt gives you the time dilation.
 
bcrowell said:
I have a treatment in my SR book, http://www.lightandmatter.com/sr/ , in ch. 5. You don't need GR or the Schwarzschild metric.
You do if the expression involves the Schwarzschild radius.
 
Bill_K said:
You do if the expression involves the Schwarzschild radius.

It can be expressed in terms of the gravitational potential.
 
Bill_K said:
The way you have it written, the formula applies only to the Schwarzschild solution, where R is the Schwarzschild radius. In that case, the way to derive it is to write down the Schwarzschild metric, and look at g00. For a motionless test particle (dr = dθ = dφ = 0),

ds2 = (1 - 2GM/Rc2) dt2,

and from this, ds/dt gives you the time dilation.
I don't understand what you mean by "##R## is the Schwarzschild radius". I would expect that ##R## is just the radial coordinate, in his notation. But I agree with the 'derivation'. Just assume Schwarzschild metric, and then choose a stationary test object outside the event horizon, then you get the time dilation formula. And this is often used as an approximation for the general relativistic equation for the time dilation due to planets, I think. So it's a good metric to use in most of the simple textbook problems on GR time dilation.

bcrowell said:
I have a treatment in my SR book, http://www.lightandmatter.com/sr/ , in ch. 5. You don't need GR or the Schwarzschild metric.
pretty cool. I always assumed that anyone that posts on physicsforums wouldn't have time to write a book along with doing their full-time job. hehe. you must have good time-management skills. In the discussion of gravitational potential in chapter 5, I'm not totally sure what you mean though. I understand
\frac{\Delta f}{f} \approx -g \Delta x
But then I don't understand how that implies the next equation:
\frac{\Delta f}{f} \approx - \Delta \Phi
Ah wait, duh, I get it now. I thought that this triangle was the Laplacian (and so the right-hand-side would always equal zero in free space). But it is a Delta symbol again. OK that makes sense. haha, sorry, I only realized it was meant to be a Delta symbol when I typed out the equation here, and I realized I wrote a Delta and thought "uh... of course it's a Delta, not a Laplacian".

Anyway, that's kinda interesting. Although I suppose it only works in the Newtonian limit. So in comparison to the Schwarzschild solution, we must choose a radius which is much greater than the event horizon. But in the Schwarzschild solution, we can choose any radius (except the radius of the event horizon, or r=0).
 
BruceW said:
Anyway, that's kinda interesting. Although I suppose it only works in the Newtonian limit.

It's true in general, not just in the Newtonian limit, that df/f=-d\Phi. If you integrate both sides, you get f\propto e^{-\Phi}, which holds for any static spacetime, including the Schwarzschild spacetime.
 
BruceW said:
I don't understand what you mean by "##R## is the Schwarzschild radius". I would expect that ##R## is just the radial coordinate, in his notation.
If R is some other radial coordinate and not the Schwarzschild radial coordinate, then the formula is false.
 
Bill_K said:
If R is some other radial coordinate and not the Schwarzschild radial coordinate, then the formula is false.

Are you sure about that? I'm reading the formula as applying to any radius larger than the Schwarzschild radius. At the Schwarzschild radius the formula produces infinite red-shift.
 
  • #10
I think Bill_K does mean "Schwarzschild radial coordinate" as meaning any radius larger than the Schwarzschild radius. I think he is saying "Schwarzschild radial coordinate" to distinguish it from any other choice of radial coordinate that we could make. (I don't know of any other choices though. But I'm sure you could make up some non-useful type of radial coordinate, maybe like the cylindrical radial coordinate).
 
  • #11
bcrowell said:
It's true in general, not just in the Newtonian limit, that df/f=-d\Phi. If you integrate both sides, you get f\propto e^{-\Phi}, which holds for any static spacetime, including the Schwarzschild spacetime.
cool. Is ##\Phi## one of the Hansen potentials? (the mass potential). I just read this on the wikipedia page for stationary spacetime. That's some pretty interesting stuff. So.. it seems like the Newtonian gravitational potential has a (perfect?) analogue in stationary spacetimes in GR. It's not often that a concept carries on so nicely from Newtonian gravity to GR.
 
  • #12
BruceW said:
So.. it seems like the Newtonian gravitational potential has a (perfect?) analogue in stationary spacetimes in GR.

As long as you're in a space-time with a time-like Killing field ##\xi^{\mu}## you can always define a gravitational potential by ##\Phi = \frac{1}{2}\ln (-\xi_{\mu}\xi^{\mu})##. It's clear why this makes sense physically as observers following orbits of ##\xi^{\mu}## are at rest in the gravitational field, or equivalently the asymptotic Lorentz frame if the space-time is asymptotically flat, and their 4-acceleration is simply ##a^{\mu} = \nabla^{\mu}\Phi## which is (the negative of) the gravitational acceleration of freely falling particles with respect to the stationary ("Copernican") frames defined by ##\xi^{\mu}##. If ##\xi_{[\gamma}\nabla_{\mu}\xi_{\nu]} \neq 0## then we can also write down Newtonian analogues of the centrifugal and Coriolis accelerations of freely falling particles relative to the stationary frames.
 
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