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Calculating proper time using schwartzchild metric

by demonelite123
Tags: metric, proper, schwartzchild, time
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demonelite123
#1
Mar3-12, 11:02 PM
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I am using the schwartzchild metric given as [itex] ds^2 = (1 - \frac{2M}{r})dt^2 - (1 - \frac{2M}{r})^{-1} dr^2 [/itex], where I assume the angular coordinates are constant for simplicity.

So if a beam of light travels from radius r0 to smaller radius r1, hits a mirror, and travels back to r0, I am trying to find how much proper time has passed for an observer fixed at r0. So far, i have that this path can be parametrized by r = r0 and t = x, where x is just my parameter. Therefore, r' = 0 and t' = 1. Using the formula for arc length, i have that the proper time is given by [itex] \int \sqrt{1 - \frac{2M}{r_0}} dx [/itex].

this is where i am stuck as i am having trouble determining the limits of my integral. can someone give me a hint or two in the right direction? thanks
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bcrowell
#2
Mar3-12, 11:14 PM
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Your integrand doesn't have any variable in it. The [itex]r_0[/itex] shouldn't be inside the integral; it should relate to a limit of integration.

You could try setting [itex]ds^2=0[/itex] and then separating variables and integrating to get a relation between r and t for the light beam.
PeterDonis
#3
Mar3-12, 11:15 PM
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First of all, you don't really need the extra parameter x; as far as the observer fixed at r0 is concerned, he's just traveling from t0, the time when he emits the light beam, to t1, the time when it returns to him. So you could just write the integral as:

[tex]\tau = \int_{t_{0}}^{t_{1}} \sqrt{1 - \frac{2M}{r_{0}}} dt[/tex]

But the integrand doesn't depend on t, so you can just factor it out, and that makes the integral trivial:

[tex]\tau = \sqrt{1 - \frac{2M}{r_{0}}} \left( t_{1} - t_{0} \right)[/tex]

Which, of course, should make you realize that the real focus of the problem is determining the coordinate time interval t1 - t0. The way to do that is to focus, not on the worldline of the observer fixed at r0, but on the worldline of the light beam. There are two segments to it (the one from r0 inward to r1, and the one from r1 back outward to r0), but they are mirror images, so to speak, so they should take equal coordinate time to traverse. So figuring out the coordinate time for one is sufficient. That's where I would recommend focusing your efforts. The key fact you need, in addition to what you've already posted, is that the light beam's worldline is null; that is, the interval ds^2 along the light beam's worldline is zero.

pervect
#4
Mar4-12, 02:20 AM
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Calculating proper time using schwartzchild metric

Well, the way I'd approach it is this:

Integrating along the path that the light takes won't give us the right answer - we want to integrate along the path that the clock takes between transmission and reception. Which is a simple path, of constant r = r0.

So we need to draw a space-time diagram with the ingoing light beam, and the outgoing lightbeam. How do we do this?

Given the line element

[tex]ds^2 = (1 - \frac{2M}{r})dt^2 - (1 - \frac{2M}{r})^{-1} dr^2p[/tex]

we know that for a light beam, ds = 0. This immediately gives us the ratio dr/dt for the light beam - which will be a function of r.

So we'll have f(r) dr = dt, where I'm too lazy to write out f(r).

Integrating this we'll get [itex]\Delta t=F(r)[/itex]. We'll have the same [itex]\Delta t[/itex] on the ingoing and outgoing null geodesic - so we double it.

This will give us the coordinate time that elapses between emission and reception. To get the proper time, we integrate along the worldline at r=r0 between the emission and reception events. dr=0 for this intergal, so we get a simple time dilation factor

[tex]ds = \int \sqrt{1 - \frac{2M}{r}} \, dt = \sqrt{1 - \frac{2M}{r}} \Delta t[/tex]
demonelite123
#5
Mar5-12, 12:56 AM
P: 219
ok so setting [itex] ds^2 = 0 [/itex], i get [itex] (1 - \frac{2M}{r})dt^2 = (1 - \frac{2M}{r})^{-1} dr^2 [/itex] or [itex] dt^2 = (1 - \frac{2M}{r})^{-2} dr^2 [/itex]. then taking the square root of both sides, i get [itex] dt = (1 - \frac{2M}{r})^{-1} dr [/itex].

now i can integrate both sides and i get [itex] t_1 - t_0 = \int_{r_1}^{r_0} \frac{r}{r - 2M}dr = (r_0 - r_1) + 2Mln(r_0 - 2M) - 2Mln(r_1 - 2M) [/itex].

would this be correct? thanks for you replies.
PeterDonis
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Mar5-12, 09:24 AM
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Quote Quote by demonelite123 View Post
now i can integrate both sides and i get [itex] t_1 - t_0 = \int_{r_1}^{r_0} \frac{r}{r - 2M}dr = (r_0 - r_1) + 2Mln(r_0 - 2M) - 2Mln(r_1 - 2M) [/itex].

would this be correct? thanks for you replies.
As noted before, to get the final answer you need to multiply the result by 2 because the integral you have given gives the "one-way" time, and you need the "round trip" time. The integral itself looks OK to me.


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