calculating proper time using schwartzchild metricby demonelite123 Tags: metric, proper, schwartzchild, time 

#1
Mar312, 11:02 PM

P: 219

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 r_{0} to smaller radius r_{1}, hits a mirror, and travels back to r_{0}, I am trying to find how much proper time has passed for an observer fixed at r_{0}. So far, i have that this path can be parametrized by r = r_{0} 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 



#2
Mar312, 11:14 PM

Emeritus
Sci Advisor
PF Gold
P: 5,500

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. 



#3
Mar312, 11:15 PM

Physics
Sci Advisor
PF Gold
P: 5,510

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. 



#4
Mar412, 02:20 AM

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P: 7,439

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 = r_{0}. So we need to draw a spacetime 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=r_{0} 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] 



#5
Mar512, 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. 



#6
Mar512, 09:24 AM

Physics
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PF Gold
P: 5,510




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