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1PN metric vs. Schwarzchild

by andert
Tags: metric, schwarzchild
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andert
#1
Mar18-10, 02:15 PM
P: 12
I'm sure there is a simple answer to this question, but I have been looking at the first Post-Newtonian (1PN) metric (for my own research) and noticed that the time-time component of the GR metric is:

[tex]g_{00} = -1 + 2U - 2 U^2[/tex]

where U is the Newtonian potential.

The time-time component of the Schwarzchild metric, however, is

[tex]g_{00} = -1 + 2M/r[/tex].

There is no quadratic in the Newtonian potential even though this metric is an exact solution of the field equations. Is this because it is in a different gauge?
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bcrowell
#2
Mar18-10, 05:36 PM
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You can make g00 look like anything you like, just by a choice of coordinates. Suppose I set g00=f(r), where f is some arbitrary function, and I don't give you any other information about my coordinates. You can then use this equation to define an r coordinate. For instance, suppose g00 changes by a factor of 2 between r1 and r2. Then we've effectively defined r=r2 to be the location where gravitational time dilation differs by a factor of 2 from its value at r=r1. With this implicit definition of the r coordinate, we can now go ahead and infer the rest of the metric.
andert
#3
Mar19-10, 08:50 AM
P: 12
Alright, yes, coordinate system or gauge. I see that. Each of them is a in a specific gauge. Now, what is different about the gauges of the 1PN and Schwarzchild metrics specifically? The 1PN gauge is a harmonic one. So if I were to take a static spherically symmetric field, I would have the 1PN time-time component,

[tex] g_{00} = -1 + 2M/r - 2(M/r)^2[/tex]

Is the idea that, in this special case, we can make a change of gauge (coordinates) to eliminate the quadratic term but in the general case of many bodies (with a sum over masses) we cannot?


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