# 1PN metric vs. Schwarzchild

by andert
Tags: metric, schwarzchild
 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: $$g_{00} = -1 + 2U - 2 U^2$$ where U is the Newtonian potential. The time-time component of the Schwarzchild metric, however, is $$g_{00} = -1 + 2M/r$$. 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?
 Emeritus Sci Advisor PF Gold P: 5,441 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.
 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, $$g_{00} = -1 + 2M/r - 2(M/r)^2$$ 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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