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Source term of the Einstein field equation

+ 4 like - 0 dislike
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I tried this question on the Physics SE in November and didn't get a response. Let's see how it does over here!

My copy of Feynman's "Six Not-So-Easy Pieces" has an interesting introduction by Roger Penrose. In that introduction (copyright 1997 according to the copyright page), Penrose complains that Feynman's "simplified account of the Einstein field equation of general relativity did need a qualification that he did not quite give." Feynman's intuitive discussion rests on relating the "radius excess" of a sphere to a constant times the enclosed gravitational mass $M$: for a sphere of measured radius $r_{\mathrm{meas}}$ and surface area $A$ enclosing matter with average mass density $\rho$ smoothly distributed throughout the sphere, $$\sqrt{\frac{A}{4π}}−r_{\mathrm{meas}}=\frac{G}{3c^2}\cdot M,$$ wherein $G$ is Newton's gravitational constant, $c$ is the speed of light in vacuum, and $M=4π\rho r^3/3$. I don't know what $r$ is supposed to be, but it's presumably $\sqrt{\frac{A}{4\pi}}$. Feynman gratifyingly points out that $\frac{G}{3c^2}\approx 2.5\times 10^{−28}$ meters per kilogram (for Earth, this corresponds to a radius excess of about $1.5$ mm). Feynman is also careful to point out that this is a statement about average curvature.

Penrose's criticism is: "the 'active' mass which is the source of gravity is not simply the same as the energy (according to Einstein's $E=mc^2$); instead, this source is the energy plus the sum of the pressures". Damned if I know what that means -- whose pressure on what?

So, taking into account Penrose's criticism but maintaining Feynman's intuitive style, what is the active mass $M$?

This post has been migrated from (A51.SE)
asked Jan 21, 2012 in Theoretical Physics by Kernel (125 points) [ no revision ]
Cross-listed from: http://physics.stackexchange.com/q/16725/2451

This post has been migrated from (A51.SE)

1 Answer

+ 3 like - 0 dislike

I'll try and answer in an intuitive way as best I can (as you asked for on the crosslink).

The relation between the true, or physical, surface area of a sphere with radius $r_{meas}$ and the surface area one expects from standard Euclidean space is a measure (as you say) of the average curvature (more precisely it is a measure of the scalar curvature $R$: http://en.wikipedia.org/wiki/Scalar_curvature). In GR this curvature is generated by "mass-energy" (again I'm sure you know this), or the stress-energy tensor $T_{\mu \nu}$. The above relation that Feynman gives for a average mass density $\rho$ is valid if you are talking about cold matter or "dust" where in the average rest frame of the matter, there are no internal motions (i.e. pressures). If there are internal motions of the matter then there is kinetic energy in addition to rest mass energy which you should intuitively expect to also contribute to curvature. This additional energy plus the rest mass energy is what I think you mean by "active" mass.

So it seems to me that Penrose's criticism is really that Feynman didn't use a realistic model of matter, since the original relation is true if the matter you are talking about is ultra cold.

This post has been migrated from (A51.SE)
answered Jan 21, 2012 by Kyle (335 points) [ no revision ]
I'm pondering this answer... my concern is that Penrose might have been referring to the pressure required to keep all the energy enclosed within the sphere, and the reason I am concerned is because Feynman may have intended for the active mass to inclue kinetic energy terms.

This post has been migrated from (A51.SE)
Dear @Kernel, it seems to me that you are repeating the same pre-existing opinions of yours and you didn't really pay much attention to Kyle who explains what the actual core of Penrose's complaint was, and in my opinion, correctly so. When you have a star, it has not only mass density inside; it also have a pressure because the particles are moving somewhat quickly. In GR, the very presence of pressure - inside the matter (so "pressure on what" is a totally irrelevant question here) - is affecting the properties of the gravitational field because the whole $T_{\mu\nu}$ including $p$ is RHS.

This post has been migrated from (A51.SE)
Otherwise what the mass "including the pressure" is depends on what we exactly calculate; Feynman's was an estimate. But in various situations, the mass may be modified by $\pm C \int p\,\, {\rm d}V/c^2$ where $C$ is a numerical constant. In other contexts, it's important that the influence of the pressure is nonlinear, and therefore only in higher orders, and so on.

This post has been migrated from (A51.SE)
I think someone ought to write down the precise definition of mass in general relativity that yields the above equation. It is some integral involving the energy momentum tensor but what is it precisely?

This post has been migrated from (A51.SE)
I believe that the deviation of the surface area of a sphere is determined by the scalar curvature, which is proportional to the trace of the stress energy tensor: $R = - \frac{8 \pi G}{c^4} T$. Then it depends on how you are modeling your matter.

This post has been migrated from (A51.SE)
Lubos, I didn't have a previously existing opinion, so I'm not sure what I repeated.

This post has been migrated from (A51.SE)
So I guess my concern came down to whether the active mass of an ideal gas in a container of fixed size varied in the temperature or the pressure... but they are equal up to a constant in this case. I accept the answer now.

This post has been migrated from (A51.SE)

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