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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 $M$?

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You may be interested in John Baez's wonderful introduction to general relativity which explains how the pressure comes in to it. –  Michael Brown Mar 1 '13 at 10:06
    
Thanks! I consider the following quote to be an excellent answer to my question: "Given a small ball of freely falling test particles initially at rest with respect to each other, the rate at which it begins to shrink is proportional to its volume times: the energy density at the center of the ball, plus the pressure in the direction at that point, plus the pressure in the direction, plus the pressure in the direction." Baez claims that this statement is equivalent to the usual formulation of Einstein's equation. –  Kernel Mar 8 '13 at 13:02
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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.

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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. –  Kernel Jan 22 '12 at 1:21
    
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. –  Luboš Motl Jan 22 '12 at 6:11
    
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. –  Luboš Motl Jan 22 '12 at 6:14
    
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? –  Squark Jan 22 '12 at 20:59
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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. –  Kyle Jan 22 '12 at 22:47
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