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PeterDonis said:Hi pervect, yes, this is a good point; in this particular case the pressure contribution to J is negligible (I believe--see below), but in general it might not be.
In post #57, if you have time to look, I posted a metric from MTW Box 23.2 for the interior of a static spherical object that is not a "shell", i.e., it has no hollow portion inside it. The total mass M that appears in that metric is defined in MTW as
[tex]M = \int_{0}^{R} 4 \pi \rho r^{2} dr[/tex]
I.e., M does not contain any contribution from the pressure inside the object. If I'm reading MTW correctly here, they don't intend this formula to be an approximation; it is supposed to be exact.
The formula is exact - but read the part in MTW that says that 4 pi r^2 dr is NOT a volume element.
It superfically looks like one at first glance, but isn't. 4 pi r^2 is ok, but dr needs a metric correction. Using the actual volume element, MTW also calculates the integral of rho* dV, dV being the volume element, and find that said integral is larger than the mass M. The quantity [itex]\int \rho dV[/itex] is given a name, the "mass before assembly". Because there is no compression to worry about, (the pieces are modeled as not changing volume with pressure), the only work being done by assembly is the binding energy, which you can think of being taken out of the system as you assemble it - for instance, you might imagine cranes lowering the pieces into place, and work is made available in the process.
You'll see a chart, where they tabluate the binding energy for various sizes too, as I recall.