Recent discussions on pressure as a source of gravity and the related Tolman paradox have reminded me that few people seem to be aware that even in Newtonian gravity the pressure is related to the potential energy, which I've mentioned a few times on these forums before.(adsbygoogle = window.adsbygoogle || []).push({});

This is easy to show, but a bit messy to describe. I'd like to know whether this is a known named result, and whether someone can give me a simpler way of describing it in Newtonian terms (preferably without requiring the use of GR-style tensor notation as in the Komar mass expression).

Specifically, if you have an isolated static system of masses in free space internally supported in equilibrium by any means (rods, strings, springs, balloons or whatever), then if you integrate the normal pressure component over planes perpendicular to three perpendicular axes through the whole system, the total of the three integrals is a positive quantity equal and opposite to the gravitational potential energy of the system.

To show this, you can divide up the system conceptually into point masses where each one is joined to all other point masses by lines representing their gravitational interaction. For the system to be static, the total force (including gravitational) across any plane must be zero. The gravitational force along the line between masses ##m_i## and ##m_j## is then ##-Gm_i m_j/r_{ij}^2##, so this is the contribution to the overall force across any plane cutting the line between those masses, which must be equal and opposite to the integral over that plane of the pressure. The force is the same at all points on the line between the masses. If the pressure is integrated along that line, that simply multiplies the result by ##r_{ij}## giving a contribution ##Gm_i m_j / r_{ij}## for that pair of masses, equal and opposite to the potential energy between those masses.

More generally, if the displacement vector ##(x,y,z)## between two masses is not perpendicular to the plane, the normal pressure is reduced by the cosine of the angle between the displacement line and the perpendicular to each plane, and so is the perpendicular component of the length of the line, so the contributions for three planes add up as ##x^2/r^2##, ##y^2/r^2## and ##z^2/r^2##, adding up to the same total as the original case. As these quantities add linearly for all contributing point masses, the total for the system is equal and opposite to the total potential energy of the system.

This means for example that if the system just consists of a perfect fluid held together by gravity, the gravitational potential energy is equal and opposite to the volume integral of three times the pressure.

Note that in the Newtonian case it is clear that although this pressure integral is equal and opposite to the potential energy for a static system, it cannot be identified with the potential energy, because for example if a support is removed and the system starts to collapse, the Newtonian potential energy is initially unaffected, but terms in the pressure integral can vanish immediately, at least to orders of magnitude smaller than their original values.

(For this reason, I feel that Tolman was on to something and there is something wrong either with the conventional interpretation that pressure is a source of gravity in GR or with the field equations themselves, but that's a different topic).

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# A Pressure and Newtonian potential energy

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