- #1

Jonathan Scott

Gold Member

- 2,295

- 997

Consider a flat space stress-energy tensor describing some self-contained system of masses, not necessarily static, in Cartesian (x,y,z) coordinates with gravity being treated as an acceleration field ##\mathbf{g}##, with energy density ##\rho c^2## and local directional pressure ##P_{xx}## in the ##x## direction (the negative of the normal stress ##\sigma_{xx})##, with no transverse stresses. The continuity equation for the ##x## component of momentum is then as follows:

$$\frac{\partial(\rho v_x)}{\partial t} + \frac{\partial(P_{xx})}{\partial x} = \rho g_x$$

This is equivalent to the ##x## component of the covariant divergence in the weak approximation in GR.

We can take the dot product of both sides with the vector ##-\mathbf{x}## relative to some arbitrary origin. The ##x## component of the result is as follows:

$$-x \cdot \frac{\partial(\rho v_x)}{\partial t} - x \cdot \frac{\partial(P_{xx})}{\partial x} = -x \cdot \rho g_x$$

From my previous thread, the RHS of this integrated over all three spatial dimensions is equal to the Newtonian potential energy, because if one considers the gravitational force between every pair of masses making up the collection, taking the dot product with the position means that one has overall taken the dot product of the two equal forces, of opposite sign, with the vector for the displacement between them, which is equal to the potential energy of that pair but positive. Note that the RHS is not affected by whether the object is static or accelerating. In the case where there is no pressure, the mass is in free fall and the rate of change of momentum is equal to the gravitational force. In the static case, the rate of change of momentum and the term containing the pressure integrates to the potential energy.

The term containing the pressure is not simply the pressure, but its integral is equal to the integral of the pressure over the whole object, which we can confirm as follows (with force ##F_{xx}## being the area integral of the pressure):

$$\iiint P_{xx} + x \cdot \frac{\partial(P_{xx})}{\partial x} \, dx \, dy \, dz = \iiint \frac{\partial}{\partial x} \left ( x \cdot P_{xx} \right ) \, dx \, dy \, dz = \int \frac{\partial}{\partial x} \left ( x \cdot F_{xx} \right ) \, dx = \left [x \cdot F_{xx} \right ]$$

That is, the pressure plus that term with its sign switched to positive integrates to the difference in ##x\cdot F_{xx}## between the limits of integration, which is zero when the integration extends to beyond the edge of the system of masses, so the integral of the pressure minus the integral of that term is zero over the whole system. However, the pressure is not necessarily locally equal to that term; the equality only holds for the integral over the whole system.

So the integral of the directional pressure terms (negative normal stresses) in all three dimensions over a system is equal to the integral of ##-x \cdot \partial(P_{xx})/\partial x##, and the integral of that term plus the force density term (which only appears when acceleration is present) adds up to the Newtonian potential energy of the system.

Note that the expression involving ##-x \cdot \partial(P_{xx})/\partial x## is a local equation, which says that there is a local quantity which can be integrated to give the potential energy, and that quantity is proportional to the local energy density, so it is associated with the existing mass density, not with support structures or stresses, and it cannot change abruptly (although the total potential energy can change gradually as it is converted to or from kinetic energy).

The effective energy of an object of local total energy ##E## in a Newtonian potential ##\phi = -Gm/r## is expected to be given in General Relativity by ##E \, e^{\phi/c^2}##, applying the time dilation associated with the potential. For multiple objects this counts the binding energy twice, but there is an alternative expression for the effective source strength based on the above equations which gives the correct effective binding energy (where ##\mathbf{g} = -\nabla \phi##):

$$\nabla \left ( \mathbf{x} \, E \, e^{\phi/c^2} \right ) = E \, e^{\phi/c^2} \left ( 1 - \frac{\mathbf{x}\cdot\mathbf{g}}{c^2} \right ) $$

If ##\mathbf{x}## is relative to an arbitrary location then this still does not give a specific value for the potential energy of each part (which is similar to the Newtonian case) but if it is however taken to be relative to the centre of mass of the system (taking into account the potential) then that is equivalent to the following constraint on the origin to use for ##\mathbf{x}##, which involves an expression that just appeared in the previous equation, summed for all components of the system:

$$\sum \left ( \mathbf{x} \, E \, e^{\phi/c^2} \right ) = 0$$

So far this hasn't helped me very much with my attempts to understand in what way the binding energy is reflected in the SET and in the EFE in the general dynamic case, but it does make the important point that even if the source strength in the static case is effectively increased by an amount

**equal**to the pressure, that doesn't mean that the pressure

**is**the source. The above shows that there is another quantity which provides the same correction to the effective source strength and which has the same integral as the pressure in the static case but is also valid in the dynamic case. Perhaps something like that quantity is hiding somewhere in some form of the EFE or a time derivative and gives rise to the pressure in static cases and some other term in other cases. Any constructive suggestions for how to progress from here would be gratefully received.

In my previous thread I thought that perhaps the gravitational induction linear frame-dragging effect of the accelerating matter could explain the dynamic case, but I have now ruled that out. On calculating the effect more carefully I found some additional terms which cancelled out those induction terms, and it also would not have worked for transverse accelerations, which could cancel by symmetry.