Interior Schwarzschild Metric: Pressure Dependence

[Sonderval]

I'm looking influence of pressure on the general interior Schwarzschild metric (see for example the book by Weinberg, eq. 11.1.11 and 11.1.16.
The radial component of the metric (usually called A(r)) depends only on the mass included up to radius r
$$A(r) = \left(1-\frac{ 2G M(r)}{r}\right)^{-1}$$
For the time-component, there is a rather complicated differential equation that depends on pressure; which is what I expect due to the pressure from the Energy-Momentum-Tensor.
Is there a physical reason why the pressure cancels out in the spatial part of the metric?
I can see mathematically from the derivation of the equation that it does, but would like to know whether this can be explained in any intuitive way.

 

[PeterDonis]

Is there a physical reason why the pressure cancels out in the spatial part of the metric?

I think this is because of the particular choice of coordinates being made. The form of $A(r)$ that you give actually applies to any spherically symmetric, static geometry, if you define the radial coordinate $r$ in the way it's defined for Schwarzschild coordinates, i.e., so that the area of a 2-sphere at $r$ is $4 \pi r^2$. But any other choice of coordinates will give a different form for $g_{rr}$, which AFAIK will not, in general, depend only on $M(r)$ (or equivalently on density only, not pressure).

 

[Sonderval]

@PeterDonis
I agree that the pressure dependence should not cancel in other r-coordinates.
However, I'm not totally sure I see how the purely geometrical fact that I define the r-coordinate using the surface relation (as done by Schwarzschild) causes the vanishing pressure term.
I suspect it is related to the fact that the mass defect can be calculated by the difference between "nominal volume" (using the standard sphere formula) and the actual volume (using the grr-term to integrate)., but I have no clear picture of this.

 

[PeterDonis]

I'm not totally sure I see how the purely geometrical fact that I define the r-coordinate using the surface relation (as done by Schwarzschild) causes the vanishing pressure term.

Yes, this is a better way of phrasing the question since it treats $r$ not as a coordinate but as a geometric parameter describing the (square root of the) area of the 2-spheres.

I suspect it is related to the fact that the mass defect can be calculated by the difference between "nominal volume" (using the standard sphere formula) and the actual volume (using the grr-term to integrate)

I think this is correct.

 

[George Jones]

Assume that $A$ is a function of $r$ and $p$, i.e., $A = A \left( r , p \right)$. The solution is static and spherically symmetric, and hence, by the way the coordinates are chosen, $p$ is only a function of $r$, i.e., $p = p \left( r \right)$. Since $p$ is a monotonically decreasing function of $r$, this function is invertible, and we can write $r = r \left( p \right)$. Consequently, nothing is lost by writing either $\tilde{A} \left( r \right) = A \left( r , p \left(r\right) \right)$ or $\tilde{A} \left( p \right) = A \left( r \left(p\right) , p \right)$.

In other words, because of the one-to-one relationship between $r$ and $p$, it makes sense to express $A$ as a function of either of these quantities, as opposed to both.

 

[Sonderval]

@Peter
Thanks for that.

@George Jones
I do not think this answers the question, because the same argument would apply to the tt-component of the metric which cannot be written in closed form without explicit pressure dependence. The A(r) relation is actually universal and holds for any matter or stellar model if I understand things correctly (which I may not...), whereas to solve for the tt-component ( usually called B(r)), you need a p(r)-equation.

 

[Sonderval]

So, finally I can answer my own question at least partly:
The metric inside of a mass is related to the "excess radius" (how much longer is the way through a sphere than expected from its circumference), and this is directly related to the 00-component of the Einstein-tensor. (See Feynman Lectures on Gravitation, Lecture 11.)
Since this is equal to the 00-component of the energy-momentum-tensor, no pressure is involved.