Gravity distortion at the center of a massive object?

[yuiop]

What Carroll wrote down in those notes is definitely not what Schutz has. In fact Schutz has the exact same interior solution as the one given by Padmanabhan for a constant density star. Both have $m(r) = 4\pi \rho r^3/3$,
$g_{rr} = e^{2\Lambda} = (1 - 2m(r)/r)^{-1}$,
and $e^{\Phi} = \frac{3}{2}(1 - 2M/R)^{1/2} - \frac{1}{2}(1 - 2Mr^2/R^3)^{1/2}$ where $g_{tt} = e^{2\Phi}$.

This solution of course makes perfect sense.

OK, assuming that $(1 - 2m(r)/r) = (1 - 2mr^2/R^3)$, that appears to be basically the standard interior Schwarzschild solution given by George Jones that I linked to in #10.

If that is the case then the Shutz/ standard interior Schwarzschild metric can be stated as:

d\tau^{2}=-\left( \frac{3}{2}\sqrt{1-\frac{2M}{R}}-\frac{1}{2}\sqrt{1-\frac{2Mr^{2}}{R^{3}}}\right) ^{2}dt^{2}+\left( 1-\frac{2Mr^{2}}{R^{3}}\right) ^{-1}dr^{2}+r^{2} d\Omega ^{2}

or alternatively as:

d\tau^{2}=-\left( \frac{3}{2}\sqrt{1-\frac{2M}{R}}-\frac{1}{2}\sqrt{1-\frac{2M(r)}{r}}\right) ^{2}dt^{2}+\left( 1-\frac{2M(r)}{r}\right) ^{-1}dr^{2}+r^{2} d\Omega ^{2}

Does that seem reasonable?

P.S. From the second form it is obvious that the time dilation is constant everywhere within the cavity.

 

[PAllen]

This formula is in fact the same as the formula I linked to in post #10 although it is heavily disguised. i.e. it is the standard interior Schwarzschild metric with homogeneous density.

That's true, but only at the point where A is taken to be a constant. The framework would cover A being a function of r and, and then A becomes related to density variation.

I agree that the Carroll equation contradicts the regular interior Schwarzschild metric. However, in its favour, Carroll suggests it is backed by Schutz, which is normally a reliable source. Does anyone have the Schutz text? Also, the Caroll metric clearly indicates that the spacetime inside a cavity is flat which I have seen claimed many times.

It contradicts physical plausibility. For a static solution where a potential can be introduced, gravitational time dilation is determined by potential. This metric then implies that you have to apply work to push matter down a tube through the star, and free fall would be upward. Absurd is not too strong a word for that.

Does your metric show that the spacetime inside a cavity is Minkowskian? Does your metric match the exterior vacuum Schwarzschild metric at the outer surface of the massive body?

Yes, Peter's does, as long as J(r) becomes constant inside the shell, which is the required by plausibility anyway. For the outside, J(r) has to match the exterior SC value at the outer surface.

 

[PeterDonis]

Does your metric show that the spacetime inside a cavity is Minkowskian?

Yes. Inside a cavity, $m(r) = 0$, so the only term in the line element that is not explicitly Minkowski is $g_{tt}$, and you can rescale the time coordinate to make $g_{tt} = -1$ inside the cavity.

[Edit: technically, you also have to prove that $J(r)$ is constant inside the cavity, but this is easy; see my response to PAllen.]

Does your metric match the exterior vacuum Schwarzschild metric at the outer surface of the massive body?

Yes; if you work out what $J(r)$ is at the outer surface of an isolated body surrounded by vacuum, you will obtain

$$
J(r) = 1 - \frac{2M}{r}
$$

where $M$ is the total mass of the body.

Can you for example state the time dilation when dr=dΩ=0 , m=1, the outer surface radius of the massive object is R=18/8 using units of G=c-1 when :

1) r=R=18/8
2) r=0 and
3) r = R/2 = 9/8?

Sure; this is just the case that Einstein worked out as the minimum possible radius for a body in static equilibrium. He did it using the same application of the EFE that I used in the blog post, though he didn't use the same notation that I used. (Technically, the pressure increases without bound as $r \rightarrow 0$ and $g_{tt}$ becomes zero at some point in the solution, so this case is not actually physically possible; it's a limiting case.)

Also, in your blog does s represent stress and p pressure, or does p represent density?

$p$ is radial stress, and $s$ is tangential stress. In the case of a perfect fluid, $p = s$ and the stress is normally referred to as the (isotropic) pressure.

 

[PeterDonis]

Yes, Peter's does, as long as J(r) becomes constant inside the shell

This is guaranteed by the equation for $dJ / dr$; inside the cavity, $m = 0$ and $p = 0$ so we must have $dJ / dr = 0$.