# Spacetime geometry measurements inside an extremely dense spherical shell

From http://jila.colorado.edu/~ajsh/bh/schwp.html The amount by which radial distance is "squished" is:

$$\dfrac{1}{\sqrt{ 1 - r_s/r}}$$

where the Schwarzschild radial distance (or "circumferential radius") r is the point at which measured circumference is $$2 \pi r$$ and the Schwarzschild radius is

$$r_s=2 G M/c^2$$

is the Scharzchild radius at which we get a black hole for that amount of mass ($$M$$). The amount by which time is dilated is the inverse of this (so pretty similar to the case for moving objects in flat space). Usually, this equation is integrated over dr to calculate the total distance between 2 points near a dense gravitational body.

Lets assume we have a thin spherical shell that is dense enough to make 50% time dilation immediately above the surface, which I believe comes out to

$$r=(4/3)r_s$$

The radial distance will be squished by 50% as well (though some people refer to this as "expansion" since you can fit more into the same space relative to the reference frame outside this gravitational well) and the circumference of the outside of the sphere will be

$$2\pi(4/3)r_s$$

According to Birchoff's theorem, the apparent gravitational field should be zero inside (just as it is for Newtonian gravity).

According to an answer to the question at Does a massive spherical shell expand the time inside itself?, "gravitational time dilation depends on the gravitational potential" so should be the same inside (which makes since if you think of photon red-shift as an indication of time difference).

It seems like you would still have the same length contraction inside as well (if it is just as much an effect of gravitational potential as time dilation) but it would be the same in all directions, which implies that the measured inner circumference of the sphere would be twice the outer circumference (so $$2 (2\pi(4/3)r_s)$$ ) and the inner (measured) radius would be $$2 ((4/3)r_s)$$.

Is this right?

Dustin Soodak

• Welcome to physics.SE. Please mark up your question using mathjax. That means basically putting dollar signs around it and then using LaTeX math inside. There are tutorials you can find online for how to do math in LaTeX. – Ben Crowell Nov 28 '18 at 1:30
• Got the LaTeX in...much easier than I thought it would be. – Dustin Soodak Nov 28 '18 at 18:24
• Please note that the "Schwarzschild Metric" section in the first link is incorrect. – safesphere Nov 29 '18 at 14:05
• what about it in particular? A typo in the equation for Schwarzschild Metric itself? – Dustin Soodak Dec 3 '18 at 22:41

• Just to make sure I understand this, if the circumference outside the sphere is $2\pi(4/3)r_s$, the circumference inside will be the same and the inner measured radius will be $(4/3)r_s$ but the time dilation will still be at 50%. This is the same as immediately outside where there is also 50% radial length contraction. Correct? – Dustin Soodak Nov 29 '18 at 7:28
• @DustinSoodak I haven't done the math (although it's not complicated), but intuitively, you "see" the event horizon so far away only if you hover over it using rocket engines to hold you steady. However, if you are in a free fall, then your speed approaches the speed of light and the length contraction due to the relativistic speed cancels out the gravitational length expansion. The % difference ($f(t)$ in the link, 50% for $\dot{R}=0$) for a collapsing sphere also is due to the relativistic speed of the collapse ($\dot{R}$). Frame dragging refers to rotation, but there is none in this case. – safesphere Dec 1 '18 at 1:05