Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Assuming I have a hollow shell with total mass $M$ and radius $r$. On the surface, the gravitational time dilation would be

$$\tau=t \cdot \sqrt{1-\frac{v_{esc}^2}{c^2}}$$

where

$$v_{esc} = \sqrt{\frac{2 \cdot G \cdot M}{r}}$$

but inside the shell there would be no gravitational field (Newton's shell theorem and Birkhoff's theorem).

But still, the escape velocity required to escape to infinity would be the same as on the surface, since inside the shell you could move without any accelerating or decelerating forces acting on you until you reach the surface, from where you would get pulled backwards.

So is the time dilation inside the hollow shell relative to a field free observer at infinity

  1. zero (I assume it's not) or
  2. the same as on the surface (my best guess), or
  3. something completely different?

I found some not really duplicate but related threads on the interior of black holes which did not really focus on the math, but I am more behind the calculations in terms of $M$ and $r$.

share|cite|improve this question
up vote 8 down vote accepted

For an asymptotically flat metric, the proper time measured by a "stationary" observer (defined here as one whose path through spacetime only has changing $t$, and no changing spatial coordinates) is $$ d \tau = \sqrt{ - g_{tt}} dt, $$ where $g_{tt}$ is the time-time component of the metric. For a "weak" gravitational field, this works out to be $$ g_{tt} \approx - \left( 1 + \frac{2 \Phi}{c^2} \right), $$ where $\Phi$ is the gravitational potential, defined such that $\Phi \to 0$ as $r \to \infty$. Thus, $$ d \tau = \sqrt{ 1 + \frac{2 \Phi}{c^2}} dt. $$ In this form, it is pretty obvious that the time dilation factor is the same everywhere inside the shell, since $\Phi$ is a constant inside a hollow shell (compare the electrostatic equivalent if you're not convinced of this.)

Note that your formula, in terms of the escape velocity, is equivalent to this one if you define the escape velocity at any point as "the velocity for which the object's total energy is zero." (Zero total energy means, of course, that the particle can escape to infinity.) In this case, we have $$ \frac{1}{2} m v_\text{esc}^2 + m \Phi = 0 \quad \Rightarrow \quad v_\text{esc}^2 = - 2 \Phi $$ and your result above is recovered. In this interpretation, the "escape velocity" from inside a hollow sphere would be the same as the escape velocity from the surface: if we launch a projectile inside the shell, it will travel with constant velocity until it reaches the surface of the shell; and if we open a little porthole in the shell at that point for the projectile, it's as if we launched it from the surface with that same velocity.

share|cite|improve this answer
    
1  
Would the same conclusions still hold if the gravitational field was strong instead of weak? – no_choice99 Jan 16 at 18:29
    
Just to get it right: if I place a planet with a mass and radius corresponding to a gravitational time dilation factor of x inside a hollow shell which has (without the planet inside) a time dilation of factor y on its surface the new factor on the surface of the planet which is now inside the shell would be x*y, compared to an observer at infinity? – Симон Тыран Jan 16 at 19:25
1  
@no_choice99: it turns out that for the Schwarzschild metric the weak field expression using the Newtonian potential gives the correct answer, so Michael's conclusion is correct even for objects that are massive enough to be nearly black holes. However this isn't generally true. – John Rennie Jan 17 at 11:33
1  
@no_choice99: In addition to what John said, the notion of "escape velocity" (in my last paragraph) is not related to $\Phi$ in such a simple way when $\Phi$ is large. – Michael Seifert Jan 18 at 16:26

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.