1
$\begingroup$

If you drop an object into a gravitational field, is its final velocity equal to what it would have to be in flat space in order to generate the same time dilation that you get at a given radius for an object that is stationary relative to the gravitational body (sitting on the surface in the case that it isn't a black hole)? I don't have enough GR background to do the calculation myself but this seems consistent with the effects on photons going into a gravitational well.

Here's what I've already figured out (mostly from http://jila.colorado.edu/~ajsh/bh/schwp.html)

  1. The distance toward the black hole is contracted/expanded by an amount $\dfrac{1}{\sqrt{1−r_s/r}}$ where $r$ is "circumferential radius" that you get from dividing the orbit length by $2\pi$ and $r_s=2GM/c^2$ is the Schwarzschild radius.

  2. Time dilatation relative to "Schwarzschild time" is $\sqrt{1−r_s/r}$.

$\endgroup$
1
$\begingroup$

The Schwarzschild metric in Schwarzschild coordinates $(t, r, \theta, \phi)$ shows
$ds^2 = -(1 - 2M/r) dt^2 + (1 - 2M/r)^{-1} dr^2 + r^2 (d\theta^2 + \sin^2\theta d\phi^2)$
where:
$c = G = 1$ natural units
$M$ black hole mass
$r_s = 2M$ Schwarzschild radius (event horizon)

The gravitational time dilation measured at infinity (far away from the horizon) vs. the proper time $\tau$ of a stationary observer at a radial coordinate $r$ is
$dt = (1 - 2M/r)^{-1/2} d\tau$

Let us drop an object at rest from infinity. The time symmetry allows to write
$-K_\mu p^\mu = constant = E_\infty = (1 - 2M/r) p^t$
where:
$K^\mu = \partial_t = (1, 0, 0, 0)$ time Killing vector
$p^\mu$ 4-momentum
$E_\infty = m$ energy at infinity (rest energy)
The energy of the object as measured by the stationary observer is
$E = -p_\mu u^\mu = (1 - 2M/r) (1 - 2M/r)^{-1} m (1 - 2M/r)^{-1/2} = (1 - 2M/r)^{-1/2} m $ Eq. (1)
where:
$u^\mu = (dt/d\tau, 0, 0, 0)$ stationary observer 4-velocity
Applying the equivalence principle, from special relativity we get
$E = \gamma m = (1 - v^2)^{-1/2} m$ Eq. (2)
where:
$\gamma = (1 - v^2)^{-1/2}$ Lorentz factor
By comparing Eq. (1) and Eq. (2) we have
$\gamma = (1 - v^2)^{-1/2} = (1 - 2M/r)^{-1/2}$
that is
$v = (2M/r)^{1/2}$ velocity of a free falling object (at rest from infinity) relative to a stationary observer

As you read, the Lorentz factor $\gamma$ (time dilation in Minkowski) equals the gravitational time dilation.

Note: If you want the time dilation far away from the horizon vs. the proper time of the free falling object, you have to compose the two effects.

$\endgroup$
1
$\begingroup$

Yes, this is correct. From Wikipedia:

"Time dilation in a gravitational field is equal to time dilation in far space, due to a speed that is needed to escape that gravitational field. Here is the proof.

  1. Time dilation inside a gravitational field $g$ is $t_0 = t_f \sqrt{1 - \frac{2GM}{rc^2}}$

  2. Escape velocity from $g$ is $\sqrt{2GM/r}$

  3. Time dilation formula per special relativity is $t_0 = t_f \sqrt{1-v^2/c^2}$

  4. Substituting escape velocity for v in the above $t_0 = t_f \sqrt{1 - \frac{2GM}{rc^2}}$

Proved by comparing 1. and 4."

$\endgroup$
  • $\begingroup$ Isn't there a way for the formatting to indicate that you have copied & pasted directly from your reference? $\endgroup$ – D. Halsey Dec 1 '18 at 2:58
  • $\begingroup$ @D.Halsey There are many different ways of marking up the text like quotes, italics, bold, a colored text or background, etc. In this case though a markup like that would look too busy and counterproductive for the numbered list. So instead I have started with a clear reference, so there is no ambiguity. If you know a way of marking up the text without making it harder to read, e.g. a thin box around the whole thing, then I'd be happy to make an edit. In the meanwhile I've added double quotes. $\endgroup$ – safesphere Dec 1 '18 at 3:13
  • $\begingroup$ It is surprising that the formula for escape velocity is the same in GR as for Newtonian mechanics when you simply replace the normal flat-space radius with GR "circumferential radius". $\endgroup$ – Dustin Soodak Dec 5 '18 at 22:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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