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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 $\sqrt{1−r_s/r}$.

safesphere
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Dustin Soodak
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2 Answers2

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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."

safesphere
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1

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.

Michele Grosso
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