# Relationship between freefall velocity time dilation and gravitational time dilation in a Schwarzschild metric

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.

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 $$\sqrt{1−r_s/r}$$.

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."

• Isn't there a way for the formatting to indicate that you have copied & pasted directly from your reference? Commented Dec 1, 2018 at 2:58
• @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. Commented Dec 1, 2018 at 3:13
• 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". Commented Dec 5, 2018 at 22:54

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.