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My concern relates to Einstein's rock-down-a-well thought experiment. If Einstein drops a rock, it picks up velocity and gains in total energy. If an assistant at the bottom of the well converts the rock into a photon by $E=mc^2$ it will have more energy than if Einstein had done that atop the well. When the assistant shines the photon back up to Einstein, he must find that the energy has decreased, or he will be able to violate the conservation of mass and energy by forming a bigger rock. This is one way to show that time must slow with elevation in order to exactly offset energy gained from gravitational potential energy conversion.

So what happens if the assistant is right outside a black hole? In standard theory, the rock arrives with a less than infinite KE, yet there can be any ratio of time speed we want right up to infinity. That is, convert the rock into a photon, and shine the photon back to the origin, and you can red-shift it to as small an energy as you want, and eliminate mass and energy from the universe.

Is this a violation of conservation of mass and energy?

Ralph Berger
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  • Why do you think time dilation could be anything you want near the event horizon? – Adrian Howard May 18 '21 at 00:10
  • "it arrives to the horizon with the kinetic energy of $K=mc^2$" That makes the problem worse. You've made the gravitational potential energy vanish. Anyway, the assistant is in a place of slow time. He is the one who thinks that there is a lot of energy there to make a energetic photon. It is only when the photon returns to the source that it is redshifted to next-to-nothing. – Ralph Berger May 18 '21 at 15:08
  • "Why do you think time dilation could be anything you want near the event horizon?" A lot of comments on the web, here and elsewhere, but I am suspecting that is wrong. Einstein said it, too, of course, writing in 1939 “a clock kept at this place would go at the rate zero.” At https://physics.stackexchange.com/questions/48026/in-general-relativity-gr-does-time-stop-at-the-event-horizon-or-in-the-centra#:~:text=If%20you're%20sitting%20outside,stops%20at%20the%20event%20horizon, it states "the clock slow as it approaches the horizon and you'll see it running slower and slower" forever. – Ralph Berger May 18 '21 at 15:17
  • @safesphere - If, for the rock, $E=\frac{dt}{d \tau}mc^2$ and if $\frac{dt}{d \tau}$ approaches infinity at the event horizon, will the assistant not see that the rock is approaching lightspeed? That is really at the root of my question. I am told that objects do not reach lightspeed at the event horizon, but your comment and my thinking imply to me that, in the assistant's reference frame hovering right outside the event horizon, the only possible velocity is light speed. – Ralph Berger May 18 '21 at 17:36
  • @safesphere I am happy with that, so long as it asymptotically approaches light speed. Some people use $v = c\sqrt{r_s(1/r - 1/R)}$ to argue that the speed is well below lightspeed. I am pretty sure that equation is not useful because one cannot define both r and R from the same reference. I have convinced myself that the proper equation is $v = c\sqrt{r_s(1/r - 1/(r+\gamma_g^3(R-r))}$ where $\gamma_g$ is the gravitational time dilation. – Ralph Berger May 20 '21 at 20:13

2 Answers2

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The time dilation factor applied to the emitted photon exactly compensates the factor by which the total rock energy has increased by falling at any $r$ and there is no energy conservation problem.

Let's make it simple and assume the rock is originally at rest at infinity, with energy $E_\infty = mc^2$, and falls radially inwards.

A static observer (at $r> r_s$ of course), will observe the speed of the rock as it passed to be $c\sqrt{r_s/r}$. The rock energy (in the local inertial frame of the static observer) will be $$E = \gamma mc^2 = \left(1- \frac{r_s}{r}\right)^{-1/2} mc^2\ .$$

If the energy of the rock is turned into a photon of frequency $\nu_e= E/h$ and emitted radially outward, then the photon will be redshifted. The factor by which it is redshifted (according to an observer at infinity) is given simply by the Schwarzschild metric as $$\nu_{\rm obs} = \nu_e \left( 1 - \frac{r_s}{r}\right)^{1/2} = \frac{mc^2}{h}\ .$$ Thus the photon has an energy of $E_\infty$ and energy is conserved.

ProfRob
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  • I agree with that! Here's my dilemma. I drop the rock from 3rs, say. I believe it will reach lightspeed at rs (or at least, asymptotically approach it). Then all is well. As it approaches the event horizon, time dilation goes to infinity and the kinetic energy goes to infinity, and reversing the path leads to the original conditions. But if, as some say, the object reaches an area of infinite time dilation due to gravity at less than lightspeed, then your formula will allow me to redshift back to $v_{obs}$ as close to zero as I want, thus making the object vanish. – Ralph Berger May 20 '21 at 20:06
  • that is, I contend and you seem to support me, that it is a necessary condition that anything that reaches the event horizon must be at lightspeed, even it it is dropped in with zero velocity from a meter away. – Ralph Berger May 20 '21 at 20:08
  • @RalphBerger it doesn't matter where you drop it from, a similar energy-conserving proof can be done and yes, the speed approaches $c$ for all cases. The dilation for $\nu_{\rm obs}$ changes appropriately if the observer is not at infinity. – ProfRob May 20 '21 at 22:12
  • "The speed approaches c for all cases." Thank you very much! I have been in arguments with others, including Roy Patrick Kerr, about that. – Ralph Berger May 20 '21 at 22:24
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In standard theory, the rock arrives with a less than infinite KE, yet there can be any ratio of time speed we want right up to infinity

These two things, the amount of KE gained and the amount of time dilation are not independent, as you seem to imply. The further you fall the more KE you gain and the more gravitational time dilation you undergo. The two are tied together. The time dilation is always the precise amount needed to compensate for the KE so that energy is conserved. This can be calculated, and it is not trivial but it is straightforward.

Dale
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  • That is my understanding and I agree (although I actually think it is easy to calculate). But then - if time dilation is infinite at the event horizon, doesn't that mean KE must approach infinity, hence velocity must approach lightspeed? The thing that I think violates conservation of energy is when I hear people say that the velocity is less than lightspeed when reaching the event horizon even as time dilation due to gravity is infinite. I think those people are wrong and you and I are right. – Ralph Berger May 20 '21 at 19:57
  • The coordinate speed is less than c at any location outside of the horizon. The Schwarzschild coordinates don’t exist at the horizon. It doesn’t really make sense to talk about the speed at the horizon. – Dale May 20 '21 at 20:43
  • Hence the phrase "approach lightspeed". If right outside the horizon, and a signal is emitted, we can figure its velocity by the redshift (due partly to gravity, but also due to doppler), so the velocity still has meaning. The time dilation due to KE and GPE for a drop from 1 meter to a 10,000 m black hole is: – Ralph Berger May 20 '21 at 22:11
  • The time dilation due to KE and GPE for a drop from 1 meter to a 10,000 m black hole is, 1e-5m from the horizon, 97,071 and 10,000, respectively. At 1e-9m, it is 3.16e6 for both effects (differs by 0.06%) as velocity has hit within 0.000015 m/s of lightspeed. I have a paper under review, just checking here if I am missing something. If any physicist want to vet, improve, and co-author, it might end up an interesting addition to a publications list. – Ralph Berger May 20 '21 at 22:20