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As I understand the mass of an object doesn't increase in a gravitational field according to general relativity. It just follows a geodesic, its worldline.

Now imagine a small marble falling straight down the gravitational field of a supermassive black hole. And imagine a photon falling down a similar trajectory next to it, in a parallel worldline. According to time dilation its energy will increase the further it enters the viscinity of the gravitating black hole. The energy increase is observed as an increase in the photon's frequency.

Now, as the energy increases in the photon due to gravitational time dilation, its energy will eventually be so large that it should collapse into a black hole of its own (imagining the gravitational field that is attracting it is almost infinitely large).

My question is: why doesn't the same thing happen to the marble that follows a parallel geodesic into the gravitational field of the black hole?

And if the marble does turn into a black hole, firstly why, and secondly does it do it at exactly the same altitude as the photon? And as what exactly do we identify the energy increase in the marble, like frequency is to the photon?

Níck Aguiar Alves
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Apsteronaldo
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2 Answers2

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And imagine a photon falling down a similar trajectory next to it, in a parallel worldline.

The marble and photon cannot follow parallel word lines because no local observer ever measures the speed of the marble to be the speed of light. Any local observer will see the photon moving at c and the marble moving at less than c.

Now, as the energy increases in the photon due to gravitational time dilation, its energy will eventually be so large that it should collapse into a black hole of its own (imagining the gravitational field that is attracting it is almost infinitely large).

My question is: why doesn't the same thing happen to the marble that follows a parallel geodesic into the gravitational field of the black hole?

Neither the photon nor the marble become a black hole in their own right. There is an easy way to convince yourself of this. Consider two particles co-moving next to each at a high speed (relative to you the observer) parallel to each other in flat space. The Lorentz transform of transverse force says the gravitational force (or any other transverse force) is less than the force measured in the rest frame of the particles, by a factor of gamma. (Apologies for using gravitational force in a special relativity equation, but the transform of transverse force really does apply to any transverse force. If it did not, we would be able to detect the ether by observing gravitationally interacting objects moving relative to us). Even if you want to consider relativistic mass increase to be a real thing, there is no net increase in the attracting force of the two particles. The marble is simply a bunch of particles and there is no additional force collapsing it.

And if the marble does turn into a black hole,

It doesn't.

And as what exactly do we identify the energy increase in the marble, like frequency is to the photon?

The kinetic energy of a particle in relativity is defined differently to the Newtonian equation. See the calculation of the relativistic kinetic energy in the answer I posted in another question.

We can identify the increase of energy of the marble as kinetic energy pretty much the way we do in Newtonian physics and for both the marble and the photon the loss in potential energy balances the increase in kinetic energy and there is no net gain in energy.

From the comments:

The formulas are correct and they don’t include the “potential energy”, because there is no “potential energy” in General Relativity. It doesn’t exist. The increase in the kinetic energy is due to the mass defect since mass is variable in General Relativity. So your reference to the Newtonian “potential energy” is incorrect. – safesphere

Noether's theorem that relates conservation laws to symmetries, suggests that in special case of the Schwarzschild metric, the metric has a time translational symmetry (does not change with time) and that in that case energy is conserved. See equation (5) of this paper.

Now the Schwarzschild metric is an idealised simplified version of our universe and does not allow for the fact that there are other moving gravitational bodies present and dark energy continually changing the geometry of the spacetime and things get much more complicated. Whether energy is conserved in the general case is hotly debated by the experts with the general consensus that it is not, but there are counter arguments by other experts. See this article outlining some of the main arguments and counter arguments by Philip Gibbs.

KDP
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Your premises are wrong, neither the photon nor the marble turns into a black hole of its own. A black hole needs to be smaller than its Schwarzschildradius in its own rest frame, which is not the case in your scenario. You are also only considering the positive kinetic energy, but the negative potential energy also counts.

The total energy of the marble is $\rm E= m \ g_{tt} \ \dot{t} = m c^2 \ \sqrt{(1-r_s/r)/(1-v^2/c^2)}$, which is a conserved quantity in free fall, so the marble only contributes its rest energy to the system if it freefalls towards the central mass from infinity. A marble hovering stationary close above the Schwarzschildradius on the other hand contributes almost nothing to the total energy.

Yukterez
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