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What happens to gravity and spacetime when mass turns to energy?

Physics Asked on September 12, 2020

What will happen to the distorted space and time around a mass when it is converted into energy?

Will it go back to its original configuration (i.e. with $0$ gravity)?
Or does space time oscillate? Or is there something else that happens?

3 Answers

Okay, this can get a bit tricky if I just throw the full science at you outright, so what I'm going to do is start with a very unrealistic physical situation and then move on to full realism. So just bear with me.

Unrealistic: let's suppose you took a lump of matter and converted it all into light and that you have a system of magic mirrors (that's the technical term) to bounce that light around and effectively keep it contained to the same volume as the original mass. So far so good? Great. Now it's boring math time:

The mass density of matter is simple to write down. Let's call it $rho$. If we change this to pure energy, then we can write down the energy density:

$$rho=frac{1}{2c^2}left(epsilon_0E^2+{1over{mu_0}}B^2right)$$

The terms in this equation correspond to the energy density of the electric and magnetic fields ($E$ and $B$ respectively) in EM waves and the $1over{c^2}$ out front converts energy to mass to make the equation valid.

Why did I just tell you that? Am I that boring? Am I going mad? Is it actually useful? The answer is both.

In General Relativity (GR), we describe gravity pretty well using the Einstein Field Equations. I won't bore you with the description of those. Short story is that one side describes gravity warping spacetime and the other side describes all things that generate gravity (yeah, that's an interpretation. Sue me if you don't like it). On the latter side, we have something called the Stress-Energy tensor. In this tensor, one of the terms represents the sum Energy density of matter and radiation. The matter component is just $rho$ and the radiation component is the right side of the above equation. Why is this important? Well, if you have only a lump of mass at first and then change it to radiation, the equality above means that the amount of gravity generated doesn't change! (cue gasp of astonishment)

So if you could contain the energy, nothing would change with the gravity.

Realistic: So you have a lump of matter and convert it to energy. Well, being massless, energy has this weird preference of travelling as fast as it can (I know, what a loser). It's also notoriously hard to fully contain. So, unless the object you convert to energy was a black hole, odds are high that the energy will escape your system. What does this mean for the gravity? To an observer outside the light cone, nothing. Mind you, they won't observe the mass being gone either so it's just a regular Tuesday for them. To you, the mass and/or energy will no longer be where it was. In fact, if you're not destroyed by the massive energy release during the conversion, you'll probably be inside of a spherical shell of light. And we all know that inside a spherical shell there's no gravitational force.

So as your light moves outwards, what moves with it is a gravity wave (a rather outstanding one at that). Ahead of the wave, normal gravity; behind the wave, zero gravity.

Doomsday device: Okay, you accept everything so far. Let's say you can convert the mass to energy and, instead of radiation outwards, you can have it all point inwards towards the center of the formerly-existing mass. For a sufficiently massive object, you could get enough photons compressed into a small enough volume (no smaller than the largest photon wavelength across) that it would create a black hole. Once done, this black hole would be fixed (barring evaporation) and could continue to destroy anything nearby (try this at parties to amuse your friends!). It's essentially the ultimate doomsday weapon. No longer do you have to compress the matter itself (durn fermions make that annoying), you can just convert it to energy and let it compress itself. Tada! Instant black hole to engulf anyone foolish enough to question your authority.

Correct answer by Jim on September 12, 2020

To answer this, lets use a single atom as an example. Essentially, the energy contained in that atom (bonds between the sub-atomic particles in the atoms nucleus) radiates away at the speed of light away when released, mostly in the form of Gamma radiation. the gravitational field would radiate out in exactly the same way, since mass is essentially energy stored in the bonds between the particles that make up the nucleus of the atom.

This means the mass really was just stored energy, which is released when those bonds break. It would cause a ripple in the gravitational field. Of course, there being nothing were the atom used to be, there's nothing to attract anything there.

Answered by Russell King on September 12, 2020

There are several ways to convert mass into energy. The mass of a star e.g. decreases slowly due to nuclear fusion processes. In this case energy is radiated away which corresponds to a slight decrease of mass. As the mass of the star remains spherical symmetric during this process, so does its gravitational field and thus the spacetime around the star will not be distorted.

The complete conversion of mass into energy is possible by particle-antiparticle annihilation. Perform this process in a ideal box (withstanding the heat) the weight of the box will not change and neither does the spacetime.

Distortions of spacetime are caused by asymmetric processes (more technically if the mass quadrupole moment of a system changes over time) as during the merger of black holes, supernovae explosions and the like.

Answered by timm on September 12, 2020

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