# How is energy conserved in this scenario?

There are two objects A and B of equal mass (m). They are laying in close proximity to each other. Object A is then moved further from object B, to a distance r. Since there is gravity between these objects, hence energy is utilized for this purpose. This energy is thereafter stored in both these objects as gravitational potential energy, and should be available for utilization (according to the law of conservation of energy).

Object B is later converted into energy, which is equal to mc². Now that object B is gone (converted into energy), what happened to the gravitational potential energy of object A? Has it just vanished because object B is no longer present, and hence there is no longer any gravitational pull which was present earlier.

How is the total energy of this closed system conserved?

• It is better to think of the equation $E =mc^2$ as a true equality rather than a conversion. Commented Jun 14, 2015 at 17:58
• First of all, one can't simply "convert an entire object to energy". That's ruled out by the laws of thermodynamics and microscopic conservation laws. Secondly, mass-energy is one concept, not two and the gravitational interaction associated with it is the same no matter what the "object" is made of. That's one of the main points of the equivalence principle. Commented Jun 14, 2015 at 17:58
• If you split a larger nucleus into a smaller one, the effect remains the same i.e. some (albeit extremely little) mass is converted into energy and is no longer active in the gravitational equation between the objects. So even during radioactive decay, some matter is converted into energy, and hence is "lost", unless explained. Commented Jun 14, 2015 at 18:03

It is better to think of the equation $E =mc^2$ as a true equality rather than a conversion. Mass is energy.

If one has that mindset, then it is intuitive that energy has a gravitational field. A hot cup of tea weighs more than a cold one.

• But the photons produced from the energy conversion would dissipate in (apparently) random directions. If they all had directed right towards object A, then it would make more sense. Commented Jun 14, 2015 at 18:05
• @YoustayIgo Photons have a gravitational field too. Energy has a gravitational field. Commented Jun 14, 2015 at 18:09
• Right. But before being converted into photons, the matter was all in one place, exerting one-directional gravitational force on object A. After being converted into energy, the photons will disperse in random (each angle having equal probability of photon dispersal) directions and their overall effect would be cancelled out by each other (a photon dispersed to the left side of object A will cancel out the gravitational effect of the photon dispersed at the right side of it). Commented Jun 14, 2015 at 18:22
• @YoustayIgo: If you disperse all the atoms of object B, then the gravity of these atoms will also mostly cancel out. No difference. Commented Jun 14, 2015 at 23:19
• Yes. That's what my point is. What happened to the energy we used in displacing the objects? Apparently this energy is lost when some of object B is converted into photons. Commented Jun 16, 2015 at 7:18

By "converting to energy," I'm assuming you mean converting part of the mass of the system to photons. These photons will quickly zoom off to large distances, making them irrelevant gravitationally. The original gravitational potential energy was caused by the original masses A and B warping spacetime. This warping is not instantaneously changed when mass B is converted to photons, but propagates at the speed of light (as far as we can tell from theory and indirect measurements). The propagation of the change of spacetime warping is a wave that carries energy away from the system. Any rapid change in the distribution of mass and energy in a section of space will cause gravitational waves to be emitted.

Before the annihilation of B, you have mass-energy and gravitational potential energy. Afterwards, you have kinetic energy (of the photons), mass-energy (of the remaining mass A), and gravitational wave energy.

Gravitational waves have been indirectly observed by the decay of orbiting neutron stars. The emission of these waves saps energy from the binary stars' orbits and causes them to slowly fall towards each other. There are several large projects around the world that are using lasers to attempt a detection of gravitational waves from supernovas and colliding black holes. LIGO is the most well-known currently running experiment of this type.

• With much confusion and reluctance, I am placing this answer as the solution. Although I am not 100% sure to what you are intending to mean, but it looks better than the other answers and I don't want to look like a douchebag, not accepting any answer. Commented Jun 20, 2015 at 11:03
• I'm sorry for writing a confusing answer. Feel free to ask questions so I can clarify. Commented Jun 21, 2015 at 4:43

I would need to look into this more, but I believe gravity couples to energy rather than the Lorentz invariant mass $m$. If this is true, then it doesn't matter if you view the mass of particle B as mass or energy.

• The issue with this is, that when the energy transforms into photons and those restless kids run away in all directions, the total energy-mass sum left with object B is lesser than before (since the photons have escaped). So what happened to the gravitational potential energy of both objects? Is it lost? Commented Jun 20, 2015 at 11:00
• As mentioned by @Mikael Fremling, if the photons are running away then since energy is conserved the photons must lose kinetic energy as they gain gravitational potential energy. When photons lose energy they are red shifted. Energy conservation is a fundamental consequence of time translation invariance, so it is one of the most basic requirements for any theory of an isolated system. Energy conservation is never violated for a system obeying time translation invariance.
– Ian
Commented Jun 26, 2015 at 21:46

You have to think of what happens when object $B$ is "converted to energy". Since there is no such thing as raw energy (dark energy excluded) $B$ has to be transformed into other particles and or kinetic energy. As many of the other comment have mentioned the kinetic energy also has a gravitational pull, so initially after the "conversion" nothing has really happened.

For sake of argument, say that $B$ was transformed into photons (the closest thing you can come to "pure" energy). These photons will now disperse and as they do so the energy stored in the gravitational potential disappears.

What conserves energy here is that the photons will be red-shifted (and thus loosing energy) as they climb the gravitational well around object $A$.

• Is the red shift perspective, your idea or a proven scientific fact? Commented Jun 16, 2015 at 7:16
• @YoustayIgo:The redshift is a real thing. en.wikipedia.org/wiki/Gravitational_redshift Commented Jun 17, 2015 at 5:18
• Yes I am aware of the concept. My question was, whether photons moving through a gravitational field undergo redshift or is it that photons emitted from a BODY MOVING AWAY produces photons which, when viewed collectively, show redshift? Commented Jun 17, 2015 at 6:06
• @YoustayIgo: The phenomena should be valid for individual photons escaping a gravitational potential. However, your remark is correct that photons emitted from a moving body also exhibit red-shift, but this is simply the ordinary Doppler effect. Also, you can not change the wavelength of light by adding more photons, only the strength. Commented Jun 17, 2015 at 10:00

Here you have to be clear how you are looking at the energy. Of course all energy can take the form mass times c2. In this particular case of object A and B we have a total energy equal to mc2 where m is mass of A plus Mc2 where M is mass of B. With this you subtract the gravitational potential energy since it is attractive. You then get the system total energy. If you take away from this the energy Mc2 plus the gravitational potential energy you will be left with the mc2 part of energy. In other words taking away Mc2 alone is not sufficient for energy conservation.

• This is hard to follow, and I don't think it answers the question.
– Ian
Commented Jun 15, 2015 at 5:11