Why does mass increase when gravitational potential energy increases?

I saw a solved example in a book (Concepts of Physics by H.C. Verma, volume 2), where there is a body near surface of the earth, the problem is to calculate the increase in mass of the body when it is lifted 1 meter on the surface of earth. The book assumes that the increase in potential energy goes to the mass of the body, and applies Einstein's formula. I am attaching a link to the screenshot of the book. Check question number 7. (Somehow the uploader on this site says file type not supported)

However, the potential energy can also be thought of as integral over all space of the function $\frac { -g_1 \cdot g_2}{4\pi G}$, where $g_1$ and $g_2$ are the gravitational field of the earth and the body, respectively, and the integral can be shown to be equal to the interaction in potential energy (there is one similar problem with electrostatic potential energy in Introduction to Electrodynamics by Griffiths).

Why do we assume that the increased energy is stored as increase in mass of the body, instead of being stored in the field?

Also, why do we assume that only the mass of the body increases, not that of earth?

• Would be good to see some context from the book. It could just be sloppy language. Einstein's equation tells us that if you increase the internal energy of an object, you will also increase its mass - hence the $E=mc^2$ equation. In this situation, the increase in mass should correspond to an increase in the total mass of the Earth-body system treated as a single bound object. Commented May 2, 2018 at 18:49
• @enumaris I added a link to the question Commented May 2, 2018 at 19:24
• The text seems intentionally ambiguous to me. It says "the increase in mass", not "the increase in mass of the body" like your question seems to state. I see nothing in the text's question or answer that is incompatible with assigning the increased mass to the entire system. Commented May 2, 2018 at 21:25

Mass is a tricky concept in general relativity. Since it would take too long to explain this in full, I will just make a few relevant remarks.

First, suppose we have a spherical planet. The mass $$M$$ of the planet can be measured by measuring the periods of orbiting bodies. If we now let an object fall onto the planet, then the mass of the resulting planet will be $$M' = M + E/c^2$$ where $$E$$ is the total energy of the body which fell, including both its rest energy and its kinetic energy.

For the next experiment, suppose we have an object of mass $$m$$ sitting on the surface of the planet of mass $$M$$. The total mass of the system is then $$M + m + w$$ where $$w$$ is a term (negative) accounting for the binding energy (I am not going to calculate it). If we now raise the mass $$m$$ then this requires some energy to arrive in order to do the work of raising the mass. If that energy comes from elsewhere on the planet then, by energy conservation, the total energy of the overall system does not change and neither does its mass. If, on the other hand, the energy is supplied from elsewhere (e.g. it arrives in sunlight or something) then the internal energy of the planet+object goes up, and therefore so does its rest mass. This increase can be interpreted as a reduction in the binding energy $$w$$.

This is not the only way to discuss the concepts, however. When I asserted that $$m$$ is the rest mass of the object, you could ask "rest mass at what gravitational location?" It is not easy to compare masses located at two different locations. If we want to compare them we might consider first bringing one of the masses to the location of the other. But in order to do that we shall need to do some work. So where does the work go? Does it change the rest mass of the object we moved? The answer depends on how one is using the term "rest mass". As I said at the start, this is quite a tricky area. If you lower an object slowly to the ground on a rope, then you gain some energy at the top of the rope. If you lower an object slowly to a black hole on a rope, then you gain the whole rest energy of the object, and when it is released and goes into the black hole, the mass of the black hole then does not increase!

Mass corresponds to its internal energy as @enumaris has pointed out in his comment. The internal energy of a mass depends on the kinetic and potential energy of its constituents. This means that the mass as a whole doesn't change if its potential energy changes.

As an aside, mass is invariant and thus the same in all frames of reference.

• What about the energy stored in the field? It also increases Commented May 3, 2018 at 3:04
• If you mean the gravitational energy of a mass m due to an attractive mass M at distance r that is $U=-GMm/r$ which equals the work done to bring m to this point. Otherwise see energy of gravitational field
– timm
Commented May 3, 2018 at 7:42
• EDIT it should read: gravitational potential energy.
– timm
Commented May 3, 2018 at 14:25
• Yes, this potential energy is stored in the field. The integral of $\frac { -g_1 \cdot g_2}{4\pi G}$can be shown to be equal to total interaction potential energy. So why are we assuming that it is stored as mass, not in the field? This was my question. Your answer is not very clear to me Commented May 4, 2018 at 6:33
• "mass is invariant" in newtonian physics and galilean transformations, it should be "rest mass is invariant" You are right it is unnecessary to call it invariant mass, but to call it restmass is unnecessary either. Restmass and invariant mass means the same. Many people say just mass in contrast to relativistic mass which is something else.
– timm
Commented May 5, 2018 at 9:12