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If mass and energy are equivalent, and if gravitational potential energy is energy, why doesn't an object have more mass when it is at a higher altitude? Does the mass-energy equivalence work for kinetic energy only?

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    $\begingroup$ Mass has an associated energy. Not all energy actually has mass. However it is always possible to say "this is the same thing as IF it has x amount of added mass" $\endgroup$ Jun 23 at 10:11

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An object doesn't increase mass when its energy increases. The rest mass is the rest mass. Even with kinetic energy. If an object is moving, its mass doesn't increase. You can associate a mass to the system, however. So let's say two particles are vibrating in a molecule, each with mass $m_0$, then the molecule's mass wouldn't just be $2m_0$, you have to take into consideration that the particles have energy. The same is with gravitational potential energy. The object's mass doesn't increase when it goes up similar to how it doesn't increase when it moves faster.

That said, let's find out how such mass would work. Similar to how we assumed the molecule isn't moving, but its particles are. Let's assume a large box with the vertical length 1km. If there is an object at the top of the box, it has higher potential energy, so you could say the box has more mass than if the object was on the bottom of the box.

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In the view of general relativity, gravity in reality does not have any form of well-defined potential that can be considered "energy" in the same sense as the components of the stress-energy tensor $T^{\mu \nu}$. The notion of energy as we define it is simply the volume integral of one of the elements of this tensor:

\begin{equation} E = \int T^{00} \, d\mathcal{V} \end{equation}

The effects of gravitation are simply geometry, or more aptly put, the geometric deformation of the spacetime metric $g_{\mu \nu}$. The constituents of your theory when describing a certain system (in the Lagrangian formalism) are explicitly sundered into gravitational ones and matter ones:

\begin{equation} \mathcal{L} = \mathcal{L}_{G} + \mathcal{L}_{m} \end{equation}

Any notion of energy in the classical sense that can be expressed in the stress-energy tensor is strictly part of the matter part of the Lagrangian $\mathcal{L}_{m}$.

The gravitational potential in the Newtonian picture can be seen as a small perturbation of a flat Minkowski metric (in units where $c=1$):

\begin{equation} g_{\mu \nu} = \begin{pmatrix} -1 - 2\Phi & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{pmatrix} \end{equation}

where $\Phi \ll 1$ the gravitational potential that gives the Newtonian law of gravity:

\begin{equation} \nabla ^{2} \Phi = 4\pi G \rho \end{equation}

with $\rho$ being the energy density of what causes this tiny perturbation of the metric, and which is precisely the $T^{00}$ component of the stress-energy tensor:

\begin{equation} E = \int \rho \, d\mathcal{V} \approx M \end{equation}

where the approximation occurs when we consider that the object is non-relativistic in the kinematic sense ($v \ll c$) which means its energy density is roughly equal to its matter density.

To put it more succinctly, in general relativity energy is anything that bends the spacetime, while gravity is the observable effect of that bending i.e. how it impacts the dynamics of those objects who act as sources of energy.

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Well, I would say object has less mass when at lower altitude.

As an example, it's easier to accelerate two boulders when they are close to each other. I mean accelerate to the same direction.

If those boulders are accelerated to different directions, then gravito-magnetism (not mass) makes the required forces larger.

See the accepted answer here: Special relativity and acceleration of an object falling in a direction perpendicular to the direction of motion

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  • $\begingroup$ I am afraid you are confusing mass calculated from energy (E/c^2) with mass meant as inertial mass or rest mass. When you shoot horizontally with rail-gun on Mount Everest or at sea the outlet velocity of you bullet will be the same. $\endgroup$ Jun 23 at 14:22
  • $\begingroup$ @JanGogolin When you shoot horizontally with rail-gun on Mount Everest or at sea the outlet velocity of you bullet will be the same, although the upper capacitor pack contains more energy. $\endgroup$
    – stuffu
    Jun 23 at 15:17
  • $\begingroup$ "Well, I would say object has less mass when at lower altitude." What is your object in my example, the bullet or the capacitor? Maybe you will find useful my recent answer regarding the notion of inertial mass: physics.stackexchange.com/a/714937/281096 . $\endgroup$ Jun 23 at 16:42
  • $\begingroup$ @JanGogolin Answer: Bullet. Let's say we have two coil-guns, far apart. When we move them closer together, some energy is released. We store that energy. I trust that, as you say, the velocity of fired bullets is not affected by this operation. But we can give the bullets some more speed by using the stored energy. But now the bullets have to climb uphill, so that the final speed is unchanged, as it should be. So we took mass from bullets, we accelerated low-mass bullets to extra high speeds, then a system of fast low-mass bullets became a system of slow normal-mass bullets. $\endgroup$
    – stuffu
    Jun 23 at 17:36
  • $\begingroup$ I would like to recommend you to read Evan Currie's "Odyssey One Series" (my favorite SF book), especially part "Holy Ground". They use there some kind of system you are talking about. Regarding your consideration, there is a difference between some system of interacting objects and a single one. As you certainly know the mass of hydrogen nucleus is lower as of proton and neutron disconnected. The energy of missing mass, the so-called mass defect ($\Delta E=\Delta m c^2$) is what we hope to use for energy generation by fusion. $\endgroup$ Jun 23 at 18:34
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When one mass is at some distance from another, we may say there is gravitational potential energy in the system. But where is this energy physically located? It is not inside one mass or the other. Rather, it is located in the gravitational field.

If we adopt the perspective of general relativity, we do not have to speak of a gravitational field, but we can say that spacetime carries this energy. But the physical interpretation of gravitational energy is quite subtle. It is accounted for via the non-linearity of the Einstein field equation.

To help think about this you can notice the following: the total gravitation presented by two close-together masses is less than when they are further apart. To be precise about what I mean by this, consider the case where we have a spherical mass $M_1$ and, around it, a spherical shell of mass $M_2$. Let mass $M_1$ have radius $a$ and let the internal radius of the shell be $b$, with $b$ comfortably larger than $a$. The gravitational effects at some radius $r > b$ can be measured or calculated. Now let $b$ get smaller by allowing the shell to shrink without extracting any matter or energy. It helps to have a precise model in order to track what is happening, so let's suppose that $M_2$ consists of a large number of heavy rocks held apart by rods. To make $b$ get smaller, we let all the rods get shorter, allowing them to be compressed.

In this scenario, as $b$ gets smaller, the gravitational effects at the given $r$ (indicated by things like the period of a circular orbit, the acceleration needed to stay in one place, and the time dilation) do not change. This is associated with a theorem named after Birkhoff. But in this thought-experiment, as $b$ gets smaller there is some elastic potential energy building up in the rods which are keeping $M_2$ from collapsing. So some of the gravitation is owing to this elastic potential energy, rather than to just the rest mass of the rocks of $M_2$. If we now extract that elastic potential energy and carry it away then, after it has been removed, the gravitational effects at $r$ are smaller than they were. So in this end configuration there is the same amount of matter (i.e. the rocks) but they cause a smaller gravitational effect when those rocks are lower down (i.e. nearer to the central mass $M_1$).

In a similar way, if one now supplies energy so as to lift the rocks up again, then the gravitational effects at some given $r > b$ will get larger. In this way the intuition which motivated the question is largely vindicated, but the whole thing is quite subtle when gravitation is involved. In order to speak about mass and energy in general relativity one has to be quite careful to say how they are defined whenever the system in question is large enough that its own gravitation has to be considered.

A good intuition can also be obtained by using the concept of binding energy. The total inertial mass of any bound object is smaller than the sum of the inertial masses of its constituents taken one at a time, and this remains true whether the object is bound together by electromagnetic or gravitational (or other) forces.

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