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Would it be true to say a person is less massive on the surface of the Earth than they are in space because they have less gravitational potential energy? Would the Earth also be reduced in mass slightly?

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    $\begingroup$ In General Relativity, mass depends on the frame. As I look at you from space, your mass next to me is slightly larger than your mass down on the Earth, while your mass at the black hole horizon becomes zero. Yet as you look at yourself, your own mass is always the same. Your question does not specify the frame of reference thus prompting six misleading answers. Can you please clarify the question? $\endgroup$
    – safesphere
    Jun 25, 2020 at 18:43
  • $\begingroup$ If mass varies depending on frame of reference would this not violate conservation of mass? $\endgroup$ Jun 26, 2020 at 11:08
  • $\begingroup$ There is no such a law as “conservation of mass”, so mass is not required to conserve. There is a law of conservation of energy. This law is not violated, because the total energy remains the same. The energy that was a part of the mass is transferred to the kinetic energy of falling. Mass is frame invariant in Special Relativity without gravity, but not in General Relativity. $\endgroup$
    – safesphere
    Jun 26, 2020 at 15:15
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    $\begingroup$ @DerekSeabrooke Having a conservation law doesn't imply invariance across frames. For example, in special relativity, energy is conserved but its value depends on the frame of reference. There is nothing wrong here because it's still conserved in each frame. This is also true of momentum. Mass, in special relativity, is both conserved and invariant. So it neither changes with time nor does it change depending on your frame of reference. $\endgroup$
    – user87745
    Jun 26, 2020 at 18:08
  • $\begingroup$ @AVS Yes, the mass is both conserved and invariant in special relativity. It's invariant for the obvious reason that it's a Lorentz scalar. The four momentum is conserved in special relativity and thus, it's norm, the mass (squared) is also conserved. Of course, you need to talk about the mass of the whole system if you want to talk about this conservation, just like with any other conservation such as the conservation of energy. Neither mass nor energy is generically conserved for a part of the system, but it is conserved for the whole system. $\endgroup$
    – user87745
    Jun 26, 2020 at 18:52

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The crux of your question doesn't change if we replace "the Earth and the apple" system with a system of two (electrically) oppositely charged particles. I'll do so because I would need to appeal to the mass of the combined system consisting of both the bodies and it is a tricky thing to discuss the mass of extended bodies in the presence of gravity. So, we will ignore gravity and transfer the responsibility of introducing potential energy to the electrostatic interactions between the said charged particles.

So, let's suppose that both the particles have mass $m$ (when they are free, i.e., infinitely far away from each other) and charges $\pm q$. Now, for simplicity, let's also suppose that the particles have a size of radius $r/2$. The idea is that they are not elementary particles, they are rather macroscopic spheres made out of normal material.

First, imagine a system of these particles where the two particles are as close to each other as possible, i.e., pulled in contact with each other owing to their electrostatic attraction (and basically form a bound state). When we talk about the mass of this combined system, we do need to talk about the energy of this system in its rest frame. This energy is $mc^2+mc^2-k\frac{q^2}{r^2}=2mc^2-k\frac{q^2}{r}$. Thus, the mass of the system is $2m-k\frac{q^2}{rc^2}$. So, the mass of the system is a little less than the sum of the masses of its constituents when they were free. This is a general result. The mass of the hydrogen atom is a bit smaller than the sum of the masses of a free proton and a free electron.

However, when we speak of a single particle in the system, its energy is still going to be just $mc^2$ in its rest frame. The reason is that the potential energy is not really stored within any of the particles. It is distributed in the electric fields. So, the mass of one of the particles is still $m$ (the answer to your question). When we talk of the whole system, what we are really calculating is not the energy located within the two particles but the energy of the configuration of two charged particles including the energy stored in the fields that exist along with them. That potential energy stored in the fields is what changes the calculus when we talk of the system as a whole.


Now, despite the issues I raised about mass in the presence of gravity, since the gravitational field of the Earth is pretty weak, basically, a similar calculation would hold for the mass of the Earth and the apple system as well. In short, the mass of the system would be $M+m-\frac{GMm}{rc^2}$ while the masses of the Earth and the apple would continue to be $M$ and $m$ respectively when considered individually.

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    $\begingroup$ What about something with strong gravity like a blackhole? $\endgroup$ Oct 21, 2019 at 14:14
  • $\begingroup$ @DerekSeabrooke A blackhole doesn't necessarily imply strong gravity but in any case, the basic idea is the same. The potential energy is not stored within a particular particle. Rather, it is in the fields--or in terms of GR, it is in the curvature of spacetime. So, no, basically, it wouldn't change the mass of an individual particle. $\endgroup$
    – user87745
    Oct 22, 2019 at 14:19
  • $\begingroup$ I just revisited this and realized something. This calculation cannot possibly be complete! It would imply that the mass of the system would become infinitely negative if the distance r between them became zero. $\endgroup$ Jun 19, 2020 at 7:20
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    $\begingroup$ @DerekSeabrooke You're absolutely right. However, that's an issue with the whole of Newtonian gravity, if you talk about forces between particles sitting on top of each other, it gives you infinities. This simply means that our theory is not valid at those distances. $\endgroup$
    – user87745
    Jun 19, 2020 at 13:31
  • $\begingroup$ @dvij-d-c is that really what it's saying, or is it saying that gravity becomes repulsive at small distances? Otherwise it would seem that the formula given is incomplete and I'd like to know what the complete version is. $\endgroup$ Jun 20, 2020 at 6:08
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A few general comments first. The proper mass of any object is a property characterizing the object, and is invariant of its spatial location with respect to other objects. You might be confused between the weight of an object and its mass, where the weight is proportional to the gravitational force acting on the object. The person in your question has the same mass on the earth and in space, whereas his/her weight on earth and in far space differs because the gravitational potential energy of the person-earth system is different for the two cases.

I will derive a mathematical expression for the gravitational potential energy for the person-earth system. This derivation of the potential energy is valid for any massive spherical object, including a person-black hole or a person-star system as well. Here the mass of the sphere is $M$ and mass of the person is $m$. In the last equation of the answer I have derived the expression for the gravitational potential energy between the person and the earth/spherical object with zero angular momentum, and you can see that as the distance $r$ increases, the potential energy goes to zero. As discussed, the mass is invariant, whereas the potential energy decreases, and hence the weight of the person is zero if $r$ is very large. However close to the surface the weight is non-zero. This is the resolution to your question.

Now I will derive the potential energy. Please see the final equation if the following discussion is a bit technical. The metric outside any such spherical massive object is given by the Schwarzchild metric:

$$ c^2 {d \tau}^{2} = \left( 1 - \frac{r_{\rm s}}{r} \right) c^{2} dt^{2} - \frac{dr^{2}}{1 - \frac{r_{\rm s}}{r}} - r^{2} d\theta^{2} - r^{2} \sin^{2} \theta \, d\varphi^{2}, $$

where $r_{\rm s}$ is the Schwarzchild radius given by $r_{\rm s} = \frac{2GM}{c^{2}}$. Using this metric we will write the geodesic equation, which is the equation of motion of the person situated at a distance $r$ outside the massive object, such that $r$ is larger than the spherical massive object's radius. We integrate the equation of motion and arrive at the expression for the energy of the system.

$$ \frac{1}{2} m \left( \frac{dr}{d\tau} \right)^{2} = \left[ \frac{E^2}{2 m c^2} - \frac{1}{2} m c^2 \right] + \frac{GMm}{r} - \frac{ L^2 }{ 2 \mu r^2 } + \frac{ G(M+m) L^2 }{c^2 \mu r^3}, $$

In writing the equation above I have fixed various constants of motion. I now assume that that there is no angular momentum in the system, i.e. $L=0$. Finally I can read off the gravitational potential energy of the person-earth system as: $$ V(r) = -\frac{GMm}{r} $$

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  • $\begingroup$ @A.V.S. The complete G.R. corrections are given in my second equation, am I missing anything? I set $L=0$ afterwards. In fact, the last term in equation 2 is what causes orbits to precess. $\endgroup$
    – Bruce Lee
    Jun 24, 2020 at 16:38
  • $\begingroup$ @A.V.S. I should have mentioned that I am considering the person as a test object, which anyways should be clear from the context. Also the case I am considering is spherically symmetrical. Also it should be clear that I want to derive an expression for the gravitational potential energy, so I ignored free fall/static on earth question. $\endgroup$
    – Bruce Lee
    Jun 24, 2020 at 17:09
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    $\begingroup$ @A.V.S. The asker commented on the initial answer about black holes and was interested in the potential energy only, so I used the Schwarzchild metric. So I answered the problem in this fashion. I think the person as a test particle approximation is justified here. Of course if you want an answer to your problem, you can go ahead, solve the same and write your answer without any "trickery". $\endgroup$
    – Bruce Lee
    Jun 24, 2020 at 17:14
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I assume that when you say that mass increase in regions of higher gravitational potential, you think that mass increases due to energy. This is the famous result from Special Relativity: $$m = \frac{E}{c^2}$$ But, there is a catch: you cannot use this in this situation. Here's why:

Imagine Einstein throws an apple to Newton? What is the energy of the apple? Special relativity say: $$E_{total} = E + \frac{1}{2} mv^2; E = mc^2$$ $$\therefore{} E_{total} = mc^2 + \frac{1}{2} mv^2$$

Let's give the $E$ term an arbitrary name: innate energy. Now, if Einstein throws the apple faster, the kinetic energy will increase and the total energy will increase by the same amount. But the innate energy will not change. Therefore, whatever alteration to the energy of the apple Einstein does, it does not change the innate energy. In fact, the mass of the apple will remain constant.

Sometimes, when you consider different reference points or perspectives, you may see that an object losses or gains some energy (by for example, giving or taking in light), and the kinetic energy does not change. Where does this energy, call it $E'$ come from? That is when you can use $$\Delta m = \frac{E'}{c^2}$$ To compensate for the extra incoming or outgoing energy, the object losses or gains some mass $\Delta m$.

Coming back to your situation: Gravitational potential energy is just another sort of energy that increases the total energy, not the 'innate' energy.

So, you can say: $$E_{total} = mc^2 + mgh$$

Let's do a thought experiment: imagine you keep a balls of same mass at two heights $h_1$ and $h_2$. The total energies for the two situations are: $$E_1 = mc^2 + mgh_1$$ $$E_2 = mc^2 + mgh_2$$

If $h_2 > h_1$, then $$E_2 > E_1$$ but $$mc^2 = mc^2; \therefore{} m = m$$

What we called the 'innate' energy is the same in both situations and also the mass is the same. So, even if you keep an object on the Earth or the space, the mass of the object can never increase or decrease because of the gravitational potential.

Answering your second question, yes, if you take away a part of the Earth into space, yes the mass of the Earth system decreases. The mass of the Earth is not only due to the planet itself; the objects on it matter too. When you take a bit of mass to space, the mass of the Earth, planet and stuff on it, goes down. But this is way too small to be detected and could cause no difference whatsoever to the gravity or the potential of the Earth.

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    $\begingroup$ +1, this answer is correct as well. $\endgroup$
    – Bruce Lee
    Jun 23, 2020 at 19:15
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In Newtonian theory mass of an object is an intrinsic property so it does not depend on anything let alone potentials. What you might be alluding to is the mass energy equivalence of special relativity which being only a special case of a more general model, doesn't include gravity.

Also potentials in classical physics are arbitrary to an extent because the dynamics is given by their derivative (gradient). So in Newtonian gravity potential of spherically symmetric body is by convention taken as negative and than one has to take negative gradient. But if you define it as positive and take the gradient it's the same. You can also add any constant, it will dissaper by the gradient.

Theoretically energy is a tricky thing to define In great generality, so it's not really sensible to ask such general questions about it without more technical details.

Practically such en effect would be impossible to measure.

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No, the mass of the object doesn't change with its location in a gravitational field.

Suppose for example a small body with a very elongated eliptical orbit around the sun. If we neglect the perturbation of the planets and other orbital bodies, its orbit can be determined knowing its position and velocity in a given instant. And it is independent of its mass because:

$$\mathbf a = \frac{\mathbf F}{m} = \frac{\partial^2 \mathbf r}{\partial t^2} = \frac{-GMm\mathbf r}{mr^3} = \frac{-GM\mathbf r}{r^3}$$

Knowing the mass of the sun, the differential equation is only a function of $\mathbf r$. That is why the orbit is determined if we know the bondary conditions $\mathbf r_0$ and $\mathbf v_0$.

It is possible because the inertial mass $m$ in the denominator is the same as the gravitational mass $m$ in the numerator. Or at least its ratio is constant, and it is embeded in the constant G.

If the mass changed with the gravitational potential, it should be the gravitational mass. There is no reason for the inertial mass (relation $\mathbf F$ and $\mathbf a$) be affected.

In that case, orbits could not be calculated so precise, and in the way they are.

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Mass is proportional to internal energy in the rest frame by $E=mc^2$. For a massive body in an external potential the internal energy thus its mass does not change. For a bound system of two bodies, such as Earth and Moon, their mutual potential energy, as well their kinetic energy, do contribute to the internal energy of the total system hence to the mass of the total system.

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    $\begingroup$ +1: Can the downvoter explain? This is a perfectly correct answer :/ $\endgroup$
    – user87745
    Jun 23, 2020 at 22:13
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Would it be true to say a person is less massive on the surface of the Earth than they are in space ...?

Yes in the frame of the Earth (or in the frame of a distant observer); no in the frame of the person.

Would the Earth also be reduced in mass slightly?

Yes in the frame of the person (or in the frame of a distant observer); no in the frame of the Earth.

Mass is the internal energy of the object that is always the same in the proper frame of this object (assuming no other changes like temperature, etc.). In general, energy is frame dependent. This has no effect on mass in Special Relativity without gravity where mass is invariant. However, the situation is different in General Relativity where mass does depend on the frame of reference.

Imagine a weightless ideal mirror box filled with light. It is well known that the mass of this box is $m=E/c^2$ where $E$ is the energy of the light (in the frame of the box).

99% of the mass of the ordinary baryonic matter consists of the energy of virtual gluons. Just like photons of light, gluons are massless particles moving with the speed of light. So ordinary massive objects (like ourselves) are not conceptually too much different from the mirror box analogy.

Now let’s move the mirror box from space to the Earth and see if its mass changes. The energy of each photon is $E_{\gamma}=hf$ where $f$ is the frequency and $h$ is the Planck constant. Frequency depends on the time dilation. If we observe the box up close, our time moves at the same rate as the time of the box, so the frequency, energy, and mass are the same. However, if we observe the box on tbe Earth from space, we would see that it’s time is dilated relative to ours and consequently the frequency, energy, and mass are smaller.

On the Earth, the mass defect is very small, but scientifically measurable. In a stronger gravity, such as near a black hole, the mass defect can be dramatic. For example, since the time dilation at the horizon is infinite, the mass of a falling object, as observed from afar, approaches zero as the object approaches the horizon. Yet in the frame of the object, its own mass always remains the same.

Would it be true to say a person is less massive on the surface of the Earth than they are in space because they have less gravitational potential energy?

It depends on the way you look at it, but generally yes. The gravitational potential energy is essentially the binding energy that defines the mass defect. They are two equivalent ways to look at the same physics. You just need to be very careful with the choice of the frame of reference, as described.

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  • $\begingroup$ What definition of mass are you using for GR, in particular, how do you express it in terms of stress energy tensor? $\endgroup$
    – user87745
    Jun 27, 2020 at 14:58
  • $\begingroup$ While there is no general definition of mass in GR, mass still can be defined in many specific cases. For example, the ADM mass for asymptotically flat spacetimes, Komar mass for the fields with the asymptotic time translation symmetry, etc. Certainly we can sufficiently define mass to answer this question as well. The stress-energy tensor is the density of the energy-momentum. The “invariant” mass is the length of the total energy-momentum four-vector of the system (e.g. a person). While the tensor is Lorentz covariant, in GR, neither the total energy-momentum, nor the “invariant” mass is. $\endgroup$
    – safesphere
    Jun 27, 2020 at 21:38
  • $\begingroup$ Mass is an invariant in gr. Energy-momentum is a vector and covariant, but by the same token it is defined only in the tangent space at a point. For a distant observer, it appears as a coordinate dependent quantity (e.g. time dilation), but that observer should correct using the metric to find the actual energy momentum. I don't know any other sensible way to define the terms. $\endgroup$ Jul 9, 2020 at 6:42
  • $\begingroup$ @ÁrpádSzendrei Mass of a body is its inner energy consisting of internal movements and interactions. They are processes happening in time, so their energy depends on the rate of time in the same sense as the energy of an emitted photon depends on the time dilation. In a remote frame, time stops at the horizon and so do all processes and interactions that constitute body’s inner energy. Body’s mass becomes zero, but the total energy is conserved, because the kinetic energy proportionally increases. Thank you! $\endgroup$
    – safesphere
    Sep 12, 2020 at 1:56
  • $\begingroup$ @safesphere do you think this is correct: physics.stackexchange.com/questions/580908/… $\endgroup$ Sep 20, 2020 at 16:32

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