Does E=mc^2 apply to gravitational potential energy?

[snoopies622]

TL;DR Summary

does the mass of a falling object increase, decrease, or stay the same?

I'm reading Schutz's A First Course In General Relativity and in chapter 5 he discusses an idealized experiment in which an object is dropped from a tower, then turned into a photon and sent back up to its original height.

In classical mechanics we would say that as the object falls it loses potential energy and gains equal kinetic energy so at every moment its total energy is constant. In relativity, if an object gains energy its mass increases according to [itex] \Delta m = \Delta E / c^2 [/itex]. Does this apply to both kinetic and potential energy so that — as in classical mechanics — the object's mass remains constant as it falls?

On the one hand, energy can't just disappear and as I understand relativistic mechanics, [itex]E=mc^2[/itex] applies to potential energy as well as kinetic. On the other hand, the whole point of the tower thought experiment is that the object has more mass[energy] after it has fallen to its lower position.

 

[Dale]

Relativistic mass is a concept that has been largely discarded. When physicists today just say "mass" they mean the invariant mass, not relativistic mass.

 

[Ibix]

Gravitational potential energy (to the extent it can be defined in relativity) isn't a property of the body, but of its interaction with the gravitational field. It's very difficult to pin down "where" gravitational energy is, but you can loosely think of it as a property of the field rather than the body.

So the total energy of the body is higher when it's moving relative to a hovering observer than when it isn't. Probably best not to think of that as mass, though, since GR is formulated in terms of invariants.

 

[Vanadium 50]

I think what people are saying is that your question is a hodgepodge of Newtonian mechanics, SR and GR. It's probably unanswerable in its present form. You might think of rephrasing.

By the way, if you are 5 chapters into Schutz and are still thinking in terms of relativistic mass and energy as "localizable stuff", you are likely to have an uphill slog.

 

[Nugatory]

In relativity, if an object gains energy its mass increases according to Δm=ΔE/c2.

You'll hear that a lot, but it is not quite right, and the inaccuracy can lead to all sorts of misconceptions. It is more accurate to say that if a system with zero total momentum gains energy its mass (necessarily rest mass, because we're talking about a system with zero momentum) will increase according to ##\Delta{m}=\Delta{E}/c^2##. Because ##c^2## is the conversion factor between measuring energy in units of kilograms and in units of Joules (or whatever other units you choose), this statement is basically a tautology: If a system gains energy it gains energy.

Does this apply to both kinetic and potential energy so that — as in classical mechanics — the object's mass remains constant as it falls?

Here the system of interest is not the object, but the object and the earth. If we were to enclose the entire earth/object system in a single huge box, we would find that the mass of the box and everything in it remained constant: the kinetic energy of the object goes up as it falls, the potential goes down by the same amount, the total mass/energy inside the box is unchanged.

On the other hand, if we were to reach into the box and apply a force to lift the object away from the Earth (strictly speaking we would have to apply an equal and opposite force to the Earth to maintain the condition of zero total momentum) then we're adding energy to the system from outside by doing work on it, and we would find that the mass of the box will increase accordingly.

But note that there is no way of associating this mass increase with the Earth or the object and saying that the mass of of either has increased. It's a property of the system as a whole, the box and its contents.

 

[snoopies622]

Thanks all, I appreciate your efforts.

Let me bring this down to Earth: Suppose I have a labratory which includes a tower, and I drop an object from the top of the tower, and — using measuring devices that are at rest relative to the tower — I measure the object's mass as it falls. It sounds like, based on

So the total energy of the body is higher when it's moving relative to a hovering observer than when it isn't.

that I would in fact measure an increase? (It's fine with me if we call this energy instead of mass.)

 

[Ibix]

Suppose I have a labratory which includes a tower, and I drop an object from the top of the tower, and — using measuring devices that are at rest relative to the tower — I measure the object's mass as it falls. It sounds like, based on

that I would in fact measure an increase?

No, because "mass" means invariant mass, which doesn't change by definition. If you measure the relativistic mass, then yes that will increase by definition.

 

[Dale]

It's fine with me if we call this energy instead of mass.

Then please call it energy instead of mass.

 

[PeterDonis]

I would in fact measure an increase?

What are you measuring, and how are you measuring it? Saying...

using measuring devices that are at rest relative to the tower — I measure the object's mass as it falls

...is pretty vague. How are you going to measure the mass of something that's moving relative to the measuring device?

 

[Vanadium 50]

I measure the object's mass as it falls.

How do you do this? Surely not by grabbing it and putting it on a scale.

 

[pervect]

Thanks all, I appreciate your efforts.

Let me bring this down to Earth: Suppose I have a labratory which includes a tower, and I drop an object from the top of the tower, and — using measuring devices that are at rest relative to the tower — I measure the object's mass as it falls. It sounds like ... that I would in fact measure an increase? (It's fine with me if we call this energy instead of mass.)

The energy of a moving object is higher than the energy of an identical stationary object. Since you specified "measuring" it, you might want to sketch out how you propose to do your measurment. Alterntively, you can treat it as a question about defintions, rather than experiment, and say that by definition, if you have two identical objects, one moving, and one stationary, the moving object has more energy.

Note that this implies that the energy of an object is not solely the property of the object - you can't assign an energy to an object until you know whether it's moving or not, which means you need to specify the frame of reference of an object before you can know it's energy.

 

[Nugatory]

which means you need to specify the frame of reference of an object

Somewhere between a quibble and scratching a pet peeve... but should that not be “the frame of reference you have chosen to use”?

 

[snoopies622]

This reminds me of a question I asked here years ago: If I put a mousetrap on a scale and then set the mousetrap, would it weigh more? The answer came back: In principle, yes.

 

[PeterDonis]

If I put a mousetrap on a scale and then set the mousetrap, would it weigh more? The answer came back: In principle, yes.

Yes, because you setting the mousetrap requires reaching in from the outside and doing work, which adds energy to the mousetrap, for the reasons @Nugatory gave in post #5.

Consider an alternative scenario: you have a mousetrap with a charged battery and a battery-operated mechanism that sets the mousetrap, controlled by a timer. Before the timer triggers the mechanism, you put the mousetrap, including the battery-timer-mechanism, on a scale. When the timer triggers and the mechanism sets the mousetrap, the weight recorded by the scale does not change, because the energy used to set the mousetrap is now coming from something internal to the system--the battery--instead of being put in from outside.

 

[jartsa]

Let me bring this down to Earth: Suppose I have a labratory which includes a tower, and I drop an object from the top of the tower, and — using measuring devices that are at rest relative to the tower — I measure the object's mass as it falls.

Let's say that at the top of a tower a pea is turned into light, which is sent down to the middle of the tower, where the light is turned into a potato, which is then turned into light, which is sent down to the ground floor of the tower, where the light is turned into a melon.

Let's say that the locally measured inertial mass of the pea is one gram, the locally measured inertial mass of the potato is 100 grams, and the locally measured inertial mass of the melon is 10 kg.

(A more complete description of the thought experiment in the book would be nice)

 

[PeroK]

Let's say that at the top of a tower a pea is turned into light, which is sent down to the middle of the tower, where the light is turned into a potato, which is then turned into light, which is sent down to the ground floor of the tower, where the light is turned into a melon.

Let's say that the locally measured inertial mass of the pea is one gram, the locally measured inertial mass of the potato is 100 grams, and the locally measured inertial mass of the melon is 10 kg.

Is that from a first course in physics by Monty Python?

 

[PeroK]

Thanks all, I appreciate your efforts.

Let me bring this down to Earth: Suppose I have a labratory which includes a tower, and I drop an object from the top of the tower, and — using measuring devices that are at rest relative to the tower — I measure the object's mass as it falls. It sounds like, based on
that I would in fact measure an increase? (It's fine with me if we call this energy instead of mass.)

Here's the real issue. If you were ready for the Schutz book, you would already know that some older texts use the concept of relativistic mass while modern texts generally do not. You would have a reasonable grasp of why this is and - although ultimately it's a matter of taste - the advantages of avoiding relativistic mass.

You would also be able to laugh at the pop-science obsession with relativistic mass as a cornerstone of SR.

That you are unsure about this suggests that you need significantly more study of SR before you tackle GR.

 

[vanhees71]

That's really interesting. I thought that the very inconvenient theory of "relativistic mass" is only used in some old-fashioned textbooks on SR. You can get away with this confusing concept in SR and that's why unfortunately it's still used even in some newer textbooks.

It's new to me that this concept is also used in GR textbooks. I cannot conceive how you get away with this, using non-covariant concepts in GR. Does Schutz really use "relativistic mass" in GR? I cannot believe it, before I've seen it!

 

[Ibix]

Does Schutz really use "relativistic mass" in GR?

No, at least not on page 112 which discusses this "drop a massive particle, convert it to light and shine it back up" way of deducing the existence of gravitational redshift. He uses "rest mass" and "total energy". Nor does the term appear in the index.

 

[Vanadium 50]

Does Schutz really use "relativistic mass" in GR?

He does not. He does say "rest mass" in one place.

 

[vanhees71]

I'd have been very surprised by the opposite. I've no clue how to use a concept like relativistic mass in GR, as it seems very difficult, if not dangerous, to work with non-covariant objects. One must not forget that the general covariance is a gauge symmetry, and to introduce objects with unclear definitions of how to transform under diffeomorphisms of the coordinates is thus very dangerous.

 

[pervect]

I'd have been very surprised by the opposite. I've no clue how to use a concept like relativistic mass in GR, as it seems very difficult, if not dangerous, to work with non-covariant objects. One must not forget that the general covariance is a gauge symmetry, and to introduce objects with unclear definitions of how to transform under diffeomorphisms of the coordinates is thus very dangerous.

I don't have Rindler's textbook, but I've read it in the past, in response to some other posts by some long-gone (banned from PF for incorrect claims) relativistic mass enthusiasts.

From memory, I believe Rindler was fond of talking about mass as the ratio of 3-acceleration to 3-force. Rindler used his concept of mass to motivate the stress-energy tensor, and he'd describe the effect as a rod under pressure as having more "mass" (by his defintions) depending on the longitudinal pressure.

Understanding the 4-dimensional stress-energy tensor - or at least accepting it well enough to work with it - is a key requirement to proceed in General Relativity, one that I believe many students struggle with. Textbooks, for the most part, don't really try to motivate the stress-energy tensor, though they all present it and show how to use it.

I do remain convinced that it is both simpler and more helpful to students to introduce other 4-dimensonal objects, such as the 4-dimensional 4-velocity and the 4-force, before going on to the 4 dimensional stress-energy tensor. But of course students and readers have their own ideas. I tend to believe that most students who refuse to learn about 4-vectors cripple themselves, but there is little I can do about it other than to suggest that they not do that.

The tie into "relativistic mass" here is that the 4-vector approach naturally leads to the invariant mass, which is the "length" of the 4-vector, and that "relativistic mass" in the 4-vector formalism is just one of the compnents of the 4-vector, namely the energy component of the energy-momentum 4-vector.

Using relativistic mass is in my mind associated with not having learned or refusing to learn about 4-vectors. Presumably this seems easier to the student at the time, but I believe in the end it hurts their understanding. Anyway, to my mind, the whole relativistic mass issue is a speedbump on the way to understanding the stress-energy tensor. It's not so important in and of itself, because the whole idea of mass is going to wind up being replaced by other notions anyway. At least it well if things go well. Sometimes things do not go well.

Einstein's field equations do not describe gravity in terms of "mass", but in terms of the stress-energy tensor.

Anyway, there is a small possibility that someone who simply cannot let go of "relativisitc mass" might be helped by Rindler's old textbook. But in general I think that the best thing to do is to learn the 4-vector approach.

 

[jartsa]

In classical mechanics we would say that as the object falls it loses potential energy and gains equal kinetic energy

Let's first put a photon in a box at the top of a tower and then lower the box down to the bottom of the tower. Then let's shine the photon back up and lift the box back up.

A box that had gravitating energy ##E_{photon} + E_{box}## was lowered.

A box that had gravitating energy ## E_{box}## was lifted

Clearly the agent that lowered and lifted, gained some net energy in the process.

Now, as I want to stay in the mainstream, I must say that during the lowering the lowering agent took energy from the box-photon-planet system. And when the photon climbed back up the photon lost energy to the box-photon-planet system.

Now I have a problem: Should I say that when the photon climbed back up it gained potential energy, or should I say that it did not gain potential energy?? Oh yes, it is the box-photon-planet system that gains potential energy, not the photon.

So, does the problem go away, if objects do not lose or gain potential energy when they climb or fall?

 

[PeterDonis]

A box that had gravitating energy ##E_{photon} + E_{box}## was lowered.

A box that had gravitating energy ## E_{box}## was lifted

What do you mean by "gravitating energy"?

By that I mean, show us the actual math that backs up your claims here.

 

[PeterDonis]

Clearly the agent that lowered and lifted, gained some net energy in the process.

How?

 

[Dale]

So, does the problem go away, if objects do not lose or gain potential energy when they climb or fall?

I think that the problem goes away if you remember to account for the stress in the box.

 

[PeterDonis]

does the problem go away, if objects do not lose or gain potential energy when they climb or fall?

You can't make a perceived problem go away by violating the laws of physics.

I think that the problem goes away if you remember to account for the stress in the box.

I think the problem goes away if the hidden inconsistencies in the scenario are exposed and corrected.

First, what happens when you let the photon out of the box at the bottom? The photon can't just sit at rest at the bottom until you decide to send it back upward. You have to point it upward when you release it from the box. That means you effectively are giving it an upward impulse, because something has to absorb the recoil. So the photon is being "thrown" back upward. That means we have to add energy to it; but we haven't included the source of that energy in our analysis. We need to include that energy source at the bottom and make it part of the total system.

Second, what about the raising/lowering mechanism? It is an external thing that interacts with the system, but that means that for a complete energy accounting we need to include the energy it captures, i.e., we need to make it part of the total system.

With the above issues corrected, here is what the scenario looks like:

We start with a box with energy ##E_\text{b}## and a photon with energy ##E_\text{p}##, at height ##h##.

We put the photon in the box, and slowly lower box + photon to height ##0##. This extracts energy ##U = U_\text{b} + U_\text{p} = \left( E_\text{b} + E_\text{p} \right) g h / c^2## and stores it in the raising/lowering mechanism (a flywheel, a battery, whatever).

We let the photon out of the box, releasing it directly upward and absorbing the recoil. This requires an energy source that can provide energy ##U_\text{p} = E_\text{p} g h / c^2##, in order to provide the opposing momentum necessary to absorb the recoil.

We raise the box back upward to height ##h##. This requires energy ##U_\text{b} = E_\text{b} g h / c^2## from the raising/lowering mechanism. That leaves energy ##U_\text{p} = E_\text{p} g h / c^2## stored in the raising/lowering mechanism.

So, for a total energy accounting: we started with energy ##E_\text{b} + E_\text{p} + U_\text{p}##, i.e., the energy of the box and the photon, plus the energy stored in the recoil-absorbing mechanism at the bottom.

We end up with energy ##E_\text{b} + E_\text{p} + U_\text{p}##, where energy ##U_\text{p}## has been transferred from the recoil-absorbing mechanism at the bottom to the raising/lowering mechanism at the top.

So total energy is conserved, if we properly account for all of it.

(Note that all of these energies are energies at infinity. A really proper technical accounting would include the redshifting/blueshifting of energy with height, and would thus relate the energies at infinity above to locally measured energies at the two heights. Such an analysis would show that the locally measured energy provided by the recoil mechanism at the bottom is slightly larger, numerically, than the locally measured energy stored in the raising/lowering mechanism at the top. But when translated back into energies at infinity, they are both the same.)

 

[Dale]

So the photon is being "thrown" back upward. That means we have to add energy to it

I don’t think that is necessary. You merely need to change its momentum, not its energy. Impulse is required, not work.

A spherical shell under tension has less mass than a spherical shell not under tension. When the light is in the box the box itself weighs less. So raising the box requires more energy than obtained from lowering it.

Plus, it may require work to put the light back in the box.

 

[PeterDonis]

You merely need to change its momentum, not its energy.

That may be true; but if we don't change its energy at the bottom, then when it gets back to the top, its energy at infinity will be smaller than what it was at the start (because the photon will redshift as it rises); it will not be ##E_\text{p}## any more, but ##E_\text{p} - U_\text{p}##. So the energy accounting will still balance; the total energy before will be ##E_\text{b} + E_\text{p}##, and the total energy after will be ##E_\text{b} + \left( E_\text{p} - U_\text{p} \right) + U_\text{p}##. In other words, the net effect will be to transfer energy ##U_\text{p}## from the photon to the raising/lowering mechanism (instead of transferring it from some energy source at the bottom to the raising/lowering mechanism). So there is still no problem, as long as we correctly analyze the entire process.

 

[PeterDonis]

A spherical shell under tension has less mass than a spherical shell not under tension. When the light is in the box the box itself weighs less. So raising the box requires more energy than obtained from lowering it.

I don't think this is correct. The energy accounting for photon out of the box vs. photon in the box should be:

Photon out of the box: photon energy ##E_\text{p}## plus box energy ##E_\text{b}##.

Photon in the box: photon energy ##E_\text{p}## plus box energy ##E_\text{b}## plus pressure exerted by photon on box wall ##p## minus tension in box wall ##p##. As long as we treat the box with the photon inside as static, the pressure/tension terms cancel and the total energy is unchanged.

In other words, putting the photon into the box, or taking it out of the box, can be done, in the idealized case, without any work having to be done either by the environment on the photon/box, or by the photon/box on the environment, as long as we consider everything to happen at constant altitude.

 

[Dale]

The energy accounting for photon out of the box vs. photon in the box should be:

Photon out of the box: photon energy Ep plus box energy Eb.

I agree it should be that way, but the whole point was to treat the box and the light separately. If you do it that way then you have to consider that the box is different in those cases

In any case, there are now at least three possible explanations, so it is not clear that there actually is any problem. And the conservation of energy makes it clear that there is not a problem.

 

[vanhees71]

I don't think this is correct. The energy accounting for photon out of the box vs. photon in the box should be:

Photon out of the box: photon energy ##E_\text{p}## plus box energy ##E_\text{b}##.

Photon in the box: photon energy ##E_\text{p}## plus box energy ##E_\text{b}## plus pressure exerted by photon on box wall ##p## minus tension in box wall ##p##. As long as we treat the box with the photon inside as static, the pressure/tension terms cancel and the total energy is unchanged.

In other words, putting the photon into the box, or taking it out of the box, can be done, in the idealized case, without any work having to be done either by the environment on the photon/box, or by the photon/box on the environment, as long as we consider everything to happen at constant altitude.

It depends a bit on, what you mean by "photon in a box". Take a box with ideally reflecting walls. Then you have the usual cavity modes. A "photon in the box" means that the em. field state is the corresponding one-photon Fock state, i.e., one of the eigenmodes is occupied by one photon with energy ##\hbar \omega## (as measured in the rest frame of the box). The total mass of the box then is ##m_{1\gamma}=m_0+\hbar \omega/c^2>E_{0\gamma}/c^2=m_{\text{empty box}}##.

 

[PeterDonis]

The total mass of the box then is ##m_{1\gamma}=m_0+\hbar \omega/c^2>E_{0\gamma}/c^2=m_{\text{empty box}}##.

Yes, and that is equivalent to saying that no work needed to be done in either direction to put the photon in the box, because the mass of the box is just the sum of the empty box mass plus the mass-equivalent of the photon's energy. If work needed to be done in either direction to put the photon in the box, the mass of the box with the photon in it would be different from what you give, by the amount of work done.

 

[vanhees71]

Of course you need the energy ##\hbar \omega## to excite the corresponding mode.

 

[jartsa]

So you guys have solved my perceived problem in three different ways. But I did not mean to say that there is a problem. I meant to say: "does this solve your problem @snoopies622".

Snoopies622 thought that a falling object loses potential energy and gains kinetic energy, which would keep total energy of a falling object constant, which would be a problem, because the energy of a falling object may be turned into light, which redshifts when it climbs.

A simplified version of Snoopie622's problem: Falling light blueshifts (gains energy), falling rock has constant energy.

How do we correctly solve that problem?