Gravitational potential energy, a thought experiment

  • #36
Ibix said:
The energy in your case is already in the system and already accounted for in the mass measurement. So the mass does not increase.

Again, look up Oppenheimer-Snyder collapse for an analytically tractable case.
I will have to read the original paper. I will try to find time for it and will come back on this hopefully enlightened.
 
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  • #37
Lok said:
"Given a large box with our Sun and Sagitarius A* at their respective distance one from the other, with the tangential speed of the Sun equal to zero and free falling towards Sag A*, how much does that box weigh? And how much would that box weigh when the Sun falls via gravity close enough to touch Sag A*?"
Even without doing any calculations, we know one thing: if we assume that no matter or radiation crosses the boundary of the box, then the externally measured mass of the box is constant, regardless of what processes take place inside it.

That said, the calculations for this case are pretty simple if we assume that there are no other objects involved besides the Sun and Sag A (i.e., the box only contains those two objects), and that at the start, the gravitational potential energy between the Sun and Sag A is negligible (i.e., we take it to be zero). Then the starting total mass is ##m + M##, where ##m## is the mass of the Sun and ##M## is the mass of Sag A.

To analyze the end state, we make one more assumption, that ##m << M##, so that we can take the center of mass frame of the system to be the frame in which Sag A is at rest. In that frame, at the end state (just before the Sun falls through the horizon of Sag A), the Sun has kinetic energy ##K##, but the gravitational potential energy of the system is now ##- K##. (This calculation is a simple one if we treat the Sun as a test object in the Schwarzschild spacetime geometry around Sag A.) So the kinetic and gravitational potential energies cancel and the total mass is still ##m + M##.
 
  • #38
Lok said:
In the initial state the weight of the box should be M+m.
In the final M+m*1.24
No. See my previous post. You appear to have been confused by assuming there must have been some kind of potential energy stored in the initial state. That's not the case. The non-negligible gravitational potential energy is in the final state, and it is negative--just negative enough to cancel out the Sun's kinetic energy in the final state.
 
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  • #39
Ibix said:
You can do the calculation exactly for something like Oppenheimer-Snyder collapse.
Oppenheimer-Snyder collapse is not really relevant here; this scenario is simpler. The Sun's mass is small enough compared to the mass of Sag A that we can treat the Sun as a test body in the spacetime geometry created by Sag A, which we can assume to be Schwarzschild.
 
  • #40
Ibix said:
The energy in your case is already in the system and already accounted for in the mass measurement.
This way of putting it might be misleading. The actual solution is just what Newtonian physics would tell you: the sum of kinetic and gravitational potential energy is constant for free-fall motion under gravity. That sum is zero at the start, so it must also be zero at the end. So the thing that would need to be "accounted for in the mass measurement" is zero.
 
  • #41
PeterDonis said:
the gravitational potential energy between the Sun and Sag A is negligible (i.e., we take it to be zero).
I cannot make this assumption. Gravity might be small at these distances but adds up to significant values fast.
 
  • #42
Lok said:
I cannot make this assumption.
Remember, this is only the case at the start. If you read my posts again, you will see that I explicitly say the gravitational potential energy is not negligible at the end (indeed, it can't be since it cancels out the Sun's kinetic energy).
 
  • #43
PeterDonis said:
No. See my previous post. You appear to have been confused by assuming there must have been some kind of potential energy stored in the initial state. That's not the case. The non-negligible gravitational potential energy is in the final state, and it is negative--just negative enough to cancel out the Sun's kinetic energy in the final state.
There is a difference between the Sun close to Sag A* with zero KE and Sun with 24% Solar mass equivalent KE towards Sag A*. One enters the BH with more speed, how that translates is beyond this topic.
Yet in a regular matter collision, higher KE translates to higher internal pressures/temperatures and more chances of creating actual rest mass matter via pair production or fusion above Fe. So KE transforming into matter. This latter part is only to overstep heat as measureble mass issue.
 
  • #44
PeterDonis said:
Remember, this is only the case at the start. If you read my posts again, you will see that I explicitly say the gravitational potential energy is not negligible at the end (indeed, it can't be since it cancels out the Sun's kinetic energy).
I do not think we speak of the same concept. As the gravitational potential of a free falling body is to the best of my knowledge at infinty from the gravitational attractor.
 
  • #45
Lok said:
I do not think we speak of the same concept.
Quite possibly. This is why most threads claiming to be "thought experiments" tend to run into trouble. People seem to think a "cute" story makes things either to understand when the reverse is true.

You could have said, "I have two objects with masses m and M (M >> m) initially at rest in a box..." and we could have gone from there. Simple and clear.

As has been pointed out, if nothing enters or exits the box, its mass does not change. As m nears M, it gains kinetic energy and loses exactly the same amount of potential energy (the same can be said for M). This is the case for Newtonian mechanics as well.
 
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  • #46
Lok said:
There is a difference between the Sun close to Sag A* with zero KE
Which is irrelevant to the OP's scenario. The total mass of the system in that case would be less than ##m + M##.

Lok said:
and Sun with 24% Solar mass equivalent KE towards Sag A*.
And at an altitude just above the horizon of Sag A. That is the case under discussion in this thread.

Lok said:
in a regular matter collision, higher KE translates to higher internal pressures/temperatures and more chances of creating actual rest mass matter via pair production or fusion above Fe. So KE transforming into matter. This latter part is only to overstep heat as measureble mass issue.
None of this is relevant unless some of the energy produced escapes to infinity, i.e., outside the box the OP specified (most likely as radiation, less likely as actual mass ejected). I explicitly ruled out this possibility in my post. Of course in real scenarios these things will happen, but I think discussion of them is premature until the OP understands the simpler idealized case that is under discussion.
 
  • #47
Lok said:
I do not think we speak of the same concept.
I do not think you have read my posts carefully enough.

Lok said:
As the gravitational potential of a free falling body is to the best of my knowledge at infinty from the gravitational attractor.
Is zero at infinity. And at a very large distance, as I took you to be assuming for the Sun and Sag A at the start, it is negligible. That is the starting condition I assumed in my posts, and that I took you to be assuming in the OP of this thread.

For objects not at infinity (or a large enough distance for the gravitational potential energy to be negligible), the gravitational potential energy is negative, and becomes more negative the closer you get to the gravitating body (in this case Sag A).

(Note that different normalizations are possible for gravitational potential energy for other scenarios, but the normalization where it goes to zero at infinity is the one that is correct for this scenario.)
 
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  • #48
Vanadium 50 said:
Quite possibly. This is why most threads claiming to be "thought experiments" tend to run into trouble. People seem to think a "cute" story makes things either to understand when the reverse is true.

You could have said, "I have two objects with masses m and M (M >> m) initially at rest in a box..." and we could have gone from there. Simple and clear.

As has been pointed out, if nothing enters or exits the box, its mass does not change. As m nears M, it gains kinetic energy and loses exactly the same amount of potential energy (the same can be said for M). This is the case for Newtonian mechanics as well.
Does that KE have any weight? Is there any scenario in wich it could?
 
  • #49
Lok said:
Does that KE have any weight?
That is not a well-defines question. Two observers, even in Newtonian physics, disagree about how much kinetic energy a body has.
 
  • #50
Vanadium 50 said:
That is not a well-defines question. Two observers, even in Newtonian physics, disagree about how much kinetic energy a body has.
An observer at rest to the initial positions of both the Sun and BH for example.
 
  • #51
PeterDonis said:
Is zero at infinity.
Ah I meant to say is greatest at infinity (or the farther the freefall starts).
Why would it be zero at infinity, or the greater the distance? (to avoid the singularities of infinity in general)
 
  • #52
PeterDonis said:
Which is irrelevant to the OP's scenario. The total mass of the system in that case would be less than ##m + M##.And at an altitude just above the horizon of Sag A. That is the case under discussion in this thread.None of this is relevant unless some of the energy produced escapes to infinity, i.e., outside the box the OP specified (most likely as radiation, less likely as actual mass ejected). I explicitly ruled out this possibility in my post. Of course in real scenarios these things will happen, but I think discussion of them is premature until the OP understands the simpler idealized case that is under discussion.
The initial conditions had a mass of m+M, then the Sun gets accelerated and attains enough energy to rival it's own mass. How much does the box weigh?
There are more solutions to this. Either the initial mass is not m+M and GPE has actual mass.
Or KE ads no mass even though it can and should.
Or the mass of the box changes in time.
Or some other solution.
 
  • #53
Lok said:
It should'nt be unrelated, as anything that gives 24% more mass from a single interaction of a small part of this galaxy is not nothing.
Be aware, this should be 24% less mass, not more.

The process is this: the sun starts out far away. As it falls it gains KE and loses PE. When it is in the low PE high KE state the mass of the system is unchanged. The sun can collide with other objects, breaking apart, and thermalizing its KE. When it is in the low PE high thermal energy state the mass is still unchanged. The resulting hot mass can radiate energy away. Only after the radiation has left the box is the mass decreased. At that point it is a low PE and low thermal energy state. Only then is the mass lower.

There is a limit to how much the mass can drop, but if I recall correctly 24% is definitely possible.
 
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  • #54
Lok said:
I meant to say is greatest at infinity (or the farther the freefall starts).
Yes.

Lok said:
Why would it be zero at infinity, or the greater the distance?
You could define the value at infinity to be anything you like; if you define it to be anything except zero, you will just be adding an irrelevant constant that drops out of the analysis (because that constant cannot appear in the actual externally measured mass of the system, which is an observable and can't depend on what conventions you adopt). So it's easiest to just define the value at infinity to be zero.
 
  • #55
Lok said:
The initial conditions had a mass of m+M, then the Sun gets accelerated and attains enough energy to rival it's own mass. How much does the box weigh?
There are more solutions to this.
Not as you've stated the problem, no, because you have stated that the box is isolated. As has already been pointed out, if the box remains isolated the whole time (nothing goes in or out), then its externally measured mass cannot change. That is the only valid solution.
 
  • #56
PeroK said:
Are you sure this applies in GR?
I don’t see that this is a GR question. It’s basically about energy conservation and how we choose the zero point of total energy.
SR was dragged in only because OP has to convert energy to mass before they can add it to ##m+M## to work with the total mass/energy of the system. But because ##m## and ##M## are both constant there’s no need to do that; they could just be working with the sum of the potential and kinetic energy. Comparing ##m+M+E_K+E_P## when one or the other energy terms is zero doesn’t show us anything we won’t see just by looking at ##E_K+E_P##.

[edit to add - I posted this before it was fully baked, then finished it, which is what the exchange with @Vanadium 50 below was about]
 
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  • #57
Nugatory said:
I don’t see that this is a GR question.
It's not even an SR question.
 
  • #58
Vanadium 50 said:
It's not even an SR question.
Quite right…. Your post crossed my edit intended to complete my thought.
 
  • #59
PeroK said:
Are you sure this applies in GR? In terms of conservation of invariant mass?
In this particular case, yes. We have an isolated system, and its externally measured mass is constant.

PeroK said:
I thought that more generally we have conservation of stress-energy.
We have local conservation of stress-energy in GR, but that does not come into play here as we are talking about the externally measured mass of an isolated system. That is a global quantity.

PeroK said:
There is additional stress-energy - although even that is not so clear cut, as the gravitational field is not part of the stress-energy tensor, but described separately.
Which means there is no additional stress-energy. The various "gravitational pseudo-tensors" in the literature are not stress-energy. And in this case they don't tell us anything useful that we can't get from the much simpler analysis that has already been done in this thread.
 
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  • #60
Lok said:
it's own mass. How much does the box weigh?
I suggest the following:

Clear up the concepts:
##m## - mass of the Sun
##M## - mass Sag. A
##(t,r,\theta,\phi)## -spherical coordinates centered at Sag. The coordinates ##\theta## and ##\phi## are constant for the Sun.

Weight is the force that occurs when you place a mass in a gravitational field.
You shouldn't use it because then you would need a third object in whose gravitational field these two (the Sun and Sagittarius A) would be.

Potential energy, kinetic energy and work are three terms that are closely related.

For potential energy, you should choose one of two conventions.
One is that the potential energy is maximal at the beginning and decreases towards zero, and the other is that it is zero at the beginning and becomes more and more negative.

As the Sun falls towards Sag and it does some work (##W_{ork}=F_{orce}\cdot d_{istance}##), its potential energy decreases and its kinetic energy increases.

Try to analyze what is happening using:

1. Classical Newtonian mechanics when velocities are small compared to the speed of light and when space-time is approximately flat

2. Special relativity when the speed of the Sun is not negligible compared to the speed of light, but the Sun is not close to Sag A (space-time of significant curvature).
The rest masses of both objects will remain the same, but you should pay attention to the relativistic mass of the Sun.
Relativistic mass is not "real" mass, but only the body's resistance to acceleration (in this case, the gravitational force of Sag A)
The sun will accelerate less and react more sluggishly to the external force than in the case of lower speeds of Newtonian mechanics.

3. General relativity.
##r_s=\frac{2GM}{c^2}## the Schwarzschild radius (the event horizon) of Sag A
Because the coordinates ##\theta## and ##\phi## are constant for the Sun you can simplify the Schwarzschild metric:
$$ds^2=-(1-\frac{r_s}{r})c^2dt^2+\frac{dr^2}{1-\frac{r_s}{r}}+\cancel{r^2(d\theta^2+\sin^2 \theta d\phi^2)}$$
 
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  • #61
Bosko said:
Weight is the force that occurs when you place a mass in a gravitational field.
You shouldn't use it because then you would need a third object in whose gravitational field these two (the Sun and Sagittarius A) would be.
This is a valid point; I think everyone commenting in the thread has assumed that the OP actually meant "invariant mass of the system" instead of "weight", but it is good to be precise.

Bosko said:
Classical Newtonian mechanics when velocities are small compared to the speed of light and when space-time is approximately flat
This obviously won't apply at the end state of the scenario, where neither of these things are true.

Bosko said:
Special relativity when the speed of the Sun is not negligible compared to the speed of light, but the Sun is not close to Sag A (space-time of significant curvature).
This won't tell us anything useful. In fact no dynamical analysis is required at all; the question can be answered purely in terms of energy conservation. As has already been done in the thread.

Bosko said:
Relativistic mass is not "real" mass, but only the body's resistance to acceleration (in this case, the gravitational force of Sag A)
This is wrong. "Acceleration" due to gravity is not proper acceleration; it is not felt as a force, and therefore relativistic mass as "inertia" (resistance to an applied force) does not apply.

Bosko said:
General relativity.
This is implied by the use of a black hole in the scenario, yes. However, in this particular case the analysis looks exactly the same as a Newtonian analysis using energy conservation.
 
  • #62
PeterDonis said:
This obviously won't apply at the end state of the scenario, where neither of these things are true.
Of course you are right but OP is talking about masses, but in fact he is interested in speed, kinetic energy stored as relativistic mass and the energy of the collision that will happen at the end.
I thought he should practice Newtonian mechanics and then move on to more complex relativistic formulas.
PeterDonis said:
This is wrong. "Acceleration" due to gravity is not proper acceleration; it is not felt as a force, and therefore relativistic mass as "inertia" (resistance to an applied force) does not apply.
It is also true, IMO, but you speak from the point of view of GR and I from a static observer in SR relative to Sag A
PeterDonis said:
This is implied by the use of a black hole in the scenario, yes. However, in this particular case the analysis looks exactly the same as a Newtonian analysis using energy conservation.
Of course, but reading the OP, it seems to me that he is still interested in final speed and kinetic energy.
 
  • #63
Bosko said:
kinetic energy stored as relativistic mass
Is, as I have already said, irrelevant to the scenario under discussion (and relativistic mass is an outdated concept anyway, as you will see if you search for relevant PF Insights articles and previous threads).

Bosko said:
the energy of the collision that will happen at the end
There is no collision. Sag A is a black hole; the Sun just falls through its horizon.

Bosko said:
I thought he should practice Newtonian mechanics and then move on to more complex relativistic formulas.
As I have already said, there is no need to dynamically analyze the situation at all. The analysis that is actually needed looks the same in Newtonian mechanics as it does in GR. You can't do the analysis in SR (see further comments below).

Bosko said:
a static observer in SR relative to Sag A
There is no such thing in this scenario since Sag A is a black hole and the spacetime around it is curved, and the Sun does not stay a very large distance away for the entire scenario so you cannot ignore the curvature of the spacetime.

Bosko said:
reading the OP, it seems to me that he is still interested in final speed and kinetic energy.
Only because he didn't understand when he posted the OP that the Sun's kinetic energy is exactly canceled by negative gravitational potential energy, so the externally measured mass of the system is constant even though the Sun's kinetic energy changes during the scenario.
 
  • #64
Ibix said:
The energy in your case is already in the system and already accounted for in the mass measurement. So the mass does not increase.

Again, look up Oppenheimer-Snyder collapse for an analytically tractable case.
Nifty little paper with nice heart warming historical notations.
I see where the similarity begins, but they do not treat mass and potential gravitational energy at all and are more concerned with the physicality of the geometry as far as i understood it.
 
  • #65
PeroK said:
Heat is internal energy that is effectively frame independent. That contributes to the rest mass or invariant mass of an object. And, that contributes to its gravitational mass.

The object's KE is entirely frame dependent. There is always a frame of reference in which the centre of mass of the object is at rest. You can't simply take that KE to be gravitating mass. That's why you need to be a lot more precise about "mass-energy equivalence". And, especially, once you go beyond SR into GR and cosmology.
There is an argument to be made that heat is randomly distributed KE vectors in a body.
 
  • #66
Bosko said:
Of course, but reading the OP, it seems to me that he is still interested in final speed and kinetic energy.
Not really interested in final speed and real value of KE as long as it is >0 and can in theory be counted as a gravitationally observable mass.
 
  • #67
PeterDonis said:
There is no collision. Sag A is a black hole; the Sun just falls through its horizon.
I used the Sun and Sag A* as the initial bodies to underline the massive amount of energy that such a system holds as potential energy. It could have been an electron and a star, were the collision can easily produce rest mass, but then I would get quantum arguments of fields holding whatever mass somewhere, I wanted to avoid this and make it cosmological because IMO the amount of mass that comes about is of dark matter proportions.
 
  • #68
PeterDonis said:
Only because he didn't understand when he posted the OP that the Sun's kinetic energy is exactly canceled by negative gravitational potential energy, so the externally measured mass of the system is constant even though the Sun's kinetic energy changes during the scenario.
I still do not understand this. In the OP I used the as in Wiki weirdly formulate value for Gravitational potential energy.
U=-GMm/R, where R is the smallest distance that can be attained by the masses, in my case the radius of Sag A* or close by, and that is the final state, it does not matter what happens after.
Thus I would not state that KE cancels out GPE, but rather one transforms into another.
So either the extra 24% mass is a real thing, and this leaves me wondering where it is distributed in the initial state.
Or it is a figment of equations and there should be no difference in outcome whether there is or there isn't KE in the final state.
 
  • #69
Nugatory said:
I don’t see that this is a GR question. It’s basically about energy conservation and how we choose the zero point of total energy.
SR was dragged in only because OP has to convert energy to mass before they can add it to ##m+M## to work with the total mass/energy of the system. But because ##m## and ##M## are both constant there’s no need to do that; they could just be working with the sum of the potential and kinetic energy. Comparing ##m+M+E_K+E_P## when one or the other energy terms is zero doesn’t show us anything we won’t see just by looking at ##E_K+E_P##.

[edit to add - I posted this before it was fully baked, then finished it, which is what the exchange with @Vanadium 50 below was about]
Still would those ##m+M+E_K+E_P## energy terms add to the mass of the box?
If yes, how is that mass distributed in the initial state as the final state is less of an issue.
 
  • #70
Bosko said:
Clear up the concepts:
##m## - mass of the Sun
##M## - mass Sag. A
##(t,r,\theta,\phi)## -spherical coordinates centered at Sag. The coordinates ##\theta## and ##\phi## are constant for the Sun.
I used the Sun Sag A* masses and radii as found on Wiki. But they do not matter as long as they are non zero, assuming initial state distance is greater than the final state.
Bosko said:
Relativistic mass is not "real" mass, but only the body's resistance to acceleration (in this case, the gravitational force of Sag A)
I agree up to a point as KE does have a measurable mass via e.g. heat (non uniform energy vectors) can produce rest mass via nuclear or pair production (thus I am a bit liberal with their use interchangeably) so having uniform KE in one specific direction does not seem to me to be different.
IMO the moving Sun should have more gravitationally measurable mass.

I am going for dark matter of course, as that 24% is only the contribution of GPE of the SMBH Sag A* and the total GPE should account for all masses and distances in the Milky way once the Sun ends up in the final state somewhere in the center.

As a side note I find it interesting to calculate this for more mundane celestial bodies.
So Moon Earth system ends up 6.9e-10% (basically a rounding error).
Earth Sun system ends up as 6.36e-10% (less of a basic rounding error).
So if true, this physically shows up only for some pretty extreme cases of BH or galaxies.
 

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