B Gravitational potential energy, a thought experiment

[Nugatory]

Hi Nugatory, glad to see you are still on PF. And this horridly uncalculable mass is acounted for in the dark matter visible to gravitational mass difference?

Dark matter is completely unrelated.

(and "still on PF"? Are you thinking of someone else?)

 

[berkeman]

(and "still on PF"? Are you thinking of someone else?)

LOL, he hasn't been here for the last 8 years, so maybe you were one of the folks he remembers best from way-back-then.

 

[Lok]

(and "still on PF"? Are you thinking of someone else?)

I did miss out for a bit from PF and even though I'm not someone of note I do remember you schooling me and others on many topics.

Dark matter is completely unrelated.

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.
I find it that gravitational potential energy can easily surpass baryonic matter in a galaxy. Would this assesment be even true, as I never heard it mentioned anywhere.

 

[Nugatory]

LOL, he hasn't been here for the last 8 years, so maybe you were one of the folks he remembers best from way-back-then.

@Lok OK I see - now it makes sense.
Although if you check the member lists you'll find a lot of other people who have been around even longer than me.... Welcome back, and I hope you still find the forum helpful and fun.

 

[PeroK]

That’s where you’re going wrong. The total mass of a multi-body system is not generally equal to the sum of the masses of the individual bodies; there is additional mass from the energy associated with their interaction.
So the quick answer tois “yes”. This assumes no energy radiated away or added to the two-star system and some other simplifying assumptions.

Are you sure this applies in GR? In terms of conservation of invariant mass? I thought that more generally we have conservation of stress-energy. 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.

Is it clear that the local, SR concept of invariant mass extends to non-local, curved spacetime?

 

[Lok]

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.

 

[PeterDonis]

"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 $m + M$.

 

[PeterDonis]

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.

 

[PeterDonis]

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.

 

[PeterDonis]

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.

 

[Lok]

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.

 

[PeterDonis]

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).

 

[Lok]

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.

 

[Lok]

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.

 

[Vanadium 50]

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.

 

[PeterDonis]

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 $m + M$.

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.

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.

 

[PeterDonis]

I do not think we speak of the same concept.

I do not think you have read my posts carefully enough.

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.)

 

[Lok]

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?

 

[Vanadium 50]

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.

 

[Lok]

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.

 

[Lok]

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)

 

[Lok]

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.

 

[Dale]

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.

 

[PeterDonis]

I meant to say is greatest at infinity (or the farther the freefall starts).

Yes.

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.

 

[PeterDonis]

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.

 

[Nugatory]

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]

 

[Vanadium 50]

I don’t see that this is a GR question.

It's not even an SR question.

 

[Nugatory]

It's not even an SR question.

Quite right…. Your post crossed my edit intended to complete my thought.

 

[PeterDonis]

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.

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.

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.

 

[Bosko]

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:
$$ds^2=-(1-\frac{r_s}{r})c^2dt^2+\frac{dr^2}{1-\frac{r_s}{r}}+\cancel{r^2(d\theta^2+\sin^2 \theta d\phi^2)}$$