# Gravitational potential energy, a thought experiment

• B
• Lok
Lok
TL;DR Summary
A simple physics problem to understand the nature of mass by falling celestial bodies.
Hi PF, long time no see. Hope you are all well.

Recently I have come into a mental conundrum of a cosmological physical nature.
After doing some napkin calculations about the energy of celestial bodies and transforming them into mass via E=mc^2 I've found that said energy is by no means small.
Therefore I would like to post a though experiment.

"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*?"

I treated it classically and found that the Sun would attain enough kinetic energy to equal ~24% of it's mass, and while I am sure relativistic effects play a massive part, 24% is hardly something to ignore or cancel out.
Basically I assume that the box in the initial state should weigh exactly as much as the final state as measured via lensing from a distant enough observer, yet the final state has 24% more of a solar mass. Was the final state mass hidden in the initial state as potential energy? How?
Data used:
mSun=1.9885e+30 kg
MSagA*=8.544e+36 kg
R Sag A*=2.59e+10 m
Energy of Sun close to Sag A*=4.378e+46 J
excess mass= 4.86e+29 kg
Classical energy formula of U=-G*(M*m)/R

I am sorry if this is similar to other threads and if someone could help me understand this better I would be very grateful.
Cheers!

You know that mass of a system (I'll use special relativistic definition, and set c=1) is: ##m=\sqrt{E^2-p^2}##? That is, momentum also plays role.

Note that ##E = mc^2## is part of SR (Special Relativity - flat spacetime) and only applies locally in curved spacetime. In particular, the gravitational mass of an object is not equivalent to its so-called relativistic mass.

The gravitational potential energy is not itself part of the Einstein stress-energy tensor. See, for example:

https://en.wikipedia.org/wiki/Stress–energy_tensor#Gravitational_stress–energy

The following is required to explain your scenario.

https://en.wikipedia.org/wiki/Stress–energy–momentum_pseudotensor

weirdoguy said:
You know that mass of a system (I'll use special relativistic definition, and set c=1) is: ##m=\sqrt{E^2-p^2}##? That is, momentum also plays role.
Sure, in the initial state momentuum is zero. In the final state it has a value. Was mass created during falling?

PeroK said:
Note that ##E = mc^2## is part of SR (Special Relativity - flat spacetime) and only applies locally in curved spacetime. In particular, the gravitational mass of an object is not equivalent to its so-called relativistic mass.

The gravitational potential energy is not itself part of the Einstein stress-energy tensor. See, for example:

https://en.wikipedia.org/wiki/Stress–energy_tensor#Gravitational_stress–energy

The following is required to explain your scenario.

https://en.wikipedia.org/wiki/Stress–energy–momentum_pseudotensor
Kinetic energy has a measurable mass afaik. A hot object should weigh more than a cold one by virtue of kinetic energy stored within.
The reverse of this problem is possible. Sun close to Sag A* should weigh M+m. And to lift the Sun to 26000 ly you require 24% of the Sun's mass as kinetic energy.

Lok said:
in the initial state momentuum is zero. In the final state it has a value.
In the center of mass frame, the total momentum is zero all the time, isn't it?

Hill said:
In the center of mass frame, the total momentum is zero all the time, isn't it?
Yes.

Lok said:
Sure, in the initial state momentuum is zero. In the final state it has a value.
In the final state the momentum (which is a vector) remains zero.
Lok said:
Was mass created during falling?
Kinetic energy is not equivalent to mass. In a Newtonian treatment of the problem, there was initially GPE, which was converted to KE; and overall momentum and total mechanical energy are conserved.

In a GR treatment, the calculations are more complicated - see the above references.

PeroK said:
In the final state the momentum (which is a vector) remains zero.

Kinetic energy is not equivalent to mass. In a Newtonian treatment of the problem, there was initially GPE, which was converted to KE; and overall momentum and total mechanical energy are conserved.

In a GR treatment, the calculations are more complicated - see the above references.
The final state as I defined it is right before the collision, where one can measurably see the star moving at a certain speed, sure the BH moves too. I am not trying to state that momentum is not conserved, only to know the weight of the box in the initial and final state.
Is it the same?

PeroK said:
In the final state the momentum (which is a vector) remains zero.

Kinetic energy is not equivalent to mass. In a Newtonian treatment of the problem, there was initially GPE, which was converted to KE; and overall momentum and total mechanical energy are conserved.

In a GR treatment, the calculations are more complicated - see the above references.
This is my problem.
In the initial state the weight of the box should be M+m.
In the final M+m*1.24 (even tough the 1.24 is a napkin calc).
Was the 24% mass previously present?

Lok said:
I am not trying to state that momentum is not conserved, only to know the weight of the box in the initial and final state.
Is it the same?
Yes, assuming you mean mass and not weight, and up to any energy radiated away by gravitational waves. Roughly speaking, the kinetic energy came from the gravitational potential, so the mass is unchanged (it wasn't m+M to start with edit: see my next post).

You can do the calculation exactly for something like Oppenheimer-Snyder collapse.

Last edited:
Ibix said:
Yes, assuming you mean mass and not weight, and up to any energy radiated away by gravitational waves. Roughly speaking, the kinetic energy came from the gravitational potential, so the mass is unchanged (it wasn't m+M to start with).

You can do the calculation exactly for something like Oppenheimer-Snyder collapse.
M+m is the observable while the final state would be a truly gravitationally active value IMO.

Lok said:
The final state as I defined it is right before the collision, where one can measurably see the star moving at a certain speed, sure the BH moves too. I am not trying to state that momentum is not conserved, only to know the weight of the box in the initial and final state.
Is it the same?
I suspect you are influenced by the idea of relativistic mass. This is still prevalent in popular science books and videos, although it has largely disappeared from SR as an academic subject. There is a FAQ on this:

https://www.physicsforums.com/insights/what-is-relativistic-mass-and-why-it-is-not-used-much/

In any case, relativistic mass is no use in General Relativity.

If you learn physics from popular science sources, then it's difficult to analyse your own thought experiments, because you have been led astray with simplified ideas that, ultimately, may not stand up to scrutiny.

PeroK said:
I suspect you are influenced by the idea of relativistic mass. This is still prevalent in popular science books and videos, although it has largely disappeared from SR as an academic subject. There is a FAQ on this:

https://www.physicsforums.com/insights/what-is-relativistic-mass-and-why-it-is-not-used-much/

In any case, relativistic mass is no use in General Relativity.

If you learn physics from popular science sources, then it's difficult to analyse your own thought experiments, because you have been led astray with simplified ideas that, ultimately, may not stand up to scrutiny.
So an object would gravitationally lense in a similar fashion, from a distant observation point, similarly whether stationary or if it has enough kinetic energy to total 24% of it's mass?

Here's an example. Imagine a ball dropped from 5 metres above the Earth's surface that takes 1 second to fall and hit the ground. As measured locally on Earth.

Now consider the scenario in a frame of reference where the Earth and ball are moving away at relativistic speed, at right angles to the direction the ball falls. There is no length contraction in this direction, so the ball still falls 5 metres. But, the local clock runs slow as measured in the reference frame where it is moving. So, the ball takes longer to fall as measured in this frame. I.e. gravitational acceleration is less.

However, if you take relativistic mass literally, then both the Earth and the ball have a greater mass in this frame and the Earth's gravity should be greater. So, the object should fall faster. Which contradicts the calculation above.

This is a simple thought experiment to demonstrate that relativistic mass is a dodgy concept, which must be used with care - if it is used at all.

phinds
Lok said:
So an object would gravitationally lense in a similar fashion, from a distant observation point, similarly whether stationary or if it has enough kinetic energy to total 24% of it's mass?

Lok
PeroK said:
Here's an example. Imagine a ball dropped from 5 metres above the Earth's surface that takes 1 second to fall and hit the ground. As measured locally on Earth.

Now consider the scenario in a frame of reference where the Earth and ball are moving away at relativistic speed, at right angles to the direction the ball falls. There is no length contraction in this direction, so the ball still falls 5 metres. But, the local clock runs slow as measured in the reference frame where it is moving. So, the ball takes longer to fall as measured in this frame. I.e. gravitational acceleration is less.

However, if you take relativistic mass literally, then both the Earth and the ball have a greater mass in this frame and the Earth's gravity should be greater. So, the object should fall faster. Which contradicts the calculation above.

This is a simple thought experiment to demonstrate that relativistic mass is a dodgy concept, which must be used with care - if it is used at all.
If that energy can be transformed into heat as in not hitting a massive black hole but something more mundane. Would that heated matter not have a measurably higher mass?

Lok said:
M+m is the observable while the final state would be a truly gravitationally active value IMO.

The gravitational mass will be m+M (or very nearly - it will be slightly lower given the specified initial conditions) at all times. If the two objects coalesce into one, the heat generated in the collision would be radiated away into space and this would result in a total mass less than m+M. However, in this case one of the objects is a black hole, so no energy is radiated and the mass remains m+M at all times. Oppenheimer-Snyder collapse remains a good example to look up.

PeroK said:
I don't recall that. It's easy enough to work out, though, at least if the lens is moving perpendicular to the line from source to receiver. Work in the rest frame of the lens and everything is normal. Then remember that, in this frame, the source is moving fast so there will be some Doppler beaming affecting the receiver's measures of the brightness of the source, and the receiver is moving fast so will see aberration.

In short, I'd expect a perfect Einstein ring to flatten into an ellipse as seen by someone who says the lens is moving, but the major axis ought to be the same as if it weren't moving.

Lok said:
If that energy can be transformed into heat as in not hitting a massive black hole but something more mundane. Would that heated matter not have a measurably higher mass?
No. The heated matter would have the same m+M mass. It would cool, and then would have a lower mass.

Ibix said:

The gravitational mass will be m+M (or very nearly - it will be slightly lower given the specified initial conditions) at all times. If the two objects coalesce into one, the heat generated in the collision would be radiated away into space and this would result in a total mass less than m+M. However, in this case one of the objects is a black hole, so no energy is radiated and the mass remains m+M at all times. Oppenheimer-Snyder collapse remains a good example to look up.
And if the collision speed is relativistic and ordinary matter involved instead of a BH pair production would be able to generate non-relativistic mass for a while.

Ibix said:
I don't recall that.
It was a thread about the trajectory of a particle travelling at relativistic speed and why you couldn't use the Newtonian approximation. Something like that.

PeroK said:
It was a thread about the trajectory of a particle travelling at relativistic speed and why you couldn't use the Newtonian approximation. Something like that.
I am not so disturbed by my ignorance of the true value of the energy.
Say it is not the Sun, be it a stationary electron crashing gravitationally into a star. It's kinetic energy can lead to pair production and avoid the relativistic mass confusion.

Lok said:
If that energy can be transformed into heat as in not hitting a massive black hole but something more mundane. Would that heated matter not have a measurably higher mass?
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.

Lok said:
I am not so disturbed by my ignorance of the true value of the energy.
The energy is clearly defined in terms of GPE and KE. But, this approach does not extend to GR, where we have not mass but a stress-energy tensor. And, including the GPE in this stress-energy is complicated, as already highlighted.

You can't have it both ways. If you want a relativistic scenario in terms of a collision between a black hole and a star, you can't work in terms of classical mass, KE and Newton's laws; you have to use the stress-energy tensor and the Einstein Field Equations: those represent the laws of physics in GR.

The Newtonian approximation is useful in many cases - like modelling the solar system - but it can only be taken so far.
Lok said:
Say it is not the Sun, be it a stationary electron crashing gravitationally into a star. It's kinetic energy can lead to pair production and avoid the relativistic mass confusion.
A high-energy electron would be absorbed and contribute to the mass of the star. The complication is when you use gravity itself to accelerate the electron from a long way away. Then, the full relativistic picture becomes somewhat complicated. And, you need to find a way to model GPE in a GR framework. Which is:

https://en.wikipedia.org/wiki/Stress–energy–momentum_pseudotensor

PeroK said:
The energy is clearly defined in terms of GPE and KE. But, this approach does not extend to GR, where we have not mass but a stress-energy tensor. And, including the GPE in this stress-energy is complicated, as already highlighted.

You can't have it both ways. If you want a relativistic scenario in terms of a collision between a black hole and a star, you can't work in terms of classical mass, KE and Newton's laws; you have to use the stress-energy tensor and the Einstein Field Equations: those represent the laws of physics in GR.

The Newtonian approximation is useful in many cases - like modelling the solar system - but it can only be taken so far.

A high-energy electron would be absorbed and contribute to the mass of the star. The complication is when you use gravity itself to accelerate the electron from a long way away. Then, the full relativistic picture becomes somewhat complicated. And, you need to find a way to model GPE in a GR framework. Which is:

https://en.wikipedia.org/wiki/Stress–energy–momentum_pseudotensor
As long as there is some excess energy IMO that can translate into excess mass (even rest mass) in the final state via various means. Mass that appeared at some point in time or was a gravitational measurably observable from the beginning.

Lok said:
As long as there is some excess energy IMO that can translate into excess mass (even rest mass) in the final state via various means.
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.

Lok
Lok said:
This is my problem.
In the initial state the weight of the box should be M+m.
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.
Was the 24% mass previously present?
is “yes”. This assumes no energy radiated away or added to the two-star system and some other simplifying assumptions.

Lok
Lok said:
Was the 24% mass previously present?
Yes.

A similar example a wind-up-car, which accelerates to relativistic speeds using energy stored internally in a spring.

In contrast to particles which are accelerated by energy supplied from the outside, the "apparent inertia" or "relativistic mass" of the wind-up-car doesn't increase, because it's there from the start due to internally stored energy. The same wind-up-car would initially accelerate faster, if it wasn't wound up, put pushed with the same external force.

Lok
Nugatory said:
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.
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?

Lok said:
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?)

Nugatory said:
(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
Nugatory said:
(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.

Nugatory said:
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.

berkeman said:
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.

berkeman and Lok
Nugatory said:
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?

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