B Gravitational potential energy, a thought experiment

[Lok]

I have no idea what you’re trying to ask here, but I do know from the timestamps that you spent no more than fifteen minutes thinking before posting, and that’s not enough.

True, sorry for the hasty reply.
I mean to say that in your reference you would measure a higher total gravity of the moving Sun but also a slower time for clocks in the moving reference frame. And for everyone to experience the same outcomes these 2 effects can cancel out, or should. IMO.

Conversely, if you see time being slower in a Sun moving past you, would you be able to calculate it's gravity from observing an object free falling into it? And if gravity is slower by virtue of relative motion, would the Sun have less gravity acting upon your reference frame?
I hope this made some sense.

 

[Dale]

More in comparison to m+M of initial state, as in the astronomically visually observable mass of the 2 body system at rest

I don’t know how I can be more clear about this. It is less, not more. It starts at M+m and remains M+m until radiation leaves the box. When radiation leaves the box the mass becomes less than M+m. At no point is it ever more.

As in the inital state has a mass of m+M observably, and m+M+KE in the final.

This is false. At no point is the mass ever greater than m+M. After the radiation leaves the box it becomes m+M-E where E is the energy of the radiation that leaves.

I assume it is not created instantaneously

There is no delay. The mass is M+m the entire time until the radiation leaves at which point it is without delay M+m-E.

 

[Lok]

I don’t know how I can be more clear about this. It is less, not more. It starts at M+m and remains M+m until radiation leaves the box. When radiation leaves the box the mass becomes less than M+m. At no point is it ever more.

This is false. At no point is the mass ever greater than m+M. After the radiation leaves the box it becomes m+M-E where E is the energy of the radiation that leaves.

I explicitly avoided radiation leakeage by stopping the problem right before the merger. And ask what the weight of the box is initially versus Sun close to A*.
My gripe is with the system in the final state having a mass of m+M+ a big chunk of Kinetic energy that does something.
And I do concur KE having mass is an assumption that is most evident at an impact, but that would mean mass is generated at impact, which makes just as little sense if trying to conserve the mass in the box.

 

[Dale]

I explicitly avoided radiation leakeage by stopping the problem right before the merger.

Then there is never any change in mass. Not at any time. The only way for the mass in the box to change is for something to leave the box.

that would mean mass is generated at impact

Mass is never generated in this scenario. In other related scenarios, the box will only gain mass if something enters from outside and it will only lose mass if something exits from inside. No internal state change of any kind changes the mass.

I hope that is sufficiently clear to avoid misunderstanding.

 

[Lok]

Then there is never any change in mass. Not at any time. The only way for the mass in the box to change is for something to leave the box.

Mass is never generated in this scenario. In other related scenarios, the box will only gain mass if something enters from outside and it will only lose mass if something exits from inside. No internal state change of any kind changes the mass.

I do understand the mass conservation problem, but this does not explain what happens to m, M and the 24% solar mass equivalent energy in the final state so that it would show up as a total mass of m+M of the initial.

 

[Dale]

this does not explain what happens to m, M and the 24% solar mass equivalent energy in the final state so that it would show up as a total mass of m+M of the initial

Does this explain it for you? If not, please ask about the part of the explanation that is confusing:

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.

 

[Lok]

Does this explain it for you? If not, please ask about the part of the explanation that is confusing:

Dale's:
So box starts out as m and an unspecified hard wall plus gravity from another source, Sun starts to attain kinetic energy losing PE, thermalizes it by hitting the wall, and creates actual rest mass via heat, fusion and pair production. What is the mass of the final thermalized Sun?

 

[Dale]

creates actual rest mass via heat, fusion and pair production

The mass does not change at any point. Heat, fusion, and pair production do not change the mass inside the box.

What is the mass of the final thermalized Sun?

To determine that you would have to draw a box around the sun and keep track of anything that enters or leaves that box. If nothing enters or leaves then the mass would remain m. But of course in that case there would be no way for the KE to thermalize.

 

[PeroK]

I do understand the mass conservation problem, but this does not explain what happens to m, M and the 24% solar mass equivalent energy in the final state so that it would show up as a total mass of m+M of the initial.

It may be worth reading this:

https://en.wikipedia.org/wiki/Mass_in_general_relativity

I think it says what I was trying to say nearly a hundred posts ago!

 

[PeterDonis]

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.

That is correct. The scenario they are analyzing is different from the scenario you are asking about. Their paper, as I have already said, does not give any useful information about your scenario.

 

[PeterDonis]

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.

Again you are missing the key point: in the starting configuration of the system, it holds zero gravitational potential energy. The total mass is just $m + M$; there is no extra "potential energy" term in there.

It is true that as the Sun falls, it gains kinetic energy, and to compensate it loses gravitational potential energy, i.e., as I have said, its gravitational potential energy becomes negative, and gets more negative as it falls. But viewing this as somehow using gravitational potential energy that was "stored" in the initial configuration is not helpful, because it makes you think there is some kind of issue involved with the total mass just being $m + M$. There isn't. Once you properly understand this, the question you think you are asking simply evaporates.

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

That's wrong. The formula does not depend on the "smallest distance", it depends on the actual distance. At the start it is zero because the actual distance is effectively infinity. At the end it is a negative value that exactly cancels the Sun's kinetic energy.

I would not state that KE cancels out GPE, but rather one transforms into another.

And your insistence on looking at it this way is why this thread keeps going on--the correct answer is staring you in the face and you are refusing to accept it.

either the extra 24% mass is a real thing, and this leaves me wondering where it is distributed in the initial state.

Which, again, is because you are refusing to accept the correct answer that is staring you in the face: the total mass is just $m + M$, always, and what happens internally can't change it because the system is isolated. There is no "extra mass" anywhere.

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.

It is impossible for the Sun to not have KE in the final state. As I have already said, there is only one valid solution to this scenario. Talking as though there were other possibilities is simply wrong. And this error appears to be part of what is confusing you and keeping you from accepting the obvious correct answer.

would those $m+M+E_K+E_P$ energy terms add to the mass of the box?

Yes, because $E_P = - E_K$ so $m + M + E_K + E_P = m + M$. This is the obvious correct answer that you are refusing to accept.

If yes, how is that mass distributed in the initial state as the final state is less of an issue.

In the initial state, $E_K = E_P = 0$. Which is just a special case of $E_P = - E_K$; the latter is always true in this scenario.

KE does have a measurable mass via e.g. heat

Sure, but in this scenario that is irrelevant, because PE exactly cancels out KE and there is no net contribution to the total mass. For KE to appear as "extra mass" in the total mass of the system you have to look at a different scenario in which PE and KE do not exactly cancel. And for such a scenario the total mass would not be $m + M$, even at the start (although even here calling the KE "extra mass" can be misleading, since the total mass will be less than $m + M$). If you want to discuss such scenarios, you can start a new thread to ask about one (for example, you could consider having the Sun in a circular orbit around Sag A at some finite radius).

I am going for dark matter of course

That is because you keep talking as if there were some "extra mass" in this scenario, when there is none.

 

[PeterDonis]

Not the gravitationally lensed mass, as my assumption is that includes the 24% extra mass somewhere somehow.

It doesn't. There is no "24% extra mass" anywhere in the system as measured from outside.

if the Sun can attain a higher mass by transforming GPE into KE which can be transformed into rest mass

If you want to consider the Sun's extra KE as contributing to the total mass of the system, you also must consider the Sun's negative GPE as contributing to the total mass of the system--and the two contributions exactly cancel in this scenario. So there is no "extra mass".

Basically, you are trying to consider the KE as "extra mass" but ignore the negative GPE as a negative contribution to the mass. That's wrong. You have to consider both.

where was that mass in the initial state?

Nowhere; it doesn't exist in this scenario. See above.

IMO this problem is a contradiction already.

It's not. You are just refusing to analyze it correctly. See above.

I strongly encourage anybody to do the most basic calculation

No such calculation is needed to give the obvious correct answer I gave above, and have given already multiple times in this thread.

 

[PeterDonis]

How can the initial geometry be described by a single mass parameter?

Please, please, let's not derail the thread any further with irrelevant confusions about the Oppenheimer-Snyder model. The fact is that that scenario can be described with a constant external mass parameter: this follows from the fact that the scenario is spherically symmetric and asymptotically flat and an obvious application of Birkhoff's theorem. But discussion of that is off topic here; we're having enough trouble with the even simpler scenario the OP has posed.

 

[Vanadium 50]

I strongly encourage anybody to do the most basic calculation and see how much GPE is in a as OP system or galactic if one has available data and modeling software. It is above my current setup.

Are you saying that you cannot do this calculation yourself? Yet you are sure we're wrong? That's a tough, tough position to hold.

Further, the actual numbers are unnecessary. You did disregard ny suggestion to eschew the cute stories, but had you followed it, we could discuss this all symbolically.

I have no idea what you’re trying to ask here, but I do know from the timestamps that you spent no more than fifteen minutes thinking before posting, and that’s not enough.

If you're not thinking about what we are writing, why are we writing it? Skimming and reacting will not teach you anything. You have to spend the time thinking.

I suggest you stop posting for some period of time - a few hours, a day, whatever it takes - and reread the entire thread, thinking about every response carefully. If after the endn of this period, there is something you still do not understand, post as short and concise a question as you can about it, with minimal filler.

 

[Ibix]

we're having enough trouble with the even simpler scenario the OP has posed.

I'm not sure I agree that the OP's scenario is simpler, but will leave it for now. Happy to discuss in another thread.

 

[PeterDonis]

I'm not sure I agree that the OP's scenario is simpler

It's simpler in that no computation whatever is required to get the correct answer; at most (if you want more than just applying the obvious fact of conservation of energy for an isolated system) you just need to read it off a simple one line formula that has already been posted.

 

[Lok]

It's simpler.

Negative energy. I cannot unsee that one.
Enough! A nice day to you all.

 

[PeterDonis]

Negative energy. I cannot unsee that one.

Negative gravitational potential energy. That has been a useful concept in physics for several centuries now.

 

[Lok]

Negative gravitational potential energy. That has been a useful concept in physics for several centuries now.

Negative gravitational potential energy. That has been a useful concept in physics for several centuries now.

The thing you create negative mass with.
I now understand.
Thank you!

 

[Nugatory]

Negative energy. I cannot unsee that one.
Enough! A nice day to you all.

Negative potential energy is one of the great simplifying concepts in classical physics - for example, you’ll find it very helpful when deriving Kepler’s Laws and solving orbital motion, which physics students will be doing no later than their first year of college. If you are not completely comfortable with the concept you really need to back up and review elementary classical mechanics before you try to take on relativity.

 

[Dale]

Negative energy. I cannot unsee that one.
Enough! A nice day to you all.

See https://en.wikipedia.org/wiki/Gravitational_energy
$$U=-\frac{GMm}{r}$$ Since $G$, $M$, $m$, and $r$ are all positive, $U$ is negative. This is ordinary Newtonian physics.

 

[pervect]

The thing you create negative mass with.
I now understand.
Thank you!

I'm not so sure you do understand - I rather suspect you do not, from your comments. I haven't read the entire thread in detail, as it is quite long, I've only skimmed it.

In Newtonian theory, if you have a planet, and you "dissassemble" it (my own term for lack of anything better) by cutting it into pieces and moving them all very far away from each other, this process requires energy.

This implies that the total energy of an assembled planet is lower than the energy of the disassembled one. This is called the "binding energy".

The total energy of a bound system is lower than the total energy of its parts.

Note that a planet orbiting a sun (or the sun orbiting Sag A) can also be considered a bound system in the same manner.

The gravitational binding energy of a system is the minimum energy which must be added to it in order for the system to cease being in a gravitationally bound state. A gravitationally bound system has a lower (i.e., more negative) gravitational potential energy than the sum of the energies of its parts when these are completely separated—this is what keeps the system aggregated in accordance with the minimum total potential energy principle.

Note particularly the remarks in wiki that the binding energy is negative.

This is standard usage - the potential energy of a bound system is negative because the energy of the bound system is lower than the energy of the unbound system.

As for how things actually work in GR (or even SR), I'd say it's too advanced for this thread.

I will say that SR has a concept of something called the "stress energy tensor", which GR borrows, and uses instead of "mass".

GR does also have several concepts of the mass of a system (which are derived from the stress-energy tensor) such as the "big three", Komar mass, ADM mass, and Bondi mass. The fact that it has three of them rather than one is a foreboding that things here are not simple.

However, other than dropping these names, I fear that a detailed discussion of these issues is beyond the scope of the thread.

 

[Lok]

See https://en.wikipedia.org/wiki/Gravitational_energy
$$U=-\frac{GMm}{r}$$ Since $G$, $M$, $m$, and $r$ are all positive, $U$ is negative. This is ordinary Newtonian physics.

I said that now I understand that negative mass cancels the positive one from KE. Problem solved.
In all honesty, how old is this mess?

 

[Dale]

In all honesty, how old is this mess?

About 300 years.

 

[Lok]

Negative potential energy is one of the great simplifying concepts in classical physics - for example, you’ll find it very helpful when deriving Kepler’s Laws and solving orbital motion, which physics students will be doing no later than their first year of college. If you are not completely comfortable with the concept you really need to back up and review elementary classical mechanics before you try to take on relativity.

About 300 years.

I appreciate the honesty. Best of luck as this will crash eventually.

 

[PeterDonis]

The thing you create negative mass with.

No. The total mass of any system is still positive.

Best of luck as this will crash eventually.

What do you mean by this? If you think the concept of negative gravitational potential energy is somehow going to "crash", I strongly suggest that you think again.

 

[Bosko]

In all honesty, how old is this mess?

1666 was a turning point for England. The outbreak of the Great Plague, the eruption of the Second Dutch War and the devastating Great Fire of London all hit the country in quick succession and with devastating consequences.

Isaac Newton took refuge in the countryside.
It is not known exactly what he was doing under that tree, but they say that an apple fell on his head.
In the following year and a half, the foundations of Newton's theory of gravitation were created.

I like another consequence of the same year 1666 > Iron Maiden - The Number Of The Beast

 

[Bosko]

I said that now I understand that negative mass cancels the positive one from KE.

I recommend that you start the calculations first using Newtonian mechanics.
You calculate the speed of the Sun after some time, the kinetic energy, how long it takes to travel a certain distance...
Interesting questions are:
How long will it take the Sun to reach Sag A?
What will be its speed at that moment?
...
When you clear everything up, you move on to the special theory of relativity and compare the results.

Then you do the same using the general theory of relativity.

 

[Bosko]

I appreciate the honesty. Best of luck as this will crash eventually.

If you use Newtonian mechanics, the potential and the kinetic energy will have the same value and the opposite sign.
You can start from
$$-E_P=\frac{GMm}{r}=\frac{mv^2}{2}=E_K$$
and calculate speed for various distances (r).

Then you can fix this formula a bit because at the beginning $E_P$ is small but not quite zero and the speed ( $v$ ) is zero so $E_P$ is also zero. ( The above formula is valid for falling from infinity )

 

[PeterDonis]

If you use Newtonian mechanics, the potential and the kinetic energy will have the same value and the opposite sign.

This is also true in GR for this scenario.