Is Gravity Affected by its Own Potential Energy?

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The discussion centers on the relationship between gravity and potential energy, particularly in the context of the Earth-moon system. It is suggested that potential energy, which is tied to gravitational fields, may have mass and thus complicates calculations of gravitational interactions. Participants debate whether potential energy should be considered a separate mass or if it inherently contributes to the mass of the system. The conversation touches on the complexities of calculating gravitational effects, especially when introducing additional bodies, and the challenges posed by the non-linear nature of gravity in general relativity. Ultimately, the discussion highlights the theoretical intricacies involved in understanding gravitational interactions and potential energy.
  • #31
1mmorta1 said:
(Since no experts are chiming in...)

Weird isn't it? They normally jump on a chance to correct people.

Anyway, I'm fairly certain you don't need motion to have energy. Motion is just a form of it. Energy is also contained in fundamental (rest) masses, as well as fields (strong, weak, electromagnetic, gravitational). That's all I can think of -- maybe there's some more.
 
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  • #32
That sounds correct to me, in fact I know your correct. I just wanted to at least bring up the fact that there exists another relativistic formula for calculating the energy of an object(Which is why photons don't have infinite mass).
Did you see how many views this page has? Maybe we've opened a can of worms that the experts don't want to be a part of.
 
  • #33
Speaking of lack of participation, did you see my post on the next step after the higgs(its on this board)? No one has anything to say on that either(maybe no one likes us) but I'd be interested on any knowledge/opinions you have on the matter.
 
  • #34
It's really the same formula (and will give you the same result), but in a different form. For example, a photon has mass. Just not rest mass. I don't know why they've made the terms so confusing, but I remember being confused by it in high school.
 
  • #35
I was researching my original post a little more, and found that energy is indeed carried away from systems in the form of gravity waves. Most physicists believe these to exist, however, they are having a very hard time detecting them because the measurements must be so precise.
 
  • #36
Your a better man than I am, I didn't study general relativity in high school ;) Keep in mind that you can't use the first formula for objects in motion though. Duh...you know that(its just me being picky about them being the same formula)
That would be a nice solution to our dilemma, it would mean there's no need to calculate "extra" mass because its being carried away from the system. I'll have to look into these.
I have an uncle who worked on a project for NASA, he was a cryogenics engineer, detecting gravity waves by building the most precise gyroscopes ever created(and the most perfectly spherical objects in the universe)
 
  • #37
Oh really? That sounds pretty cool! Ask him about it.
 
  • #38
Ha. Its hard to get a hold of the guy these days. He's always working...I don't even know where he lives anymore, it was NASA's gravity probe b. Yuck...engineering. Don't think I could do it.
 
  • #39
  • #40
Runner 1 said:
how one goes about doing these calculations.

Potential energy cannot gravitationally attract itself.

If you want to calculate the gravitational attraction force of the Earth for the moon, then select your gravitational model and do the calculation. Now move the moon out to infinity and this will give you the potential energy gain of the moon in its original orbit. This potential energy has no place in this particular gravitational force calculation.

If you want to know the earth-moon pair gravitational force to the sun, then you take can include the earth-moon gravitational attraction into this calculation. In this case, the sun-moon force and the sun-earth force calculated separately will not give you the same answer as the earth-moon-potential energy composite attraction force to the sun. The two calculations will only differ very slightly. The gravitational potential energy of the earth-moon pair is very small compared to the mass of the material of those two bodies.
 
  • #41
? said:
Potential energy cannot gravitationally attract itself.

If you want to calculate the gravitational attraction force of the Earth for the moon, then select your gravitational model and do the calculation. Now move the moon out to infinity and this will give you the potential energy gain of the moon in its original orbit. This potential energy has no place in this particular gravitational force calculation.

If you want to know the earth-moon pair gravitational force to the sun, then you take can include the earth-moon gravitational attraction into this calculation. In this case, the sun-moon force and the sun-earth force calculated separately will not give you the same answer as the earth-moon-potential energy composite attraction force to the sun. The two calculations will only differ very slightly. The gravitational potential energy of the earth-moon pair is very small compared to the mass of the material of those two bodies.

This is your response on how? Thank you, but did you read the rest of the thread...we are trying to figure out HOW, not looking for the conceptual process. Can you back the information above with some equations and directions?
 
  • #42
? said:
Potential energy cannot gravitationally attract itself.

I don't believe this statement is true. I could be wrong though. Have you got a source (like a textbook) perhaps?
 
  • #43
Runner 1 said:
I don't believe this statement is true. I could be wrong though. Have you got a source (like a textbook) perhaps?
Did you read about mass deficit in the link given by DaleSpam?
I think it explains the problem very clearly.
DaleSpam said:
1) The mass of a bound system is, in general, different from the mass of the individual components. There is a good Wikipedia article:
http://en.wikipedia.org/wiki/Binding_energy#Mass_deficit

Potential energy does not add or subtract energy from the system. If just specifies how much of the energy that the system has already can be radiated away.
 
  • #44
1mmorta1 said:
This is your response on how? Thank you, but did you read the rest of the thread...we are trying to figure out HOW, not looking for the conceptual process. Can you back the information above with some equations and directions?

Perhaps I am wrong and you are right. I did read the thread and I thought I was answering the original question.

By the way, my original statement "Now move the moon out to infinity and this will give you the potential energy gain of the moon in its original orbit." is in error. The correct statement would be "Now move the moon down to the Earth and this will give you the potential energy gain of the moon in its original orbit."

As long as the OP wants to consider energy in gravitational calculations, I would suggest adding the kinetic energy of the moon in its orbit around the Earth when calculating the moon-earth-potential energy combination gravitational attraction for the sun. And you might as well throw in the kinetic energy of the Earth rotating on it's axis too.
 
  • #45
zonde said:
Did you read about mass deficit in the link given by DaleSpam?
I think it explains the problem very clearly.

Yes, I did read it. That's why I followed it up with a question.


zonde said:
Potential energy does not add or subtract energy from the system. If just specifies how much of the energy that the system has already can be radiated away.

I never said anywhere that potential energy added or subtracted additional energy to/from the system. What I am saying is that, assume for a second that gravitational fields have no relationship to energy at all. They just lock objects into place. If you calculate the system dynamics of the Earth and the moon now, it will be DIFFERENT than how it is in reality.

Take the spring example again. Let's assume potential energy does NOT obey Einstein's formula -- meaning it does not contribute to the gravitational attraction of the spring. This means a compressed spring will weigh the same as an uncompressed spring. This is obviously not true -- there is a difference. By analogy, I'm trying to quantify that sort of difference with the Earth and moon system.
 
  • #46
? said:
Perhaps I am wrong and you are right. I did read the thread and I thought I was answering the original question.

By the way, my original statement "Now move the moon out to infinity and this will give you the potential energy gain of the moon in its original orbit." is in error. The correct statement would be "Now move the moon down to the Earth and this will give you the potential energy gain of the moon in its original orbit."

As long as the OP wants to consider energy in gravitational calculations, I would suggest adding the kinetic energy of the moon in its orbit around the Earth when calculating the moon-earth-potential energy combination gravitational attraction for the sun. And you might as well throw in the kinetic energy of the Earth rotating on it's axis too.

It is true that if we were looking at the total mass of the system we would have to observe that. We were looking into the mass of the system with regards to the original masses of the Earth and moon plus the mass of gravitational potential energy, and runner1 is curious as to whether or not the new mass considered in the system would result in more gravity potential, and thus more potential energy...in a recursive process.

Runner1, i think in taking the equations for finding gravitational energy potential(double check mine to make sure they're right), and using our original conversion rate of e = mc2(I should never have brought up relativistic mass), and converting that potential energy to mass which is added to the mass of the Earth and the Moon, we have answered our questions.

(In this instance though, we are ignoring the rest of the universe and the fact that the system "gravitates" that energy.)
 
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  • #47
Runner 1 said:
I don't believe this statement is true. I could be wrong though. Have you got a source (like a textbook) perhaps?

If two compressed springs are sitting in open space, the potential energy's stored in each spring will gravitationally attract each other (and the mass of the material in each spring). But a single compressed spring in open space does not attract itself, in a kind of endless increasing spiral.
 
  • #48
Runner 1 said:
Yes, I did read it. That's why I followed it up with a question.

I never said anywhere that potential energy added or subtracted additional energy to/from the system. What I am saying is that, assume for a second that gravitational fields have no relationship to energy at all. They just lock objects into place. If you calculate the system dynamics of the Earth and the moon now, it will be DIFFERENT than how it is in reality.
I don't understand your question.

Runner 1 said:
Take the spring example again. Let's assume potential energy does NOT obey Einstein's formula -- meaning it does not contribute to the gravitational attraction of the spring. This means a compressed spring will weigh the same as an uncompressed spring. This is obviously not true -- there is a difference. By analogy, I'm trying to quantify that sort of difference with the Earth and moon system.
No, when you compress the spring you put "real" energy into the spring. This "real" energy of course contribute to the mass. So it's not potential energy that contributes to the mass.
Now if you release the spring you can take away some "real" energy from the spring. But if you release the spring and let it oscillate eventually converting energy into the heat then the energy is still in the spring and it's mass wouldn't change.
 
  • #49
1mmorta1 said:
It is true that if we were looking at the total mass of the system we would have to observe that. We were looking into the mass of the system with regards to the original masses of the Earth and moon plus the mass of gravitational potential energy, and runner1 is curious as to whether or not the new mass considered in the system would result in more gravity potential, and thus more potential energy...in a recursive process.
This is wrong. Gravitational potential energy is negative. You have less mass in the system after system has released energy that is equivalent to gravitational potential energy.
 
  • #50
If gravity effected gravity, this interactive gravitational energy for a particular mass would have to accumulate by some function to the energy of that mass. This would mean that this interaction of gravity with itself would change that thing's effective gravitational mass; but would not change that thing's inertial mass.

The strong equivalence principle holds that moving things within a gravitational field should only be influenced by their position and not be influenced by what they might be made of or how much matter comprises them.

This suggests that gravity affecting gravity violates the strong equivalence principle?
 
  • #51
After reading, I'm actually still not sure what it is this thread is about :rolleyes: But, it sounds like perhaps you are interested in the Weyl Tensor. It describes the curvature due intrinsically to the gravitational field. Here is the Wiki article for your convenience.

http://en.wikipedia.org/wiki/Weyl_tensor"
 
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  • #52
Guys, the reason why you are arguing is because gravitational energy can't be localized. You can only talk about the total energy of the system. You can't split it up into "matter energy" and "field energy" (or, put another way, there are too many inequivalent ways to try to do this--hence all your arguing). This is a standard topic covered in GR books. The most challenging part about learning GR is learning the right questions to ask.

That said, the physical intuition in the OP is sound. If you view the theory perturbatively off of flat spacetime then there is a sense in which this "iterative" procedure described in the OP does take place. In fact there is a famous derivation (really more of a plausibility argument) due to Feynman whereby you begin with a linear equation and add "gravity gravitates" to boostrap your way up to the full nonlinear Einstein equations.
 
  • #53
Sam Gralla said:
If you view the theory perturbatively off of flat spacetime then there is a sense in which this "iterative" procedure described in the OP does take place. In fact there is a famous derivation (really more of a plausibility argument) due to Feynman whereby you begin with a linear equation and add "gravity gravitates" to boostrap your way up to the full nonlinear Einstein equations.

There we go! That's exactly what I wanted to know. What are these equations called? (I don't know anything about general relativity -- is it the same thing?)
 
  • #54
Sam Gralla said:
Guys, the reason why you are arguing is because gravitational energy can't be localized. You can only talk about the total energy of the system. You can't split it up into "matter energy" and "field energy" (or, put another way, there are too many inequivalent ways to try to do this--hence all your arguing). This is a standard topic covered in GR books. The most challenging part about learning GR is learning the right questions to ask.

I don't think this is true... The Ricci tensor, or equivalently the stress-energy-momentum tensor, is precisely the object which describes the energy not due to the gravitational field, However, a separate object exists which does include that energy.

See here http://en.wikipedia.org/wiki/Stress-energy-momentum_pseudotensor"
 
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  • #55
Runner 1 said:
There we go! That's exactly what I wanted to know. What are these equations called? (I don't know anything about general relativity -- is it the same thing?)

if you want to ask gravity, and relativity questions, you should study GR, since this is exactly what it covers.
 
  • #56
jfy4 said:
I don't think this is true... The Ricci tensor, or equivalently the stress-energy-momentum tensor, is precisely the object which describes the energy not due to the gravitational field, However, a separate object exists which does include that energy.

See here http://en.wikipedia.org/wiki/Stress-energy-momentum_pseudotensor"

Because it's not a true tensor, the stress-energy tensor isn't coordinate independent (and IIRC it even depends on the gauge). In any event, it won't give you a unique answer for the distribution of energy, because it's not a true tensor and hence not coordinate independent.

So Sam's answer is spot-on, and very well written.
 
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  • #57
jfy4 said:
if you want to ask gravity, and relativity questions, you should study GR, since this is exactly what it covers.

Well, I didn't realize it was a GR question. My post just got moved to this forum.
 
  • #58
pervect said:
Because it's not a true tensor, the stress-energy tensor isn't coordinate independent (and IIRC it even depends on the gauge). In any event, it won't give you a unique answer for the distribution of energy, because it's not a true tensor and hence not coordinate independent.

So Sam's answer is spot-on, and very well written.

I'll have to do some looking into that, I'm sorry to doubt, but wiki gives an explanation as to why what you said isn't true, and I need more time because of my naivete to be sure.

Thanks all the same.

EDIT: also, I guess I'm a little confused by your post since you segued with "because". Are you saying that the stress-energy-psudeo-tensor does reflect an object that represents the energy of the gravitational field. It seems like you wrote "the SEM-pseudotensor is that object 'because'..." but your post seems to object to what I said. Could you clarify what you meant up above.
 
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  • #59
Perhaps you are talking about gravitational waves. Gravitational waves carry energy just like electromagnetic waves, and it is gravitating.
 
  • #60
Sam Gralla said:
Guys, the reason why you are arguing is because gravitational energy can't be localized.
And why do you think that there is any "gravitational energy" at all?

Sam Gralla said:
You can only talk about the total energy of the system. You can't split it up into "matter energy" and "field energy" (or, put another way, there are too many inequivalent ways to try to do this--hence all your arguing).
If you can attribute all of the total energy of the system to "matter energy" plus radiation energy without anything left then there is no reason to think that there anything like "field energy".
 

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