Is Gravity Affected by its Own Potential Energy?

[1mmorta1]

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)

 

[Runner 1]

Oh really? That sounds pretty cool! Ask him about it.

 

[1mmorta1]

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.

 

[Dale]

Wow, this thread really ran amok.

OK, in the OP there are a couple of valid points:
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

2) Gravity gravitates. This is because the Einstein field equations are non-linear:
http://archive.ncsa.illinois.edu/Cyberia/NumRel/EinsteinEquations.html

 

[?????]

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.

 

[1mmorta1]

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?

 

[Runner 1]

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?

 

[zonde]

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.

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.

 

[?????]

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.

 

[Runner 1]

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.

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.

 

[1mmorta1]

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.)

 

[?????]

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.

 

[zonde]

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.

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.

 

[zonde]

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.

 

[bahamagreen]

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?

 

[jfy4]

After reading, I'm actually still not sure what it is this thread is about 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"

 

[Sam Gralla]

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.

 

[Runner 1]

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?)

 

[jfy4]

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"

 

[jfy4]

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.

 

[pervect]

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.

 

[Runner 1]

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.

 

[jfy4]

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.

 

[ZealScience]

Perhaps you are talking about gravitational waves. Gravitational waves carry energy just like electromagnetic waves, and it is gravitating.

 

[zonde]

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?

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".

 

[Sam Gralla]

And why do you think that there is any "gravitational energy" at all?

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".

There is such a thing as "gravitational energy" because gravitational waves carry it. You can precisely characterize the rate at which an isolated system is losing energy to gravitational-wave emission. What you can't do is say that so much of the energy was at one place in spacetime and so much of the energy was elsewhere. or whether the energy "came from" matter or field energy. (Of course, you could try to make such a statement using a "psuedotensor" as somebody brought up, but somebody else could come along with another inequivalent pseudotensor and claim that *that* one was the true gravitational energy. What everyone will agree about is the total energy in the system as well as its rate of change.) People do use "stress-energy pseudotensors" in computations, but that's mainly for convenience in performing some specific computation. All claims about a "true" local gravitational energy density have disappeared from the literature at this point. (There is a community still looking for "quasi-local" gravitational energy, but that doesn't seem to work very well either.)

You may be trying to call this sort of energy "radiation energy" so you don't need "field energy". But there are reasons for saying that even a non-radiating system has "gravitational binding energy" (that can't be localized). For example, in the Newtonian limit of GR the energy/mass of an isolated system can be written as an integral over the mass density of the matter. But as soon as you go back to GR (or just a post-Newtonian correction), this property is lost. What is the other contribution? It makes sense to think of it as energy in the field. But there's no way to write it as an integral over matter plus an integral over field, so again it makes sense to say "gravitational energy exists but can't be localized".

 

[pervect]

Do you happen to have MTW's textbook, "Graitation"? I know they have a good section on the topic of why you can't localize the energy of a gravitational field.

I'm not sure how much more clear I can be, but I'll try saying it again.

Covariance is an important physical principle. It boils down to saying that measurements made by different observers represnt the same underlying reality.

The sort of covariance we need for relativity is Lorentz covariance. Any four-vector, regardless of whether it is (time, distance) or (energy, momentum) must transform via the lorentz transforms to have a physical meaning that's independent of the coordinate system.

If you don't have covariance, your quantity cannot be defined in an observer independent way.

Pseudotensors, in the sense used in General Relatiavity (i.e. the energy pseudotensor you refer to) do not define energy in a way that's independent of the observer.

Some people have remarked, with some merit, that the "pseudotensors" in GR are really just non-tensors.

The thing that makes energy pseudotensors useful at all is that while they don't offer an observer-compatible defintion of energy, the total energy computed via them will transform properly given the proper conditions (usually asymptotic flatness).

So the pseudotensors themselves do not offer any physically meaningful way to localize energy because different observers will, as other posters have remarked, not have compatible views of how the energy is distributed.

They do allow you to come up with a total energy that everyone agrees on, however. (And they aren't the only method of doing it, there's a proof I think that the pseudotensor definition of energy matches the Bondi defiition).

 

[zonde]

There is such a thing as "gravitational energy" because gravitational waves carry it. You can precisely characterize the rate at which an isolated system is losing energy to gravitational-wave emission.

Isolated system can't loose energy.
Actually we don't have truly isolated systems so in order to model such a system we can imagine system in environment that mirrors all the outward interactions of the system with equivalent inward interactions.
This means that if the system is emitting gravitational waves then equivalent waves are directed toward the system from environment. And I would say that if gravitating system can emit these waves then it can absorb the same waves unless you have some strong arguments why this shouldn't be so.

What you can't do is say that so much of the energy was at one place in spacetime and so much of the energy was elsewhere. or whether the energy "came from" matter or field energy. (Of course, you could try to make such a statement using a "psuedotensor" as somebody brought up, but somebody else could come along with another inequivalent pseudotensor and claim that *that* one was the true gravitational energy. What everyone will agree about is the total energy in the system as well as its rate of change.) People do use "stress-energy pseudotensors" in computations, but that's mainly for convenience in performing some specific computation. All claims about a "true" local gravitational energy density have disappeared from the literature at this point. (There is a community still looking for "quasi-local" gravitational energy, but that doesn't seem to work very well either.)

What I say is that mass is moving down in gravitational potential when the system emits gravitational waves. So it's mass that is loosing this energy. The same way if you would try to reverse this process it's mass where you have to put this energy to move it upwards in gravitational potential.

You may be trying to call this sort of energy "radiation energy" so you don't need "field energy". But there are reasons for saying that even a non-radiating system has "gravitational binding energy" (that can't be localized). For example, in the Newtonian limit of GR the energy/mass of an isolated system can be written as an integral over the mass density of the matter. But as soon as you go back to GR (or just a post-Newtonian correction), this property is lost. What is the other contribution? It makes sense to think of it as energy in the field. But there's no way to write it as an integral over matter plus an integral over field, so again it makes sense to say "gravitational energy exists but can't be localized".

I have problems with that. "Gravitational binding energy" is negative energy. So we have missing energy not excess energy.

 

[jfy4]

Thanks pervect,

your helpful as always. I do have that book and I take a look at that section.