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lordoftheselands

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lordoftheselands

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So it is always with any energy generation scheme that "uses" gravitational potential energy from something moving down. Look to whatever lifted the thing up in the first place.

Since this is in relativity (at time of writing) it's worth noting that conservation of energy doesn't necessarily apply in all spacetimes. It works in Eartylhbound situations, though.

- #3

PeterDonis

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No, it doesn't. Objects moving solely under the influence of gravity are in free fall, with zero proper acceleration.gravity accelerate things

Not at all. For example, water flowing downhill is not gaining any energy in total. It is converting gravitational potential energy to kinetic energy.there must be a consense that it adds energy to things

- #4

Feynstein100

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I think that's what the OP was referring to. Had the earth not been there, the water wouldn't have any gravitational potential energy. Imagine an object outside the earth's gravitational field. It doesn't have any energy. But place it just inside the field and it now gains kinetic energy which it didn't have before.It is converting gravitational potential energy to kinetic energy.

Ah but wait. You have to move it to get it inside the field and in doing so, you give it kinetic energy. Still, it does speed up once it enters the field (assuming it gets closer to the earth), meaning there is a net gain of energy. And assuming conservation of energy holds true, this extra energy has to come from somewhere, in this case the earth's gravitational field.

One possible resolution to this problem is to say that the earth's gravitational field extends to infinity and thus the object was never outside it, and always had potential energy. However, this feels like cheating, and I can't help but wonder if there's a better resolution.

- #5

russ_watters

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- #6

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Since the gravitational field extends to infinity, it's hard to imagine such a place.Imagine an object outside the earth's gravitational field.

- #7

Lluis Olle

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Being "Objects" by definition "any substance that has mass and takes up space by having volume", would that mean that you found an easy analytical solution to the two body problem in GR?No, it doesn't.Objectsmoving solely under the influence of gravity are in free fall, with zero proper acceleration.

- #8

Feynstein100

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I literally said that in the last part of my comment. Did you even read it?Since the gravitational field extends to infinity, it's hard to imagine such a place.

- #9

Dale

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The gravitational field goes to infinity. There is no "outside" and no "just inside". All you can do is move from a region of one potential to a region of another potential.Imagine an object outside the earth's gravitational field. It doesn't have any energy. But place it just inside the field and it now gains kinetic energy which it didn't have before.

No net gain of energy. There is a loss of PE and a conversion of that PE into the same amount of KE.Still, it does speed up once it enters the field (assuming it gets closer to the earth), meaning there is a net gain of energy.

Do you think that you and the universe are childhood friends playing a game where you get some say in the rules? We have no say in the rules. And there is no possible way for the universe to “cheat” since the rules are just descriptions of whatever the universe does, so whatever it does is by definition a rule and not a cheat.One possible resolution to this problem is to say that the earth's gravitational field extends to infinity and thus the object was never outside it, and always had potential energy. However, this feels like cheating

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- #10

Dale

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Why should it be an easy analytical solution? Do you think that the laws of nature preclude true statements about difficult solutions or numerical solutions?Being "Objects" by definition "any substance that has mass and takes up space by having volume", would that mean that you found an easy analytical solution to the two body problem in GR?

- #11

PeterDonis

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Not due to the earth, but the earth is not the only gravitating mass in the universe.Had the earth not been there, the water wouldn't have any gravitational potential energy.

There is no such place, strictly speaking. The earth's field gets weaker as you get farther away from it, but it never vanishes completely. In practical terms, we don't see this because there are other masses in the universe whose fields get much larger than the earth's at distances far enough away from the earth.Imagine an object outside the earth's gravitational field.

- #12

Vanadium 50

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There's always the Whistler quote "Two plus two continue to make four, despite the whine of the amateur for three or the cry of the critic for five."Do you think that you and the universe are childhood friends playing a game where you get some say in the rules?

- #13

Feynstein100

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Not quite what I hand in mind 😂 I simply meant that at this point, it feels like you're forcing the math to confirm your preconceived notions rather than seeing where it actually leads.Do you think that you and the universe are childhood friends playing a game where you get some say in the rules? We have no say in the rules. And there is no possible way for the universe to “cheat” since the rules are just descriptions of whatever the universe does, so whatever it does is by definition a rule and not a cheat.

In particular, I find the notion of the gravitational field extending to infinity problematic because a) there are no infinities in real life, and b) it's too newtonian of a view. Perhaps the math is the same in GR, but I find the idea that a mass here can affect things on the other side of the universe to be completely antithetical to the concept of locality, on which SR and GR are based.

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It's the maths that says the field extends to infinity. We didn't pull the notion out of nowhere.I simply meant that at this point, it feels like you're forcing the math to confirm your preconceived notions rather than seeing where it actually leads.

Prove it.there are no infinities in real life,

I don't think you really understand what "locality" means in this context. And again, you're saying that the maths is inconsistent with the maths, which is obviously nonsense.it's too newtonian of a view. Perhaps the math is the same in GR, but I find the idea that a mass here can affect things on the other side of the universe to be completely antithetical to the concept of locality, on which SR and GR are based.

- #15

PeterDonis

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That concept just means that interactions propagate at the speed of light. Changes in spacetime geometry due to changes in the configuration of masses meet that requirement in GR. There is no limit on the distance over which such changes can propagate.I find the idea that a mass here can affect things on the other side of the universe to be completely antithetical to the concept of locality, on which SR and GR are based.

- #16

Dale

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Well, yes. We force the math to conform to our experimental data. It isn’t the math “leading” anything. It is the data. We see what the universe actually does and then write the rules to reflect the facts.it feels like you're forcing the math to confirm your preconceived notions rather than seeing where it actually leads

GR contains Newtonian gravity. And far away from any mass is precisely where GR matches Newtonian gravity.it's too newtonian of a view.

Although, in GR’s FLRW cosmology, energy is not conserved, but not for the reasons mentioned here. There is no “outside of the gravitational field” even there.

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Amongst everything else, you are confusing two different ideas of infinite. You cannot have two points an infinite distance apart. But, there is no limit to the possible distance between two points.a) there are no infinities in real life

- #18

Nugatory

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”Extending to infinity” is indeed problematic, but the problem is in the very common but imprecise natural language wording. There’s no infinity involved when we state it differently: if the gravitational potential is non-zero at a finite distance ##R## from the gravitating body and ##\Delta R## is some small number, then the gravitational potential will be smaller but still non-zero at the finite distance ##R+\Delta R##.In particular, I find the notion of the gravitational field extending to infinity problematic because a) there are no infinities in real life, and b) it's too newtonian of a view.

Informally, there is no hard edge detectable even with arbitrarily precise instruments.

- #19

Lluis Olle

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You're absolutely right, as always.Why should it be an easy analytical solution? Do you think that the laws of nature preclude true statements about difficult solutions or numerical solutions?

I think I just "condensed" the remark too much.

What I tried to mean is that if "an object" (which let me say that in physical "terminology" would mean any object, even a piece of a neutron start so to speak), is in

Obviously I don't know, but I thought that to calculate a Geodesic from the GR equations is something "mechanical" (I would guess tedious), but easy in the sense that if you know what you're doing, you follow the recipe and you get that Geodesic trajectory.

- #20

jbriggs444

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On the contrary. When the earth is present, the water hasHad the earth not been there, the water wouldn't have any gravitational potential energy.

Make sure that you are comparing apples with apples. Do not change the zero point of your potential energy calculation when you coallesce the Earth.

- #21

DaveC426913

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Or the Python quote:There's always the Whistler quote "Two plus two continue to make four, despite the whine of the amateur for three or the cry of the critic for five."

"...shalt thou count to [four], no more, no less. [Four] shall be the number thou shalt count, and the number of the counting shall be [four]. [Five] shalt thou not count, neither count thou [three], excepting that thou then proceed to [four]. [Six] is right out.

- #22

Lluis Olle

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I guess that it comes from the mass, as E=mc

By the way, the other day I saw I video which was something like this: the same way that the binding energy from the proton-electron is part of the mass of one atom, the binding gravitational energy for two massive objects has in itself a mass, and that mass has been demonstrated to obey the strong equivalence principle up to a number of decimal places (in a nutshell, accelerates the same way as the other mass).

- #23

PeterDonis

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Videos are usually not valid references here, but the only way we can know for sure is for you to give a link to it.the other day I saw I video

A negative mass--it makes a negative contribution to the total mass of the system.he binding gravitational energy for two massive objects has in itself a mass

Yes.that mass has been demonstrated to obey the strong equivalence principle up to a number of decimal places

- #24

PeterDonis

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Don't guess. When you are trying to help someone understand something here, you need to first understand it well enough yourself that you don't have to guess.I guess that it comes from the mass

In fact this answer is not correct, as other posts in this thread will show.

- #25

PeterDonis

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No, that's not what "objects" meant in the statement of mine that you quoted.Being "Objects" by definition "any substance that has mass and takes up space by having volume"

No, because "objects" in the statement of mine that you quoted has a much more limited meaning than you think.would that mean that you found an easy analytical solution to the two body problem in GR?

- #26

PeterDonis

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This is only an exact statement for test objects, but test objects by definition have zero effect on the spacetime geometry. So they are not "bodies" in the sense of finding a solution to the "two-body problem".if "an object" (which let me say that in physical "terminology" would mean any object, even a piece of a neutron start so to speak), is infree falland "proper acceleration" is zero (which for an "object" with a measurable size and volume I don't know very much what it means), thenit follows a Geodesic trajectory.

There has been considerable theoretical work on the topic of how good an approximation it is to assume that objects which are

It isObviously I don't know, but I thought that to calculate a Geodesic from the GR equations is something "mechanical" (I would guess tedious), but easy in the sense that if you know what you're doing, you follow the recipe and you get that Geodesic trajectory.

- #27

Lluis Olle

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The referenced video is in Spanish. Talks about the retroreflectors left in the Moon, and how they where used to check for the strong equivalence in GR.Videos are usually not valid references here, but the only way we can know for sure is for you to give a link to it.

- #28

Lluis Olle

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The concept of "the energy of gravity" is a little bit ambiguous, but don't tell me is that is not mass related!In fact this answer is not correct, as other posts in this thread will show.

- #29

Nugatory

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No one is trying to tell you that . What they are saying is that "It comes from the mass" is incorrect, which shouldn't surprise you as you were uncertain enough to preface that statement with "I guess"but don't tell me is that is not mass related!

- #30

PeterDonis

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Even in English it would probably not be the best reference to give. See below for better ones. It only took me a few minutes of web searching to find the references below; that is something you could (and should) have done.The referenced video is in Spanish.

This is the Lunar Laser Ranging experiment. The Wikipedia article gives a reasonable overview and a good number of references:Talks about the retroreflectors left in the Moon, and how they where used to check for the strong equivalence in GR.

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

This appears to be a good paper on the use of lunar laser ranging to test the equivalence principle:

https://arxiv.org/pdf/gr-qc/0507083.pdf

- #31

Vanadium 50

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The OP seems to have departed, and he seems to have a history of doing so. (Looking at past threads) Looking at those self-same threads, I suggest that the thread be relabeled as B-level.

The thread has been taken a little off-track by someone trying to help the OP, but not entirely sure. In this case, it really is best to check things out before posting. If questions remain, you can always start a new thread.

Now, a question for the OP (if he's still around). Why do you think this is a relativity question? Why and how do you think the answer is different in Newtonian gravity? And if the answer is "they are the same", we shuld probably answer this classically and not drag the additional compications of relativity into it.

- #32

PeterDonis

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Fair point. Done.I suggest that the thread be relabeled as B-level.

- #33

russ_watters

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Note that the concept of gravitational potential energy doesn't really require gravitational force extending to infinity. It's just the integral of gravitational force with distance, and if that force ever drops to zero a million light years away or whatever, it has basically no impact on the gravitational potential energy (it would be too small of a deviation to be noticeable). So even if the first sentence of that prior post were true, the second still wouldn't be.In particular, I find the notion of the gravitational field extending to infinity problematic because a) there are no infinities in real life...

[prior post]

Imagine an object outside the earth's gravitational field. It doesn't have any energy.

- #34

PeterDonis

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Since this thread was posted in the relativity forum, it should be noted that this is not quite how things work in relativity. The statement of Feynstein100 that you quoted can certainly be criticized (and I and others have done so elsewhere in this thread), but I'm not sure trying to do so along these particular lines will work.Note that the concept of gravitational potential energy doesn't really require gravitational force extending to infinity. It's just the integral of gravitational force with distance

- #35

Dale

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That is a really good point. Even if we hypothesize a conservative force with a finite range, the region outside the force would be an equipotential region. The potential would be defined, even outside the finite range of the force.So even if the first sentence of that prior post were true, the second still wouldn't be

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