Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

A Where is potential curvature stored in GR?

  1. Feb 20, 2017 #1
    It is commonly said that if you lift an object above the earth it gains potential
    energy equal to mgh (m=mass, g=gravitational acceleration, h=height), suggesting
    that the potential energy is in the lifted mass.

    This cannot be. Consider the case of two perfectly rigid spheres, isolated in space, and touching one another, held together only by gravity.
    Imagine an outside source provides the energy to lift up one of them and then
    insert a rigid, weightless rod between them to prevent them from falling together.
    The situation is symmetrical.
    If the potential energy is in the spheres and only in the spheres, then it must be
    divided equally between them, not concentrated in the smaller, because there is no
    smaller one.

    One answer I have seen is that the potential energy is in the "field" (or system or configuration), but where in the field is it?
    Is it at the midpoint between the two spheres? Is it distributed throughout the
    field? And if so, how is it distributed?
    The potential energy must be somewhere, or we have energy that is not contributing to the gravitational field, don't we?

    I don't see how GR gets rid of the problem, although it would use a different
    vocabulary. There should be some additional curvature of spacetime due to the "potential curvature" of spacetime that was created by the lifting, right?
    Or is there no potential curvature, just curvature immediately realized as the lifting is done?
     
  2. jcsd
  3. Feb 20, 2017 #2

    PeterDonis

    User Avatar
    2016 Award

    Staff: Mentor

    No, this is not necessary. For systems in which the concept is well-defined at all (it isn't for all systems), the potential energy is a property of the system as a whole. It is not localized anywhere.

    In GR only stress-energy contributes to the gravitational field.
     
  4. Feb 20, 2017 #3

    Dale

    Staff: Mentor

    What if the potential is not only in the spheres, but also in the whole spacetime?
     
  5. Feb 20, 2017 #4

    Jonathan Scott

    User Avatar
    Gold Member

    Although conservation of energy is a fundamental principle in Newtonian gravity, the concept of gravitational energy, including potential energy, is problematic and not usually very useful in GR. For a start, the very existence of such energy depends on the frame of reference, in that for example an object which is freely falling and hence has constant energy from one point of view may be accelerating and exchanging kinetic energy for potential energy from another point of view.

    In standard GR, the source of a gravitational field is the non-zero terms of the stress-energy tensor. This means that matter and electromagnetic fields can give rise to gravitational source effects, but gravitational energy cannot. (This approach however makes it rather difficult to explain how the black hole merger detected by LIGO emitted a huge amount of energy in the form of gravitational waves).

    One way to get back the usual Newtonian conservation laws is to use a "pseudotensor" to represent a local effective gravitational energy density as seen from a specific frame of reference. For weak fields and a static or slowly-changing system, the gravitational energy density can be set to ##g^2/8\pi G## where ##g## is the magnitude of the Newtonian gravitational field. This is equivalent to the "Einstein pseudotensor". The effective energy of each source is then reduced by the potential due to all other sources, so every source has its own energy reduced by the total potential energy (causing a reduction by twice the potential energy overall), but at the same time the gravitational energy density when integrated over all space adds up to a positive equivalent of the potential energy (by a similar process to the Coulomb energy for electromagnetism). This means that the overall "energy" of the system is reduced relative to the energy of the parts by its internal potential energy.

    However, that model clearly uses a different concept of "energy" from the GR stress-energy tensor, in that the total energy of the sources appears to be reduced by twice the potential energy, which is corrected by the positive energy in the gravitational field, but in GR the field is not a gravitational source.

    Edit: I should have been a bit more specific about the gravitational energy density. What matches the potential energy density is the difference between the densities of the separate gravitational fields and the density of the total field. That is, if the field due to mass 1 is ##\mathbf{g_1}## and that due to mass 2 is ##\mathbf{g_2}##, then the potential energy is the difference between ##(\mathbf{g_1})^2/8\pi G + (\mathbf{g_2})^2/8\pi G## and ##(\mathbf{g_1}+\mathbf{g_2})^2/8\pi G## which is ##2\mathbf{g_1}.\mathbf{g_2}/8\pi G## or equivalently ##\mathbf{g_1}.\mathbf{g_2}/4\pi G##. (For a point mass, the gravitational energy density expression increases without limit close to the mass, so it is not a useful concept in that case).
     
    Last edited: Feb 20, 2017
  6. Feb 20, 2017 #5
    The same problem arises even in Newtonian gravity doesn't it? If, for example, I am in free fall with an elevator passenger-space containing
    Einstein, from my point of view the elevator is not accelerating, but if I am on the earth on the geodesic of the elevator, it may whack me on the head with considerable kinetic energy.

    Difficult or impossible? I remember reading Logunov's criticism of Einstein's quadrupole formula (I think), which Logunov claimed was correct, but went outside the realm of GR! I didn't understand (or is it remember?) his criticism very well. I know Logunov is generally dismissed in the West as "fringe", meaning "somewhat better than a crackpot", but that doesn't mean all his criticisms of GR are wrong, (not that I'm ascribing that view to you personally).

    I will have to do some substantial study to figure out what this means.

    Thank you for your comments. If or when I understand what they mean, I may have some more questions.
     
  7. Feb 21, 2017 #6

    Jonathan Scott

    User Avatar
    Gold Member

    It's true that energy is different from different viewpoints in Newtonian gravity and special relativity too. However, in Newtonian gravity the location of potential energy is not important because it is not expected to affect the gravitational source strength.

    I don't personally know of any satisfactory resolution of the explanation of how gravitational waves transfer energy and momentum within GR. There is no doubt that they do, as proved by LIGO. The calculation of the flow of energy and momentum in the distant field is a standard GR calculation based on a linear approximation. However, some maintain that this is not "energy" or "momentum" in the sense that applies at points where the stress-energy tensor is non-zero, and use terminology in which the energy and momentum does not flow but is rather recreated at the interaction points, explicitly violating Newtonian conservation laws.

    In GR itself, the self-gravitation of an object means that the potential energy of each part of it is reduced by the time dilation factor, which again would mean that the total energy relative to the local energy of the components from which the object is built is decreased by twice the potential energy. This would suggest that the overall source strength of an object would be reduced by twice its internal potential energy. However, in GR there is an additional source strength contribution from pressure within the object. In a static situation, this is positive but exactly equal to the potential energy, so again it balances out, so the overall gravitational source strength is equivalent to the sum of the energies of the component parts minus the potential energy. The static expression including the pressure term is known as the "Komar mass".

    The same pressure term is present in Newtonian gravity. If you consider an isolated pair of masses held apart in a static configurations by rods or similar, then if you integrate the pressure over a plane perpendicular to the line between the masses, you get the force across that plane, equal to ##G m_1 m_2 / r^2## for any plane between the masses. If you integrate that over the distance ##r## between the masses, you get the potential energy between that pair of masses, ##G m_1 m_2 / r ##. As force adds as a vector, if you add any number of additional masses and integrate over the whole system using any three perpendicular planes, you get a positive quantity matching the total (negative) potential energy of that configuration. Note that this still applies even if the configuration contains springs, balloons or whatever; if it is static overall, the net force (including gravitational) through any plane must be zero.

    Note however that this only works for a static situation. Although the pressure integral is equal to the potential energy, it does not seem right to assume that the potential energy is located where the pressure is located. If one considers the Newtonian case and what happens if a rod is disconnected, then the pressure across that plane drops to zero (at up to the speed of sound in the rod), but it seems implausible that the gravitational source strength of the system as a whole would be abruptly affected as a result. (The energy equivalent of the pressure term which "vanishes" in that case cannot be matched with any mechanical energy in the springiness or motion of the rod because the maximum mechanical energy in that form can easily be shown to be much smaller than the potential energy of the configuration). The same applies in GR; Richard Tolman pointed out that this leads to the paradoxical situation that if a large spherically symmetrical object suddenly temporarily changes pressure inside, due to some sort of transient structural change, this suggests that its gravitational source strength should temporarily change, even though the total energy enclosed has remained constant, which seems quite implausible. This "Tolman Paradox" is quite disturbing and its resolution remains an active subject of research.
     
  8. Feb 23, 2017 #7

    PeterDonis

    User Avatar
    2016 Award

    Staff: Mentor

    The general resolution is to carefully define exactly what is locally conserved in GR. The local conservation law in GR is that the covariant divergence of the stress-energy tensor is zero. Since it's the covariant divergence, it involves the metric as well as the stress-energy tensor itself. So changes in the metric, such as a gravitational wave, have to be taken into account in assessing the local conservation law in a given small region of spacetime. Once that is taken into account, it is seen that what you are calling "transfer of energy and momentum" by a gravitational wave is actually a change, due to a changing metric (which is what a gravitational wave is, a propagating change in the metric), in how a local observer sees the various pieces of the stress-energy tensor. For example, if a GW passes through an object and heats it up, the increase in the object's energy density as seen by a locally comoving observer is exactly balanced by the change in the metric induced by the GW, so that the covariant divergence of the SET remains zero.

    All of this is rather complicated mathematically and hard to visualize intuitively, which is why a simpler heuristic of GWs "carrying energy" by analogy with other kinds of waves (such as EM waves) is often used in such cases. This heuristic works as long as you keep in mind that it's a heuristic and is not intended to be rigorous. The rigorous treatment requires actually writing down the covariant divergence equation and evaluating it for the GW case.

    Do you have any recent references?
     
  9. Feb 23, 2017 #8

    Jonathan Scott

    User Avatar
    Gold Member

    I understand the general idea about the problem about energy and momentum in the gravitational wave, but since the analogy of "carrying energy and momentum" works in the same way as for EM waves, that suggests to me that the GR definition of "energy" might not be as useful as the Newtonian one, or might need to be split into different forms to reflect energy as seen relative to a coordinate system (probably pseudotensor) and energy as seen locally.

    Here's a fairly recent one which admittedly seems to reach a very controversial conclusion. The references include the previously discussed 2005 paper by J Ehlers et al.
    Title: On the relativistic formulation of matter
    Author: Vishwakarma, Ram Gopal
    http://adsabs.harvard.edu/abs/2012Ap&SS.340..373V
     
  10. Feb 23, 2017 #9

    PeterDonis

    User Avatar
    2016 Award

    Staff: Mentor

    Definitely controversial, yes. I don't think the resolution of Tolman's Paradox is considered an open issue by most physicists in the field of GR.

    Btw, the harvard.edu link to this paper appears to be paywalled. An arxiv link to the paper is here:

    https://arxiv.org/abs/1204.1553

    Yes, but note that the only subsequent papers to cite this one are by the same author.
     
  11. Feb 24, 2017 #10
    I think these paragraphs from Wald's General relativity are pertinent here:

    "However, the interpretation of equation ##\nabla^{a}T_{ab}=0## is altered now. A family of observers is represented by a unit timelike vector field ##v^a## . If one could find such a vector field which is covariantly constant, i .e .,##\nabla_{a} v_{b}=0##-or for which merely##\nabla_{(a}v_{b)}=0##-then we would have ##\nabla^a(T_{ab}v^b) = 0## . Applying the curved spacetime version of Gauss's law (see appendix B), we again would obtain strict conservation of energy in the form (4 .2 .18) for the energy-momentum four-vector##J_a = -T_{ab}v^b## measured by the observers represented by ##v^b##. However, in curved spacetime in general one no longer can find a ##v^a## satisfying## v^av_a = -1## and ##\nabla_{(a}v_{b)} = 0## . (Indeed, the equation ##\nabla_{(a}v_{b)} = 0## is Killing's equation and holds if and only if ##v^a##generates cone-paramete r group of isometries [see appendix C] .) Thus the argument fails that equation ##\nabla^{a}T_{ab}=0##implies strict energy conservation . Physically, this makes sense because the gravitational tidal forces can do work on the fluid and may increase or decrease its locally measured energy . However, if one considers a spacetime region of dimension small compared with radii of curvature, then, physically, the tidal forces can do little work and the energy of the fluid should be approximately conserved . But over this small spacetime region it is possible to find vector fields with ##\nabla_{b}v^a≈0##, and thus equation##\nabla^{a}T_{ab}=0## does yield approximate conservation of energy as measured by these observers. Thus, equation ##\nabla^{a}T_{ab}=0## may be interpreted as a local conservation of material energy over small regions of spacetime ."

    This seems to restrict the local conservation to small regions compared with the specific radii of curvature considered. I read your concern about explanatory resolution of the local conservation issue as wondering with respect to what local curvature are the in principle arbitrarily large regions gravitational waves travel compared .
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Where is potential curvature stored in GR?
  1. GR and Curvature (Replies: 3)

  2. Curvature in GR (Replies: 10)

Loading...