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Definition of energy in general relativity

  1. Oct 9, 2009 #1

    tom.stoer

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    I searched for related discussions but could not find one; so I hope that somebody can give me hint:

    I know that the definition of energy E[V] as a volume integral of T°°(x) over a spacelike slice = 3-volume V is not (always) possible in general relativity. The basic reason is that the usual current conservation dj=0 is replaced by the covariant equation DT=0 (D is the covariant derivative; T is the energy momentum tensor). Therefore the usual trick making use of integration by parts does no longer work because additional terms (not vanishing on the boundary) are present. In addition the requirement that the integral E[V] must be the 0-component of a 4-vector would no longer hold.

    As the energy E[V] is no longer well-defined, global energy conservation becomes meaningless in general relativity.

    Now I have the following questions:
    1) is there a quantity E* that can serve as "energy" in a more general sense and that corresponds to E in special limiting cases (like flat Minkowskian spacetime)?
    2) what are the most general conditions (for the metric g) such that E[V] is well-defined, conserved (dE/dt=0) and has the correct transformation properties?
    3) is it possible to define the energy of the whole universe E[universe] if V="universe" has some boundary at infinity, or has no boundary at all?
    4) is it possible to define E[universe] in some special cases, e.g. FRW, dS, AdS etc.?

    Thanks for your help
    Thomas
     
  2. jcsd
  3. Oct 9, 2009 #2

    atyy

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    Last edited by a moderator: Apr 24, 2017
  4. Oct 9, 2009 #3
    Forget about energy in GR. There is no translation invariance, there is no energy conservation and even definition. All additive conservation laws are lost in GR.
     
  5. Oct 9, 2009 #4

    tom.stoer

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    Thanks for the links!

    So that means that one can construct something like energy even in GR, but one has to use rather cumbersome objects like pseudotensors and non-covariant expressions. This seems to be the reason that the definition (and the value) of energy depend on the fundamental object used.

    In addition it seems strange that one has to use the equations of motion in some cases. And as expected in some cases the boundary terms are not uniquely defined.

    What does that mean for my questions 3 and 4? Is it possible to apply these definitions for the whole universe?
     
  6. Oct 9, 2009 #5
    Although it's probable neither what you are looking for, nor sufficiently enlighening, I believe thermal energy density and heat flux density form a Lorentz invariant vector. This would be sufficent to obtain a continuity equation that is connection free; *d*dq=0.
     
  7. Oct 9, 2009 #6
    Alternatively, if it makes sense to talk about pv=(E,-p) as a continuous field (does it?), then by the same means, *d*dpv=0 . And also, d2pv=0 .
     
  8. Oct 9, 2009 #7

    Hurkyl

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    Don't the Hodge duals depend on the connection?
     
  9. Oct 9, 2009 #8
    If they do I've wasted a great deal of time using a false assumption. I can't prove it, nor find support at this time. It's a very good question. I'll look for the answer.
     
  10. Oct 10, 2009 #9

    tom.stoer

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    In the above mentioned papers the idea was always to calculate E from some expression starting with T (and related objects); but all the expressions were in the very end related to the metric g.

    If I want to define the energy for the whole universe, I have to add me add another question

    5) Is it possible to define energy for the gravitational field w/o referring to matter?

    Let me explain why this is interesting: assume that we start with a radiation dominated universe for which we can calculate "energy". At a later stage we may have a universe that is dominated by black holes. But black holes are vacuum solutions, that means T=0 identically. Therefore it must be possible to show that the energy can be always expressed via g. For the black holes this is the only way to calculate E which of course must be identical to the E calculated in an earlier stage via T. So the energy flows from the radiation into the gravitational field! I think something like that must be kept in mind when one discusses (eternal) inflation starting from a vacuum fluctuation. The overall energy of the multiverse is exactly zero, but locally positive or negative energy regions (bubbles, "universes") may exist.
     
  11. Oct 10, 2009 #10
    Hurkyl, in n dimensions, the Hodge duality operator by definition takes a p-form and returns an (n-p)-form. Since the exterior derivative acting on a form is a coordinate independent expression, so is the exterior derivative of it's Hodge dual.
     
    Last edited: Oct 10, 2009
  12. Oct 11, 2009 #11

    haushofer

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    A time ago I read these pieces of Bertschinger, especially #5 (Symmetry transformations, the EH-action and gauge transformations), but I must say that these notes confuse me. Especially when it comes to "dynamical symmetries" and "non-dynamical" symmetries, which also has to do with the fact that I never quite understood the exact relationship between active and passive transformations.

    Maybe I should open a topic about this text of Bertschinger,

    http://web.mit.edu/edbert/GR/gr5.pdf
     
    Last edited by a moderator: Apr 24, 2017
  13. Oct 11, 2009 #12

    Hurkyl

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    But the definition of the operator itself depends on the metric tensor.

    If (x,y) are the standard coordinates on the Euclidean plane, then three values of the Hodge dual associated with the standard metric are:
    *dx = dy
    *dy = -dx
    *d(2x+y) = -dx + 2dy​

    However, if we use a metric tensor were {d(2x), dy} is an orthonormal basis, then the same three values for the associated Hodge dual would be
    *dx = (1/2) dy
    *dy = -2 dx
    *d(2x+y) = -2dx + dy​
     
  14. Oct 11, 2009 #13
    Yes, * is a function of the metric.

    The beauty of the d and * operators is that you can do all your symbol manipulation using the Minkowski metric of Riemann normal coordinates, and the expressions are equally true under general coordinate transforms as long as your variables are forms. Of course the manifold has to be orientable and other stuff.

    Where A and B are forms, p+q=n, and n is the dimension of the manifold,

    [tex]A_{\mu_1 \mu_2 ... \mu_q} = \frac{1}{p!} {\epsilon ^{\nu_1 \nu_2 ... \nu_p}}_{\mu_1 \mu_2 ... \mu_q} B_{\nu_1 \nu_2 ... \nu_p} [/tex]

    Here, epsilon is not to be confused with the Levi-Civita symbol, but is the Levi-Civita tensor. The Levi-Civita symbol isn't a true tensor (called a tensor density) except in Rn Euclidean space. Calling [itex]\tilde{\epsilon}[/itex] the Levi-Civita symbol,

    [tex]{\epsilon_{\nu_1 \nu_2 ... \nu_n}} = \sqrt{|g|} \tilde{\epsilon} _{\nu_1 \nu_2 ... \nu_n}[/tex]

    |g| is the absolute value of the determinate of the metric, which is where the metric dependancy you noticed comes from.
     
  15. Oct 12, 2009 #14
    If a timelike Killing field exists in the spacetime, then you can relate a conserved quantity similar to energy with it, something along the lines of
    [itex] E = T_{\mu} \frac{dx^{\mu}}{d\lambda} [/itex].

    I think (but you might want to check from a reliable source) that all spherically symmetric spacetimes have timelike Killing vectors, and therefore some kind of conserved quantity you can relate to energy.

    Also, I think in GR you usually say that the energy of the entire universe is exactly zero, as the gravitational potential energy is negative, and balances out everything inside.

    This also something you might want to look into: http://www.scholarpedia.org/article/Arnowitt-Deser-Misner_energy
     
  16. Oct 12, 2009 #15

    tom.stoer

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    The timelike Killing vector is nothing else but the generalization of time-displacement symmetry in ordinary Lagrangian mechanics. Therefore it is a very restrictive concept. The main problem with it is not that w/o a timelike Killing vector energy is not conserved, but that w/o a timelike Killing vector energy cannot be defined in that way.

    As far as I know only static spacetimes have timelike Killing vectors. dS and AdS no not have timelike Killing vectors, only a maximum number of spacelike ones.

    Your statement "... in GR ... the energy of the entire universe is exactly zero, as the gravitational potential energy is negative, and balances out everything inside" is something I heard quite often, but w/o any single line of maths. There are two reasons:
    1-4) the definition of (conserved) energy based on the energy-momentum tensor
    5) the definition of the energy of the gravitational field
    Number 5) because it was questions 5) in one of the last posts.

    In order to make progress towards the exact meaning of "gravitational potential energy is negative" we need a definition of that energy; I was not able to find any.
     
    Last edited: Oct 12, 2009
  17. Oct 12, 2009 #16

    Haelfix

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    The best discussion of global energy conservation in the GR context that I can recall, was in Wald's textbook, which I don't have handy atm. Otherwise MTW has a decent section as well.

    There are several different notions of energy in general and they are only valid for subclasses of possible spacetimes (like static cases, or ones with sensible asymptotics). Bondi energy/ADM energy and the like, which you see a lot in the literature are for the latter case, Komar energy for the former.

    Generally when people tlak about the energy of the whole universe, they are talking about a closed FRW spacetime where its zero under most definitions (for compact S^3 * R topology).

    http://en.wikipedia.org/wiki/Mass_in_general_relativity
     
    Last edited: Oct 12, 2009
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