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Conservation of Energy

  1. Aug 23, 2008 #1


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    I have seen it stated that GR doesn’t always maintain the axiom of the conservation of energy. However, I haven’t been able to find any general explanation of why this is the case without getting drawn into the complexity of Killing vectors etc. So my initial questions are:

    - Is it true that GR doesn’t maintain the conservation of energy?
    - Is there a simple illustrative example?
    - Is it a fringe issue under the extremes of spacetime curvature, i.e. gravity?

    As a related, but somewhat tangential issue, it is sometimes pointed out that potential energy is not a concept used by general relativity. However, most of us will probably have been introduced to the concept of gravity using the Newtonian concept of potential energy and gravitational force. Equally, the classical concept of the conservation of energy is often described in terms of just 2 fundamental forms of energy, i.e. kinetic and potential. So my next question is:

    - Can the GR theory of gravity still be accurately transposed into the ‘concept` of potential energy and gravitational force, e.g. [tex] F=GMm/r^2[/tex], even though this may not now be the preferred approach under GR?

    By way of an example, classic effective potential is defined in the context of the conservation of total energy in the form of kinetic and potential energy, i.e.

    [tex]E_T=1/2m(v_r^2+v_o^2) - \frac{GMm}{r}[/tex]

    Where [tex][v_r][/tex] is the radial velocity, while [tex][v_o][/tex] is the orbital velocity. If we only consider a circular orbit with no radial component of velocity, the effective potential is stated to be:

    [tex]Veff=1/2mv_o^2 - \frac{GMm}{r}[/tex]

    However, a very similar form can be derived from the Schwarzschild metric, which is an accepted GR solution of Einstein’s field equations given the usual caveats. While the derivation is not reproduced here, the equivalent solution is given as:

    [tex]Veff=1/2mv_o^2 - \frac{GMm}{r} \left( 1 + \frac{v^2}{c^2} \right)[/tex]

    The implication seems to be that as the orbital velocity increases towards [c], an additional relativistic factor comes into play to maintain the orbit, i.e. as [tex][v_o \rightarrow c][/tex] the factor approaches 2. It is not clear whether this factor of 2 explains the anomaly in the classical and GR equations for light deflection?

    Classical: [tex]\theta = \frac{2GM}{rc^2}[/tex] as opposed to GR: [tex]\theta = \frac{4GM}{rc^2}[/tex]

    However, the only point of all this was simply to ask whether the GR theory of gravity can still be accurately presented in terms of potential energy?

    P.S. PF Admin, please note that the thread title keeps resetting after preview!
  2. jcsd
  3. Aug 23, 2008 #2


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    A good place to start would be here.
  4. Aug 23, 2008 #3


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    I have only quickly read the link provided,
    it seems very clear on the question, i.e.
    Is Energy Conserved in General Relativity?,
    but less clear on the answer, but maybe I just missed it.
    Thanks anyway.
    Last edited: Aug 23, 2008
  5. Aug 24, 2008 #4


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    Any other thoughts on this topic based on #1?
    Going, going.....nearly gone!
  6. Aug 25, 2008 #5

    Jonathan Scott

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    That link covers the answer fairly thoroughly and I think it might help to read it a little more carefully. As it says at the start of the linked page:

    In special cases, yes. In general -- it depends on what you mean by "energy", and what you mean by "conserved".​

    The Newtonian concept of "potential energy" appears at first glance to work in GR as well when describing the motion of a test mass in the field of a central body, in that the effective rest energy change due to time dilation in a static field matches the potential energy, and the potential energy plus (relativistic) kinetic energy remain constant for motion in a static field.

    However, I was surprised to find that it doesn't appear to work if you also consider the energy change of the source mass, or consider two similar masses orbiting around one another, even in a linearized weak field approximation. In that model, each of the objects apparently experiences an equal decrease in rest energy as it approaches the other (because of time dilation due to the other object) matching the Newtonian potential energy, so the rest energy of the system changes by twice the amount in the Newtonian model (where the potential energy is a property of the configuration of the system, not of the individual objects). However, the kinetic energy of the system only increases by the Newtonian kinetic energy, equal to the Newtonian potential energy, so this doesn't seem to add up.

    In gravitational Quantum Field Theory, some people apparently assume that the energy of the field increases by the same amount, preserving the total energy, and that works very nicely from a mathematical point of view, giving a result very much like the Maxwell energy density of the field in electromagnetism. However, GR says that there is no energy in a vacuum regardless of the strength of the field, so that appears to mean either that QFT isn't compatible with GR or that the definition of "energy" is not the same in these two cases.

    Given that the Newtonian concept of conservation of energy allows us to calculate complex gravitational interactions of many bodies to very high accuracy in non-relativistic situations, it seems surprising to me that there should be a problem with the concept in such a simple two-body situation in a weak GR approximation.
  7. Aug 26, 2008 #6


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    Response to #5:

    Hi Jonathan,
    You are probably right; I will go back and review the link in more detail. I guess I had become a little impatient to find some sort of `simple` summary of the ‘state of the conservation axioms’ under GR. Many references seem to allude to issues, but I was finding in difficult to latch on to any specific example that did not quickly get ‘lost’ in tensor notation. Hence my starting point:
    However, I wasn’t totally sure about the context of your following example:
    Can this example be described by the Schwarzschild metric? If you solve this metric for a free-falling radial path, the velocity [v] depends on `who` does the measuring, e.g.

    [1] Distant Observer [v] = [tex]c\left(1-\frac{Rs}{r}\right)\sqrt{\frac{Rs}{r}}[/tex]

    [2] Onboard Observer [v] = [tex]c\sqrt{\frac{Rs}{r}}[/tex]

    Where [tex]Rs = 2GM/c^2[/tex], which allows [2] to be written as:

    [3] Onboard Observer [v] = [tex]\sqrt{\frac{2GM}{r}}[/tex]

    What is a bit surprising is that [3] can be derived from a classical conservation of energy assumption based on the non-GR concept of potential energy, e.g.

    [4] [tex]E_K = 1/2mv^2 = \frac{GMm}{r} = E_P[/tex]

    I say `surprising` because [tex]1/2mv^2[/tex] is only a valid approximation of kinetic energy when [v<<c], i.e. non-relativistic. I outlined the example for a relativistic orbit in #1. Anyway, I probably just need to sit down and do a bit more reading, but would welcome any additional pointers into this subject. Thanks.
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