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Conserved quantities for massless particles (Schwarzschild)

  1. Jan 22, 2016 #1
    If a metric admits a Killing vector field ##V ## it is possible to define conserved quantities: ## V^{\mu} u_{\mu}=const## where ## u^{\mu}## is the 4 velocity of a particle.

    For example, Schwarzschild metric admits a timelike Killing vector field. This means that the quantity ##g_{\mu 0} u^{0}## is conserved. Thus since the metric is asymptotically flat, and since ##u^{0}=\gamma## I can assume that the conserved quantity is the energy per unit mass.

    Since the metric has a timelike Killing vector field, I can say that there is another conserved quantity: for a massive particle that quantity is the angular momentum.

    But how can I extend this construction to massless particles? How can I say that energy and angular momentum of a photon are conserved in Schwarzschild spacetime?
     
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  3. Jan 22, 2016 #2

    PeterDonis

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    The fact that the Killing vector field is timelike does not mean the worldline of the particle for which the conserved quantity is defined has to be timelike. The actual construction uses the 4-momentum of the particle, not the 4-velocity; 4-momentum is well-defined for a massless particle.

    Often when discussing massive particles the 4-velocity is used for convenience, so that the units of the conserved quantity are "energy per unit mass" instead of "energy", as you say. But that's a convenience; it's not the fundamental quantity. The fundamental quantity is energy, i.e., ##V^\mu P_\mu##, where ##P_\mu## is the 4-momentum of the particle.
     
  4. Jan 22, 2016 #3

    PeterDonis

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    Also, the angular momentum is associated with a spacelike KVF, not a timelike one (the timelike KVF is only associated with the energy). More precisely, the angular momentum about a particular axis is conserved if rotations about that axis generate a (spacelike) KVF for the spacetime. In Schwarzschild spacetime, rotations about any axis generate a spacelike KVF, so angular momentum about any axis is conserved. In Kerr spacetime, by contrast, only rotations about the axis of the rotating black hole generate a spacelike KVF, so only angular momentum about that axis is conserved.

    The same remarks as I made for energy in my previous post also apply to angular momentum: the conserved quantity is fundamentally defined using the 4-momentum, not the 4-velocity, so it is well-defined for a massless particle. As a convenience, the 4-velocity for a massive particle can be used to define "angular momentum per unit mass"; but again, that's a convenience, not the fundamental quantity.
     
  5. Jan 22, 2016 #4
    Ok, Thanks. Where can I find a proof of what you say, namely that the fundamental conserved quantity is ## V^{\mu} P_{\mu}## and not ##V^{\mu}U_{\mu}##?
     
  6. Jan 22, 2016 #5

    PeterDonis

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    What sort of "proof" are you looking for? These things aren't a matter of "proof"; they're a matter of choosing the best definitions. Defining the conserved quantity as ##V^\mu P_\mu## means that the same definition automatically applies to both massive and massless particles. Defining it as ##V^\mu u_\mu## means that it only applies to massive particles, which leads directly to the problem you posed in the OP, of what definition to use for massless particles. The answer to that problem is to pick the better definition--the one that applies equally to both massive and massless particles.

    To put it another way: for a massive particle, both ##V^\mu P_\mu## and ##V^\mu u_\mu## are valid conserved quantities. But for massless particles, only the first is; the second isn't well-defined. So you have two choices: you can say that ##V^\mu u_\mu## is the fundamental conserved quantity for massive particles, but ##V^\mu P_\mu## is for massless particles; or you can say that ##V^\mu P_\mu## is the fundamental conserved quantity for both, and ##V^\mu u_\mu## is a useful convenience for massive particles. Both choices are logically consistent; but the second one is easier to work with, so it's the one that is standardly used.
     
  7. Jan 22, 2016 #6

    bcrowell

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    This doesn't seem quite right to me. Of course, each person is entitled to make their own choice of postulates and definitions. But in every presentation I've ever seen, a Killing vector is defined as a solution of the Killing equation, and conservation of ##V_a U^a## or ##V_a P^a## is a theorem with a proof. I also don't think it's true that one of these is inherently more fundamental than the other. Conservation of ##V_a P^a## is more *general* than conservation of ##V_a U^a##, in the context of GR. On the other hand, Killing vectors and geodesics are purely geometrical topics studied in a much broader context than GR.

    If you first do a purely geometrical proof that ##V_a U^a## is conserved, then it automatically holds in GR for ##V_a P^a## as well. In the case of a massive particle this is trivial. The massless case follows by taking the limit of the massive case.

    If you take it as a postulate that the stress-energy tensor is divergenceless, ##\nabla_a T^{ab}=0##, then a straightforward computation shows that the quantity ##P^a=T^{ab}V_b##, interpreted as a density of energy-momentum, also has zero divergence. Application of Gauss's theorem then shows that the flux of energy-momentum through a closed surface is zero. In the limit where the density of energy-momentum is concentrated close to a single world-line, this implies that the energy-momentum of a test particle is conserved. There's a more detailed discussion in this style on p. 62 of Hawking and Ellis, including a discussion of angular momentum.
     
  8. Jan 22, 2016 #7
    But these conservation laws are not obvious. A mathematical proof is needed. What do you mean by "they're a matter of choosing the best definition"?
    I mean, it can be proved that if a metric admits a KVF one can define conserved quantities. I know how to prove that ##U_{\mu} V^{\mu}## is conserved along the geodesic of a particle (and it is quite simple: it suffices to contract the geodesic equation with the 1 form associated to the KVF and to use Killing equation Then the expression ##\frac{d}{d \lambda}(U_{\mu} V^{\mu}) =0 ## comes quite easily ). In this sense I'm asking for a proof that ##\frac{d}{d \lambda}( V^{\mu} P_{\mu})=0##.
     
  9. Jan 22, 2016 #8

    George Jones

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    Doesn't this proof work for any geodesic, including lightlike and (non-physical) spacelike geodesics.
     
  10. Jan 22, 2016 #9

    bcrowell

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    Which proof? We've posted 3 of them.
     
  11. Jan 22, 2016 #10

    PeterDonis

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    It's the same proof as for ##V^{\mu} U_{\mu}##. The proof does not depend on any properties of ##U_\mu## except that it satisfies the geodesic equation, and ##P_\mu## also satisfies that equation, so the proof works for it as well.
     
  12. Jan 22, 2016 #11

    bcrowell

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    That doesn't work unless you make some sort of argument about limits as in my #6. If you don't take a limit, then for m=0 the normalization of the velocity vector is undefined, and in any case multiplying it by m would mean multiplying by zero.
     
  13. Jan 22, 2016 #12

    PeterDonis

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    I was misinterpreting somewhat the question he was asking. I certainly agree that the constants of the motion can be established by rigorous proofs.
     
  14. Jan 22, 2016 #13

    PeterDonis

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    But you don't need to approach it that way. You can prove that ##V^a P_a## is conserved directly, using Killing's equation and the geodesic equation, without having to assume that ##P_a## is timelike. You just need to have an affine parameterization along the geodesic; you don't have to assume that the affine parameterization is proper time (or proper distance for a spacelike geodesic--the proof works for those as well), or that the tangent vector has a well-defined normalization.
     
  15. Jan 22, 2016 #14

    bcrowell

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    OK. Or you could do what I did in the second half of #6.
     
  16. Jan 23, 2016 #15
    Ok, thanks everyone.
    But i've never seen the geodesic equation written in terms of momentum. On the Internet I can only find something about deriving the geodesic equation from ##\nabla _{\nu}T^{\mu \nu} =0##.

    Moreover, for massive particles the conserved quantity in S. metric is: ##\frac{dt}{d \tau} = \frac{E}{1-\frac{2m}{r}}## in geometrical units. I've read that it is possible to define the same quantity for massless particles, just by using an affine parameter and defining it properly, but I can't do that.
     
  17. Jan 23, 2016 #16

    PeterDonis

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    The geodesic equation, in its most general form, is written in terms of a vector that is tangent to the worldline (and an affine parameterization of the worldline). In many presentations this vector is assumed to be the 4-velocity, but this is not necessary; all that is necessary is that it is tangent to the worldline. (Also, you don't have to assume that the affine parameter is proper time, which is what you are doing when you use the 4-velocity as the tangent vector.) The 4-momentum vector is tangent to the worldline for both massive and massless particles. (For massive particles, it is just the 4-velocity times ##m##, so it is easy to see that any equation satisfied by the 4-velocity is also satisfied by the 4-momentum. But if you're trying to see how the equation works for massless particles, it's better to start with the 4-momentum and not make any assumptions about ##m##.)

    Carroll's lecture notes on GR have a good discussion of how to derive the geodesic equation using the concept of parallel transport and tangent vector, without making any particular choice of affine parameter, in Chapter 3 (starting on p. 65, leading up to equation 3.47).

    See Carroll, Chapter 7, starting on p. 173 and leading up to equation 7.46, which is valid for both massive and massless particles and includes both the energy and angular momentum constants of the motion. (The discussion of radial null curves and different coordinate charts from p. 182 on is also interesting.)
     
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