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I Meaning of constants of motion in General Relativity

  1. Jun 15, 2017 #1
    Hello,
    I have a mess in interpretation of constants in description of movement in GR.

    First of all I define Lagrangian ##l=1/2g_{\mu\nu}u^{\mu}u^{\nu}##, and I would like to talk about axial smyetric spacetime (for example Kerr black hole) ##l(r,\theta)##. l is independent from ##t## and ##\phi## i.e. ##\frac{dl}{d\dot{t}}=const.=-E=u_t## and ##\frac{dl}{d\dot{\phi}}=const.=L=u_{\phi}##.
    One can say, ##E## is energy of test particle measured by static observer at infinity and ##L## is angular momentum of test particle measured by static observer at infinity.
    Then I may define a localy non rotating frame (LNRF) as frame at some ##r## and has ##L=0## (this frame rotates only thanks a dragging by rotating central body).

    1) Are my definition and everything above alright?
    2) If I won't be a static observer at infinity and I will measure the energy (##u_t##) of a test particle. Is true that I do not find a value ##-E##? For what and why it will be different?
    3) Will I find the value ##-E## for ##u_t## if I will be in LNRF? Or is there some conection with LNRF and measuring ##u_t=-E##?
    4) Does it have some conection with killing vectors? For example if I move in direction of some killing vector I will measure same things as static observer at infinity or something like that?
     
  2. jcsd
  3. Jun 15, 2017 #2

    PeterDonis

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    They look OK as far as they go, but I would recommend consulting a reference such as the Matt Visser paper linked to below, which will help with standardized definitions and notation:

    https://arxiv.org/abs/0706.0622

    Yes. (Note also that your sign convention for the energy at infinity appears to be nonstandard; usually it is defined as ##+E##.)

    Because the static observer measuring the energy is not at infinity. If you want to think about ##E## as a constant of the motion for the test object, that's fine, but the static observer's measurement of the energy doesn't just depend on the properties of the test object. It also depends on the properties of the observer, and in particular on his 4-velocity. The 4-velocity of a static observer not at infinity is different from the 4-velocity of a static

    observer at infinity; that is why his measurement of the energy of the test particle is different.

    In general I don't think so. See below.

    Only in the sense that the measurement of the test particle's energy depends on the 4-velocity of the observer measuring it. Also, it depends on the particular trajectory being followed by the test particle, i.e., on the test particle's angular momentum ##L##, as well as ##E##.

    Yes. Can you see what it is? Does it help to have the hint that the reason why the Lagrangian is independent of ##t## and ##\phi## has something to do with Killing vectors?
     
  4. Jun 22, 2017 #3
    Thank you very much for long answer. I probably try to calculate energy for some typical observers at some random places and I guess I will find what is happening. If you know about some good text where solve such things please give some link.

    I'm sorry, I forgot to mention, that I use signature of metric in this way (-,+,+,+). Maybe nonstandard but I started with that so I continue :-)

    Killing vectors are connected with symmetries of the spacetime and also I may say that "values" of metric coefficients are constant in this direction. True?

    Theorem of Noether connects symmetries and constants of motion. A independence of the Lagrangian on ##t## and ##\phi## says that there are symmetries.
    Well, now I probably start to understand. So if I'm talking about constants of motion and if I will move in direction of some Killing vector I'm still measure same value of these constants. But NOT same as at infinity (if I'm not starting in infinity). Do you agree?

    Now LNRF observer is moving only in direction of Killing vectors - true?
     
  5. Jun 22, 2017 #4

    PeterDonis

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    Yes.

    This is a bit vague, but the general answer is that any property of the metric coefficients, which are expressed in terms of a particular coordinate chart, will depend on your choice of coordinates.

    A better statement is that the local spacetime geometry remains constant along integral curves of any Killing vector field. That statement is independent of any choice of coordinates.

    Yes, and to each one corresponds a constant of geodesic motion. (The qualifier is crucial--see below.)

    No. You're mixing up two different things.

    If you move along an integral curve of a Killing vector field, the local spacetime geometry stays the same. (Note that this "motion" might not actually be possible in terms of an observer following such an integral curve; that is only possible if the Killing vector field is timelike. In Kerr spacetime outside the static limit, the KVF corresponding to the ##t## coordinate is timelike. But the KVF corresponding to the ##\phi## coordinate is not; it's spacelike. So no observer can actually move along the integral curves of the ##\phi## KVF.)

    If you move on a geodesic worldline, then for each KVF, there will be a constant of your motion. Note that your worldline itself will be timelike, but it will not, in general, be an integral curve of any KVF. But that doesn't matter for constants of the motion: the constant of the motion is just the inner product of your 4-momentum with the appropriate Killing vector. For example, in Kerr spacetime we have two KVFs, which we can call ##T^\mu## and ##\Phi^\mu##. Then, for an observer with 4-momentum ##P_\mu## moving on a geodesic worldline, there will be two constants of the motion: ##E = T^\mu P_\mu## and ##L = \Phi^\mu P_\mu##.

    No. See above.
     
  6. Jun 23, 2017 #5
    The velocity going into the energy and angular momentum is measured relative to the locally nonrotating observer (the ZAMO), and given by

    [tex]\rm E_{total}= E_{kin}+E_{pot}+E_{rest} = \sqrt{\frac{\Delta \ \Sigma}{(1-v^2) \ ((a^2+r^2)^2-a^2 \Delta \sin ^2 \theta)}} + \Omega \ L_z = const[/tex]

    for the total energy,

    [tex] \rm {L_z = \frac{v_{\phi} \ \bar{\omega}^2 \ \sin \theta}{\sqrt{1-v^2}}} = const[/tex]

    with the radius of gyration

    [tex] \rm \bar{\omega} = \sqrt{\frac{\left(a^2+r^2\right)^2-a^2 \Delta \sin ^2(\theta )}{a^2 \cos ^2 \theta +r^2}}[/tex]

    for the axial and

    [tex] \rm p_{\theta} = \frac{v_{\theta} \ \sqrt{\bar{\omega}^2+z^2}}{\sqrt{1-v^2}}[/tex]

    for the polar component of the angular momentum. So the kinetic component of the energy is always the special relativistic one:

    [tex]\rm E_{kin}=\frac{1}{\sqrt{1-v^2}}-1[/tex]

    where v is measured by a local and locally stationary probe. An observer at infinity does not measure

    [tex]\rm v=\sqrt{v_{r}^2+v_{\phi}^2+v_{\theta}^2}[/tex]

    but rather

    [tex]\rm v'=\sqrt{\dot x^2+\dot y^2+\dot z^2}[/tex]

    (overdot is differentiation by proper time τ) with

    [tex] \rm x = \sqrt {r^2 + a^2} \sin\theta \ \cos\phi \ , \ y = \sqrt {r^2 + a^2} \sin\theta \ \sin\phi \ , \ z = r \cos\theta \quad[/tex]

    using natural units with

    [tex]\rm G=M=c=1[/tex]

    The two shorthand terms Δ and Σ are

    [tex]\rm \Sigma =a^2 \cos ^2 \theta +r^2 \ , \ \ \Delta =a^2+r^2-2 r[/tex]

    Also see arxiv.org/pdf/gr-qc/0101023.pdf page 5, arxiv.org/pdf/1601.02063.pdf page 2 and Wikipedia: Kerr equations of motion
     
    Last edited: Jun 23, 2017
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