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Some questions on GR

  1. Aug 25, 2004 #1
    a)Can we add extra terms with covariant derivative equal to 0 so the coupling constant in relativity is dimensionless?.

    b)for a metric gab that is time independent,can we define an "energy" term E so dE/dt is conserved?...

    c)from the Wheller-De Witt equation can we construct a linear differential equation time dependent and first order in functional derivatives respect to gab?..
     
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  3. Aug 25, 2004 #2

    DW

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    Yes. It is called the energy parameter. For a time independent metric the following is a timelike killing vector:
    [tex][T^{\mu }] = \left[\begin{array}{cc}1\\0\\0\\0\end{array}\right][/tex]
    And the conserved energy parameter will be
    [tex]\frac{E_{cons}}{mc} = g_{\mu}_{\nu}T^{\mu }U^{\nu }[/tex]
    which reduces to
    [tex]\frac{E_{cons}}{mc} = g_{0}_{\nu}U^{\nu }[/tex]
    If the metric also happens to be diagonal, then this reduces further to
    [tex]E_{cons} = g_{0}_{0}\frac{dt}{d\tau}mc^{2}[/tex]
    And in special relativity where [tex]g_{0}_{0} = \eta _{0}_{0} = 1[/tex] this reduces to
    [tex]E_{cons} = \frac{dt}{d\tau}mc^{2}[/tex]
    [tex]E_{cons} = \gamma mc^{2}[/tex]
     
    Last edited: Aug 25, 2004
  4. Aug 26, 2004 #3

    pervect

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    We've been discussing b) quite a bit. Assuming you want the total energy of a gravitating system, the answer is yes, as long as the system is embeded in an asymptotically flat space-time with a vacuum at infinity.

    Because you have a time-independent metric, there will be a time-like Killing vector, Ka. It turns out that the killing vector, and the volume element of the space time, are all you need to compute the energy.

    The energy inside a volume enclosed by a two-sphere S, assumed to be orthogonal to the timelike Killing-vector will be

    [tex] \frac {1} {8 \pi} \int_S \epsilon_{abcd} \nabla^c K^d [/tex]

    where [tex] \epsilon [/tex] is the levi-civita tensor normalized so as to be a volume element, and Kd is a Killing vector of the space-time.

    K is normalized so that Ka Ka approaches -1 at infinity. (That's where the requirement for an asymptotically flat space-time comes from).

    Note that the above intergal is integrating a two-form (a fully anti-symmetric tensor) over a 2 sphere

    Because the metric is time-indepenent, there are no gravitational radiation terms to worry about.

    This result was derived in Wald, "General Relativity", pg 289 (it's not particularly obvious why it's true from my post, but I can't realistically reproduce an entire textbook chapter in a post).

    It's instructive to actually work this out for the Schwarzschild metric.

    K^d turns out to be a unit vector in the time direction

    [tex] \nabla^c K^d [/tex] turns out to have two components of value +/-M/r^2 in the 'rt' and 'tr' slots. It's anti-symmetric

    After multiplication by the volume element, one is left with two components of value [tex] 2 \sqrt{sin(\theta)^2} [/tex] in the [tex] '\theta \phi' and '\phi \theta' [/tex] slots.

    Integrating this two-form over a two-sphere gives the end result is that the mass M of a Schwarzschild black hole of mass M is M
    :smile:
     
  5. Aug 26, 2004 #4

    Garth

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    A little further explanation is required to make sense of this sentence, which otherwise appears a tautology.
    I think we have to explain:
    i) It applies to all Schwarzschild solutions and not just that of a Black Hole.
    ii) that the first mass M is that derived by Kepler at great distance from the gravitating body. This is the value inserted in the gravitational potentials of the metric.
    iii) the final M is the total energy of matter and gravitation of the system.
     
  6. Aug 26, 2004 #5

    pervect

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    Yes, the formula is valid for solutions other than the Scwharzschild metric - it is valid for any metric that is asymptotically flat, with a vacuum at infinity, and that has a time-like killing vector. The existence of a time-like killing vector is equivalent to the original requirement that the metric coefficients be independent of time.

    Working out the expression mainly confirms that the formula, which I've posted straight from the textbook without the supporting development, gives the right answer for the special case of the Schwarzschild metric for the total mass of the system. (I've been using the terms mass and energy somewhat interchangably here, hopefully this hasn't confused anyone). I'm assuming that it is already known that the quantity M in the Schwarzschild metric is equivalent to the mass of the system, which is a black hole, without going into any more details.
     
  7. Aug 29, 2004 #6
    -And for a given metric with line element ds^2=gabdxadxb then would be a basis in wich this metric is diagonal and ds^2=ha(dxa)^2

    -If exist an Energy term then we can make H=E=ih/dt and have a Schroedinguer type equation for gravity HPhi(x)=EPhi(x)
     
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