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Is black hole entropy invariant?

  1. Apr 20, 2008 #1
    Hawking gave the black hole entopy equation:

    (1) [tex]S_{BH} = {k c^3 \over 4G \hbar} A[/tex]

    where A is the surface area of the black hole event horizon. All the other factors on the right hand side of the equation are constants.

    From the point of view of an observer moving relative to the black hole the spherical event horizon appears to be an oblate spheriod with its shotest radius parallel to the relative motion. The surface area of an oblate spheriod varies in a non linear way with respect to the contraction of one of the radii and it is certainly not invariant under Lorentz tranformation. This contradicts almost text on relativistic thermodynamics that almost universally accept that entropy is a Lorentz invariant.

    Perhaps the situation can be rectified by substituting the the equation for the volume (V)of a sphere into the equation?

    The equation would then be:

    (2) [tex]S_{BH} = {k c^3 \over 4G \hbar} \times {3 \over R} \times V = {k c^3 \over 4G \hbar} \times {3 \over R} \times {4 \pi R_x R_y R_z \over 3}[/tex]

    By assuming that the relative motion is along the x axis and by assuming the undefined R is the radius parallel to the relative motion the equation becomes:

    (3) [tex]S_{BH} = {\pi k c^3 \over G \hbar} \times { R_y R_z } = {A_{tranverse} \over 2L_p^2}[/tex]

    where Ax is the transverse cross sectional area of the event volume and Lp is the Planck Area. This formulation is invariant under Lorentz transformation as all length measurements are transverse.

    Since the Radii along the y and z axes remain unaltered under transformation we can substitute the values for the Schwarzschild radii :

    (4) [tex]S_{BH} = {\pi k c^3 \over G \hbar} \times { 4 G^2 M^2 \over c^4 }[/tex]

    which simplifies to

    (5) [tex]S_{BH} = {4 G \pi k \over \hbar c} \times {M}^2 = 4 \pi k {M^2 \over M_p^2}[/tex]

    where M is the mass of the black hole and Mp is the Planck mass.

    Equation (5) appears not to be invariant although it can be noted that equation (1) can be expressed in similar way using Planck units as:

    (6) [tex]S_{BH} = {k \over 4 } {A \over L_p^2} [/tex]

    so maybe we can assume invariance is assured by the Planck constants transforming in the same way as area and mass respectively? Was the assumption that the "constants" in equation (1) really are constants, misplaced?
  2. jcsd
  3. Apr 23, 2008 #2
    This is really interesting kev. But here's the problem: although something like "contraction" should appear in Schwarzschild spacetime, it will be determined by the Schwarzschild metric, not by the Lorentz transformations.

    The Lorentz transformations are rotations of Minkowski (flat) spacetime, and map inertial observers to inertial observers. But observers around a Schwarzschild blackhole are not "inertial"; they are in freefall in a curved spacetime. So contraction won't be determind by the Lorentz transformations.

    However, Hawking radiation is just radiation, so it travels along timelike geodesics. What you might do is this:

    Pick the geodesic along which you'd like to make your observation (for example, it might be an observer in a stable circular orbit). Then consider a congruence of geodesics on escape trajectories away from the black hole, all of which intersect your observer -- these will represent your radiation. I suspect that in many situations, you have correctly predicted that the observer will *see* the congruence to have contracted as compared to what is *seen* from the perspective of coordinate time.

    However, this observer will *judge* the amount of Hawking radiation just like every other observer will judge it, using the formula you have given, since he understands the geometry of spacetime.

    Another thought: Schwarzschild spacetime is asymptotically flat. So as you move arbitrarily far away from the blackhole, the analysis you do above becomes more and more correct. Unfortunately, if you are really that far away, the change in entropy due to contraction effect would be too small to measure.

    I've always wondered why it is that black hole entropy can be *identified* with classical entropy. They of course follow very similar laws; however, they are defined completely differently. Do we really know that our intuitions about should carry over to intuitions about the other? -- Perhaps there is good reason to do this, I simply don't know.

  4. Apr 27, 2008 #3


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    [tex]c[/tex], [tex]G[/tex], and [tex]\hbar[/tex] are dimensionful constants which are assumed to have the same values within any inertial frame. [tex]S_{BH}[/tex] is a dimensionless ratio of two angular momentums which co-vary over any coordinate transformation, so it is invariant.

    These are dimensionful constants which are defined based on a system of units that is itself defined to be the same within every inertial frame, but the system of units in one inertial frame appears time-dilated and length-contracted from the perspective of another inertial frame.
    Last edited: Apr 27, 2008
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