Re: Energy extracted from a mass falling into a black hole

On Oct 26, 11:40 pm, Jim Black <fmla...@xxxxxxxxxxxxxxxx> wrote:
On Sun, 26 Oct 2008 09:20:36 -0700 (PDT), Edward Green wrote:
On Oct 25, 8:45 am, Jim Black <fmla...@xxxxxxxxxxxxxxxx> wrote:
On Fri, 24 Oct 2008 17:40:23 -0700 (PDT), Edward Green wrote:
On Oct 24, 2:28 am, Jim Black <trams...@xxxxxxxxx> wrote:
Eric Gisse wrote:
On Oct 23, 1:23 pm, moro...@xxxxxxxxxxxxxxxxxxxxxxx (Michael Moroney)
wrote:
Out of curiosity:

If you were able to lower a 1 kg mass from infinity into a black hole,
how much energy could you theoretically extract before it reaches the
event horizon? (assume 1 solar mass black hole) I know how to calculate
the Schwartzchild radius for a black hole, and how to use Newtonian
mechanics to calculate the energy difference between that and infinity,
but I figure GR and things like gravitational redshift (if energy was
released from the mass as photons) would make such calculations
meaningless.

The event horizon makes evaluating such quantities hard, as things
like energy cease to have practical meaning there. You have to
understand that energy is not a well defined quantity in general
relativity.

This is true in general, but a static black hole is invariant under a
translation in time, so Noether's theorem can be applied to get the
conserved energy quantity he wants.  For a test mass m near a
Schwarzschild black hole

ds^2 = (1 - 2GM/rc^2) dt^2 - (1/c^2) [dr^2 / (1 - 2GM/rc^2) + r^2
d(theta)^2 + r^2 sin^2(theta) d(phi)^2],

we have the action S = int -mc^2 ds.  We avoid giving time a special
role by parametrizing the trajectory with the arbitrary parameter u
instead of t:  S = int -mc^2 ds/du du.  Differentiating with respect
to dt/du and t, we get the Euler-Lagrange equation

(d/du) [ -(1 - 2GM/rc^2) mc^2 dt/ds ] = 0.

This indicates the energy is

(1 - 2GM/rc^2) mc^2 dt/ds,

which evaluates to

sqrt(1 - 2GM/rc^2) mc^2

for an object at rest.  This puts the answer to his question -- for a
nonrotating, electrically neutral black hole -- at mc^2, as one might
expect.

That's interesting... and if all the rest energy radiates away, what
is left to fall into the black hole?

I'd expect that if all this energy was extracted, the black hole wouldn't
gain any mass from the infalling object.  But this won't happen -- to get
the object's energy arbitrarily close to zero, whatever is slowing the
object's fall would have to bring it to a stop arbitrarily close to the
event horizon, which would require an arbitrarily large force.  Otherwise
the object will carry some kinetic energy in.

Nonetheless we could cast our "arbitraries" the other way around: by
using a large enough, but finite, force, one can extract an
arbitrarily large fraction of the potential energy before letting go
of the weight.

Say our aribtrarily large fraction is .9 . Does this mean the black
hole only gains 1/10 of the mass/energy?

I would guess so.

Maybe the factor of mc^2 is
essentially a coincidnece, and, if released from infinity and allowed
to fall, a test mass carries in  twice mc^2, one factor being its rest
mass, and the second being its KE.

Consider a spherically symmetric shell of matter with mass m falling into a
black hole with mass M from a great distance with no energy radiated away..
Both the metric inside and outside the shell are vacuum solutions to the
Einstein equation; by their spherical symmetry and Birkhoff's theorem, they
must be Schwarzschild solutions.  If the shell of matter is sufficiently
far from the black hole, Newtonian gravity applies there, and by Gauss's
law, the exterior solution must have mass M+m.

We can extend this argument to the case where some energy is radiated away
by enclosing everything in a radiation-absorbing shell with initial mass
m3.  As above, the solution outside the second shell must be a
Schwarzschild solution with mass parameter M+m+m3.  We wait until the
radiation from the infalling matter is absorbed.  If it has energy E, the
mass of the absorbing shell is now m3+E/c^2.  Applying Gauss's law again,
the final solution inside the absorbing shell must have mass M+m-E/c^2.

Well, how about a purely Newtonian thought experiment then: a
spherically symmetric shell of matter collapses symmetrically,
converting gravitational potential energy into mass-energy. It seems
to me we either conclude the gravity increases externally or that
gravitational potential energy gravitates, or else kinetic energy does
not gravitate. All possibilities sound wrong!

I notice in a purely Newtonian world we have an infinity of potential
energy relative to collapse to a mass point, so evidently potential
energy does not gravitate. This infinity seems to be "dressed" in a
black hole.


We can extend to non-spherically-symmetric cases if we can show that the
flux of the gravitational field vector through a closed surface at a
sufficiently large distance is conserved.  Here, we'd need the distance
large enough that the surface encloses everything, and also large enough
that Newtonian gravity works and we can even sensibly talk about a
gravitational field.  Locality suggests this will be true, but I'm not
quite convinced yet, because locality doesn't hold in Newtonian mechanics..
So this may require a GR-compatible definition of the flux, and a GR-based
proof it is conserved.

.

Relevant Pages

  • Re: What is gravity, I have no clue
    ... so all things warp, and warping is what gravity is, warping around ... mass in the fabric ... energy, and the formula does not work. ... Why is it that spiral galaxies have a lot of dark matter and elliptic ...
    (sci.astro)
  • Re: "Einsteins Unfortunate Legacy #2"
    ... revealed to be a force resulting from the release of mass energy due to ... However, as to the assertion re gravity quoted above, I would like to ... work titled "The Quantum as a Physical Entity". ... the energy of expansion and contraction ...
    (sci.physics.relativity)
  • Re: Nobel Prize for David Thomson?!
    ... force will invariably stress the aether to ... >> having zero mass, no electric charge, and an indefinitely long lifetime. ... So indeed, a photon IS energy. ...
    (sci.physics.particle)
  • Re: Nobel Prize for David Thomson?!
    ... force will invariably stress the aether to ... >> having zero mass, no electric charge, and an indefinitely long lifetime. ... So indeed, a photon IS energy. ...
    (sci.physics)
  • Re: Nobel Prize for David Thomson?!
    ... force will invariably stress the aether to ... >> having zero mass, no electric charge, and an indefinitely long lifetime. ... So indeed, a photon IS energy. ...
    (sci.physics.relativity)