Re: Mass of particles in GR field
- From: a student <of_1001_nights@xxxxxxxxxxx>
- Date: Sat, 10 Feb 2007 15:42:49 +0000 (UTC)
On Feb 10, 6:05 am, harry <harald.vanlintelButNotT...@xxxxxxx> wrote:
I am also interested to hear the correct answer; but the above cannot be it,
as you say to use "the time measured by a clock falling with the particle" -
thus for a local observer. Of course, proper mass does not change, by
definition.
The question was "wrt the observer" who was suggested to be not taking part
with that motion ("by definition he can't be with the particle").
No, all observers will see the same rest mass, since it is an
invariant as noted above. However, any observer at a fixed location
above the horizon will never actually see the particle reach the
event horizon, or fall through it (since nothing can escape from the
horizon or from within it).
Also that doesn't really answer the question if according to that observer
the particle's rest mass tends to zero.
The 'rest mass' and the `proper mass' are the same quantity, m - and
the same for *all* observers in GR. This is because
m = sqrt{ g_uv p^u p^v }
is a scalar on the worldline of the particle (where p^u is the 4-
momentum), irrespective of who calculates it in whatever coordinate
system. It further follows from the geodesic equation that m is a
*constant* along the worldline. Similarly, if the worldline x^u is
parameterised by some parameter p, then the quantity T defined via
dT := sqrt{ g_uv (dx^u / dp) (dx^v / dp) } dp
is a scalar, with the same value for any observer (for an agreed
'origin' T=0), not just a co-moving observer.
Perhaps the original post does not mean 'rest mass', but instead the
'contravariant energy'
E = p^0,
i.e., the zero-component of the 4-momentum ? Or perhaps what is
meant is the 'covariant energy'
E' = p_0 = g_0v p^v
(the latter is actually the more logical definition of 'energy' in
this case). Each of these can change in different frames, of course,
as they are not scalars.
One can calculate these for a radially infalling particle in the
Schwarzschild metric, using
p^u = m dx^u / dT
and the known geodesic solution (eg, Weinberg, "Gravitation and
Cosmology, sec. 8.4) in the usual coordinates:
dx^0 / dT = dt / dT = 1/(k B),
dx^r / dT = dr /dT = - (1/k) sqrt{ 1 - k^2 B },
where k is a positive constant of the motion and B denotes g_00 . If
the particle starts at rest at radius r*, then
k =1 / sqrt{ B*} = 1 / sqrt{ 1 - 2GM/r* } .
For a photon, k=0.
In particular, it follows, noting g_0r = 0, that
E = m / (kB) ,
and
E' = m/k .
Thus the covariant energy is a constant of the motion, while the
contravariant energy diverges as the particle approaches the
Schwarzschild radius (since B -> 0).
Finally, if the original post was referring to E or E' for a photon
emitted radially outward by the particle as it fell, then this may
also be calculated from the above (with k -> 0, m/k -> 1), giving
E = 1/B, E' = 1 .
Hence the contravariant energy E decreases as the photon moves
outwards, corresponding to a redshift.
.
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