Re: Mass of particles in GR field

On Feb 6, 7:23 pm, "Rich L." <ralivings...@xxxxxxxxxxxxx> wrote:
In QM energy is related to frequency by h_bar. In Relativity mass is
related to energy. A particle at rest can be represented by a wave
function exp(-i*w_0*t) where w_0 is the frequency corresponding to the
rest mass of the particle. GR predicts a gravitational red shift. I
assume that this red shift applies to this rest mass frequency as
well. Therefore, if I'm thinking about this correctly, the rest mass
of a particle deep in a gravitational well (wrt the observer) is
reduced. Is this correct, or am I missing a key concept?

Even in flat space, the wave function exp(-i*w_0*t) only applies to
the particle at rest, and for a general inertial frame one has instead
exp(i k.x - i w_0 t),
Thus w_0 corresponds to the energy, E, of the particle, rather than
the rest mass (and k corresponds to the momentum, p). The rest mass
is given by m^2 = E^2 - p^2 in all inertial frames (units c=1), and so
if the particle is not at rest then both E and |p| must increase.

To bring in gravity and GR, the generalisation of rest mass for a
freely falling particle is
m^2 = g_uv p^u p^v ,
where g_uv is the metric tensor (signature +---), and p^u is the 4-
momentum. Here, if the spacetime coordinate of the particle is x^u,
and the time measured by a clock falling with the particle (i.e., in
the rest frame of the particle) is T, then the 4-velocity is given by
dx^u / dT = p^u / m = p^u / sqrt{ g_uv p^u p^v }.
The geodesic equation implies that m is a constant of the motion.
Thus, the rest mass doesn't change at all while falling to the
horizon.

If this is correct, then by the time such a particle reaches the event
horizon it's apparent mass (as seen by the observer higher in the
well) will be reduced to zero (since g_00 goes to zero at that
point). If the particle was allowed to fall freely to this point, I
assume all the rest mass energy has been converted to kinetic energy.
Of course in the frame comoving with the particle it would always have
the original rest mass.

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).


.

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