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?

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.

I'm also not quite sure what it means to the observer that the rest
mass of the particle has been reduced. How would he measure this,
since by definition he can't be with the particle. Would this be
observable, or is this "mass reduction" meaningless in terms of
observations?

On Feb 6, 7:23 pm, "Rich L." 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).

On Feb 6, 3:23 am, "Rich L." 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?

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.

I'm also not quite sure what it means to the observer that the rest
mass of the particle has been reduced. How would he measure this,
since by definition he can't be with the particle. Would this be
observable, or is this "mass reduction" meaningless in terms of
observations?

This is observed in the Pound-Snider experiment. There is
more inertial coupling to the subatomic constituants of an
iron-salt near the earth's surface than a similar structure
at the top of a tower. The proximity to the earth's surface
lowers the nuclear resonant frequency enough that atomic
absorption of light emitted from the tower-top is inhibited.

http://en.wikipedia.org/wiki/Mossbauer_spectroscopy
http://link.aps.org/abstract/PRL/v13/p539

--L.B.Okun
http://arxiv.org/abs/physics/9907017

Sue...

"Sue..." wrote:
On Feb 6, 3:23 am, "Rich L." 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?

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.

I'm also not quite sure what it means to the observer that the rest
mass of the particle has been reduced. How would he measure this,
since by definition he can't be with the particle. Would this be
observable, or is this "mass reduction" meaningless in terms of
observations?

This is observed in the Pound-Snider experiment. There is
more inertial coupling to the subatomic constituants of an
iron-salt near the earth's surface than a similar structure
at the top of a tower. The proximity to the earth's surface

?? Why should 'proximity to the earth's surface' be relevant? 'Deeper
in the earth's gravitational field' I would understand.

lowers the nuclear resonant frequency enough that atomic
absorption of light emitted from the tower-top is inhibited.

http://en.wikipedia.org/wiki/Mossbauer_spectroscopy
http://link.aps.org/abstract/PRL/v13/p539

--L.B.Okun
http://arxiv.org/abs/physics/9907017

- Dushan Mitrovich

This is one of those cases where everything depends on conventions,
frames of reference and coordinate systems, but most answers are right
from someone's point of view.

In a local inertial frame of reference, mass and energy are related
via E = m c^2. However, a coordinate system from which gravity looks
like a force is not inertial, and in such a coordinate system c
depends slightly on the potential. This makes it a bit tricky to talk
about what happens to energy and mass as seen from a distance.

In the simple case of a static central mass described in isotropic
coordinates, the potential can be approximated as a single scalar
factor, equal to (1-GM/rc^2) in a weak semi-Newtonian approximation.
This scale factor (let's call it Phi) determines the rate of a
standard clock and the size of a standard ruler at a distance r from
the central mass, so c varies as Phi^2.

When a small object falls towards the central mass, as in the
Newtonian model, its total energy as measured in the isotropic
coordinates remains constant, but potential energy is turned into
kinetic energy. The potential energy varies as Phi, and is equal to m
c^2 where m is the mass of the test object (which remains fixed in its
own frame of reference). Since c varies as Phi^2, this means that
relative to the isotropic coordinate system, the rest mass varies as
Phi^-3, which means it actually INCREASES with decreasing potential,
although the corresponding rest energy decreases.

For most purposes, quantities measured in energy units are more useful
than those measured in mass units; energy is conserved, and is closely
related to frequency which can be compared at a distance.

As an object falls towards the event horizon, then its calculated rest
energy does decrease towards zero. However, the overall total energy
and momentum (taking into account the movement of the black hole
towards the object) are unaffected, so this has no physical effect
that could be observed from a distance.

"a student" wrote in message
oups.com...
On Feb 6, 7:23 pm, "Rich L." 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.

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").

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).

Also that doesn't really answer the question if according to that observer
the particle's rest mass tends to zero.

Regards,
Harald

On Feb 10, 6:05 am, harry 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.

On Feb 9, 2:05 pm, (Dushan Mitrovich) wrote:
"Sue..." wrote:
On Feb 6, 3:23 am, "Rich L." 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?

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.

I'm also not quite sure what it means to the observer that the rest
mass of the particle has been reduced. How would he measure this,
since by definition he can't be with the particle. Would this be
observable, or is this "mass reduction" meaningless in terms of
observations?

This is observed in the Pound-Snider experiment. There is
more inertial coupling to the subatomic constituants of an
iron-salt near the earth's surface than a similar structure
at the top of a tower. The proximity to the earth's surface

?? Why should 'proximity to the earth's surface' be relevant? 'Deeper
in the earth's gravitational field' I would understand.


Gravity is maxmum at the surface and that is how the experiment
was performed. That is a good question tho. I am not aware
of anyone repeating the experiment down a well or mine shaft.

It is well known that pendulum clocks speed up whether moved
above or below the surface but that isn't quite a fair comparison.

GR (or Schwartzchild geometry) gets pretty vague without
a free-space path, making singularities event horizions and
other spooky stuff. I would think a mine shaft expeiment
might clear some of that up.

The concept is: "gravity there, makes the inerta here", and
you test the concept by moving an inertial oscillator closer to the
a gravitational source. Sure enough, GPS clocks are slower
down here.

A bullet follows a curved trajectory on the surface of the earth
and a straight trajectory at the earths center or in deep space
so my guess is that an atomic clock would speed up with with
either altitude or depth.

Whether the clock's oscillator exploits a ridgid frame or
an eliptical path to conserve momentum would also seem
an important factor.

I still would like to see a mine-shaft experiment
if you know of one.

Sue...

lowers the nuclear resonant frequency enough that atomic
absorption of light emitted from the tower-top is inhibited.

http://en.wikipedia.org/wiki/Mossbauer_spectroscopy
http://link.aps.org/abstract/PRL/v13/p539

--L.B.Okun
http://arxiv.org/abs/physics/9907017

- Dushan Mitrovich-

On Feb 9, 2:05 pm, (Dushan Mitrovich) wrote:
"Sue..." wrote:
Correction

It is well known that pendulum clocks *slow down* whether moved
above or below the surface but that isn't quite a fair comparison.

Sue...

This answer makes a lot of sense to me. I do want to challeng one
point however for clarification:

On Feb 9, 1:05 pm, Jonathan Scott wrote:
...
When a small object falls towards the central mass, as in the
Newtonian model, its total energy as measured in the isotropic
coordinates remains constant, but potential energy is turned into
kinetic energy. The potential energy varies as Phi, and is equal to m
c^2 where m is the mass of the test object (which remains fixed in its
own frame of reference). Since c varies as Phi^2, this means that
relative to the isotropic coordinate system, the rest mass varies as
Phi^-3, which means it actually INCREASES with decreasing potential,
although the corresponding rest energy decreases.

Here you suggest that in calculating m_0*c^2 you should use the
apparent speed of light at the mass as observed by the observer. Why
should you use this speed instead of the observers own local speed of
light? My (unthinking) inclination was that the observer should use
his own value of c, which of course is the universal constant. It had
not occured to me to use the distant mass' apparent speed of light.
What is the rational for one over the other?

An observer local to the test mass (at rest with it) would calculate a
"rest mass frequency" of m_0*c^2/h_bar. If this frequency is real and
observable, the distant observer would see it red or blue shifted by
phi (in Jonathans terminology). He would thus calculate a mass of
m_0*phi*c^2/h_bar. The observer would interpret this frequency to
correspond to a rest mass of m_0*phi. By the argument I gave in my
last post this would give a constant apparent frequency as the
particle falls (that is, the mass "frequency" plus the kinetic energy
frequency). Can you point out the flaw in this argument?

For most purposes, quantities measured in energy units are more useful
than those measured in mass units; energy is conserved, and is closely
related to frequency which can be compared at a distance.

I like this point of view. It is actually the one I'm trying to
develop (or understand why it doesn't work). I'm trying to better
understand the relationship between time and energy. I'm inclined to
consider the frequency to BE the energy, but I'm not sure that is a
useful or even an idea consistent with observed physics. I think it
is curious that there is time dilation in a gravitational field, but
not in an electric field. Also QM seems to imply that a particle in a
potential well has constant frequency as it orbits in and out,
irrespective of the nature of the force creating that potential.
Ultimately I'd like to understand how these three facts are compatible
and consistent, but I'm working on it one small bit at a time.

As an object falls towards the event horizon, then its calculated rest
energy does decrease towards zero. However, the overall total energy
and momentum (taking into account the movement of the black hole
towards the object) are unaffected, so this has no physical effect
that could be observed from a distance.
This is as it seems to me. Does this mean that since the mass goes to
zero at the event horizon that the particle (formally with mass) can
now go AT the speed of light? I'm not sure what significance this
has, if any. For one thing, based on your argument which makes sense
to me, the apparent speed of light at the event horizon goes to zero
there.

Thanks for the comments!