Mass of particles in GR field

[Rich L.]

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?

[a student]

Post by Rich L.
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.

Post by Rich L.
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).

[harry]

Post by a student

Post by Rich L.
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").

Post by a student

Post by Rich L.
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

[a student]

Post by harry
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").

Post by a student
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.

[a student]

Post by a student

Post by harry
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").

Post by a student
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
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.

[Sue...]

Post by Rich L.
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...

[Dushan Mitrovich]

Post by Sue...

Post by Rich L.
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.

Post by 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

[Sue...]

Post by Dushan Mitrovich

Post by Sue...

Post by Rich L.
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...

Post by Dushan Mitrovich

Post by 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-

[Sue...]

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...

[Rich L.]

On Feb 10, 11:11 pm, "Sue..." <***@yahoo.com.au> wrote:
..

Post by Sue...
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.

Actually, pendulum clocks run more slowly above or below the surface
because their rate (the frequency of the pendulum swing) is
proportional to the acceleration of gravity.

Post by Sue...
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.

Actually the modern point of view is that inertia (which is really
conservation of momentum) is a consequece of the "symmetry"
represented
by conservation of momentum and the relativity principle. See, for
example,
the start of the article on Wikipedia:
http://en.wikipedia.org/wiki/Principle_of_relativity

Post by Sue...
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.

The speed of clocks depends on the location of the clock wrt the
observer.
A clock at the center of the earth would run normal speed if you were
standing
by the clock. If you were on the surface of the earth (or anywhere
except by
the clock, for that matter) the clock would appear to run slower
because it
would be lower in the "potential well". This is clearly predicted by
the
Schwarzschild metric for any object (earth or black hole). Likewise,
for
a clock on the surface of the earth, an observer at the center of the
earth
would see that clock running faster than if he were standing next to
it. At
the same time an observer at the top of a tall building looking down
at the
same clock would see that clock running slower. These effects have
been
measured experimentally to good precision.

[Sue...]

..> Gravity is maxmum at the surface and that is how the experiment

Post by Sue...
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.

Actually, pendulum clocks run more slowly above or below the surface
because their rate (the frequency of the pendulum swing) is
proportional to the acceleration of gravity.

This is corrected in separate post.

Post by Sue...
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.

Actually the modern point of view is that inertia (which is really
conservation of momentum) is a consequece of the "symmetry"
represented
by conservation of momentum and the relativity principle. See, for
example,

http://en.wikipedia.org/wiki/Principle_of_relativity

That is inconsistant with Einsteins 1920 paper and
1923 Nobel lecture.
(Note Wiki's references to 1905 paper.)

http://www.bartleby.com/173/
http://nobelprize.org/nobel_prizes/physics/laureates/1921/einstein-lecture.html

Post by Sue...
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.

[RichL lines rewrapped] :

<< The speed of clocks depends on
the location of the clock wrt
the observer.A clock at the
center of the earth would run
normal speed if you were standing
by the clock. If you were on the
surface of the earth (or anywhere
except by the clock, for that matter)
the clock would appear to run slower
because it would be lower in the
potential well". This is clearly
predicted by the Schwarzschild metric
for any object (earth or black hole).
Likewise, for a clock on the surface of
the earth, an observer at the center of
the earth would see that clock running
faster than if he were standing next to
it. At the same time an observer at
the top of a tall building looking down
at the same clock would see that clock
running slower. These effects have been
measured experimentally to good precision. >>

You don't spectify the nature of the clocks
but the difference in Pound-Rebka and
Pound-Snider attest to the *bad* precision.

"On the Interpretation of the Redshift in a Static
Gravitational Field"

http://arxiv.org/abs/physics/9907017

Sue...

[Tom Roberts]

Post by Rich L.

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

Actually, pendulum clocks run more slowly above or below the surface
because their rate (the frequency of the pendulum swing) is
proportional to the acceleration of gravity.

You both seem to think that a "pendulum clock" consists only of the
pendulum, its support, and the attached counter mechanism. This is
false. The _clock_ inherently includes the earth (common word usage
notwithstanding). So the above discussion does not make sense -- the
_clock_ cannot be "moved" without changing its timekeeping mechanism, so
you really have two DIFFERENT clocks, not one that was "moved".

Attempting to discuss the "gravitational time dilation" of a pendulum
clock does not make sense, because that effect applies only to "small"
clocks, and a pendulum clock cannot be "small" (here "small" means "does
not significantly affect the geometry").

Tom Roberts

[Sue...]

Post by Tom Roberts

Post by Rich L.

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

Actually, pendulum clocks run more slowly above or below the surface
because their rate (the frequency of the pendulum swing) is
proportional to the acceleration of gravity.

You both seem to think that a "pendulum clock" consists only of the
pendulum, its support, and the attached counter mechanism. This is
false. The _clock_ inherently includes the earth (common word usage
notwithstanding). So the above discussion does not make sense -- the
_clock_ cannot be "moved" without changing its timekeeping mechanism, so
you really have two DIFFERENT clocks, not one that was "moved".
Attempting to discuss the "gravitational time dilation" of a pendulum
clock does not make sense, because that effect applies only to "small"
clocks, and a pendulum clock cannot be "small" (here "small" means "does
not significantly affect the geometry").

The rate varies above and below the point of maximum
gravitational force so it certainly makes sense to discuss it.

A Pound-Snider type experiment down a mine shaft would
make even more sense but no one seems to know of one.
A torsion pendulm in submarine?
An atomic clock in a well ?

Sue...

Post by Tom Roberts
Tom Roberts

[harry]

Post by Sue...

Post by Tom Roberts

Post by Rich L.

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

Actually, pendulum clocks run more slowly above or below the surface
because their rate (the frequency of the pendulum swing) is
proportional to the acceleration of gravity.

You both seem to think that a "pendulum clock" consists only of the
pendulum, its support, and the attached counter mechanism. This is
false. The _clock_ inherently includes the earth (common word usage
notwithstanding). So the above discussion does not make sense -- the
_clock_ cannot be "moved" without changing its timekeeping mechanism, so
you really have two DIFFERENT clocks, not one that was "moved".
Attempting to discuss the "gravitational time dilation" of a pendulum
clock does not make sense, because that effect applies only to "small"
clocks, and a pendulum clock cannot be "small" (here "small" means "does
not significantly affect the geometry").

The rate varies above and below the point of maximum
gravitational force so it certainly makes sense to discuss it.
A Pound-Snider type experiment down a mine shaft would
make even more sense but no one seems to know of one.
A torsion pendulm in submarine?
An atomic clock in a well ?
Sue...

A torsion pendulum is almost certainly not accurate enough, but contrary to
a standard pendulum it doesn't include the earth in its operating mechanism,
thus in principle it would be a valid candidate. And an atomic clock would
be OK for sure.

Harald

[Sue...]

Post by harry

Post by Sue...

Post by Tom Roberts

Post by Rich L.

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

Actually, pendulum clocks run more slowly above or below the surface
because their rate (the frequency of the pendulum swing) is
proportional to the acceleration of gravity.

You both seem to think that a "pendulum clock" consists only of the
pendulum, its support, and the attached counter mechanism. This is
false. The _clock_ inherently includes the earth (common word usage
notwithstanding). So the above discussion does not make sense -- the
_clock_ cannot be "moved" without changing its timekeeping mechanism, so
you really have two DIFFERENT clocks, not one that was "moved".
Attempting to discuss the "gravitational time dilation" of a pendulum
clock does not make sense, because that effect applies only to "small"
clocks, and a pendulum clock cannot be "small" (here "small" means "does
not significantly affect the geometry").

The rate varies above and below the point of maximum
gravitational force so it certainly makes sense to discuss it.
A Pound-Snider type experiment down a mine shaft would
make even more sense but no one seems to know of one.
A torsion pendulm in submarine?
An atomic clock in a well ?
Sue...

A torsion pendulum is almost certainly not accurate enough, but contrary to
a standard pendulum it doesn't include the earth in its operating mechanism,
thus in principle it would be a valid candidate. And an atomic clock would
be OK for sure.

Might a spinning superconductor be more accurate
than any of those ?

http://www.esa.int/SPECIALS/GSP/SEM0L6OVGJE_0.html
http://einstein.stanford.edu/

Sue...

Post by harry
Harald-

[Rich L.]

Post by Tom Roberts

Post by Rich L.

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

Actually, pendulum clocks run more slowly above or below the surface
because their rate (the frequency of the pendulum swing) is
proportional to the acceleration of gravity.

You both seem to think that a "pendulum clock" consists only of the
pendulum, its support, and the attached counter mechanism. This is
false. The _clock_ inherently includes the earth (common word usage
notwithstanding). So the above discussion does not make sense -- the
_clock_ cannot be "moved" without changing its timekeeping mechanism, so
you really have two DIFFERENT clocks, not one that was "moved".
Attempting to discuss the "gravitational time dilation" of a pendulum
clock does not make sense, because that effect applies only to "small"
clocks, and a pendulum clock cannot be "small" (here "small" means "does
not significantly affect the geometry").
Tom Roberts

I agree with your criticisim of pendulum clocks, to a point. You
certainly do have to consider the whole system, which includes the
massive body it is near. I also don't like it as a model for a clock
because it is conceptually complex and it doesn't make it perfectly
clear how fundamental the gravitational time dilation really is.

I prefer to think about clocks made from meter sticks with mirrors at
the ends and timing the propagation of a light pulse between the
mirrors. Since the speed of light is a universal constant (a
postulate of Relativity), at least locally in the case of GR, I would
argue that this is a much better clock to be using for these
arguments.

On the other hand, and this is where I disagree with your post, in
principle, the pendulum clock should show the effect as well. If it
did not, one could distinguish between a gravitational accelleration
and an inertial acceleration, which would violate another postulate of
Relativity.

[Jonathan Scott]

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.

[Rich L.]

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 <***@vnet.ibm.com> wrote:
..

Post by Jonathan Scott
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?

Post by Jonathan Scott
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.

Post by Jonathan Scott
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!

[Jonathan Scott]

Post by Rich L.
...
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?
...

In general, when you make observations in a particular coordinate
system, you need to ensure that you consistently use the same coordinate
system. The standard value of c is effectively just a way of converting
between units. However, the actual value of c at a remote location also
needs to take into account the coordinate system. The speed at which
light appears to move at that point is effectively a physical
measurement. (Of course, if you use a coordinate system which isn't
even isotropic, such as Schwarzschild coordinates, then the speed of
light isn't even the same in all directions, and the mapping of simple
physical quantities between local and background coordinate systems
becomes much less intuitive, requiring tensors).

In this specific case, you will find for example that physical laws
relating to energy and angular momentum work correctly when
expressed consistently in either coordinate system, but become
meaningless when the wrong value of c is used.

Post by Rich L.
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?

In energy or frequency terms, this is correct (although the process
of "observing" a distant frequency is complicated in practice by the
need to interpret the effects of the Doppler shift relative to the
assumed coordinate system). The only error I can see is the
sentence where the frequency is interpreted as a rest mass without
taking into account the local value of c.

Post by Rich L.
... 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.

As you approach the event horizon, it is the limiting ratios of values
which are important, as in ordinary calculus. Locally, free fall
objects suffer from unlimited tidal forces, but otherwise physics is
fairly normal.

If you want to talk about what happens AT the event horizon, then I'm
sorry but this is outside my area of expertise, mostly because the maths
is so absurd that so far I'm not convinced that black holes even exist
(so I also believe that the GR field equations aren't quite right too).
One thing I remember is that in "Kruskal coordinates" it is possible to
describe the point of view of a falling observer even past the point of
crossing the event horizon! There are plenty of articles on the
internet about black holes.

Jonathan Scott

[Igor Khavkine]

Post by Jonathan Scott

Post by Rich L.
...
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?
...

In general, when you make observations in a particular coordinate
system, you need to ensure that you consistently use the same coordinate
system. The standard value of c is effectively just a way of converting
between units. However, the actual value of c at a remote location also
needs to take into account the coordinate system. The speed at which
light appears to move at that point is effectively a physical
measurement.

You have to be really careful when talking about "the actual value of
c
at a remote location". In my preferred convention, c=1 everywhere and
everywhen. You may have a different preference. All the more reason to
state precisely what you mean. A few months ago I wrote at length
about
what is meant by "speed of light" and how using that term in a curved
space-time can get you into trouble if you're not careful. That post
is: news:***@h48g2000cwc.googlegroups.com
http://groups.google.com/group/sci.physics.research/msg/5b598d3b6e4fbce7

Post by Jonathan Scott
(Of course, if you use a coordinate system which isn't
even isotropic, such as Schwarzschild coordinates, then the speed of
light isn't even the same in all directions, and the mapping of simple
physical quantities between local and background coordinate systems
becomes much less intuitive, requiring tensors).

Forget Schwarzschild. According to your seeming definition of "speed
of
light", it's not isotropic even if you use spherical spatial
coordinates in good old ordinary Minkwoski space-time. You're right,
that is confusing. All the more reason not to adopt this definition.
Setting c to 1, or some other convenient constant (which amounts to
redefining your units of measurement) is much simpler, whether you use
tensors or not. Incidentally, using tensors does not require
abandoning
your physical intuition. One's intuition simply needs to adapt through
practice in using tensors.

Post by Jonathan Scott
In this specific case, you will find for example that physical laws
relating to energy and angular momentum work correctly when
expressed consistently in either coordinate system, but become
meaningless when the wrong value of c is used.

Right. All the more reason not to tie the value of c to any coordinate
system.

Post by Jonathan Scott

Post by Rich L.
... 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.

As was explained in another post in this thread. The zero "mass" of a
massless particle is not the same as the observer dependent "mass" of
a
particle that can be said to go to zero at the horizon. Sorry, as long
as you have non-zero "mass" (in the geometrically invariant way),
you're constrained to travel at sub-light speeds, even at or past the
event horizon.

Post by Jonathan Scott
As you approach the event horizon, it is the limiting ratios of values
which are important, as in ordinary calculus. Locally, free fall
objects suffer from unlimited tidal forces, but otherwise physics is
fairly normal.

The event horizon is not a special place in space time. That is, you
cannot perform a local measurement to figure out whether you've passed
the event horizon or not. The event horizon is defined as the boundary
of the region from which no particle can escape to infinity. This
literally means that you can't know whether a particular space-time
point is inside the horizon unless you release a bunch of particles
from it and check whether some of them escape to infinity. However,
it's intuitively obvious that this is not a local measurement. This
fact becomes even more clear when the "observer at infinity" is
properly defined.

In particular, the tidal forces are quite finite at the event horizon.
They only blow up at the singularity. There is a good description of
what happens when you fall into a black hole in the Usenet Physics
FAQ,
including a short description of the event horizon:

http://math.ucr.edu/home/baez/physics/Relativity/BlackHoles/fall_in.html

Post by Jonathan Scott
If you want to talk about what happens AT the event horizon, then I'm
sorry but this is outside my area of expertise, mostly because the maths
is so absurd that so far I'm not convinced that black holes even exist
(so I also believe that the GR field equations aren't quite right too).
One thing I remember is that in "Kruskal coordinates" it is possible to
describe the point of view of a falling observer even past the point of
crossing the event horizon! There are plenty of articles on the
internet about black holes.

There is nothing absurd about the mathematics describing black holes,
just as there's nothing particularly unintuitive about tensors. It is
possible to describe physics on either side of the horizon in several
different ways (one of which is choosing a coordinate system singular
at it, such as the Kruskal one). With all the resources describing and
explaining how to work with general relativity (both in print and
online), there is no reason that a dedicated student, with time and
guidance, couldn't grasp at least some of these descriptions enough to
make quantitative statements about the physics involved.

As to the validity of GR, they are quite good explaining a variety of
physical phenomena (planetary, astrophysical, and cosmological). And
that's all one could really ask of a theory. Of course it doesn't
explain all phenomena; in that sense you could say that it's not quite
right. But then you could say the same about any theory, as long as
there are any yet unexplained phenomena out there. One should be
specific with criticisms.

Igor

[c***@physics.ucdavis.edu]

Jonathan Scott <***@vnet.ibm.com> wrote:

[...]

Post by Jonathan Scott
As you approach the event horizon, it is the limiting ratios of values
which are important, as in ordinary calculus. Locally, free fall
objects suffer from unlimited tidal forces, but otherwise physics is
fairly normal.

Tidal forces are not infinite for a freely falling observer near, or
even passing through, an event horizon. In fact, for a large black
hole they can be very small; a person falling through a million-
Solar-mass black hole horizon would feel stresses of less than one
atmosphere. It's only at the singularity that the tidal forces
diverge.

What *is* true is that the tidal forces diverge for an observer who
is not freely falling, but is attempting to remain at rest at the
horizon.

Steve Carlip