| Topic: |
Science > Physics |
| User: |
"The Ghost In The Machine" |
| Date: |
11 Jun 2005 11:00:07 AM |
| Object: |
Photon Energy/Gravitational Field |
This is probably a dumb question but I just want to confirm this.
Assume one is standing on a nonrotating planet with (rest)
mass M, radius r, lightspeed c = 2.99792458 * 10^8 m/s,
universal gravitational constant G = 6.674215 * 10^-11
m^3/(kg s^2), and one is observing a distant light source.
(We neglect the mass of the light source for now.)
If the light source is radiating at a certain energy E0
and modeled as a classical particle of some mass m, then one
can work out that the energy gained by that light as it falls
onto the Earth's surface is
dE = M*m*G/r
(It turns out that this is expressed in a slightly different form at
http://hyperphysics.phy-astr.gsu.edu/hbase/gpot.html#gpt , so
I'm on good ground here.)
Now, if one naively replaces m by E/c^2, one gets
E = E0 + M * G * E0/(c^2 * r)
or if one prefers
f = f0 + M * G * f0/(c^2 * r)
since E = h * f (where h here is Planck's Constant = 6.626 * 10^-34 J s).
If one takes a very small h << r, then one can work out
f1 = f0 + M * G * f0/(c^2 * (r + h))
= f0 * (1 + M * G/(c^2 * (r + h)))
f2 = f0 + M * G * f0/(c^2 * r)
= f0 * (1 + M * G/(c^2 * r))
where h in *these* formulae is the height of the Harvard Tower, in
meters. (ASCII is such a pain...)
http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/gratim.html#c2
Note that the above webpage does the same replacement of m by E/c^2
so I'm probably on the right track.
We massage to express f2 in terms of f1:
f2 = f1 + M * G * f0/(c^2 * r) - M * G * f0/(c^2 * (r + h))
= f1 + f0 * M * G * h / (c^2 * r * (r + h))
= f1 + f1 * M * G * h / (c^2 * r * (r + h) * (1 + M * G/(c^2 * r)) )
= f1 + f1 * M * G * h / (c^2 * ((r*(r+h)) + (r+h)*M * G/(c^2) ) )
Since h << r we can replace r+h by r, yielding
f2 =~ f1 + f1 * M * G * h / (c^2 * ((r^2) + r*M * G/(c^2) ) )
=~ f1 + f1 * M * G * h / (c^2 * r * ((r) + M * G/(c^2) ) )
=~ f1 + f1 * M * G * h / (r * (c^2*r + M * G ) )
At this point we might want to use the more convenient g = M * G/r^2,
so:
f2 =~ f1 + f1 * g * h / ((c^2 + g ) )
=~ f1 + f1 * g * h / (c^2 + g)
=~ f1 + f1 * g * h / c^2
=~ f1 * (1 + g * h / c^2)
since g << c^2 also, at least here on Earth -- black holes are
another thing entirely.
My naive assumption is proven consistent (if I've done this correctly),
but is it correct? Did I miss anything hyper-obvious?
That's the question.
Related questions might be what happens when one factors in the
Earth's rotation, orbital perturbations, the Sun's influence,
the mass of the distant star, etc. These I more or less know
the answers to already. :-)
--
#191,
It's still legal to go .sigless.
.
|
|
| User: "Old Man" |
|
| Title: Re: Photon Energy/Gravitational Field |
11 Jun 2005 01:57:42 PM |
|
|
"The Ghost In The Machine" <ewill@sirius.athghost7038suus.net> wrote in
message news:0cern2-shh.ln1@sirius.athghost7038suus.net...
[ ....]
f2 =~ f1 * (1 + g * h / c^2)
since g << c^2
No: g * h << c^2
also, at least here on Earth -- black holes are another thing
entirely.
My naive assumption is proven consistent (if I've done this
correctly), but is it correct?
Stick with consistent. Ghost's assumption,
m_photon = E_photon / c^2
is sufficient to an incomplete set of observations. Ghost hasn't
shown it to be a necessary assumption. In general, the assumption
leads to erroneous predictions. The assumption,
p_photon = E_photon / c
is consistent with all observations.
[Old Man]
#191,
.
|
|
|
| User: "The Ghost In The Machine" |
|
| Title: Re: Photon Energy/Gravitational Field |
11 Jun 2005 05:00:04 PM |
|
|
In sci.physics, Old Man
<nomail@nomail.net>
wrote
on Sat, 11 Jun 2005 13:57:42 -0500
<P4Gdne1aHKawqTbfRVn-oQ@prairiewave.com>:
"The Ghost In The Machine" <ewill@sirius.athghost7038suus.net> wrote in
message news:0cern2-shh.ln1@sirius.athghost7038suus.net...
[ ....]
f2 =~ f1 * (1 + g * h / c^2)
since g << c^2
No: g * h << c^2
A reasonable point, and true mostly because h << r anyway.
also, at least here on Earth -- black holes are another thing
entirely.
My naive assumption is proven consistent (if I've done this
correctly), but is it correct?
Stick with consistent. Ghost's assumption,
m_photon = E_photon / c^2
is sufficient to an incomplete set of observations. Ghost hasn't
shown it to be a necessary assumption. In general, the assumption
leads to erroneous predictions. The assumption,
p_photon = E_photon / c
is consistent with all observations.
That it is, but I did say m_photon is naive anyway. I'd frankly
have to look up Einstein's field equations to get a less
naive assumption model. Hyperphysics doesn't seem
to have it:
http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html
and
http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/conrel.html
is a bit simplistic when it comes to relativity (it doesn't
factor in momentum for energy, for example).
http://scienceworld.wolfram.com/physics/EinsteinFieldEquations.html
describes the field equations, but in rather abstruse terms;
I was hoping for something a lot simpler. :-)
My main objective is to find the frequency ratio between
a photon emitted in deep space and that photon captured
by something on Earth in a known gravitational field.
Perhaps you have something along the lines of a webpage that
might have a more rigorous definition of this? Unfortunately,
I don't have a link handy to Hafele and Keating's original
thesis, though
http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/airtim.html#c4
does match the Tower formula.
[Old Man]
#191,
--
#191,
It's still legal to go .sigless.
.
|
|
|
| User: "Old Man" |
|
| Title: Re: Photon Energy/Gravitational Field |
12 Jun 2005 08:33:55 PM |
|
|
"The Ghost In The Machine" <ewill@sirius.athghost7038suus.net> wrote in
message news:lv1sn2-pam.ln1@sirius.athghost7038suus.net...
In sci.physics, Old Man
<nomail@nomail.net>
wrote
on Sat, 11 Jun 2005 13:57:42 -0500
<P4Gdne1aHKawqTbfRVn-oQ@prairiewave.com>:
"The Ghost In The Machine" <ewill@sirius.athghost7038suus.net> wrote in
message news:0cern2-shh.ln1@sirius.athghost7038suus.net...
[ ....]
f2 =~ f1 * (1 + g * h / c^2)
since g << c^2
No: g * h << c^2
A reasonable point, and true mostly because h << r anyway.
also, at least here on Earth -- black holes are another thing
entirely.
My naive assumption is proven consistent (if I've done this
correctly), but is it correct?
Stick with consistent. Ghost's assumption,
m_photon = E_photon / c^2
is sufficient to an incomplete set of observations. Ghost hasn't
shown it to be a necessary assumption. In general, the assumption
leads to erroneous predictions. The assumption,
p_photon = E_photon / c
is consistent with all observations.
That it is, but I did say m_photon is naive anyway. I'd frankly
have to look up Einstein's field equations to get a less
naive assumption model. Hyperphysics doesn't seem
to have it:
http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html
and
http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/conrel.html
is a bit simplistic when it comes to relativity (it doesn't
factor in momentum for energy, for example).
http://scienceworld.wolfram.com/physics/EinsteinFieldEquations.html
describes the field equations, but in rather abstruse terms;
I was hoping for something a lot simpler. :-)
My main objective is to find the frequency ratio between
a photon emitted in deep space and that photon captured
by something on Earth in a known gravitational field.
Perhaps you have something along the lines of a webpage that
might have a more rigorous definition of this? Unfortunately,
I don't have a link handy to Hafele and Keating's original
thesis, though
http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/airtim.html#c4
does match the Tower formula.
Hartle does it neatly via time dilation and the Equivalence
Principle. It's easily done using the Schwarzschild space-
time metric.
[Old Man]
[Old Man]
#191,
--
#191,
.
|
|
|
|
| User: "Zigoteau" |
|
| Title: Re: Photon Energy/Gravitational Field |
12 Jun 2005 03:41:05 PM |
|
|
Hi, Ghost,
I can't claim great superiority over your level, but I don't think you
should do it that way. What you've done is to mix Newtonian and GR
concepts. If you want to know what General Relativity predicts, you've
got to do things the GR way, which means in the first case finding the
variation of the metric tensor over the region of space you're
interested in.
I go along with Old Man's statement that you should ignore photon mass
m and stick with p and E. Even better, divide by Planck's constant to
get the temporal frequency f and spatial frequency k of the photon.
Then I'd use the correct metric for a spherically-symmetrical planet,
with the unit of distance the light-second for simplicity. In the
weak-field case appropriate for an earth-sized planet, you have
g_ij =
[1-2*V 0 0 0 ]
[0 -1-2*V 0 0 ]
[0 0 -1-2*V 0 ]
[0 0 0 -1-2*V ]
where V is the potential of Newtonian gravity, i.e in more usual units
G*M*c^-2*r^-1 .
Notice that here the coordinate 'speed of light' is not constant.
That's OK. GR is based on the idea of using absolutely arbitrary
coordinate systems, so that the time-like coordinate x_0 does not have
to be the time measured by an observer at the surface of the planet,
and g_11/g_00 does not have to equal c^2. The form of the metric I have
written down is the standard one, and has the desirable property that a
geodesic shifted wrt x_0 is still a geodesic. The proper times and
distances which would be measured by an observer are worked out later,
using the metric tensor at the vantage point of that observer. Now
set up your light beam with frequency f shining down from infinity to
the observer at the planet surface. Each cycle of the beam follows a
geodesic. Because geodesics can be shifted in x_0, the observer sees
each successive cycle at intervals delta x_0 = 1/f . However, because
g_00 is not unity, the proper time between cycles is not 1/f but
sqrt(1-2*V) ~ 1-V, i.e. the radiation has been blue-shifted.
You can see that energy is very closely linked to time. Since in
relativity time is not absolute, but can vary from observer to
observer, conservation of energy becomes a can of worms.
Cheers,
Zigoteau.
.
|
|
|
| User: "Traveler" |
|
| Title: Re: Photon Energy/Gravitational Field |
12 Jun 2005 05:16:03 PM |
|
|
In article <1118608865.379219.158090@f14g2000cwb.googlegroups.com>,
"Zigoteau" <zigoteau@yahoo.com> wrote:
You can see that energy is very closely linked to time. Since in
relativity time is not absolute, but can vary from observer to
observer, conservation of energy becomes a can of worms.
Well, time cannot change, by definition. So this interpretation by
relativists is both wrong and hopelessly misguided.
I would say rather that energy is closely related to the speed of
processes, like that of clocks. We can then obtain different temporal
intervals. So clocks slow down, not because time dilates (a very
harmful misnomer), but because of energy conservation principles.
It's about time physicists stop thinking in terms of spacetime.
Spacetime is an obsolete concept. It hides the true nature of things.
There is neither time nor space. There is only position and change of
position (motion). Time and space are abstract concepts derived from
motion.
Louis Savain
The Silver Bullet: Why Software Is Bad and What We Can Do to Fix it
http://users.adelphia.net/~lilavois/Cosas/Reliability.htm
.
|
|
|
| User: "Zigoteau" |
|
| Title: Re: Photon Energy/Gravitational Field |
13 Jun 2005 05:01:50 AM |
|
|
Hi, Louis,
You can see that energy is very closely linked to time. Since in
relativity time is not absolute, but can vary from observer to
observer, conservation of energy becomes a can of worms.
Well, time cannot change, by definition.
Well, I define everything you say to be wrong.
So this interpretation by
relativists is both wrong and hopelessly misguided.
I would say rather that energy is closely related to the speed of
processes, like that of clocks. We can then obtain different temporal
intervals. So clocks slow down, not because time dilates (a very
harmful misnomer), but because of energy conservation principles.
It's about time physicists stop thinking in terms of spacetime.
Spacetime is an obsolete concept.
Would that be since August 6, 1945?
It hides the true nature of things.
There is neither time nor space.
Well that's a good thing then, because it means that you are not a
waste of anything, and replying to you is not wasting anything.
There is only position and change of
position (motion). Time and space are abstract concepts derived from
motion.
Cheers,
Zigoteau.
.
|
|
|
|
|
| User: "The Ghost In The Machine" |
|
| Title: Re: Photon Energy/Gravitational Field |
12 Jun 2005 05:00:07 PM |
|
|
In sci.physics.relativity, Zigoteau
<zigoteau@yahoo.com>
wrote
on 12 Jun 2005 13:41:05 -0700
<1118608865.379219.158090@f14g2000cwb.googlegroups.com>:
Hi, Ghost,
I can't claim great superiority over your level, but I don't think you
should do it that way. What you've done is to mix Newtonian and GR
concepts. If you want to know what General Relativity predicts, you've
got to do things the GR way, which means in the first case finding the
variation of the metric tensor over the region of space you're
interested in.
I go along with Old Man's statement that you should ignore photon mass
m and stick with p and E. Even better, divide by Planck's constant to
get the temporal frequency f and spatial frequency k of the photon.
Then I'd use the correct metric for a spherically-symmetrical planet,
with the unit of distance the light-second for simplicity. In the
weak-field case appropriate for an earth-sized planet, you have
g_ij =
[1-2*V 0 0 0 ]
[0 -1-2*V 0 0 ]
[0 0 -1-2*V 0 ]
[0 0 0 -1-2*V ]
where V is the potential of Newtonian gravity, i.e in more usual units
G*M*c^-2*r^-1 .
Notice that here the coordinate 'speed of light' is not constant.
That's OK. GR is based on the idea of using absolutely arbitrary
coordinate systems, so that the time-like coordinate x_0 does not have
to be the time measured by an observer at the surface of the planet,
and g_11/g_00 does not have to equal c^2. The form of the metric I have
written down is the standard one, and has the desirable property that a
geodesic shifted wrt x_0 is still a geodesic. The proper times and
distances which would be measured by an observer are worked out later,
using the metric tensor at the vantage point of that observer. Now
set up your light beam with frequency f shining down from infinity to
the observer at the planet surface. Each cycle of the beam follows a
geodesic. Because geodesics can be shifted in x_0, the observer sees
each successive cycle at intervals delta x_0 = 1/f . However, because
g_00 is not unity, the proper time between cycles is not 1/f but
sqrt(1-2*V) ~ 1-V, i.e. the radiation has been blue-shifted.
You can see that energy is very closely linked to time. Since in
relativity time is not absolute, but can vary from observer to
observer, conservation of energy becomes a can of worms.
Cheers,
Zigoteau.
Ah, it figures. A very interesting starting point regarding tensors.
Is there a website somewhere that goes into some more details on this?
--
#191,
It's still legal to go .sigless.
.
|
|
|
| User: "Zigoteau" |
|
| Title: Re: Photon Energy/Gravitational Field |
13 Jun 2005 04:01:04 AM |
|
|
Hi, Ghost,
Ah, it figures. A very interesting starting point regarding tensors.
Is there a website somewhere that goes into some more details on this?
What exactly do you want? Do you know anything at all about tensors
and/or general relativity? There's always the Wolfram website:
http://mathworld.wolfram.com/Tensor.html
Wikipedia has a long article with some math:
http://www.algebra.com/algebra/about/history/General-relativity.wikipedia
Visiting a library might be a better approach.
Cheers,
Zigoteau.
.
|
|
|
| User: "Zigoteau" |
|
| Title: Re: Photon Energy/Gravitational Field |
13 Jun 2005 10:14:11 AM |
|
|
Hi, Ghost,
The Schwarzschild solution of the field equations of GR is derived in
polar coordinates in:
http://scholar.uwinnipeg.ca/courses/38/4500.6-001/Cosmology/Schwarzschild_Metric2.htm
Cheers,
Zigoteau.
.
|
|
|
|
|
|
|
| User: "Nick" |
|
| Title: Re: Photon Energy/Gravitational Field |
12 Jun 2005 02:05:51 AM |
|
|
In GR energy is conserved.
.
|
|
|
| User: "T Wake" |
|
| Title: Re: Photon Energy/Gravitational Field |
12 Jun 2005 05:07:31 AM |
|
|
"Nick" <macromitch@yahoo.com> wrote in message
news:1118556599.462638.20640@z14g2000cwz.googlegroups.com...
In GR energy is conserved.
I like the way your posts "seem" like deep, philosophical, thoughts - when
in reality they are pretty vacuous. Its things like this which make people
think you are a chat bot.
What happened to your original thoughts and your ability to reason things
into context?
.
|
|
|
|
|
| User: "Y.Porat" |
|
| Title: Re: Photon Energy/Gravitational Field |
12 Jun 2005 01:16:09 AM |
|
|
if you say among the others that energy has mass
i support you
i also used it in my smallest photon mass definition
all th ebest
Y.Porat
------------------------------------
.
|
|
|
|
|
|
|
Related Articles |
|
|
