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Science Forum Index » Physics - Research Forum » Gravitational Redshift from Energy Conservation
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| Murat Ozer |
 Posted: Sat Oct 21, 2006 5:23 am |
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I would like to ask for your guidance on a problem that I have always
had difficulty with whenever I tried to understand it. It's the
argument for the gravitational redshift derived from energy
conservation presented in textbooks such as by MTW and Schutz.
Consider a particle of rest mass m at a height H above the origin 0.
Let us choose the reference level for potential energy as the level H,
where the total energy of the particle is mc^2. Let the particle fall
to the origin, where its total energy is still mc^2 because the
kinetic energy 1/2*mv^2=mgH cancells with the negative potential
energy relative to H. Now let the particle be annihilated at 0
converting its energy to a photon of the same total energy. Let the
photon travel upward in the gravitational field to H, at which point it
is converted to a particle of rest mass m. What is the total energy
of the created particle? Measuring its potential energy from the
same level H (according to which it is 0), it has to be mc^2.
Is not this right? So, according to this picture the photon cannot
loose energy as it climbs up in the gravitational field from 0 to H.
Hence, no redshift.
Next, let us analyse the problem by choosing the potential energy
reference as the origin 0 (as done by MTW and Shutz). When the
particle is at its initial position H at rest, its total energy is
mc^2 + mgH. When it falls to 0, its total energy is also mc^2 + mgH,
which is also equal to the energy of the created photon upon
annihilation of the particle. When the photon travels upward
to the point H and is converted to a particle of rest mass m, what is
the total energy of the created particle? Since we are taking the
potential energy reference as the origin, the particle must be created
at rest and with an energy of mc^2 + mgH. Hence, no shift.
But, MTW (p. 187) and Shutz (p.119) say that if the photon arrives
at H with a total energy of mc^2 + mgH, it can be converted to a
particle of rest mass m, and there would be an excess energy of mgH.
To avoid this excess energy, the photon must loose energy in the
gravitational field and arrive at H with energy mc^2. However,
mc^2 is the rest energy of the particle. Since we are measuring
the potential energy from the origin its potential energy is mgH
and this must be added to its total energy. Saying that the
particle's total energy is mc^2 is equivalent to shifting the
potential energy reference level from 0 to H. This cannot be
allowed because one must use the same level for potential energy
from the beginning to the end in a problem.
So, what is the error in my thinking above?
Thank you,
Murat Ozer |
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| John C. Polasek |
| Posted: Thu Nov 02, 2006 4:45 pm |
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On Tue, 24 Oct 2006 22:24:09 +0000 (UTC), Oh No
<NotI@charlesfrancis.wanadoo.co.uk> wrote:
Quote: Thus spake Murat Ozer <Murat.H.Ozer@gmail.com
I would like to ask for your guidance on a problem that I have always
had difficulty with whenever I tried to understand it. It's the
argument for the gravitational redshift derived from energy
conservation presented in textbooks such as by MTW and Schutz.
Consider a particle of rest mass m at a height H above the origin 0.
Let us choose the reference level for potential energy as the level H,
where the total energy of the particle is mc^2. Let the particle fall
to the origin, where its total energy is still mc^2 because the
kinetic energy 1/2*mv^2=mgH cancells with the negative potential
energy relative to H. Now let the particle be annihilated at 0
converting its energy to a photon of the same total energy. Let the
photon travel upward in the gravitational field to H, at which point it
is converted to a particle of rest mass m. What is the total energy
of the created particle? Measuring its potential energy from the
same level H (according to which it is 0), it has to be mc^2.
Is not this right? So, according to this picture the photon cannot
loose energy as it climbs up in the gravitational field from 0 to H.
Hence, no redshift.
You can't say no redshift; when the photon is created, it has some
energy corresponding to the KE of the falling particle. When the photon
rises back to H, it loses this energy and is correspondingly red
shifted.
Regards
It seems fair to say that relativity seems to have no good idea what
causes gravitational redshift. This constant recourse to falling
particles that are converted at will to photons and back again raises
the grave suspicion that the photon is believed to be affected by the
gravitational field just as a massive particle is. It isn't.
Calculating gh/c^2 and applying it as needed for lower frequency and
longer wavelength (up top) with constant c is not clean physics.
The fact is that the atomic clock/radiator runs at a lower frequency
down in gravity and as the wave rises, the frequency remains the same,
unaffected by energy gh. When tested by a higher clock the radiated
frequency will compare as lower. On the way up the wavelength
stretches because c also was lowered down in the well and accelerates
out of the well.
GPS relativistic calculations depend on clocks slowing in gravity, and
when clock slowing is combined with frequency loss (above) , the
total result is double redshift. (Low clock is low by k, frequency up
top loses k, top clock with 0k measures 2k).
The frequency of the emitted wave starts out low and remains so on the
way up.
John Polasek |
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