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Science Forum Index » Physics - Research Forum » Gravitational redshift (potential well) query?
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| WG |
 Posted: Sat Apr 12, 2008 3:30 am |
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If you take a sphere of a given Radius (R) and Density (D) and have a
photon travel out from the center, it will experience a gravitational
redshift according to Zg=4.19GDr^2/c^2 (where r is the distance
traveled from the center and is less than R at this point, r < R ).
This is a standard photon climbing out of a potential well problem
taken from physics textbooks.
{To better visualize it, lets say we are at the center of the earth
and shoot a photon back towards the surface through a open shaft}
Lets assume the sphere is embedded in a background density of D1.
{Which in the earth case would be the density of the universe, (i.e.
critical Density), ignoring local lumpiness} Now distance R or
density D1 do not come into play here due to Gausses theorem. This is
reflected in the equation since neither is present. In other words we
should be able to vary R and D1 to any value and the equation should
still hold true.
Now my question is this:
Would we still get a redshift as D1 approaches D and eventually
equals D?
The answer appears to be "yes it has to" since the photon can never
know about (i.e. feel the effects of) values D1 or R due to Gausses
theorem!
But this gives rise to an interesting problem if D=D1.
You may see it already.
Bill Gilmour wgilmour@i-zoom.net |
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| Greg Egan |
| Posted: Sun Apr 13, 2008 8:49 am |
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On Apr 12, 9:30=A0pm, WG <wgilm...@i-zoom.net> wrote:
Quote: If you take a sphere of a given Radius (R) and Density (D) and have a
photon travel out from the center, it will experience a gravitational
redshift according to Zg=4.19GDr^2/c^2 =A0(where =A0r is the distance
traveled from the center and is less than R at this point, =A0 r < R ).
This is a standard photon climbing out of a potential well problem
taken from physics textbooks.
{To better visualize it, lets say we are at the center of the earth
and shoot a photon back towards the surface through a open shaft}
Lets assume the sphere is embedded in a background density of D1.
{Which in the earth case would be the density of the universe, (i.e.
critical Density), ignoring local lumpiness} =A0Now distance R or
density D1 do not come into play here due to Gausses theorem. This is
reflected in the equation since neither is present. In other words we
should be able to vary R and D1 to any value and the equation should
still hold true.
Now my question is this:
Would we still get a redshift as D1 approaches D and eventually
equals D?
The answer appears to be "yes it has to" since the photon can never
know about (i.e. feel the effects of) values D1 or R due to Gausses
theorem!
But this gives rise to an interesting problem if D=D1.
You may see it already.
Bill Gilmour =A0wgilm...@i-zoom.net
You're mixing Newtonian and General-Relativistic treatments, for a
problem that can only be treated precisely with GR.
If you want to use Newtonian physics, the background of density D1
can't be infinite. If you choose some finite size for it (e.g. a
large sphere centred on the smaller sphere), you'll get a result
perfectly consistent with Gauss's theorem.
If you want to use GR, then the universe can be infinite if you like,
but it must be expanding or contracting. The redshift in a perfectly
homogeneous and isotropic universe follows a simple formula: the
ratio of photon energies is the inverse of the ratio of length scales
for the universe at the times of measurement. Inserting a localised
body of higher density is a complication that would be messy to solve
for exactly, but at the transition that you're describing -- where the
extra density is only infinitesimally above the background -- it's
obvious that the change in redshift will be a continuous function of
the size of the density perturbation. |
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| Tom Roberts |
| Posted: Fri Apr 18, 2008 7:52 am |
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WG wrote:
Quote: If you take a sphere of a given Radius (R) and Density (D) and have a
photon travel out from the center, it will experience a gravitational
redshift according to Zg=4.19GDr^2/c^2 (where r is the distance
traveled from the center and is less than R at this point, r < R ).
This is a standard photon climbing out of a potential well problem
taken from physics textbooks.
{To better visualize it, lets say we are at the center of the earth
and shoot a photon back towards the surface through a open shaft}
It is more obvious what is going on if you discuss gravitational
potential rather than redshift. The redshift, of course, depends on the
gravitational potential difference between source and detector. The
reason is that there is a well-known differential equation for potential
(Poisson's equation), but no corresponding equation for redshift.
[This of course is in the context of Newtonian mechanics, using the GR
relationship between redshift and potential.]
Quote: Lets assume the sphere is embedded in a background density of D1.
{Which in the earth case would be the density of the universe, (i.e.
critical Density), ignoring local lumpiness} Now distance R or
density D1 do not come into play here due to Gausses theorem. This is
reflected in the equation since neither is present. In other words we
should be able to vary R and D1 to any value and the equation should
still hold true.
Not quite -- you need to apply a boundary condition in order to solve
Poisson's equation for the potential. The usual approach is to assume an
empty universe at spatial infinity, and assign potential=0 there. Then
everything you did makes sense, but this requires D1=0. If you insist on
D1>0, there is no physical solution to Poisson's equation throughout the
universe. That requires GR (below).
Note that Newtonian mechanics implicitly assumes Euclidean geometry, and
thus an infinite space. To consider what would happen in a
spatially-closed universe requires GR (below).
Quote: Now my question is this:
Would we still get a redshift as D1 approaches D and eventually
equals D?
The answer appears to be "yes it has to" since the photon can never
know about (i.e. feel the effects of) values D1 or R due to Gausses
theorem!
But this gives rise to an interesting problem if D=D1.
Yes, in an infinite universe with density D1, symmetry implies there can
be no redshift (or potential difference) between any pair of points
(well, this depends on there being no relative motion between the
points, whatever that means... [hic sunt dracones]). This is a
pathological (unphysical) case in Newtonian mechanics.
In GR, there is no intrinsic difficulty with an infinite universe having
D1>0, but it is expanding, and there is no simple "gravitational
potential". The relevant manifold for comparison to above (with D1>0) is
the FRW manifold with zero spatial curvature. For source and detector at
rest in the standard cosmological coordinates, the formula for redshift
is zero when the source and detector are close enough to ignore the
cosmological expansion [#]. This universe is expanding, and the redshift
is nonzero and growing with distance if they are far enough apart so the
expansion cannot be ignored (think many kiloparsecs or more in our
current universe). Now if you add the earth with an open shaft to the
center (also at rest in the cosmological coordinates), and treat it as a
small perturbation of the manifold [@], the redshift is nonzero for a
source at the center of the earth and a detector at rest anywhere else.
[#] Note also that the dragons have disappeared, because
here one can easily select appropriate coordinates and use
them to define "no relative motion".
[@] The earth IS small, on a cosmological scale.
In GR there are two other (simple) possibilities: the universe could be
spatially closed (i.e. the FRW manifolds with positive spatial
curvature), and the universe could be spatially open but have negative
spatial curvature. For them my remarks of the previous paragraph still
hold. In the first there is an additional possibility: the
spatially-closed universe has a finite duration, and during the second
half of its life it is contracting; if the cosmological
expansion/contraction cannot be ignored then during the contraction
phase there is a blueshift that grows with distance.
Tom Roberts |
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