Gravitational Redshift On Rotating Disk (and its implications)
Chalky
rim
= sqrt(1 - [v/c]^2) = sqrt(1 - [wr/c]^2)
= sqrt(1 - integral[ w^2 r]dr/[c]^2)
= sqrt(1 - 2delphi/c]^2)
where w = angular velocity, and del phi is difference in gravitational
potential energy from observer located at middle of disk.
Clearly this is only equal to the traditional gravitational Doppler
shift factor (1 - delphi/c^2), to a first approximation.
Is there any more subtle relativistic effect here that I have missed?
If not, does the general principle thus establish that the following
statement is, consequently, a general law of nature?
"The gravitational Doppler shift factor of 1 - del phi/c^2, is only
generally valid to a first approximation."
Chalky
On Jul 18, 12:07 pm, Chalky <chalkys...@bleachboys.co.uk> wrote:
> On repeating Einstein's rotating disk experiment, I find redshift of
> rim is given by 1(1 + z)
> = sqrt(1 - [v/c]^2) = sqrt(1 - [wr/c]^2)
> = sqrt(1 - 2integral[ w^2 r]dr/[c]^2)
> = sqrt(1 - 2delphi/c]^2)
> where w = angular velocity, and del phi is difference in gravitational
> potential energy from observer located at middle of disk.
>
> Clearly this is only equal to the traditional gravitational Doppler
> shift factor (1 - delphi/c^2), to a first approximation.
>
> Is there any more subtle relativistic effect here that I have missed?
>
> If not, does the general principle thus establish that the following
> statement is, consequently, a general law of nature?
>
> "The gravitational Doppler shift factor of 1 - del phi/c^2, is only
> generally valid to a first approximation."
>From a reference given at spf (http://www.teleconnection.info/rqg/
Gravitation#TheWeakFieldLimit) it seems that the Schwarzchild solution
of EFE also gives an accurate solution of 1 + z = (1 - 2 del phi/c^2)^
- 1/2, thus confirming that this formula IS more accurate, generally.
Comments?
Tom Roberts
> "The gravitational Doppler shift factor of 1 - del phi/c^2, is only
> generally valid to a first approximation."
Yes, of course. This becomes obvious when applying GR. Indeed, the
ability to model gravity using a potential phi inherently requires weak
fields and an approximation.
Tom Roberts
Ian Parker
This seems to me to be "Special" and not "General" Relativity. The
disc is not gravitational, it is constrained to spin by quantum
electrodynamics NOT by gravity/curvature of space. What we must do is
take the instantaneous Lorenz Transformation at a point on the rim.
This is quite a spectacular example. Blue rather than red shift
produces Vacuum UV/Soft X rays, but the principle is the same.
- Ian Parker
Chalky
> On 18 Jul, 12:07, Chalky <chalkys...@bleachboys.co.uk> wrote:
>
>
>
> > On repeating Einstein's rotating disk experiment, I find redshift of
> > rim
> > = sqrt(1 - [v/c]^2) = sqrt(1 - [wr/c]^2)
> > = sqrt(1 - integral[ w^2 r]dr/[c]^2)
> > = sqrt(1 - 2delphi/c]^2)
> > where w = angular velocity, and del phi is difference in gravitational
> > potential energy from observer located at middle of disk.
>
> > Clearly this is only equal to the traditional gravitational Doppler
> > shift factor (1 - delphi/c^2), to a first approximation.
>
> > Is there any more subtle relativistic effect here that I have missed?
>
> > If not, does the general principle thus establish that the following
> > statement is, consequently, a general law of nature?
>
> > "The gravitational Doppler shift factor of 1 - del phi/c^2, is only
> > generally valid to a first approximation."
>
> This seems to me to be "Special" and not "General" Relativity. The
> disc is not gravitational, it is constrained to spin by quantum
> electrodynamics NOT by gravity/curvature of space. What we must do is
> take the instantaneous Lorenz Transformation at a point on the rim.
>
>
> This is quite a spectacular example. Blue rather than red shift
> produces Vacuum UV/Soft X rays, but the principle is the same.
Your derivation indicates a misunderstanding of SR too. The Lorentz
transforms do NOT tell the high velocity observers that their local
physics is different. That would violate the Special Principle.
Furthermore it appears to violate the basic tenets of Quantum
Mechanics as well. The electron is bound by quantum uncertainty so you
can't really talk about what the electron 'sees' classically, let
alone apply the Lorentz transforms to this to decide that it will now
see local physics differently because it is moving.
Chalky
Quite. Nevertheless, GM / R in the exact Schwarzchild solution
certainly looks and smells like Newtonian potential energy, even
though we are advised to avoid stepping in it.
SCNR. ; )
Chalky
> On 18 Jul, 12:07, Chalky <chalkys...@bleachboys.co.uk> wrote:
>
> > On repeating Einstein's rotating disk experiment, I find redshift of
> > rim
> > = sqrt(1 - [v/c]^2) = sqrt(1 - [wr/c]^2)
> > = sqrt(1 - integral[ w^2 r]dr/[c]^2)
> > = sqrt(1 - 2delphi/c]^2)
> > where w = angular velocity, and del phi is difference in gravitational
> > potential energy from observer located at middle of disk.
>
> > Clearly this is only equal to the traditional gravitational Doppler
> > shift factor (1 - delphi/c^2), to a first approximation.
>
> > Is there any more subtle relativistic effect here that I have missed?
>
> > If not, does the general principle thus establish that the following
> > statement is, consequently, a general law of nature?
>
> > "The gravitational Doppler shift factor of 1 - del phi/c^2, is only
> > generally valid to a first approximation."
>
> This seems to me to be "Special" and not "General" Relativity.
It is at the interface between the 2
> The
> disc is not gravitational,
No, but an observer located on it can consider himself at rest in a
gravitational field, under the General Principle, just as the linearly
accelerating observer could too, as explicitly spelled out by the
General Postulate.
> it is constrained to spin by quantum
> electrodynamics NOT by gravity/curvature of space. What we must do is
> take the instantaneous Lorenz Transformation at a point on the rim.
>
> http://groups.google.co.uk/group/sci.physics/browse_frm/thread/55ba5d...
Then your disagreement is with Einstein.
See Einstein A. (1920) Relativity, the special and the general theory.
Ch 18 general principle
Ch 20 general postulate
Ch 23 rotating disk experiment
Koobee Wublee
> On Jul 21, 6:41 pm, Tom Roberts wrote:
> > Yes, of course. This becomes obvious when applying GR. Indeed, the
> > ability to model gravity using a potential phi inherently requires weak
> > fields and an approximation.
>
> Quite. Nevertheless, GM / R in the exact Schwarzchild solution
> certainly looks and smells like Newtonian potential energy, even
> though we are advised to avoid stepping in it.
Each solution or the metric to the Einstein field equations is unique
and describes an independent universe from all other solutions. One
such particular solution that Hilbert had discovered is the famous
Schwarzschild metric in which it is described in general as follows.
ds^2 = c^2 T dt^2 (1 =96 K / r) =96 dr^2 / (1 =96 K / r) =96 r^2 dO^2
Where
** K, T = Integration constants
** dO^2 = r^2 cos^2(Latitude) dLongitude^2 + r^2 dLatitude^2
To impose the limiting case to Newtonian mechanics, the integration
constants must be fudged into the following to do so.
** T = 1
** K = 2 G M / c^2
Of course, the (2 G M /c^2 / r) term smells like twice the Newtonian
potential. In particular, if you work out the mathematics of the
Lagrangian method confined to the Schwarzschild metric, you will find
the following Euler Lagrange equation associated with time.
(1 =96 2 U) / (1 =96 B^2) = Constant
Where
** B^2 c^2 = (dr/dt)^2 / (1 =96 2 U)^2 + r^2 (dO/dt)^2 / (1 =96 2 U)
** U = G M / c^2 / r
The constant can be interpreted as the following.
Constant = E / (m c^2)
Where
** E = Observed energy
** m = intrinsic mass of the particle
Under weak gravitation and low speed, you have the Newtonian
conservation of energy as described below.
B2 =96 2 U = constant
Or
B^2 / 2 =96 U = constant
Where
** B^2 c^2 ~ (dr/dt)^2 + r^2 (dO/dt)^2
Why is this simple mathematical result so surprising to you?
Ian Parker
> On Jul 21, 6:41 pm, Ian Parker <ianpark...@gmail.com> wrote:
>
>
>
>
>
> > On 18 Jul, 12:07, Chalky <chalkys...@bleachboys.co.uk> wrote:
>
> > > On repeating Einstein's rotating disk experiment, I find redshift of
> > > rim
> > > = sqrt(1 - [v/c]^2) = sqrt(1 - [wr/c]^2)
> > > = sqrt(1 - integral[ w^2 r]dr/[c]^2)
> > > = sqrt(1 - 2delphi/c]^2)
nal
> > > potential energy from observer located at middle of disk.
>
> > > Clearly this is only equal to the traditional gravitational Doppler
> > > shift factor (1 - delphi/c^2), to a first approximation.
>
> > > Is there any more subtle relativistic effect here that I have missed?
>
> > > If not, does the general principle thus establish that the following
> > > statement is, consequently, a general law of nature?
>
> > > "The gravitational Doppler shift factor of 1 - del phi/c^2, is only
> > > generally valid to a first approximation."
>
> > This seems to me to be "Special" and not "General" Relativity.
>
> It is at the interface between the 2
>
> > The
> > disc is not gravitational,
>
> No, but an observer located on it can consider himself at rest in a
> gravitational field, under the General Principle, just as the linearly
> accelerating observer could too, as explicitly spelled out by the
> General Postulate.
>
by Tensors. We do not observe the Universe revolving round us. If you
are on a disc you observe your surroundings rotating. You and your
surroundings have an instantanous Lorenz transformation.
I think that the confusion is due to a faulty understanding of Mach's
principle which relativity does not affect.
> > it is constrained to spin by quantum
> > electrodynamics NOT by gravity/curvature of space. What we must do is
> > take the instantaneous Lorenz Transformation at a point on the rim.
>
> >http://groups.google.co.uk/group/sci.physics/browse_frm/thread/55ba5d...
>
> Then your disagreement is with Einstein.
>
> See Einstein A. (1920) Relativity, the special and the general theory.
>
> Ch 18 general principle
> Ch 20 general postulate
> Ch 23 rotating disk experiment- Hide quoted text -
>
You do in point of fact get the same answer however you look at it. If
you apply special relativity you get a time dilatrion factor of =96(1-
v*v/c*c) If you integrate your centrifugal force, you will get =96(1-v*v/
c*c).
Something is wrong though. Of we move a mass to the rim energy will
increase. In fact if the rotation speed remains constant, energy has
to be supplied to the axle as we move towards the rim. Thinking of it
like this it is clear we don't have the "sealed" system of gravitating
bodies. We have a system which we supply energy to via the axle and
can take energy out, also via the axle when a mass moves from the rim
to the center.
- Ian Parker
Chalky
encoding. Please check that your posts are in plain ASCII to avoid
similar transmission problems. -ik ]
On Jul 23, 10:08 pm, Koobee Wublee <koobee.wub...@gmail.com> wrote:
> ds^2 = c^2 T dt^2 (1 =96 K / r) =96 dr^2 / (1 =96 K / r) =96 r^2 dO^2
> (1 =96 2 U) / (1 =96 B^2) = Constant
> ** B^2 c^2 = (dr/dt)^2 / (1 =96 2 U)^2 + r^2 (dO/dt)^2 / (1 =96 2 U)
> B2 =96 2 U = constant
> B^2 / 2 =96 U = constant
>
> Where
>
> ** B^2 c^2 ~ (dr/dt)^2 + r^2 (dO/dt)^2
>
> Why is this simple mathematical result so surprising to you?
Sorry, I don't understand what you mean by 1=96 or by 2=96 so I don't
understand what your question is.
Chalky
>
> > On Jul 21, 6:41 pm, Tom Roberts wrote:
> > > Yes, of course. This becomes obvious when applying GR. Indeed, the
> > > ability to model gravity using a potential phi inherently requires weak
> > > fields and an approximation.
>
> > Quite. Nevertheless, GM / R in the exact Schwarzchild solution
> > certainly looks and smells like Newtonian potential energy, even
> > though we are advised to avoid stepping in it.
>
> Each solution or the metric to the Einstein field equations is unique
> and describes an independent universe from all other solutions. One
> such particular solution that Hilbert had discovered is the famous
> Schwarzschild metric
Really? I thought Schwarzschild discovered the Schwarzschild metric
only months after Einstein published his gravitational field equation.
Are you claiming Hilbert did it all first?
> Of course, the (2 G M /c^2 / r) term smells like twice the Newtonian
> potential.
We seem to have our wires crossed here. I was referring to the
redshift which is (1 - 2GM/R) -!/2
This only acts like twice the Newtonian potential at the relativistic
event horizon, which is exactly the same as what we find by applying
the General Principle to the rotating disk.
Why should I be surprised by that?
Chalky
> Something is wrong though. Of we move a mass to the rim energy will
> increase.
No, the disk will slow down. (Spinning skater, ballerina
etc....Remember?)
Thus for an observer at the axis, if a sufficiently massive part of
the disk is moving radially outwards, this will cause experienced
'gravitational field' to decrease in magnitude near the origin.
If you think about it, something similar would happen if you moved a
sufficiently massive fraction of the Earth away from the Earth too.
Spaceman
> On Jul 23, 10:08 pm, Ian Parker <ianpark...@gmail.com> wrote:
>
>> Something is wrong though. Of we move a mass to the rim energy will
>> increase.
>
> No, the disk will slow down. (Spinning skater, ballerina
> etc....Remember?)
Correct,
The motion creates a path that moves outward at a spiral,
the further out it gets the slower rotations will occur even if at the
same "path speed" but in reality it is constantly slowing
unless constant energy is applied.
If no constant energy is supplied, the energy will slowly
be decreasing.
> Thus for an observer at the axis, if a sufficiently massive part of
> the disk is moving radially outwards, this will cause experienced
> 'gravitational field' to decrease in magnitude near the origin.
>
> If you think about it, something similar would happen if you moved a
> sufficiently massive fraction of the Earth away from the Earth too.
Again, Correct.
The gravitational potential strength will decrease when the distance
increases.
:)
--
James M Driscoll Jr
Spaceman
Chalky
> On 22 Jul, 01:05, Chalky <chalkys...@bleachboys.co.uk> wrote:
>
> > On Jul 21, 6:41 pm, Ian Parker <ianpark...@gmail.com> wrote:
>
> > > This seems to me to be "Special" and not "General" Relativity.
>
> > It is at the interface between the 2
>
> > > The
> > > disc is not gravitational,
>
> > No, but an observer located on it can consider himself at rest in a
> > gravitational field, under the General Principle, just as the linearly
> > accelerating observer could too, as explicitly spelled out by the
> > General Postulate.
>
> No he doesn't.
Jim Akerlund has carefully typed out the relevant text (by Einstein)
at
http://groups.google.com/group/sci.physics.foundations/browse_thread/thread/0d10c7aacfc6038e#
I suggest you read it very carefully. As I have already pointed out,
your argument contradicts Einstein.
Chalky
Since not all email gateways are 8-bit clean, and even if they are there
is not a standard 8-bit character set, MIME encoding if there are 8-bit
characters are OK and I usually change them back to 8-bit characters
which, as far as I know, show up correctly for everyone reading them.
However, encoding 7-bit characters, whether or not there are any 8-bit
characters in the post, is just plain stupid. Sorry, but that's the
only way to put it. This is a text-based newsgroup, and there is no
reason to send anything but 7-bit ASCII if that is what the text is to
begin with. In the future, we might have to start deleting posts which
have too much MIME garbage. Find out how to send plain text and do so.
-P.H.]
On Jul 23, 10:08pm, Ian Parker <ianpark...@gmail.com> wrote:
> On 22 Jul, 01:05, Chalky <chalkys...@bleachboys.co.uk> wrote:
> > > The
> > > disc is not gravitational,
>
> > No, but an observer located on it can consider himself at rest in a
> > gravitational field, under the General Principle, just as the linearly
> > accelerating observer could too, as explicitly spelled out by the
> > General Postulate.
>
> No he doesn't.
Who doesn't what? You are not making sense.
>If we are in a gravitational field we can describe it
>by Tensors.
We can describe it any way we like, that works.
> We do not observe the Universe revolving round us.
Of course we DO! (what planet are you on?)
We INFER that this is because the planet is rotating (and orbiting
the Sun). This is because we OBSERVE the Sun and the stars revolving
around us every 24 hours, and the seasons cycling every year.
> You do in point of fact get the same answer however you look at it. If
> you apply special relativity you get a time dilatrion factor of (1-
> v*v/c*c) If you integrate your centrifugal force, you will get (1-v*v/
> c*c).
Assuming ???? means sqrt, yes. That is the whole point
> Something is wrong though. Of we move a mass to the rim energy will
> increase.
Actually, if the a mass is an appreciable fraction of the disk's mass,
the disk will slow down, instead,
But, yes, the situation is not identical to the gravitational fields
_induced by mass_
You seem to be following the Kierkegaard argument that once you have
used a ladder to get to a higher point, you can throw away the ladder.
You seem to be going one step beyond though by now denying that the
ladder ever existed.
Koobee Wublee
Since not all email gateways are 8-bit clean, and even if they are there
is not a standard 8-bit character set, MIME encoding if there are 8-bit
characters are OK and I usually change them back to 8-bit characters
which, as far as I know, show up correctly for everyone reading them.
However, encoding 7-bit characters, whether or not there are any 8-bit
characters in the post, is just plain stupid. Sorry, but that's the
only way to put it. This is a text-based newsgroup, and there is no
reason to send anything but 7-bit ASCII if that is what the text is to
begin with. In the future, we might have to start deleting posts which
have too much MIME garbage. Find out how to send plain text and do so.
-P.H.]
On Jul 24, 12:25 pm, Chalky wrote:
> On Jul 23, 10:08 pm, Koobee Wublee wrote:
> > Each solution or the metric to the Einstein field equations is unique
> > and describes an independent universe from all other solutions. One
> > such particular solution that Hilbert had discovered is the famous
> > Schwarzschild metric
>
> Really? I thought Schwarzschild discovered the Schwarzschild metric
> only months after Einstein published his gravitational field equation.
> Are you claiming Hilbert did it all first?
Yes, really. See Schwarzschild's own writing below.
http://arxiv.org/abs/physics/9905030
In summary, below spacetime represents Schwarzschild's original
solution which is not to be confused as the Schwarzschild metric.
ds^2 ppp c^2 (1 =96 K / u) dt^2 =96 dr^2 (du/dr)^2 / (1 =96 K / u) =96 u^2 dO^2
Where
** K ppp Integration constant
** u^3 ppp r^3 + K^3
** dO^2 ppp cos^2(Latitude) dLongitude^2 + dLatitude^2
Notice Schwarzschild's original solution does not manifest any black
holes. It was Hilbert a year or two later that finally wrote down the
Schwarzschild metric below.
ds^2 ppp c^2 (1 =96 K / r) dt^2 =96 dr^2 / (1 =96 K / r) =96 r^2 dO^2
As you know the Schwarzschild metric does manifest black holes.
Another simple solution to the field equations is described as follows
where it also does not manifest black holes as the original
Schwarzschild's solution.
ds^2 ppp c^2 dt^2 / (1 + K / r) =96 dr^2 (1 + K / r) =96 (r + K)^2 dO^2
All these three solutions satisfy the following:
** Newtonian gravity as the limiting case
** Static
** Spherically symmetric
** Asymptotically flat
Thus, they are all valid solutions. The most acceptable solution is
the Scwharzschild metric discovered by Hilbert that manifest black
holes.
> > Of course, the (2 G M /c^2 / r) term smells like twice the Newtonian
> > potential.
>
> We seem to have our wires crossed here. I was referring to the
> redshift which is (1 - 2GM/R) -!/2
http://groups.google.com/group/sci.physics.relativity/msg/7dd4e0a10f3d816f?hlpppen
> This only acts like twice the Newtonian potential at the relativistic
> event horizon, which is exactly the same as what we find by applying
> the General Principle to the rotating disk.
http://groups.google.com/group/sci.physics.relativity/msg/f42dddd7a3dc9470?hlpppen
> Why should I be surprised by that?
Since you raised the issue, I thought you were surprised. That is
all. <shrug>
Chalky
wrote:
Yes, you are right. The gravitational potential difference *does*
decrease in both situations, irrespective of whether the potential
difference encourages or discourages that radial motion.
Thanks also for responding via the moderated newsgroup instead of via
the rogue sub-thread created by Kubee Wooblie removing SPResearch from
the loop.
It is unfortunate that Eric Gisse (for example) has not also tried
responding in this way, since this does tend to filter out a lot of
undesirable noise (and invariably supresses flames).
Chalky
> On Jul 24, 12:25 pm, Chalky wrote:
> > We seem to have our wires crossed here. I was referring to the
> > redshift which is (1 - 2GM/R)^ -1/2
>
> http://groups.google.com/group/sci.physics.relativity/msg/7dd4e0a10f3...
Your central argument in this link is that the rim is blue shifted not
red shifted (as was Ian Parker's), and, by implication, that
Einstein's special relativistic time dilation formula is upside down.
Perhaps you should show a rigorous derivation of this alleged
relativistic time contraction, and an explain why, despite this,
Einstein's formula had already been verified observationally by the
early 70s using accurate clocks on aircraft flights (MTW). This would
be more sensible than challenging me to prove that red is blue.
> > This only acts like twice the Newtonian potential at the relativistic
> > event horizon, which is exactly the same as what we find by applying
> > the General Principle to the rotating disk.
>
> http://groups.google.com/group/sci.physics.relativity/msg/f42dddd7a3d...
This link does at least present your prior formulae uncorrupted.
> > Why should I be surprised by that?
>
> Since you raised the issue, I thought you were surprised. That is
> all. <shrug>
When I first raised this question I did not know that the Schwarzchild
redshift formula was effectively identical to the rotating disk
formula if you interpret GM/R as Newtonian potential.
I am relieved rather than surprised that the General Principle works
that well, in this situation.