SR time dilation on remote objects ?

On Tue, 21 Sep 2004 18:50:52 +0100, Jonathan Silverlight
wrote:

In message , Marcel
Luttgens writes

In my model, everything is *very* close to the center of the (stable)
universe,
which is so big that a distance c/H is so much smaller than the universe
radius, that the universe could as well be considered as infinite.
If it were mathematically infinite, the Earth, and all of its points,
would of course coincide with its center. In particular, all points
of the trajectory of a light signal (emitted for instance by a galaxy)
would be at the center of a sphere of radius c/H centered on the universe's
center, and thus subject to an acceleration cH explaining its redshift.


What acceleration? I thought your universe was static.

On Mon, 20 Sep 2004 06:51:27 -0700, "N:dlzc D:aol T:com \(dlzc\)" N:
dlzc1 D:cox wrote:

Dear Marcel Luttgens:

"Marcel Luttgens" wrote in message
. com...
Bjoern Feuerbacher wrote in message
...
Marcel Luttgens wrote:
Bjoern Feuerbacher wrote in
message ...


But the point is that my formula is right, what you denied
till now.
...
In a situation and using assumptions which have *nothing at all*
to do with cosmology!!!


According to GRists.

Perhaps you are not being clear with your cosmology. Here is what you are
saying:
1) the Earth is at the periphery of all the matter in the Universe.
2) space is infinite, unbounded.
3) the Earth is everywhere on the periphery.

1 is your model of the "illusion" of expansion. 2 is the basis for your
mathematics. 3 is required since expansion is observed in all directions.
Because your model is similar to one used when "climbing out" of the core
of the Earth. And this is definitely not uniform in all directions.

David A. Smith

Gentlemen, the `universe' you discuss is 90+% invisible to us except
by inference. How can you be so sure of your `illusions'. Remove EM
radiation from the equations and a universe without atoms (No strong
or weak nuclear or electromagnetic forces) - just matter/energy in
some unknown state only detectible by gravity, results. Dark matter
and dark energy. Not defined visible in any direction. Perhaps with
a little adventitious `light' resulting from some fleeting quirk that
allowed some quarks to become `sticky' ,15 or so billion years ago
which allowed a scattering of Hydrogen candles to light. I submit your
`cosmology' is like attempting to describe a car after finding one
headlight lamp. Don't be too sure your illusions have anything to do
with reality. Perhaps the visible cosmos is as transitory as the
glimmer of a firefly.

Dear vonroach:

"vonroach" wrote in message
...
On Mon, 20 Sep 2004 06:51:27 -0700, "N:dlzc D:aol T:com \(dlzc\)" N:
dlzc1 D:cox wrote:

Dear Marcel Luttgens:

"Marcel Luttgens" wrote in message
.com...
Bjoern Feuerbacher wrote in message
...
Marcel Luttgens wrote:
Bjoern Feuerbacher wrote in
message ...


But the point is that my formula is right, what you denied
till now.
...
In a situation and using assumptions which have *nothing at all*
to do with cosmology!!!


According to GRists.

Perhaps you are not being clear with your cosmology. Here is what you
are
saying:
1) the Earth is at the periphery of all the matter in the Universe.
2) space is infinite, unbounded.
3) the Earth is everywhere on the periphery.

1 is your model of the "illusion" of expansion. 2 is the basis for your
mathematics. 3 is required since expansion is observed in all directions.
Because your model is similar to one used when "climbing out" of the core
of the Earth. And this is definitely not uniform in all directions.

Gentlemen, the `universe' you discuss is 90+% invisible to us except
by inference. How can you be so sure of your `illusions'. Remove EM
radiation from the equations and a universe without atoms (No strong
or weak nuclear or electromagnetic forces) - just matter/energy in
some unknown state only detectible by gravity, results. Dark matter
and dark energy. Not defined visible in any direction.

MOND, which has other problems does not require the "Dark Matter patch",
which then requires the "Dark Energy upgrade", to get compliance. So maybe
only 10-20% is yet to be discovered "optically".

Perhaps with
a little adventitious `light' resulting from some fleeting quirk that
allowed some quarks to become `sticky' ,15 or so billion years ago
which allowed a scattering of Hydrogen candles to light. I submit your
`cosmology' is like attempting to describe a car after finding one
headlight lamp. Don't be too sure your illusions have anything to do
with reality. Perhaps the visible cosmos is as transitory as the
glimmer of a firefly.

It is something to do, until the star drives come online...

David

Jonathan Silverlight wrote:
In message , Marcel
Luttgens writes


In my model, everything is *very* close to the center of the (stable)
universe,
which is so big that a distance c/H is so much smaller than the universe
radius, that the universe could as well be considered as infinite.
If it were mathematically infinite, the Earth, and all of its points,
would of course coincide with its center. In particular, all points
of the trajectory of a light signal (emitted for instance by a galaxy)
would be at the center of a sphere of radius c/H centered on the
universe's
center, and thus subject to an acceleration cH explaining its redshift.


What acceleration? I thought your universe was static.

Well, apparently in his universe, the matter is not accelerated, but
the photons are.

Who cares about consistency...



Bye,
Bjoern

Bjoern Feuerbacher wrote in message ...
Marcel Luttgens wrote:
"N:dlzc D:aol T:com \(dlzc\)" N: dlzc1 D:cox wrote in message news:yfB3d.207162$4o.59667@fed1read01...

Dear Marcel Luttgens:

"Marcel Luttgens" wrote in message
.com...

Bjoern Feuerbacher wrote in message
...

Marcel Luttgens wrote:

Bjoern Feuerbacher wrote in
message ...




But the point is that my formula is right, what you denied
till now.

...

In a situation and using assumptions which have *nothing at all*
to do with cosmology!!!


According to GRists.

Perhaps you are not being clear with your cosmology. Here is what you are
saying:
1) the Earth is at the periphery of all the matter in the Universe.
2) space is infinite, unbounded.
3) the Earth is everywhere on the periphery.

1 is your model of the "illusion" of expansion. 2 is the basis for your
mathematics. 3 is required since expansion is observed in all directions.
Because your model is similar to one used when "climbing out" of the core
of the Earth. And this is definitely not uniform in all directions.



Not exactly.

In my model, everything is *very* close to the center of the (stable)
universe,

How can *everything* be *very* close to the center? You are not making
sense.

And, BTW, if the Earth were near the center of a homogeneous sphere,
the light we receive from outside would be *blueshifted*, not
*redshifted*, since the light *gains* energy when it travels inwards
in such a sphere.


No, because the acceleration cH at the surface of the sphere of radius c/H
is negative, hence light leaving the center of the sphere would be redshifted.
Every point of the light trajectory is situated at the center of such a
sphere. Notice that the universe is not expanding.

Anyhow, when hypothetizing such negative acceleration cH, one gets the
redhift formula d = c/K * z/(z+1) corresponding to z = (v/c)/(1-v/c),
where v = Kd (K has the same value as the Hubble constant H).

Notice also that the formula z = (v/c)/(1-v/c) can also be found when
hypothetizing an expanding universe.



which is so big that a distance c/H is so much smaller than the universe
radius, that the universe could as well be considered as infinite.
If it were mathematically infinite, the Earth, and all of its points,
would of course coincide with its center.

No. In that case, it would have *no* uniquely defined center.


But every point could be considered as a center of the infinite universe.


In particular, all points
of the trajectory of a light signal (emitted for instance by a galaxy)
would be at the center of a sphere of radius c/H centered on the universe's
center, and thus subject to an acceleration cH explaining its redshift.

See above. This would give a blueshift, not a redshift.


It would be a redshift, because of the negative acceleration cH at the
surface of the sphere.


Bye,
Bjoern

Marcel Luttgens

vonroach wrote in message . ..
On Tue, 21 Sep 2004 18:50:52 +0100, Jonathan Silverlight
wrote:

In message , Marcel
Luttgens writes

In my model, everything is *very* close to the center of the (stable)
universe,
which is so big that a distance c/H is so much smaller than the universe
radius, that the universe could as well be considered as infinite.
If it were mathematically infinite, the Earth, and all of its points,
would of course coincide with its center. In particular, all points
of the trajectory of a light signal (emitted for instance by a galaxy)
would be at the center of a sphere of radius c/H centered on the universe's
center, and thus subject to an acceleration cH explaining its redshift.


What acceleration? I thought your universe was static.

What acceleration? I thought your universe was static.

It is static. At the surface of a sphere of radius c/H exists a negative
acceleration cH.

Marcel Luttgens

Marcel Luttgens wrote:
Bjoern Feuerbacher wrote in message ...

Marcel Luttgens wrote:


[snip]


In my model, everything is *very* close to the center of the (stable)
universe,

How can *everything* be *very* close to the center? You are not making
sense.

I notice you did not bother to answer that.


And, BTW, if the Earth were near the center of a homogeneous sphere,
the light we receive from outside would be *blueshifted*, not
*redshifted*, since the light *gains* energy when it travels inwards
in such a sphere.



No, because the acceleration cH at the surface of the sphere of radius c/H
is negative,

The acceleration *everywhere* in the sphere is negative (in the sense
of pointing to the center).


hence light leaving the center of the sphere would be redshifted.

Err, so what? You claim that we are near the center. At least some
light which reaches us has to come from regions which are farther away
from the center than we. And that light would be *blueshifted*, not
redshifted!

Actually, what we should see that the light coming from one direction
(the direction to the center) is redshifted, turning around, this re
shift becomes less and less, light coming from regions at an angle of 90
degrees to the direction should not be shifted at all, and all light
coming from regions at higher angles has to be blueshifted.


Every point of the light trajectory is situated at the center of such a
sphere.

Pardon? Why are there suddenly several spheres, one for every point
of the light trajectory?

You make less and less sense.


Notice that the universe is not expanding.

Notice that this is a totally unsupported assertion, and that you
still keep ignoring 99% of the evidence which shows that the
universe *is* expanding.


Anyhow, when hypothetizing such negative acceleration cH, one gets the
redhift formula d = c/K * z/(z+1) corresponding to z = (v/c)/(1-v/c),
where v = Kd (K has the same value as the Hubble constant H).

For the 20th time: go to a paper where actual redshifts are reported,
and compare the numbers there to the predictions of your formula.
Particulary, look at
Riess et al., Type Ia Supernova Discoveries at z1 From the Hubble Space
Telescope: Evidence for Past Deceleration and Constraints on Dark Energy
Evolution, Astrophys.J. 607 (2004) 665-687, astro-ph/0402512


Notice also that the formula z = (v/c)/(1-v/c) can also be found when
hypothetizing an expanding universe.

No, it can't. Your derivation is complete bogus.


which is so big that a distance c/H is so much smaller than the universe
radius, that the universe could as well be considered as infinite.
If it were mathematically infinite, the Earth, and all of its points,
would of course coincide with its center.

No. In that case, it would have *no* uniquely defined center.



But every point could be considered as a center of the infinite universe.

If the universe is not *really* infinite, then no, *not* every point
can be considered as the center. The gravitational field at every point
would show in which direction the *unique* center lies.


In particular, all points
of the trajectory of a light signal (emitted for instance by a galaxy)
would be at the center of a sphere of radius c/H centered on the universe's
center, and thus subject to an acceleration cH explaining its redshift.

See above. This would give a blueshift, not a redshift.



It would be a redshift, because of the negative acceleration cH at the
surface of the sphere.

At the surface of the sphere, and everywhere in its interior, the
acceleration points inwards. And hence gives a blueshift for light
which goes from there to the center.


Bye,
Bjoern

vonroach wrote in message . ..
On Mon, 20 Sep 2004 06:51:27 -0700, "N:dlzc D:aol T:com \(dlzc\)" N:
dlzc1 D:cox wrote:

Dear Marcel Luttgens:

"Marcel Luttgens" wrote in message
. com...
Bjoern Feuerbacher wrote in message
...
Marcel Luttgens wrote:
Bjoern Feuerbacher wrote in
message ...


But the point is that my formula is right, what you denied
till now.
...
In a situation and using assumptions which have *nothing at all*
to do with cosmology!!!


According to GRists.

Perhaps you are not being clear with your cosmology. Here is what you are
saying:
1) the Earth is at the periphery of all the matter in the Universe.
2) space is infinite, unbounded.
3) the Earth is everywhere on the periphery.

1 is your model of the "illusion" of expansion. 2 is the basis for your
mathematics. 3 is required since expansion is observed in all directions.
Because your model is similar to one used when "climbing out" of the core
of the Earth. And this is definitely not uniform in all directions.

David A. Smith

Gentlemen, the `universe' you discuss is 90+% invisible to us except
by inference. How can you be so sure of your `illusions'. Remove EM
radiation from the equations and a universe without atoms (No strong
or weak nuclear or electromagnetic forces) - just matter/energy in
some unknown state only detectible by gravity, results. Dark matter
and dark energy. Not defined visible in any direction. Perhaps with
a little adventitious `light' resulting from some fleeting quirk that
allowed some quarks to become `sticky' ,15 or so billion years ago
which allowed a scattering of Hydrogen candles to light. I submit your
`cosmology' is like attempting to describe a car after finding one
headlight lamp. Don't be too sure your illusions have anything to do
with reality. Perhaps the visible cosmos is as transitory as the
glimmer of a firefly.

I agree, we see only the tail of the elephant, but it is worth trying
to imagine the whole beast.

Marcel Luttgens

Marcel Luttgens wrote:
vonroach wrote in message . ..

On Tue, 21 Sep 2004 18:50:52 +0100, Jonathan Silverlight
wrote:


In message , Marcel
Luttgens writes

In my model, everything is *very* close to the center of the (stable)
universe,
which is so big that a distance c/H is so much smaller than the universe
radius, that the universe could as well be considered as infinite.
If it were mathematically infinite, the Earth, and all of its points,
would of course coincide with its center. In particular, all points
of the trajectory of a light signal (emitted for instance by a galaxy)
would be at the center of a sphere of radius c/H centered on the universe's
center, and thus subject to an acceleration cH explaining its redshift.


What acceleration? I thought your universe was static.


What acceleration? I thought your universe was static.


It is static. At the surface of a sphere of radius c/H exists a negative
acceleration cH.

And you don't think that the words "static" and "acceleration" are
contradictory?


Bye,
Bjoern

Bjoern Feuerbacher wrote in message ...
Marcel Luttgens wrote:
Bjoern Feuerbacher wrote in message ...

Marcel Luttgens wrote:

Bjoern Feuerbacher wrote in message ...


[snip]


Hint: LET and SR give the same results. And that result is *not* the
one *you* obtain.



I presume that *your* formula is Nu(o) = Nu sqrt((1-v/c)/(1+v/c)),

Yes. Nice that you finally got this.

Hint: that formula has been *tested*. See e.g. he
http://www.mathpages.com/rr/s2-04/2-04.htm
For a more detailed description of that experiment, see he
http://spiff.rit.edu/classes/phys314/lectures/doppler/doppler.html

Care to compare the predictions of your formula to the results
obtained there?


See also section IV of this article:
http://arxiv.org/pdf/physics/0408047


Oh, and while you are at it, you could also explain the transverse
Doppler effect, which also was tested experimentally, and where also
agreement with the predictions of SR was found.


I don't deny that such formula makes sense, I simply claim that
it cannot be applied to light emitters/receivers subject to
"space" expansion, because such expansion is symmetrical.



but
it would be clearer to write Nu(o) = Nu sqrt(1-v^2/c^2)/(1+v/c),
where sqrt(1-v^2/c^2) corresponds to the time dilation.

If you want, write it that way.


If vc, one is left with Nu(o) = Nu / (1+v/c), which is the Doppler
formula for sound when the sound source is receding from the
observer.

Yes, indeed. Hint: for v c, one can also write this as
Nu(o) = Nu (1 - v/c),
which is the Doppler formula for sound when the observer is moving wrt
the sound source.

It has to be *expected* that for v c, the sound formulas are reproduced!


But you should realize that light is not sound, because
of the speed limit c.

*sigh* Obviously. That's why one uses Nu(o) = Nu sqrt((1-v/c)/(1+v/c)),
not Nu(o) = Nu / (1+v/c), if you didn't notice.


No, those who use Nu(o) = Nu sqrt((1-v/c)/(1+v/c)) because of time dilation,
forget that in an expanding universe, the identical time dilations
on the emitter and the receiver cancel each other.


I keep claiming that the right formula is
Nu(o) = Nu * (1-v/c), which leads to z = (v/c)/(1-v/c), and which
is valid whether the observer or the source is receding.

With no supporting evidence, and with a calculation which is based
on "the distance changes, but let's consider that it it constant
anyway".


No time dilation factor
is needed in an expanding universe, because the light source and the
observer are simultaneously receding wrt each other.

Irrespective of a time dilation factor is needed or not, your
formula above is simply wrong. If you think otherwise, present
a derivation which is *not* based on "the distance changes, but let's
consider that it it constant anyway".


IOW, you are saying that Nu(o) = Nu / (1+v/c) sould be used when
the light source is receding from the observer.


Nu(o) = Nu * (1-v/c) should also be used when the source is receding
from the observer, because light can be considered as a train of photons
of wavelength lambda emitted at c by the source. But as the source
is moving away at v, the frequency Nu(o) at which the photons are
emitted becomes Nu * (c-v)/c for the observer.

That claim is still utter nonsense.


Realize that, for the observer, the photons are not emitted at c,
but at c-v, by the atoms of the source. Hence their frequency
is lessened by a factor (c-v)/c.


Of course, the velocity of the emitted light remains c.

Finally something I can agree with.


You claimed that your formula is used by radar systems, and cited
that page as support. But your formula does not appear on that page.
A formula appears on the page which can be *derived* from your formula
for small velocities. But that same formula can *also* be derived
from SR and LET for small velocities! So the page is no support for your
claim. It's that simple.


You could as well say that it is no support for LET/SR.

Yes, I could say that. But: so what? *You* brought the page up as
support for *your* formula. I merely pointed out that what is written
on the page can't serve to distinguish between your formula and the one
of LET/SR.


The page is interesting. It helps to show that the time dilation
effects on the source and the emitter cancel each other:

1) Absence of time dilation.

The radar emits at a frequency Nu, the car is moving away at v
from the radar.
The frequency Nu1 received by the car is Nu * (c-v)/c.
The moving car is now the source and emits at Nu1.
The frequency Nu(o) received by the radar is, according to
the *sound formula*, Nu1 * c/(c+v), hence
Nu(o) = Nu * (c-v)/c * c/(c+v)
= Nu (c-v)/(c+v)
Thus, 1+z = lambda(o)/lambda = (c+v)/(c-v), and
z = 2v/(c-v) = 2 v/c / (1-v/c)

2) There is time dilation on both objects.

Notice that, unlike the Doppler shift for sound, it does not matter
whether the source or the observer is the one in motion.
Only their relative velocity is needed when using the LET formula
Nu(o) = Nu sqrt(1-v^2/c^2)/(1+v/c).

The radar emits at a frequency Nu, the car is moving away at v
from the radar.
The frequency Nu1 received by the car is Nu sqrt(1-v^2/c^2)/(1+v/c).
The moving car is now the source and emits at Nu1.
The frequency Nu(o) received by the radar is
Nu1 sqrt(1-v^2/c^2)/(1+v/c), hence
Nu(o) = Nu sqrt(1-v^2/c^2)/(1+v/c) * sqrt(1-v^2/c^2)/(1+v/c)
= Nu (1-v^2/c^2)/(1+v/c)^2
= Nu (1-v/c)(1+v/c)/((1+v/c)(1+v/c))
= Nu (1-v/c)/(1+v/c)
= Nu (c-v)/(c+v)
Thus, 1+z = = lambda(o)/lambda = (c+v)/(c-v), and
z = 2v/(c-v) = 2 v/c / (1-v/c),

which is the same formula as the one obtained with no time dilation!



Bye,
Bjoern

Marcel Luttgens