Re: SR time dilation on remote objects ?

From: Marcel Luttgens (mluttgens_at_wanadoo.fr)
Date: 08/11/04 

Date: 11 Aug 2004 08:02:08 -0700

Bjoern Feuerbacher <feuerbac@thphys.uni-heidelberg.de> wrote in message news:<cfardg$js9$1@news.urz.uni-heidelberg.de>...
> Marcel Luttgens wrote:
> > Bjoern Feuerbacher <feuerbac@thphys.uni-heidelberg.de> wrote in message news:<cf7tar$6ab$1@news.urz.uni-heidelberg.de>...
> >
> >>Marcel Luttgens wrote:
> >>
> >>>Bjoern Feuerbacher <feuerbac@thphys.uni-heidelberg.de> wrote in message news:<ceqahd$f84$1@news.urz.uni-heidelberg.de>...
>
> [snip]
>
>
> >>>In such universe, observers A and B separated by a distance d move
> >>>*from each other* at a velocity v, which is a function of d.
> >>
> >>And of time, if you didn't notice.
> >>
> >
> >
> > Unless one assumes a steady state model. I know that such model is
> > controversial, but it is however plausible.
>
> Why is that plausible? How would one explain all the evidence in
> such a model?

This has been done by the proponents of such model.
>
>
> >>>It has been shown above that *in such situation*, A and B clocks
> >>>tick at the same rate.
> >>
> >>Your demonstration about SR is entirely irrelevant here.
> >>
> >>In cosmology, the relevant formula for time dilation is:
> >> dt(when the light was emitted) = f*dt(when the light is observed).
> >>You claim that you can show that f = 1 - but for showing that, you
> >>use an argument which seems to be based on
> >> dt(A) = f*dt(B) and dt(B) = f*dt(A),
> >>completely ignoring the *crucial* difference between the time when
> >>the light as emitted and the time when it was observed.
> >>
> >>I told you that about 5 times now. You *still* ignore that.
> >
> >
> > Hereafter is a response I just gave on the NG sci.physics.relativity.
>
> Which has *nothing* to do with my argument above!!!!!
>
>
>
> >>>Applying SR, observer A situated at a distance d from observer B
> >>>will claim that tB = tA * sqrt(1-v^2/c^2), but, in his frame of
> >>>reference, observer B is perfectly right (sic) to claim that
> >>>tB = tA * sqrt(1-v^2/c^2)!
> >>
> >>Using tB and tA here makes no sense, because
> >>1) time dilation applies to time *intervals*, so you should write
> >>dt or delta t or something like that.
> >>2) you *still* ignore the *crucial* difference between the time when
> >>the light as emitted and the time when it was observed.
>
> Did you get it this time? I bet you didn't.
>
> [snip]
>
>
> > 1) From http://www.astro.ucla.edu/~wright/cosmo_02.htm
> >
> > "Time dilation causes the proper time measured by an observer
> > to depend on the velocity of the observer..."
> >
> > "The Hubble law (v = HD) is true for all values of D, even very
> > large ones which give v > c."
> > "The distance in the Hubble law must be defined so that if A and B a
> > re two distant galaxies seen by us in the same direction, and
> > A and B are not too far from each other, then the difference
> > in distances from us, D(A)-D(B), is the distance A would measure
> > to B. But this measurement must be made "now" -- so A must measure
> > the distance to B at the same proper time since the Big Bang as
> > we see now. Thus to determine Dnow for a distant galaxy Z we
> > would find a chain of galaxies ABC...XYZ along the path to Z,
> > with each element of the chain close to its neighbors, and then
> > have each galaxy in the chain measure the distance to the next
> > galaxy at time to since the Big Bang.
> > The distance to Z, D(us to Z), is the sum of all these subintervals:
> > Dnow = D(us to Z) = D(us to A) + D(A to B) + ... D(X to Y) +
> > D(Y to Z)
> > And the velocity in the Hubble law is just the change of Dnow per
> > unit time."
>
> I know all that, at it is rather irrelevant to my argument - which
> you *keep* ignoring!
>
>
> > As v=HD, or D=v/H,
> >
> > v(now) = v(us to A) + v(A to B) + ... + v(X to Y) + v(Y to Z)
> >
> > According to SR, one should use the relativistic addition of
> > those "sub-velocities", and get a global velocity which is less
> > than c.
>
> How often do I need to tell you that the velocities in cosmology
> are only apparent and hence are not subject to the rules of SR?
>

Apparent velocities leading to apparent redshifts?

>
> > Then the claim "In particular, galaxies that are far enough away
> > from us necessarily have velocities greater than the speed of
> > light." is necessary wrong,
>
> No. What is wrong is your claim above that one has to use the velocity
> addition of SR.
>
>
> > as well as the following claims:
> >
> > "...the worldlines of the galaxies get flatter and giving velocities
> > v = dDnow/dt that are greater than c. But in special relativistic
> > coordinates the velocities are less than c."
>
> See? This says *obviously* that in cosmology, one does *not* use
> special relativistic coordinates and therefore also *not* the special
> relativistic velocity addition!
>

The cosmologists contradict themselves. Using Wright's demonstration,
I get v(now) = v(us to A) + v(A to B) + ... + v(X to Y) + v(Y to Z)
In this *analytical* case, one *has* to use the relativistic addition of velocities.
This is different from *globally* claiming that "in cosmology, one does
*not* use special relativistic coordinates and therefore also *not* the
special relativistic velocity addition!"
The *analytical* approach, used by Wright himself, demonstrates that
SR addition is pertinent, hence that galaxies that are far enough away
from us necessarily *cannot* have velocities greater than the speed of
light.
 
>
> > "But the Hubble law distance Dnow, which is measured now, of these
> > most distant galaxies is infinity (in this model). Furthermore,
> > this galaxy with infinite Hubble law distance and hence infinite
> > Hubble law velocity is visible to us, since in this model the
> > observable Universe is the entire Universe. The relationships
> > between the Hubble law distance and velocity (Dnow & v) and
> > the redshift z are given below:
> > v = HoDnow
> > Dnow = (c/Ho)ln(1+z)
> > 1+z = exp(v/c)"
> >
> > Notice "in this model",
>
> Yes, I noticed that. Why did you feel the need to quote something
> which is only true for one particular, rather unrealistic model?
>

If it is unrealistic, then why did Wright used it to demonstrate that
Dnow and galaxies's velocity can be infinite?

>
> > and also "Furthermore,
> > this galaxy with infinite Hubble law distance and hence infinite
> > Hubble law velocity is visible to us, since in this model the
> > observable Universe is the entire Universe."
> >
> > This wonderful conclusion suffices to falsify the whole reasoning.
>
> No! Why on earth do you think so?

Simply because nothing can be seen at infinite distances (unless, of
course, if the light speed is itself infinite).

>
>
> > "The predicted curve relating one distance indicator to another
> > depends on the cosmological model."
> > "Here are the technical formulae for these distances. The graphs
> > below show these distances vs. redshift for three models:
> > the critical density matter dominated Einstein - de Sitter model
> > (EdS), the empty model, and the accelerating Lambda-CDM model
> > (LCDM) that is the current concensus model."
> >
> > Three different models giving different results!
>
> Err, and what's your problem with that??? That obviously has to
> be expected!
>

The problem is that it is difficult to trust the results and conclusions
of theories using different models, especially the one which lead to
an accelerated expansion.

>
> > Which one will be the next?
>
> All evidence we have today points towards the LCDM.
>
>
>
> > 2) From http://www.astro.ucla.edu/~wright/tiredlit.htm
> >
> > "Tired light models invoke a gradual energy loss by photons as
> > they travel through the cosmos to produce the redshift-distance law."
> > This has three main problems:
> >
> > There is no known interaction that can degrade a photon's energy
> > without also changing its momentum, which leads to a blurring of
> > distant objects which is not observed. The Compton shift in
> > particular does not work."
> >
> > Not true, a cosmic negative acceleration cH doesn't change the
> > photon's momentum.
>
> And what would be the mechanism for this negative acceleration?
> *How* are the photons decelerated?
>

Assuming a stable (not expanding) universe with a cosmic negative
acceleration cHo,

A light ray of wavelength lambda is sent from a point P.
At a distance d from P, the energy loss of a photon of frequency Nu is

(hNu/c^2) * cH0 * d = hNu * (H0/c) * d,

where h is the Plank constant.

Hence, the residual energy hNu(o) of the photon at the distance d is

hNu(o) = hNu - hNu * (H0/c) * d = hNu (1 - H0*d/c), hence
Nu(o) = Nu (1 - (H0*d/c), and
lambda(o) = (1 - (H0*d/c) / lambda

Let's notice that H0*d corresponds to the recession velocity
v = H0*d assumed in an expanding universe.

Hence,

z = (lambda(o) - lambda) / lambda
  = (H0/c)d * (1+z), and

d = (c/H0) * z/(1+z)

Conclusively, the existence of a cosmic "deceleration" cH0
suffices to explain the so-called tiring of light in a stable universe.

> You ignored the word "interaction" above!
>

Not at all, the energy loss of the photon should be globally recycled
by the universe itself.

>
>
> > "The tired light model does not predict the observed time dilation
> > of high redshift supernova light curves. This time dilation is
> > a consequence of the standard interpretation of the redshift:
> > a supernova that takes 20 days to decay will appear to take 40
> > days to decay when observed at redshift z=1."
> >
> > Nobody knows what exactly happens on the SNe.
>
> We know enough for the measurements to be reliable.
>

Did you look at the error bars in
http://www.astro.ucla.edu/~wright/tiredlit.htm ?
The data spread is very big.
 
>
> > The apparent time dilation could be due to gravitational effects.
>
> Why would the observed time dilation then depend on the distance
> to the SNs?
>

If there is such time dilation.

>
>
> > "The tired light model can not produce a blackbody spectrum for
> > the Cosmic Microwave Background without some incredible coincidences.
> > The local Universe is transparent and has a wide range of
> > temperatures, so it does not produce a blackbody, which requires
> > an isothermal absorbing situation. So the CMB must have come from
> > a far away part of the Universe, and its photons will thus lose
> > energy by the tired light effect."
> >
> > From the many discussions about tired light and the CMB, that
> > took place on this or other NGs, one should be careful and not
> > automatically claim that the blackbody spectrum is not compatible
> > with a stable universe.
>
> I have so far seen no calculation in this newsgroup demonstrating
> that one could have a blackbody CMBR in a stable, static universe.
>
> Care to present such a calculation?
>

Calculations have been done long ago, which then gave better results
that those based on an expanding universe. But I didn't indeed see
recent calculations.
 
>
>
> > "The tired light model fails the Tolman surface brightness test.
> > This is essentially the same effect as the CMB prefactor test,
> > but applied to the surface brightness of galaxies instead of
> > to the emissivities of blackbodies".
> >
> > Same remark as just above.
>
> If you can explain this in a stable, static universe, feel free
> to present your calculations.
>
>
> Bye,
> Bjoern

Marcel Luttgens