Re: Classical wave theory and tired light

On 5 jan, 00:15, Timo Nieminen <t...@xxxxxxxxxxxxxxxxx> wrote:
On Sun, 4 Jan 2009 srp2...@xxxxxxxxx wrote:

On 4 jan, 18:57, Timo Nieminen <t...@xxxxxxxxxxxxxxxxx> wrote:
On Sun, 4 Jan 2009 srp2...@xxxxxxxxx wrote:

On 4 jan, 15:52, "Timo A. Nieminen" <t...@xxxxxxxxxxxxxxxxx> wrote:
On Sat, 3 Jan 2009, srp2...@xxxxxxxxx wrote:
On 3 jan, 06:01, "Timo A. Nieminen" <t...@xxxxxxxxxxxxxxxxx> wrote:
On Wed, 31 Dec 2008, srp2...@xxxxxxxxx wrote:
On 26 déc, 01:11, "Timo A. Nieminen" <t...@xxxxxxxxxxxxxxxxx> wrote:
On Sat, 13 Dec 2008, srp2...@xxxxxxxxx wrote:
Thirdly, my question was about _understanding_, not acceptance. The
Copenhagen interpretation is simple enough. It's understandable whether
one accepts it or not, and whether one accepts its basic premises or not.

Yes, very easy to understand because it does not require strict
logical
reasoning. You accept that localized trajectories for moving
electrons
can't exist as a fact, and then all becomes clear.

No, you _don't_ have to accept the assumptions as fact; you only have to
undersrtand them.

If you don't start off accepting the assumptions, why should that prevent
understanding of a theory (in general; perhaps your theory is a special
case)? Note that the understanding of a theory that successfully models
experimental/observed results can lead to acceptance of the assumptions;
the assumptions don't need to be accepted beforehand.

For example, Planck's quantisation hypothesis (and Einstein's photon
hypothesis).

The difference is that causality requires strict mastery of
formal logic, which has unfortunately been flushed down the
drain by the Copenhagen irrational drift.

That's the sad legacy of the Copenhagen school of thought:
No training in formal logics.

So you keep claiming, but you have provided no specific examples.
It's clear that you don't accept the assumptions of the
Copenhagen interpretation, but what does that have to do with the absence
of formal logic?

No logical training is required to accept the Copenhagen
interpretation.
It is the easy way out. No effort is required for simple belief.

Perhaps it wouldn't be a very useful example, since the Copenhagen
interpretation isn't a theory, doesn't claim to be a theory, and you say
you have no disagreement with the theory (i.e., quantum mechanics) that it
is an interpretation of.

But let me ask you this: do you understand the Copenhagen interpretation?

What does this say about the difference between accepting and
understanding?

It says accept that no least action trajectories can exist for moving
electrons, or else no degree.

No need to understand. Compute and shut up.

Very simple and very well known.

Anyway, why would sticking strictly to formal logic make a theory _harder_
to understand?

It is easy to understand for anyone having mastered formal logic.
Absolute requirement.

If your theory is solidly framed using formal logic, it
should be understandable even if your assumptions are
not fully accepted,

Not by anyone having no thorough training in formal logic.

or even accepted at all. One could construct, using formal logic, a theory
that assumes 2+2=5,

Wrong. Formal logic would immediately reveal the fraud.

2+2 makes 4. Period.

and if properly presented, the theory should be understandable.
Of course, it would be wrong, but it should be _understandable_.

Immediately proven false by applying formal logic.
Any time.

Understanding an error is not studying a theory.

The loss of energy of discrete photons in the model is linked to
change in direction. Straight line motion of photons involves no
possible loss of energy. Net change in direction, which occurs
whenever a localized photon's trajectory is deflected by gravity,
such as was proven by the 1919 Eddington et al. and many
other observations, gravitational lensing, and other assorted
observations, cannot possibly occur without some energy
being expended. 2nd principle of thermodynamics.

Do meteors lose energy when gravitationally deflected by planets or stars?

Depends if they are on stable closed orbits or on hyperbolic
trajectories.

If the first, then they energy is constantly replenished by the force
maintaining them on their stable orbit.

In the later case, they may lose or gain energy (velocity) as a
function
of their mass and the angle of deflection depending on the relative
direction of  motion of the body they deflect about.

For photons, since velocity is constant, and that they are always
on hyperbolic trajectories in the universe as they gravitationally
flyby, then they can only lose some energy at each flyby however
shallow.

Orbiting planets keep going, without falling into their stars, despite
continuous gravitational deflection. Why are photons so different?

Because orbiting planets are on closed and stable gravitational
orbits while photons are on hyperbolic trajectories.

Why are photons on hyberbolic trajectories so different from meteors on
hyperbolic trajectories?

First, photons are massless and have constant velocity. So their
velocity cannot possibly vary upon losing energy, thus automatic
redshifting, which is not the case for meteors.

Second, the velocity differential during photons flybys is so huge
with respect to the deflecting massive bodies that the relative
direction of motion of these deflecting masses amonts to an
infinitesimal factor relative to the energy involved in the net
change in direction.

Which would make the change in energy, and hence the redshift,
infinitesimal.

Yes. Definitely, but after millions such encounters, they all
add up to measurable amounts. the farther away they come from,
the more energy will have been lost. Not assuming this is the
only cause.

I wouldn't call it infinitesimal, since the relative speeds
we can expect vary from 10s of km/s to much, much higher.

But it doesn't answer the question - why can meteors on hyperbolic
trajectories either gain or lose energy, depending on the motion of the
deflecting body, while photon (so you claim) can only lose energy (whether
infinitesimally or otherwise)?

Simply because of the velocity differential between photon and
deflecting
body. Eventually, this will be studied. This is what the model
predicts.

A massive body, even one travelling at almost c, so that the velocity
differential is approximately the same, behaves very differently.

Really ? To my knowledge, there is no such body in existence.

If such a body existed, we would have detected it long ago
since its mass would constitute a sizeable fraction of the
mass of the existing universe.

We do detect objects, with mass (i.e., massive compared to massless
photons), with such speeds. E.g., cosmic rays. There doesn't appear to be
any need for their mass to consitute a sizeable fraction of the mass of
the existing universe. (If by mass, as you use it here, you mean
"relativistic mass", then photons have mass. If you mean proper mass, then
speed has effect and is irrelevant.)

I wonder what you point is exactly. the 3-spaces model shows that
photons can only be repeatedly deflected by intervening galaxies on
their
way to us. The overall effect is that they lose energy through net
changes in direction.

You would have to understand the model to see this.


So the velocity differential is clearly not the cause, since if it was
the cause, it would produce the same result in both cases.

From what I observe, it simply cannot exist.

So the question remains - why would photons behave so differently from
massive particles?

I already gave my answer to this. Same answer.

No, you repeated your assertion. Why would photons behave so differently?
You claim a qualitative difference, not just a quantitative difference.

Ok. Photons are massless and move at c. Since they cannot slow down,
any loss of energy they suffer as a result of net changes in direction
mandated
by the 2nd law of thermodynamics can only cause them to progressively
redshift. The farther away from us they have been emitted, the more
red
shifted they will appear to us. Not assuming that other additional
causes
might not be at play.

For that matter, why would photons always lose energy, when a classical
electromagnetic beam can either gain or lose energy, depending on the
motion of the deflecting body?

Not so. Even when considering EM energy as beams, if the beam
suffers a net change in direction, which becomes effective and
measurable after the beam has left the vicinity of the deflecting
body energy will be lost. ref gravitational lensing.

That's just wrong. The only changes in energy of an EM beam on deflection
in this case result from Doppler shifts due to motion of the deflecting
body. This follows from (i) classical electrodynamics/optics, (ii) QED,
which you claim is left untouched by your theory, and (iii) experimental
measurements, to within experimental error limits.

You said youself that any single encounter would amount to
infinitesimal difference.

You said yourself that the velocities of the deflecting bodies are
infinitesimal,

...with respect to the velocity of photons.

which implies the infinitesimal difference. I disagreed.

So I see.

We currently have no means to verify
the change for a single encounter since it is so small, particularly
since the theory has not yet been put to test.

It's easily tested for non-gravitational deflection, and is done so
regularly. Off-the-shelf practical technology.

What I observe is that the farther away photons have been emitted
the more red shifted they are.

The 3-spaces model has a clean explanation for this that
I already gave you.

Why would gravitational deflection be so very different from other
technology?

Uncompensate deflection (hyperbolic) can only be different
from stable orbiting body whose energy expenditure through
permanent deflection about the deflecting body is
permanently compensated by gravity induced energy as a
function of distace on a closed orbit.

I'm not asking about the difference between energy loss/gain on hyperbolic
trajectories vs elliptical orbits; I'm asking about your claimed
difference between energy loss/gain of photons vs any object with non-zero
rest mass on a hyperbolic trajectory.

Massive bodies on hyperbolic trajectories are likely to either
increase their velocity or diminish it depending on the relative
direction of motion of the deflecting body.

Already mentioned.

You'd need to understand the model to see this.

2nd law of thermodynamics has nothing to do with this.

Yes. There is no way to cause anything to effect a net change
in direction without some energy being expended.

What does the 2nd law of thermodynamics have to do with
direction of motion? It's about moving energy around, temperature,
and entropy.

The 2nd law of thermodynamics mandates that no change in
direction of moving bodies can occur without energy being
expended to account for that change.

dS >= dQ/T doesn't tell us anything of the sort.

Newtonian mechanics also tells us otherwise. As do relativistic mechanics
and quantum mechanics.

Total disagreement.

To quote a compact form of the 2nd law: dS >= dQ/T. Care to explain how
this (or another correct statement of the 2nd law) leads to your
conclusion?

So, no answer here. Is the answer in your book?

Yes.

What we can get from thermodynamics is that the deflecting encounters
between an ensemble of photons and an ensemble of gravitating bodies will
tend to bring them into thermal equilibrium with each other. That is, the
mean energy of the photons will approach the mean KE of the gravitating
bodies.

Given typical photon energies and typical KE of, e.g., stars, would this
tend to lead to redshift or blueshift?

Indeed, when a photon closes in on the inwards flyby trajectory it
will
of course blueshift until it reaches the closest point to the
deflecting
body, and on the outbound leg, it will have lost all of that energy
in
such a way that when it is at a distance from the deflecting body
where the deflection becomes negligible equal to the corresponding
distance it on the inbound leg when deflection began to be felt.

BUT, more energy is required to account for the net change
in direction that the photon will have suffered. That energy
can only come from its own internal complement.

That's what the model shows.

You would have to understand the model to see this.

You _claim_ that more energy is required to account for the net change in
direction. Does your model _show_ this, or does your model _assume_ this?

It shows it, and explains it.

Ok, let's summarize where we are at.

1) Understanding that elementary em particles can only travel
as localized quanta on least action trajectories is a mandatory
axiom of the model.

You say this only an unproven possibility. Just one of
thousands other possible explanations.

Firstly, I don't say it's just one of thousands (of) other possible
explanations.

Secondly, while it would be reasonable that the assumption that elementary
EM particles can only travel as localised quanta on least action
trajectories is an axiom of your model, how can the _understanding_ of
that assumption be an axiom of your model?

Understanding the idea of least action trajectories for permanently
localized photons it is not sufficient. Coming to the conclusion that
it is the only logical possibility is the requirement. Only causalists
can reach it.

How can the _reader_ of your description of your model possibly
affect your model?

They can't. But only causalist will be able to follow the logical
sequence and eventually understand how its description of
fundamental reality accounts for all observed data, including
the set currently considered unexplained, such as the
so-called anomalous acceleration of Pioneer 10 and 11 and
their so-called anomalous axial spin slowdown, and how
a new unsuspected source of practically unextinguishable
energy is now within our reach.

Thirdly, given that billiard-ball particles can be described as "localised
quanta travelling on least action trajectories", I'm interested in seeing
a more precise definition of what you mean by the above.

No billiard-ball like particles can exist in physical reality, so
the 3-spaces model doesn't describe any. It describes
permanently localized EM elementary particles moving on
least action trajectories and the consequences thereof.

2) The model shows that net change in direction of photons
mandatorily involves expenditure of energy, which is mandated
by the 2nd principle of thermo, and clearly confirmed by
the model.

You say that no such expenditure can exist, and that the
2nd principle of thermo doesn't apply.

Fine with me.

You have repeatedly asserted this about the 2nd law of thermodynamics,
without any support. Can you even give a statement of the 2nd law that has
any relevance to the change of direction of photons?

The second principle of thermodynamics says that in an
isolated system, a spontaneous change of state cannot
be produced, for any physical event possessing a certain
level of energy, unless circumstances allow this event to
reach a lower state of energy.

Ref Poincare.

Let us see some formal logic connecting dS >= dQ/T to the loss in energy
of photons on deflection, in support of your claim.

Sorry. I am not going to try summarizing the 3-spaces
model here.

Now, how are you going to be able to give any credit
or even follow any logical elaboration on this stuff that
you consider false to start with ?

If you clearly state your assumptions, write clearly, and
develop your theory/model logically, why should it be hard
to follow?

You apparently do not even accept the basic tokens
already presented. Why should I waste any time laying
more token that are certain to meet your disapproval since
they are based on those already presented and rejected ?

I am not looking for any approval on you part nor on the
part of any Copenhagen afficionado.

You think my model is not worth anything ? fine.

Others may find otherwise. Not a worry of mine.

Why would acceptance or non-acceptance of your initial
assumptions affect the understandability of your theory?

Already answered at the beginning of this post.


--
Timo Nieminen - Home page:http://www.physics.uq.edu.au/people/nieminen/
E-prints:http://espace.uq.edu.au/list/author_id/1189/
Shrine to Spirits:http://www.users.bigpond.com/timo_nieminen/spirits.html

André Michaud
.

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