In the beginning

[Michael Helland]

Contrary to popular belief, the big bang never actually happened. The
big bang is a hypothetical consequence of an expanding universe. And
for as long as I remember, I had just assumed it actually happened.

Now I have some serious doubts.

We think the big bang happened because we think the universe is
expanding. And we think the universe is expanding because we think all
the universe's galaxies are receding from ours.

We think the galaxies have actual recessional velocity proportional to
their distance.

Why exactly do we think that?

For starters, we observe redshifts in the light coming from those
galaxies.

The easiest way to explain the observation of redshifts is by assuming
that it is a Doppler effect. If the Doppler effect is causing the
redshift, then the galaxies must be moving away from us. If the
galaxies are moving away from us they must all be moving away from
each other, space must be expanding, and at some point in history,
there was a big bang.

But is that what's really going on?

The redshifts in the galaxies' light were discovered in the late
1920's, and earlier in that same decade we didn't even know that there
were galaxies outside of our own. We didn't even know what a galaxy
was.

So let's consider another possibility. Let's say, hypothetically, the
observed redshift isn't caused by the motion of the galaxy. In fact,
let's say the cause of the light's redshift has nothing to do with the
galaxy at all. The reason for light's redshift is light itself.
In the short amount of time between discovering that there is more
than one galaxy in the universe, and discovering the redshift from
distant galaxies, we did so under the assumption that since the laws
of physics governing light were the same whether you were dealing with
distances of 1 foot, or 1 mile, or to the moon, or to Neptune, or to
the nearest stars or the end of the galaxy or to the next galaxy over,
that light would behave the same way indefinitely. We assumed that
life travels forever, and has no limits to the laws that work in our
solar system and laboratories.

However, the evidence is that the farther and farther you look, that
is apparently not the case. Light essentially dies out and then
everything goes black.

It might be a good idea to ask:

What would the universe look like if light had a finite range?

At some point out in the cosmos, light would stop coming to us, which
observationally seems to be the case.)

And, unless it came to an abrupt stop, we would see effects of it
dieing out, which we do as redshifts.

Now how about making future predictions that can be validated by
observation? We’ll discover there are mature galaxies in the young
universe that defy theories on how galaxies form in the time allowed
by the big bang.

That too has been observed, and if light has a finite range, it will
continue to do so.

Beyond observing old galaxies in a young universe, the big bang has
other considerable problems explaining the evidence. Most of the
expansion of the universe had to occur super fast early on in a period
called “inflation”, otherwise the theory simply doesn’t fit the
evidence. Is dark energy real? Would we even need it in a universe
where light has a finite range?

The man who discovered the redshift never himself accepted that the
apparent recessional velocity of distant galaxies was due to an
expanding universe.

He believed it was an indication of a new principle of nature.

"… if redshift are not primarily due to velocity shift … the velocity-
distance relation is linear, the distribution of the nebula is
uniform, there is no evidence of expansion, no trace of curvature, no
restriction of the time scale … and we find ourselves in the presence
of one of the principle of nature that is still unknown to us today …
whereas, if redshifts are velocity shifts which measure the rate of
expansion, the expanding models are definitely inconsistent with the
observations that have been made … expanding models are a forced
interpretation of the observational results"
— E. Hubble, Ap. J., 84, 517, 1936

"[If the redshifts are a Doppler shift] … the observations as they
stand lead to the anomaly of a closed universe, curiously small and
dense, and, it may be added, suspiciously young. On the other hand, if
redshifts are not Doppler effects, these anomalies disappear and the
region observed appears as a small, homogeneous, but insignificant
portion of a universe extended indefinitely both in space and time."
— E. Hubble, Monthly Notices of the Royal Astronomical Society, 97,
513, 1937

I propose that the new principle of nature is the finite range of
light. Light doesn’t travel indefinitely.

One possible mathematical description of this principle I’ve
considered can be found by changing Hubble’s relationship between
apparent recessional velocity and distance, to the speed of light and
the distance to its source.

the speed of light in a vacuum = c - H0 * distance it has traveled
v = c - H0 * d

Because the speed of a wave v = frequency * wavelength, if over
cosmological distances light’s speed diminishes, either the frequency
or the wavelength has got to give. And it is an observed fact that
over cosmological distances, the frequency of light gives.
With this specific hypothesis, the range of light is Hubble’s Limit.

Back in Hubble’s day, the active exploration of ideas about the laws
of physics for the cosmos at large was interrupted by some world wars
and looking at the universe on a very different scale: inside an atom.

[Androcles]

"Michael Helland" <moby...@gmail.com> wrote in message
news:1377cb62-561d-4511...@s35g2000pra.googlegroups.com...

===============================================
Congratulations, you've begun to question dogma.

[gaby de wilde]

On Nov 20, 11:07 am, "Androcles" <Headmas...@Hogwarts.physics.November.
2011> wrote:
> "Michael Helland" <mobyd...@gmail.com> wrote in message

Welcome back on solid ground.

We hope you had a pleasant flight.

[Androcles]

"gaby de wilde" <gdew...@gmail.com> wrote in message
news:6d1ca3f3-d849-43c7...@y7g2000vbe.googlegroups.com...

================================================
The inverse square law 1/r^2 doesn't work if there is a lens or curved
mirror involved and the energy is channelled into a beam.
http://www.youtube.com/watch?v=680xwwEs7Lg

[7]

Certain types of supernova explode with a known brightness profile.
It means no matter what distance they are at, they will have
the same brightening and dimming profile. Hence it is certain to be
the same type supernova and since the peak brightness is the same
in all known supernova of this type, if the total brightness is low,
then it must be far away. The red shifts and brightness variations
are in agreement with each other. There is nothing else that
is needed to slipped in to make for allowances.
The red shifts are true reflections of absolute distances.

[richard....@comcast.net]

Perhaps redshift is evidence of the total energy concept. Perhaps
exposing certain resonant molecules to a high energy photon will
trigger energy redistribution (phase change or rise in temperature)
w/o changing photon direction. If so, redshift would be a quantized
function of molecular species.

[Sam Wormley]

On 11/20/11 2:20 AM, Michael Helland wrote:
> Contrary to popular belief, the big bang never actually happened. The
> big bang is a hypothetical consequence of an expanding universe. And
> for as long as I remember, I had just assumed it actually happened.
>
> Now I have some serious doubts.

WMAP: Foundations of the Big Bang theory
http://map.gsfc.nasa.gov/m_uni.html

WMAP: Tests of Big Bang Cosmology
http://map.gsfc.nasa.gov/m_uni/uni_101bbtest.html

No Center
http://www.astro.ucla.edu/~wright/nocenter.html
http://www.astro.ucla.edu/~wright/infpoint.html

Also see Ned Wright's Cosmology Tutorial
http://www.astro.ucla.edu/~wright/cosmolog.htm
http://www.astro.ucla.edu/~wright/cosmology_faq.html
http://www.astro.ucla.edu/~wright/CosmoCalc.html

[Michael Helland]

Is this to say that it is not allowed to have doubts about the big
bang?

[Sam Wormley]

We measure redshifts of moving lights in the laboratory, too.

>
> The easiest way to explain the observation of redshifts is by assuming
> that it is a Doppler effect. If the Doppler effect is causing the
> redshift, then the galaxies must be moving away from us. If the
> galaxies are moving away from us they must all be moving away from
> each other, space must be expanding, and at some point in history,
> there was a big bang.
>
> But is that what's really going on?
>
> The redshifts in the galaxies' light were discovered in the late
> 1920's, and earlier in that same decade we didn't even know that there
> were galaxies outside of our own. We didn't even know what a galaxy was.
>
> So let's consider another possibility. Let's say, hypothetically, the
> observed redshift isn't caused by the motion of the galaxy. In fact,
> let's say the cause of the light's redshift has nothing to do with
> the galaxy at all. The reason for light's redshift is light itself.
> In the short amount of time between discovering that there is more
> than one galaxy in the universe, and discovering the redshift from
> distant galaxies, we did so under the assumption that since the laws
> of physics governing light were the same whether you were dealing with
> distances of 1 foot, or 1 mile, or to the moon, or to Neptune, or to
> the nearest stars or the end of the galaxy or to the next galaxy over,
> that light would behave the same way indefinitely. We assumed that
> life travels forever, and has no limits to the laws that work in our
> solar system and laboratories.

Tired light -- As cosmological measurements became more precise and
the statistics in cosmological data sets improved, tired light
proposals ended up being falsified, and are now all-but absent from
current mainstream literature.

Tired Light is Still Dead
http://www.astro.ucla.edu/~wright/cosmolog.htm#News

24 Apr 2008 - Blondin et al. (2008) studied distant supernovae using
spectra to judge the age of the object during each observation. They
found an aging rate that varied with redshift z like

1/(1+z)(0.97 +/- 0.10),

compatible with the expected 1/(1+z) for expanding Universes, but 9.7
standard deviations away from the constant aging rate expected in the
tired light model.

[Michael Helland]

Two words that could mean many different things.

However, none of them change Hubble’s relationship between

apparent recessional velocity and distance, to the speed of light and
the distance to its source.

the speed of light in a vacuum = c - H0 * distance it has traveled
v = c - H0 * d

Because the speed of a wave v = frequency * wavelength, if over
cosmological distances light’s speed diminishes, either the frequency
or the wavelength has got to give. And it is an observed fact that
over cosmological distances, the frequency of light gives.

With this specific hypothesis, the range of light is Hubble’s Limit.

Tired Light is many different things, but it isn't this.

[Sam Wormley]

On 11/23/11 9:21 PM, Michael Helland wrote:
> And it is an observed fact that
> over cosmological distances, the frequency of light gives.

As does the wavelength with the speed of light remaining
constant.

[Michael Helland]

That's not true.

Wavelength is a property we deduce after measurement, after the photon
has been absorbed and re-emmited, thus using my v = c - H_0 * d,
results in v = c, which is what we observe.

Just because you don't understand anything outside of what you've been
trained doesn't mean my liberal interpretation of Hubble's Law is
wrong.

[Sam Wormley]

On 11/24/11 4:49 AM, Michael Helland wrote:
> Wavelength is a property we deduce after measurement

Astronomers measure wavelength directly.

"In the early part of the twentieth century, Slipher, Hubble and others
made the first measurements of the redshifts and blueshifts of galaxies
beyond the Milky Way. They initially interpreted these redshifts and
blueshifts as due solely to the Doppler effect, but later Hubble
discovered a rough correlation between the increasing redshifts and the
increasing distance of galaxies. Theorists almost immediately realized
that these observations could be explained by a different mechanism for
producing redshifts. Hubble's law of the correlation between redshifts
and distances is required by models of cosmology derived from general
relativity that have a metric expansion of space. As a result, photons
propagating through the expanding space are stretched, creating the
cosmological redshift".
--Ref: http://en.wikipedia.org/wiki/Redshift

[PD]

On 11/24/2011 4:49 AM, Michael Helland wrote:
> On Nov 23, 7:37 pm, Sam Wormley<sworml...@gmail.com> wrote:
>> On 11/23/11 9:21 PM, Michael Helland wrote:
>>
>>> And it is an observed fact that
>>> over cosmological distances, the frequency of light gives.
>>
>> As does the wavelength with the speed of light remaining
>> constant.
>
>
> That's not true.
>
> Wavelength is a property we deduce after measurement, after the photon
> has been absorbed and re-emmited, thus using my v = c - H_0 * d,
> results in v = c, which is what we observe.

Sorry, but that's just not so, Michael. The wavelength of light is
easily and directly measurable.

[Michael Helland]

On Nov 24, 8:47 am, PD <thedraperfam...@gmail.com> wrote:
> On 11/24/2011 4:49 AM, Michael Helland wrote:
>
> > On Nov 23, 7:37 pm, Sam Wormley<sworml...@gmail.com>  wrote:
> >> On 11/23/11 9:21 PM, Michael Helland wrote:
>
> >>> And it is an observed fact that
> >>> over cosmological distances, the frequency of light gives.
>
> >>     As does the wavelength with the speed of light remaining
> >>     constant.
>
> > That's not true.
>
> > Wavelength is a property we deduce after measurement, after the photon
> > has been absorbed and re-emmited, thus using my v = c - H_0 * d,
> > results in v = c, which is what we observe.
>
> Sorry, but that's just not so, Michael. The wavelength of light is
> easily and directly measurable.

First, a question: Doesn't only quantum energy matter in quantum
theory?

Second, when we measure it, it is after we've focused it in a lens,
after we've reflected it off a mirror, after it's gone through the
atmosphere and the IPM and the ISM.

It has been absorbed and re-emitted along the way, and thus the
distance it has been traveling * H_0 is effectively zero.

Thus its speed is c, and the frequency and wavelength are what we
observe.

I'm saying the light has velocities less than c (c - H_0 * d to be
precise) in the intergalactic voids.



This explains other observations, such as why there are mature
galaxies in the young universe.

Because the universe is indefinitely older than 13.7 billion years.

[Glyd]

I saw tv last night how colliding branes can produce same expanding
space, etc. like big bang without big bang.. interesting... so it's
not definite that once upon a time the universe with billions and
billions of galaxies were once the size of a hydrogen atom....

[PD]

On 11/25/2011 3:33 AM, Michael Helland wrote:
> On Nov 24, 8:47 am, PD<thedraperfam...@gmail.com> wrote:
>> On 11/24/2011 4:49 AM, Michael Helland wrote:
>>
>>> On Nov 23, 7:37 pm, Sam Wormley<sworml...@gmail.com> wrote:
>>>> On 11/23/11 9:21 PM, Michael Helland wrote:
>>
>>>>> And it is an observed fact that
>>>>> over cosmological distances, the frequency of light gives.
>>
>>>> As does the wavelength with the speed of light remaining
>>>> constant.
>>
>>> That's not true.
>>
>>> Wavelength is a property we deduce after measurement, after the photon
>>> has been absorbed and re-emmited, thus using my v = c - H_0 * d,
>>> results in v = c, which is what we observe.
>>
>> Sorry, but that's just not so, Michael. The wavelength of light is
>> easily and directly measurable.
>
>
> First, a question: Doesn't only quantum energy matter in quantum
> theory?

No.

>
> Second, when we measure it, it is after we've focused it in a lens,
> after we've reflected it off a mirror, after it's gone through the
> atmosphere and the IPM and the ISM.

Not so. You can measure light from a desktop source through a vacuum
pipe or through a short span of air without any intervening lens or
mirror, directly with a diffraction grating. In fact, this is routinely
done in teaching labs every year at colleges across the country.

We KNOW what we're measuring.

[richard....@comcast.net]

On Fri, 25 Nov 2011 01:48:54 -0800 (PST), Glyd <glyd...@yahoo.com>
wrote:

Since the mass of an electron has zero diameter, the mass part of
hydrogen has zero diameter. The non-mass part of hydrogen has greater
than zero diameter.

Because mass itself has zero diameter, I suppose that you could
confined a unverse worth of electrons (or the mass of electrons) into
a volume equal to a hydrogen atom's volume. Would the mass be stable?
Only energy with zero volume (like an electron) could exist in the
volume.

Note: The volume could not be superconductive, because the vacuum
between electrons would be a perfect insulator.

[Michael Helland]

The fact that people take the kind of hocus pocus you mentioned
seriously, and my suggestion that Hubble's Limit is the range of light
is mocked, pretty much tells the story of what kind of crap scientists
can be trained to believe.

[Michael Helland]

And what distance is being used?

If they aren't in the millions of light year range, then according to:

v = c - H_0 * d

H_0 * d = 0 (for non-intergalactic distances)

and

v = c

My hypothesis is identical with established theory at distances where
cosmological redshift is not observed.

[Michael Helland]

On Nov 25, 7:30 am, PD <thedraperfam...@gmail.com> wrote:
> On 11/25/2011 3:33 AM, Michael Helland wrote:
>
>
>
>
>
>
>
>
>
> > On Nov 24, 8:47 am, PD<thedraperfam...@gmail.com>  wrote:
> >> On 11/24/2011 4:49 AM, Michael Helland wrote:
>
> >>> On Nov 23, 7:37 pm, Sam Wormley<sworml...@gmail.com>    wrote:
> >>>> On 11/23/11 9:21 PM, Michael Helland wrote:
>
> >>>>> And it is an observed fact that
> >>>>> over cosmological distances, the frequency of light gives.
>
> >>>>      As does the wavelength with the speed of light remaining
> >>>>      constant.
>
> >>> That's not true.
>
> >>> Wavelength is a property we deduce after measurement, after the photon
> >>> has been absorbed and re-emmited, thus using my v = c - H_0 * d,
> >>> results in v = c, which is what we observe.
>
> >> Sorry, but that's just not so, Michael. The wavelength of light is
> >> easily and directly measurable.
>
> > First, a question: Doesn't only quantum energy matter in quantum
> > theory?
>
> No.
>
>
>
> > Second, when we measure it, it is after we've focused it in a lens,
> > after we've reflected it off a mirror, after it's gone through the
> > atmosphere and the IPM and the ISM.
>
> Not so. You can measure light from a desktop source

I was talking about light that has traveled millions of light years,
such that H_0 * d (the distance it has traveled) is > 0.

You're talking about light that has traveled 10 feet, such that H_0 *
d = 0.

When H_0 * d = 0, then according to v = c - H_0 * d, v = c.

There is no issue discrepancy between my hypothesis and established
theories at these distances.

[PD]

That's ok. We have bridge cases. That is, we have near objects in the
range of a few tens of meters to several tens of millions of miles where
we can test the very same thing.

If there were an established gap where we might say, "OK, but up in this
range something COMPLETELY different is going on," then you might have a
case. But there is not observational gap.

[PD]

In the freshman labs, perhaps a few meters.
In graduate exercises, it ranges from a few meters to a few hundred
thousand meters.
Then there are deliberate bridge measurements, ranging from a few
hundred thousand meters to several tens of millions of meters, using
astronomical instruments both on the ground and above the atmosphere.

There's no gap.

[Michael Helland]

If you plug several tens of millions of miles into my hypothesis:

v = c - H_0 * d

H_0 * 99 million miles is still 0 .. and ..

v = c

My hypothesis is consistent with established theory at those
distances.

[Michael Helland]

On Nov 25, 3:54 pm, PD <thedraperfam...@gmail.com> wrote:

> That's ok. We have bridge cases. That is, we have near objects in the
> range of a few tens of meters to several tens of millions of miles where
> we can test the very same thing.
>
> If there were an established gap where we might say, "OK, but up in this
> range something COMPLETELY different is going on," then you might have a
> case.

You are 100% correct.

And as empirical fact would have it, there is a range of distances
where something COMPLETELY different starts to happen: Hubble
redshift.


You explain it as actual recessional velocity of distant galaxies, and
from there extrapolate the big bang.

I explain it as light having a finite range.

[PD]

On 11/26/2011 5:46 AM, Michael Helland wrote:
> On Nov 25, 3:54 pm, PD<thedraperfam...@gmail.com> wrote:
>
>> That's ok. We have bridge cases. That is, we have near objects in the
>> range of a few tens of meters to several tens of millions of miles where
>> we can test the very same thing.
>>
>> If there were an established gap where we might say, "OK, but up in this
>> range something COMPLETELY different is going on," then you might have a
>> case.
>
>
> You are 100% correct.
>
> And as empirical fact would have it, there is a range of distances
> where something COMPLETELY different starts to happen: Hubble
> redshift.

No, see the problem is that WITHIN the Hubble range, we already have
independent confirmation that the SAME thing is going on as is going on
in desktop systems. That is, we can MEASURE the recessional velocity of
certain objects in the Hubble range with an *independent* means that has
nothing to do with their wavelength shift. You may want to do a small
Google search on astronomical methods for measuring velocities of
bodies. So for a good chunk of the Hubble data, we KNOW that it is
recessional velocity that is responsible for the wavelength shift, and
it corresponds EXACTLY to what is found in desktop systems.

[PD]

That's fine, Michael. You have a hypothesis that is consistent with data
in a certain narrow data range.

The prevailing model, however, is validated by data over a range ranging
from a meter or so to billions and billions of meters.

There is not normally good reason to take seriously an alternative model
that only applies to a narrow range of data, when the prevailing model
works over a much wider range of data.

[Michael Helland]

No.

I have a hypothesis which is identical to established theory where
Hubble redshift is not detected.

I have a hypothesis that is consistent with the observable data
everywhere.

Including why there are mature galaxies in the young universe:

<quote>
Until now, astronomers have been nearly blind when looking back in
time to survey an era when most stars in the Universe were expected to
have formed. This critical cosmological blind-spot has been removed by
a team using the Frederick C. Gillett Gemini North Telescope, showing
that many galaxies in the young Universe are not behaving as expected
some 8-11 billion years ago.
The surprise: these galaxies appear to be more fully formed and mature
than expected at this early stage in the evolution of the Universe.
</quote>

http://www.gemini.edu/node/74

> The prevailing model, however, is validated by data over a range ranging
> from a meter or so to billions and billions of meters.
>
> There is not normally good reason to take seriously an alternative model
> that only applies to a narrow range of data, when the prevailing model
> works over a much wider range of data.

That's right, but you've misinterpreted what I said.

My hypothesis works over all the data.

[Michael Helland]

On Nov 26, 11:00 am, PD <thedraperfam...@gmail.com> wrote:

> On 11/26/2011 5:46 AM, Michael Helland wrote:
>
>
>
>
>
>
>
>
>
> > On Nov 25, 3:54 pm, PD<thedraperfam...@gmail.com>  wrote:
>
> >> That's ok. We have bridge cases. That is, we have near objects in the
> >> range of a few tens of meters to several tens of millions of miles where
> >> we can test the very same thing.
>
> >> If there were an established gap where we might say, "OK, but up in this
> >> range something COMPLETELY different is going on," then you might have a
> >> case.
>
> > You are 100% correct.
>
> > And as empirical fact would have it, there is a range of distances
> > where something COMPLETELY different starts to happen: Hubble
> > redshift.
>
> No, see the problem is that WITHIN the Hubble range, we already have
> independent confirmation that the SAME thing is going on as is going on
> in desktop systems. That is, we can MEASURE the recessional velocity of
> certain objects in the Hubble range with an *independent* means that has
> nothing to do with their wavelength shift. You may want to do a small
> Google search on astronomical methods for measuring velocities of
> bodies. So for a good chunk of the Hubble data, we KNOW that it is
> recessional velocity that is responsible for the wavelength shift, and
> it corresponds EXACTLY to what is found in desktop systems.

Sure.

I'm not saying the Doppler shift isn't real.

I'm saying the cosmological redshift, the redshift we observe after
H_0 * d begins to be noticeable, which is predicted by the hypothesis
v = c - H_0 * d, I'm saying that specific redshift is not caused by
the Doppler effect.


It's caused by light's behavior in the large intergalactic voids, as
stated by v = c - H_0 * d

Where H_0 * d is zero, then v = c - 0

v = c

[G=EMC^2]

> >>>>> wrong.- Hide quoted text -
>
> - Show quoted text -- Hide quoted text -
>
> - Show quoted text -

Bang was a dirty word(makes a virgin blush) "Big Event much better.Big
Birth is also better than bang. Get the picture TreBert

[PD]

But it is. This is the part you don't seem to recognize.
We can put a source on a desktop and move it with a *measurable* speed
v. We can then measure the wavelength. The amount of wavelength shift is
in *precise* accord with the predictions of Doppler shift for that
velocity v.

This is validated *again* with sources hundreds of meters away, again by
direct and independent measurement of both v and wavelength.

This is validated *again* with sources tens of thousands of meters away,
with the same result.

This is validated *again* with sources of a few million meter away, with
the same result.

[richard....@comcast.net]

On Sun, 20 Nov 2011 00:20:31 -0800 (PST), Michael Helland
<moby...@gmail.com> wrote:

>
>
>Contrary to popular belief, the big bang never actually happened. The
>big bang is a hypothetical consequence of an expanding universe. And
>for as long as I remember, I had just assumed it actually happened.
>
>Now I have some serious doubts.
>
>We think the big bang happened because we think the universe is
>expanding. And we think the universe is expanding because we think all
>the universe's galaxies are receding from ours.

Doppler redshift should never have made sense to anybody. Emitted
energy - redshifted photon energy = energy imbalance.

>We think the galaxies have actual recessional velocity proportional to
>their distance.

Uniform initial big bang TEMPERATURE and composition dictate UNIFORM
velocity.

Assuming Doppler redshift does not exist, redshift would indicate
interaction between light and the media light travels through.

>However, the evidence is that the farther and farther you look, that
>is apparently not the case. Light essentially dies out and then
>everything goes black.

A low density of light emitters among light absorbers could result in
a black background. Virtually no light penetrates beyond a certain
depth in the ocean.

>It might be a good idea to ask:
>
>What would the universe look like if light had a finite range?
>
>At some point out in the cosmos, light would stop coming to us, which
>observationally seems to be the case.)
>
>And, unless it came to an abrupt stop, we would see effects of it
>dieing out, which we do as redshifts.

You are claiming light dies because of redshift? Imagine a universe
with no light absorbers. Every emitted photon will travel to
infinity. Given an infinite universe, the sky would be full of
redshifted light, instead of 'black'.

>Now how about making future predictions that can be validated by
>observation? We’ll discover there are mature galaxies in the young
>universe that defy theories on how galaxies form in the time allowed
>by the big bang.

Not just time, density. Collapse of scattered material into a star
(or galaxy) requires increasing density, not the decrease in density
inherent to an explosion.

>That too has been observed, and if light has a finite range, it will
>continue to do so.

Are you trying to say that if the universe is not expanding, young
stars will continue forming?

A fundamental flaw exists with models that use hydrogen as a star's
primary source of energy. Hydrogen has the most energy per atom. The
quantity of hydrogen consumed in a star will not equal the quantity of
hydrogen emitted, w/o replacing the energy.

In other words, during a star's life cycle, more reactions emit energy
than absorb energy. Each generation of stars reduces total energy.
Eventually, energy exhaustion must occur.

>Beyond observing old galaxies in a young universe, the big bang has
>other considerable problems explaining the evidence. Most of the
>expansion of the universe had to occur super fast early on in a period
>called “inflation”, otherwise the theory simply doesn’t fit the
>evidence. Is dark energy real? Would we even need it in a universe
>where light has a finite range?

Are you saying that universe expansion at greater than light speed
during 'inflation' is a requirement for a big bang?

>The man who discovered the redshift never himself accepted that the
>apparent recessional velocity of distant galaxies was due to an
>expanding universe.
>
>He believed it was an indication of a new principle of nature.
>
>"… if redshift are not primarily due to velocity shift … the velocity-
>distance relation is linear, the distribution of the nebula is
>uniform, there is no evidence of expansion, no trace of curvature, no
>restriction of the time scale … and we find ourselves in the presence
>of one of the principle of nature that is still unknown to us today …
>whereas, if redshifts are velocity shifts which measure the rate of
>expansion, the expanding models are definitely inconsistent with the
>observations that have been made … expanding models are a forced
>interpretation of the observational results"
>— E. Hubble, Ap. J., 84, 517, 1936
>
>"[If the redshifts are a Doppler shift] … the observations as they
>stand lead to the anomaly of a closed universe, curiously small and
>dense, and, it may be added, suspiciously young. On the other hand, if
>redshifts are not Doppler effects, these anomalies disappear and the
>region observed appears as a small, homogeneous, but insignificant
>portion of a universe extended indefinitely both in space and time."
>— E. Hubble, Monthly Notices of the Royal Astronomical Society, 97,
>513, 1937

The spectacular gets more money, press and attention.

>I propose that the new principle of nature is the finite range of
>light. Light doesn’t travel indefinitely.

Within a pure vacuum, removal of light energy during travel will
violate the conservation of energy law. All transparent molecules
within interstellar media reduce light velocity. Therefore, molecules
are candidates for reducing light energy.

>One possible mathematical description of this principle I’ve
>considered can be found by changing Hubble’s relationship between
>apparent recessional velocity and distance, to the speed of light and
>the distance to its source.
>
>the speed of light in a vacuum = c - H0 * distance it has traveled
>v = c - H0 * d
>
>Because the speed of a wave v = frequency * wavelength, if over
>cosmological distances light’s speed diminishes, either the frequency

>or the wavelength has got to give. And it is an observed fact that

>over cosmological distances, the frequency of light gives.

>With this specific hypothesis, the range of light is Hubble’s Limit.

In theory, on Earth, within transparent material, light slows, w/o
changing wavelength.

>Back in Hubble’s day, the active exploration of ideas about the laws
>of physics for the cosmos at large was interrupted by some world wars
>and looking at the universe on a very different scale: inside an atom.

Emitted energy - Doppler redshifted photon energy = energy in
violation of the conservation of energy law.

[PD]

On 11/28/2011 2:59 PM, richard....@comcast.net wrote:

>
> Doppler redshift should never have made sense to anybody. Emitted
> energy - redshifted photon energy = energy imbalance.
>

I don't know why you'd say this. You're comparing the energy as measured
in two different frames of reference. There is no law of physics that
says the energy of something should be independent of reference frame.

For example, a northboound car moving at 60 mph in a reference frame in
which a fire hydrant is at rest will have a certain energy, but the same
car is moving at 100 mph in a reference frame in which a southbound
truck is at rest will have a completely different energy. Does this mean
to you that there is an energy imbalance?

[Michael Helland]

On Nov 28, 12:59 pm, richard.desan...@comcast.net wrote:


> Imagine a universe
> with no light absorbers.  Every emitted photon will travel to
> infinity.

That's not supported by evidence.

What would the universe look like if light had a finite range?

(which is supported by the evidence, it's Hubble's Limit, where H_0 *
d = c )

[Michael Helland]

That's fine. I get that.

Galaxies have intrinsic velocities, just like everything else.

The Doppler effect is real. Ok.


That doesn't mean the Doppler effect is what causes the apparent
recessional velocity proportional to its distance as per Hubble's Law.

Hubble tried to make that clear.

Hubble never accepted that the apparent recessional velocity was
actual.

[PD]

Doppler effect doesn't CAUSE anything. It is simply the signal for
recessional velocity for very distant objects. Thus, if we see a Doppler
shift, then we know it is correlated with a recessional velocity. This
relationship is established with nearer objects (ranging from desktop to
astronomical) where we know independently the recessional velocity and
the wavelength shift and so we know the relationship holds. That is, the
relationship is TESTED and CONFIRMED in situations where we know both v
and delta(f). Thus, we can be confident that when we see delta(f) for
very distant objects, this is correlated with a recessional velocity.

The Hubble relationship is definitely taken today to mean that
recessional velocity is very real.

[Michael Helland]

Sure. I get that.

That doesn't mean the theory is right.

Hubble claimed that if the redshifts are not caused by the Doppler
effect, then we are looking at a new principle of nature.


That principle is light has a finite range.

It fits the evidence, including mature galaxies in the so-called young
universe.

[richard....@comcast.net]

On Mon, 28 Nov 2011 16:20:22 -0600, PD <thedrap...@gmail.com>
wrote:

Yes, inertial reference frames (observer, emitter, emitted object) do
apply to the KE of mass. Photons are different.

"Such particles and waves travel at c regardless of the motion of the
source or the inertial frame of reference of the observer."
http://en.wikipedia.org/wiki/Speed_of_light

In other words, the speed (thus the KE) of photons (within a vacuum)
is independent of the relative speeds of the emitter and the observer.

[richard....@comcast.net]

The Hubble limit is a limit of detection. The limit of detection is
not necessarily the limit of existence.

"In cosmology, the Hubble volume, or Hubble sphere, is the region of
the Universe surrounding an observer beyond which objects recede from
the observer at a rate greater than the speed of light, due to the
expansion of the Universe"
http://en.wikipedia.org/wiki/Hubble_volume

Big bang theory assumes big bang material expands into a vacuum. I
doubt it.

1. The radius of pre-big bang gravitational collapse is likely less
than subsequent big bang expansion.

2. Within an infinite universe, there is room for multiple big bangs.
3. An intact galaxy traveling at near the speed of light is weird,
unless material from a big bang triggered collapse of existing
material into the galaxy.

[richard....@comcast.net]

On Tue, 29 Nov 2011 07:06:22 -0500, richard....@comcast.net
wrote:

Correction: The galaxies at the current edge of the espanding big bang
are not likely composed of big bang material. Therefore, they would
not be travelling at relativistic speeds.

[PD]

On 11/28/2011 8:01 PM, richard....@comcast.net wrote:
> On Mon, 28 Nov 2011 16:20:22 -0600, PD<thedrap...@gmail.com>
> wrote:
>
>> On 11/28/2011 2:59 PM, richard....@comcast.net wrote:
>>
>>>
>>> Doppler redshift should never have made sense to anybody. Emitted
>>> energy - redshifted photon energy = energy imbalance.
>>>
>>
>> I don't know why you'd say this. You're comparing the energy as measured
>> in two different frames of reference. There is no law of physics that
>> says the energy of something should be independent of reference frame.
>>
>> For example, a northboound car moving at 60 mph in a reference frame in
>> which a fire hydrant is at rest will have a certain energy, but the same
>> car is moving at 100 mph in a reference frame in which a southbound
>> truck is at rest will have a completely different energy. Does this mean
>> to you that there is an energy imbalance?
>
> Yes, inertial reference frames (observer, emitter, emitted object) do
> apply to the KE of mass. Photons are different.
>
> "Such particles and waves travel at c regardless of the motion of the
> source or the inertial frame of reference of the observer."
> http://en.wikipedia.org/wiki/Speed_of_light

The energy of a photon is not connected to its speed. Thus you can make
no conclusion that if the speed is the same in all frames, then the
energy must be the same in all frames.

As you say, photons are different than massive objects.

(By the way, it's also true that the momentum of a photon is not
connected to its speed. You may have been told or led to believe that
kinetic energy and momentum are *defined* to be dependent on speed --
something you can assume to be true in all cases -- but if this is so,
then you were misled.)

[PD]

The issue, you may have surmised, is that some work has been done since
Hubble's day (1929). Since that time, the nature of the relationship has
been checked, using more local sources, and in fact the theory regarding
the expansion of the universe has been developed more completely, so
that the validation is MUCH more solid than it was in Hubble's day.

Moreover, "tired light" models (which yours is) have been subjected to
tests other than the Hubble constant, and unfortunately have been ruled
out by numerous observations.

There are some object lessons here:
1. Ambiguities and uncertainties at the time of seminal discoveries do
not usually translate to ambiguities and uncertainties persisting to the
present day.
2. Just because a model is consistent with one piece of data, does not
mean that the model has not been ruled out by conflicts with OTHER data.
It simply is not true that if a model gets one thing wrong, then it will
get everything wrong. The world of physics is littered with failed
models that get some things right and other things wrong -- they are
counted failures because of the latter, despite the former.

[PD]

On 11/29/2011 6:06 AM, richard....@comcast.net wrote:

>
> Big bang theory assumes big bang material expands into a vacuum. I
> doubt it.

Actually, this is not accurate. The big bang is not to be thought of as
an explosion of material from some spot in empty space, into empty
space. It's quite a bit weirder than that.

There is no outer boundary of the big bang. There is no "inside" the
material and "outside" the material. The expansion is expansion of
space(time) itself, not of material INTO space.

[Michael Helland]

On Nov 29, 8:25 am, PD <thedraperfam...@gmail.com> wrote:
> On 11/28/2011 6:04 PM, Michael Helland wrote:
>
> > On Nov 28, 3:26 pm, PD<thedraperfam...@gmail.com>  wrote:
>
> > Sure. I get that.
>
> > That doesn't mean the theory is right.
>
> > Hubble claimed that if the redshifts are not caused by the Doppler
> > effect, then we are looking at a new principle of nature.
>
> > That principle is light has a finite range.
>
> > It fits the evidence, including mature galaxies in the so-called young
> > universe.
>
> The issue, you may have surmised, is that some work has been done since
> Hubble's day (1929). Since that time, the nature of the relationship has
> been checked, using more local sources, and in fact the theory regarding
> the expansion of the universe has been developed more completely, so
> that the validation is MUCH more solid than it was in Hubble's day.
>
> Moreover, "tired light" models (which yours is) have been subjected to
> tests other than the Hubble constant, and unfortunately have been ruled
> out by numerous observations.

You use the words "tired light", and conveniently lump it in with all
sorts of things.

I use the mathematical expression v = c - H_0 * d

I don't think any tired light models express Hubble's Law that way.

> There are some object lessons here:
> 1. Ambiguities and uncertainties at the time of seminal discoveries do
> not usually translate to ambiguities and uncertainties persisting to the
> present day.
> 2. Just because a model is consistent with one piece of data, does not
> mean that the model has not been ruled out by conflicts with OTHER data.
> It simply is not true that if a model gets one thing wrong, then it will
> get everything wrong. The world of physics is littered with failed
> models that get some things right and other things wrong -- they are
> counted failures because of the latter, despite the former.

3. Scientists are smart enough to fool themselves into thinking
they've validated their theories. That was true in Ptolemy's time, and
it is true today. There is no profound difference.

[Michael Helland]

On Nov 29, 4:06 am, richard.desan...@comcast.net wrote:
> On Mon, 28 Nov 2011 15:04:16 -0800 (PST), Michael Helland
>

> <mobyd...@gmail.com> wrote:
> >On Nov 28, 12:59 pm, richard.desan...@comcast.net wrote:
>
> >> Imagine a universe
> >> with no light absorbers. Every emitted photon will travel to
> >> infinity.
>
> >That's not supported by evidence.
>
> >What would the universe look like if light had a finite range?
>
> >(which is supported by the evidence, it's Hubble's Limit, where H_0 *
> >d = c )
>
> The Hubble limit is a limit of detection.  The limit of detection is
> not necessarily the limit of existence.

My point exactly.

Light has a finite range. It is Hubble's Limit.

We don't get light beyond that. The big bang say there's nothing
there. I say it's farther than light is willing travel. The same force
that hold electrons to protons has no reason to travel 1 trillion
years.

[Sam Wormley]

On 11/29/11 3:43 PM, Michael Helland wrote:
> Light has a finite range. It is Hubble's Limit.

Ref: http://en.wikipedia.org/wiki/Hubble_volume

The boundary of the Hubble volume is known as the "Hubble limit". Per
Hubble's law, objects at the Hubble limit have an average comoving speed
of c relative to an observer on the Earth. This is significant, because,
in a universe in which the Hubble parameter was constant, light emitted
at the present time by objects outside the Hubble limit could never be
seen by an observer on the Earth. However, the Hubble "constant" is not
constant. In a decelerating Friedmann universe, the Hubble sphere
expands faster than the Universe and its boundary overtakes light
emitted by receding galaxies. In an accelerating universe, the Hubble
sphere expands more slowly than the Universe, and bodies move out of the
Hubble sphere.[1] So the Hubble limit need not define the cosmological
event horizon (that is, the boundary separating events visible at some
time or other and those that are never visible[4]), because (depending
upon the cosmological model) light emitted at earlier times by objects
outside the Hubble sphere still may eventually arrive inside the sphere
and be seen by us.[2] If, as is inferred from current observations, the
expansion of the universe is in fact accelerating,[5] then at a later
time, some objects within the Hubble limit no longer will be observed
(by us) as they are today.

[richard....@comcast.net]

On Tue, 29 Nov 2011 10:14:46 -0600, PD <thedrap...@gmail.com>

W/ or w/o differences in frame of reference, w/ or w/o a change in
light speed, Doppler redshift violates the conservation of energy law.

[PD]

I completely disagree, for the reasons I've already indicated to you.
Energy is not expected to be constant from frame to frame. There is no
law that says it should be, and it isn't. It isn't for slow, massive
objects (for which the kinetic energy is mv^2/s), and it isn't for light
(for which the energy is pc).

It would help if you acquainted yourself with some of the basics before
assuming that there is a fundamental, unsolved problem, when in fact it
isn't there.

[Michael Helland]

On Nov 29, 3:56 pm, PD <thedraperfam...@gmail.com> wrote:
> On 11/29/2011 5:46 PM, richard.desan...@comcast.net wrote:
>
>
>
>
>
>
>
>
>
> > On Tue, 29 Nov 2011 10:14:46 -0600, PD<thedraperfam...@gmail.com>
> > wrote:
>
> >> On 11/28/2011 8:01 PM, richard.desan...@comcast.net wrote:
> >>> On Mon, 28 Nov 2011 16:20:22 -0600, PD<thedraperfam...@gmail.com>
> >>> wrote:

Sean Carroll says energy isn't conserved in General Relativity due to
the redshifting of photons.

[PD]

On 11/30/2011 1:21 AM, Michael Helland wrote:

>
>
> Sean Carroll says energy isn't conserved in General Relativity due to
> the redshifting of photons.

I'm pretty sure you mangled that quote. Want to try again?

[richard....@comcast.net]

On Tue, 29 Nov 2011 10:27:59 -0600, PD <thedrap...@gmail.com>
wrote:

>On 11/29/2011 6:06 AM, richard....@comcast.net wrote:
>
>>
>> Big bang theory assumes big bang material expands into a vacuum. I
>> doubt it.
>
>Actually, this is not accurate. The big bang is not to be thought of as
>an explosion of material from some spot in empty space, into empty
>space. It's quite a bit weirder than that.
>
>There is no outer boundary of the big bang.

By definition, the Hubble limit is the outer boundary of a big bang.
http://en.wikipedia.org/wiki/Hubble_limit

>There is no "inside" the material and "outside" the material.

The big bang model considers material beyond the Hubble limit as
traveling faster than the speed of light.

I think there is a possibility that pre-big bang material may exist
outside the Hubble limit.

>The expansion is expansion of space(time) itself, not of material INTO space.

A redshifted galaxy headed toward the Hubble limit is material headed
toward outer space. A big bang model referenced to an observer on
Earth may regard the redshifted galaxy as an expansion of space
(time). By analogy, relativistic speed can change PERCEIVED length of
a yardstick (when pointed toward direction of travel, w/o changing
physical yardstick length.
http://en.wikipedia.org/wiki/Length_contraction#Basis_in_relativity

Note: Time at Earth's surface is almost the same as time at locations
w/o gravity. Perhaps, low gravity areas exist on both sides of the
Hubble limit. Similarly, high gravity or high speed inside or outside
the Hubble limit would slow time.

[Michael Helland]

"Actually, there is a field of physics in which energy is not
conserved: it's called general relativity. In an expanding universe,
as we have known for many decades, the total energy is not conserved.
Nothing fancy to do with dark energy -- the same thing is true for
ordinary radiation. Every photon loses energy by redshifting as the
universe expands, while the total number of photons remains conserved,
so the total energy decreases. An effect which has, of course, been
observed. "
http://preposterousuniverse.blogspot.com/2004/05/doubt-and-dissent-are-not-tolerated.html



"The thing about photons is that they redshift, losing energy as space
expands. If we keep track of a certain fixed number of photons, the
number stays constant while the energy per photon decreases, so the
total energy decreases. A decrease in energy is just as much a
“violation of energy conservation” as an increase in energy, but it
doesn’t seem to bother people as much. At the end of the day it
doesn’t matter how bothersome it is, of course — it’s a crystal-clear
prediction of general relativity."
http://blogs.discovermagazine.com/cosmicvariance/2010/02/22/energy-is-not-conserved/

[Michael Helland]

On Nov 30, 5:35 pm, richard.desan...@comcast.net wrote:
> On Tue, 29 Nov 2011 10:27:59 -0600, PD <thedraperfam...@gmail.com>
> wrote:

>
> >On 11/29/2011 6:06 AM, richard.desan...@comcast.net wrote:
>
> >> Big bang theory assumes big bang material expands into a vacuum.  I
> >> doubt it.
>
> >Actually, this is not accurate. The big bang is not to be thought of as
> >an explosion of material from some spot in empty space, into empty
> >space. It's quite a bit weirder than that.
>
> >There is no outer boundary of the big bang.
>
> By definition, the Hubble limit is the outer boundary of a big bang.  http://en.wikipedia.org/wiki/Hubble_limit
>
> >There is no "inside" the material and "outside" the material.
>
> The big bang model considers material beyond the Hubble limit as
> traveling faster than the speed of light.
>
> I think there is a possibility that pre-big bang material may exist
> outside the Hubble limit.

And I think there is a possibility beyond Hubble Limit, there are
galaxies.

Which means, beyond light's finite range, there are galaxies.

Which means we won't ever see those galaxies.


Redshift of light is due to the behavior of light at those distances.

Space is not expanding and the big bang never happened.

[PD]

On 11/30/2011 7:35 PM, richard....@comcast.net wrote:
> On Tue, 29 Nov 2011 10:27:59 -0600, PD<thedrap...@gmail.com>
> wrote:
>
>> On 11/29/2011 6:06 AM, richard....@comcast.net wrote:
>>
>>>
>>> Big bang theory assumes big bang material expands into a vacuum. I
>>> doubt it.
>>
>> Actually, this is not accurate. The big bang is not to be thought of as
>> an explosion of material from some spot in empty space, into empty
>> space. It's quite a bit weirder than that.
>>
>> There is no outer boundary of the big bang.
>
> By definition, the Hubble limit is the outer boundary of a big bang.
> http://en.wikipedia.org/wiki/Hubble_limit

No sir. That is the boundary of the *observable universe*, but not the
limit of the Big Bang.

>
>> There is no "inside" the material and "outside" the material.
>
> The big bang model considers material beyond the Hubble limit as
> traveling faster than the speed of light.

Yes, and there is nothing wrong with that.

>
> I think there is a possibility that pre-big bang material may exist
> outside the Hubble limit.

Anything is possible.

>
>> The expansion is expansion of space(time) itself, not of material INTO space.
>
> A redshifted galaxy headed toward the Hubble limit is material headed
> toward outer space.

No, you're not hearing this right. The redshifted galaxy is receding
from us not because it is traveling THROUGH space away from us, but
because the space itself is receding away from us.

There's an excellent Scientific American article from February 2005 that
deals with popular misconceptions about the Big Bang. I highly recommend
it for you.

> A big bang model referenced to an observer on
> Earth may regard the redshifted galaxy as an expansion of space
> (time). By analogy, relativistic speed can change PERCEIVED length of
> a yardstick (when pointed toward direction of travel, w/o changing
> physical yardstick length.

I have no idea what YOU mean by "physical yardstick length".

[PD]

Thanks for the more accurate quote. And it's right. So what was your point?

[richard....@comcast.net]

On Tue, 29 Nov 2011 17:56:17 -0600, PD <thedrap...@gmail.com>

You might be claiming blue or red shift is a function of relative
emitter speed. Let a slit (detector of the diffraction pattern from a
slit) travel at relativistic velocity while detecting photons from a
stationary emitter. The high speed of the slit will not increase or
decrease the wavelength of the photons.

Similarly, for a photon emitter traveling at relativistic speed, the
amount of energy from that converts to a photon = hc/wavelength = the
amount of energy within the emitted photon. In other words, no
redshift results from photon direction, or speed of the emitter, or
speed of the detector.


Does your reference to KE=mv^2/s apply to a particular situation?

[PD]

Oh, but it does, in the reference frame in which the slit is at rest.

Let's be clear here. We're talking about wavelengths being DIFFERENT in
two different reference frames. If you have a source and a receiver that
are in relative motion, you can ask the question what is the wavelength
in the reference frame in which the receiver is at rest or what is the
wavelength in the reference frame in which the source is at rest. Those
two answers are different, and you can't point to either one of them and
say, "Well, THIS one is the right one, and THAT one is the wrong one."

Here, the receiver may well be a diffraction grating or a slit with a
screen behind it. What will be true is that the angles of the maxima of
the diffraction pattern WILL be different if that apparatus is moving
relative to the source. This is easily verifiable with rather simple
equipment.

>
> Similarly, for a photon emitter traveling at relativistic speed, the
> amount of energy from that converts to a photon = hc/wavelength = the
> amount of energy within the emitted photon. In other words, no
> redshift results from photon direction, or speed of the emitter, or
> speed of the detector.
>
>
> Does your reference to KE=mv^2/s apply to a particular situation?

Sorry, that's a typo. KE=mv^2/2, and I described the conditions for
which that applies.

[richard....@comcast.net]

>> You might be claiming blue or red shift is a function of relative
>> emitter speed. Let a slit (detector of the diffraction pattern from a
>> slit) travel at relativistic velocity while detecting photons from a
>> stationary emitter. The high speed of the slit will not increase or
>> decrease the wavelength of the photons.
>
>Oh, but it does, in the reference frame in which the slit is at rest.
>
>Let's be clear here. We're talking about wavelengths being DIFFERENT in
>two different reference frames. If you have a source and a receiver that
>are in relative motion, you can ask the question what is the wavelength
>in the reference frame in which the receiver is at rest or what is the
>wavelength in the reference frame in which the source is at rest. Those
>two answers are different, and you can't point to either one of them and
>say, "Well, THIS one is the right one, and THAT one is the wrong one."
>
>Here, the receiver may well be a diffraction grating or a slit with a
>screen behind it. What will be true is that the angles of the maxima of
>the diffraction pattern WILL be different if that apparatus is moving
>relative to the source. This is easily verifiable with rather simple
>equipment.

You are right. A diffraction pattern might not be definitive. If you
were in a spacecraft traveling at relativistic speed, double slit
diffraction from a light source within the spacecraft would be a
function of angle between direction of the spacecraft and the
direction of the photons.

Perhaps a single one wavelength wide slit would be definitive. An
increase in slit width or a decrease in wavelength should convert a
point source of light into an interference pattern, regardless of
angle.

Are you claiming that the relativistic speed of the slit changes the
slit width, the size of the photon, or what?

Note: The slit (travelling at relativistic speed) could be a few
wavelengths away from the stationary emitter at the time the photon
passes through it.

[PD]

No. In the reference frame in which the slit is at rest (and the source
is moving), then the wavelength of the light is different, and so the
diffraction pattern is different. This is easily confirmed in measurement.

[richard....@comcast.net]

On Fri, 02 Dec 2011 08:52:34 -0600, PD <thedrap...@gmail.com>
wrote:

Are you saying that emitter velocity adds or subtracts momentum
to/from photons?

Direct emission of a photon containing the extra momentum is more
likely than redshift from the Doppler effect.

1. The Doppler effect on a relativistic photon may violate the
conservation of energy law.
2. Molecular energy may convert to a photon instantly. because
quantized molecular and atomic energy changes occur instantly. W/o a
molecule dragging a photon during emission, the Doppler effect should
not occur.

I am not advocating that photons have momentum. In the case of the
stationary emitter, the moving wavelength detector is in a frame that
should enhance the influence of photon momentum. ZERO wavelength
change (red or blue shifts) indirectly indicates zero photon momentum.
If photons indeed lack momentum, I think propagation through molecules
more likely cause redshift, than relativistic speed of an emitter.

[PD]

Momentum is frame-dependent. Do you know what this means?

>
> Direct emission of a photon containing the extra momentum is more
> likely than redshift from the Doppler effect.
>
> 1. The Doppler effect on a relativistic photon may violate the
> conservation of energy law.

No, it doesn't, as I've pointed out. Conservation of energy does NOT
mean that energy is the same in different reference frames.

> 2. Molecular energy may convert to a photon instantly. because
> quantized molecular and atomic energy changes occur instantly. W/o a
> molecule dragging a photon during emission, the Doppler effect should
> not occur.

Doppler shift has nothing to do with a source "dragging" the emitted light.

>
> I am not advocating that photons have momentum.

The fact that they have momentum is supported by DIRECT experimental
confirmation that has nothing to do with Doppler effect.

[richard....@comcast.net]

>There's an excellent Scientific American article from February 2005 that
>deals with popular misconceptions about the Big Bang. I highly recommend
>it for you.

I am surprised that you recommended this article.

1. "As the photons travel through expanding space, they lose energy
and their temperature decreases. In this way, the universe cools as
it expands, much as compressed air in a scuba tank cools when it is
released and allowed to expand."

The comparison between decompression of air and redshift is invalid.
The photon's lost energy does not convert into a different energy.
Energy due to gas compression converts into an equal amount of other
energy. For example, decompression can lower temperature; a
temperature difference is the source of energy for the Carnot cycle.


2. "a bomb … gets bigger by expanding into the space around it."
"Galaxies are not traveling through space away from us."

In other words, a bomb fragment has momentum caused by a blast; a
receding galaxy has space-time expansion, instead of momentum.

I think distant galaxies are not traveling away from us at
relativistic speeds.

As mass accelerates to relativistic speeds, Bremsstrahlung radiation,
spontaneous creation of particle pairs, high energy collision products
and other high-energy phenomena should predominate over low speed
phenomena. Receding galaxies lack high-energy phenomena. Therefore,
receding galaxies are traveling below relativistic speeds.

[richard....@comcast.net]

On Mon, 05 Dec 2011 10:47:52 -0600, PD <thedrap...@gmail.com>

Momentum has frame dependence. Galactic redshift (Scientific
American) is another story.
http://space.mit.edu/~kcooksey/teaching/AY5/MisconceptionsabouttheBigBang_ScientificAmerican.pdf

"Astronomers say that the galactic light has been redshifted. The
explanation is straightforward: As space expands, light waves get
stretched. If the universe doubles in size during the waves' journey,
their wavelengths double and their energy is halved."

In other words, DURING the light's entire travel between emitter and
detector, expansion stretches (redshifts) the light. Do you think the
energy change due to expansion during travel violates the conservation
of energy law?

Distant galaxies do not contain Bremsstrahlung or other emissions and
collision products that objects with relativistic speed normally have.
Therefore, distant galaxies have non-relativistic speed and redshift
occurs during light propagation.

>>
>> Direct emission of a photon containing the extra momentum is more
>> likely than redshift from the Doppler effect.
>>
>> 1. The Doppler effect on a relativistic photon may violate the
>> conservation of energy law.
>
>No, it doesn't, as I've pointed out. Conservation of energy does NOT
>mean that energy is the same in different reference frames.

Consider the relativistic Doppler effect
As a photon emitter and detector move apart, the time between
wavefronts at the detector increases. The difference in TIME between
wavefronts is the basis of the relativistic Doppler effect.
http://en.wikipedia.org/wiki/Relativistic_Doppler_effect#Motion_along_the_line_of_sight

Time between wavefronts is not necessarily a measure of photon energy
or wavelength. Glass can have a refractive index = 1.5, 3 wavefronts
exist in the glass for each 2 in a vacuum. A detector moving through
glass would detect the same wavelength as a detector moving through a
vacuum

>> 2. Molecular energy may convert to a photon instantly. because
>> quantized molecular and atomic energy changes occur instantly. W/o a
>> molecule dragging a photon during emission, the Doppler effect should
>> not occur.
>
>Doppler shift has nothing to do with a source "dragging" the emitted light.

If motion of a mirror (car) indeed changed wavelength of reflected
radiation, w/o changing KE of the mirror (Doppler effect,) I think
there would be a violation of the conservation of energy law. Does
Doppler radar actually use the Doppler effect?

>>
>> I am not advocating that photons have momentum.
>
>The fact that they have momentum is supported by DIRECT experimental
>confirmation that has nothing to do with Doppler effect.

By implication, you are saying that light has momentum. If light had
momentum, then light must have mass. Mass has gravity. Gravitational
redshift could occur. Gravitational redshift within a vacuum violates
the conservation of energy law (energy is lost w/o conversion to a
different energy).

I think the evidence for gravitational red/blue shift (the Pound-Rebka
experiment) ignores the influence of gravitational potential energy on
Fe itself. Quantized molecular energy levels are a function of total
energy, which includes KE and gravitational potential energy. In
other words, gravitational potential energy MUST influence the
wavelength of Fe emission and absorption. If gravity indeed directly
decreased photon energy in the Pound-Rebka experiment, what gains the
released energy?

[PD]

On 12/8/2011 6:20 PM, richard....@comcast.net wrote:

[a fair number of things, too many to all address in one response. I
will select a few]

No. Again, it's important to understand what conservation of energy
actually says, and it is confined to a particular reference frame.

>
> Distant galaxies do not contain Bremsstrahlung or other emissions and
> collision products that objects with relativistic speed normally have.

First of all, this is wrong. What do you think the source of X-rays from
an accretion disk is from?

> Therefore, distant galaxies have non-relativistic speed and redshift
> occurs during light propagation.
>>>
>>> Direct emission of a photon containing the extra momentum is more
>>> likely than redshift from the Doppler effect.
>>>
>>> 1. The Doppler effect on a relativistic photon may violate the
>>> conservation of energy law.
>>
>> No, it doesn't, as I've pointed out. Conservation of energy does NOT
>> mean that energy is the same in different reference frames.
>
> Consider the relativistic Doppler effect
> As a photon emitter and detector move apart, the time between
> wavefronts at the detector increases. The difference in TIME between
> wavefronts is the basis of the relativistic Doppler effect.
> http://en.wikipedia.org/wiki/Relativistic_Doppler_effect#Motion_along_the_line_of_sight

I think you've got a wrong picture of what's going on here. It's not
like the wavetrain is being stretched like a rubber band between emitter
and detector.

>
> Time between wavefronts is not necessarily a measure of photon energy
> or wavelength. Glass can have a refractive index = 1.5, 3 wavefronts
> exist in the glass for each 2 in a vacuum. A detector moving through
> glass would detect the same wavelength as a detector moving through a
> vacuum

Uh, no.

>
>>> 2. Molecular energy may convert to a photon instantly. because
>>> quantized molecular and atomic energy changes occur instantly. W/o a
>>> molecule dragging a photon during emission, the Doppler effect should
>>> not occur.
>>
>> Doppler shift has nothing to do with a source "dragging" the emitted light.
>
> If motion of a mirror (car) indeed changed wavelength of reflected
> radiation, w/o changing KE of the mirror (Doppler effect,) I think
> there would be a violation of the conservation of energy law. Does
> Doppler radar actually use the Doppler effect?
>>>
>>> I am not advocating that photons have momentum.
>>
>> The fact that they have momentum is supported by DIRECT experimental
>> confirmation that has nothing to do with Doppler effect.
>
> By implication, you are saying that light has momentum.

Yes, of course.

> If light had
> momentum, then light must have mass.

No. Why do you think bearing momentum requires bearing mass?

[richard....@comcast.net]

>> "Astronomers say that the galactic light has been redshifted. The
>> explanation is straightforward: As space expands, light waves get
>> stretched. If the universe doubles in size during the waves' journey,
>> their wavelengths double and their energy is halved."
>>
>> In other words, DURING the light's entire travel between emitter and
>> detector, expansion stretches (redshifts) the light. Do you think the
>> energy change due to expansion during travel violates the conservation
>> of energy law?
>
>No. Again, it's important to understand what conservation of energy
>actually says, and it is confined to a particular reference frame.

The Scientific American article made a distinction between galactic
expansion and a bomb.

"a bomb … gets bigger by expanding into the space around it."
"Galaxies are not traveling through space away from us."

http://space.mit.edu/~kcooksey/teaching/AY5/MisconceptionsabouttheBigBang_ScientificAmerican.pdf

A bomb converts chemical energy into KE. Galactic expansion converts
compression energy into redshifting photons, w/o changing KE of
galaxies. In other words, most of the redshift of distant galaxies
occurs after the photons leave the distant galaxy.

"As space expands, light waves get stretched. If the universe doubles
in size during the waves' journey, their wavelengths double and their
energy is halved."

http://space.mit.edu/~kcooksey/teaching/AY5/MisconceptionsabouttheBigBang_ScientificAmerican.pdf
The Earth can reflect a photon from a distant galaxy, the reflected
photon can return to its original reference frame (near the redshifted
galaxy,) doubly redshifted. In other words, redshift can occur during
travel within the photon's original reference frame.


If redshift indeed had frame dependence, relativistic electrons would
blueshift FEL and synchrotron radiation. Retreating electrons would
redshift IFEL. In practice, the stimulating radiation for IFEL is
close to output radiation energy, not red or blue shifted.

"One problem with SASE FELs is the lack of temporal coherence due to a
noisy startup process. To avoid this, one can "seed" an FEL with a
laser tuned to the resonance of the FEL. Such a temporally coherent
seed can be produced by more conventional means, such as by
high-harmonic generation (HHG) using an optical laser pulse. This
results in coherent amplification of the input signal; in effect, the
output laser quality is characterized by the seed. While HHG seeds are
available at wavelengths down to the extreme ultraviolet, seeding is
not feasible at x-ray wavelengths due to the lack of conventional
x-ray lasers."
http://en.wikipedia.org/wiki/Free-electron_laser

>>
>> Distant galaxies do not contain Bremsstrahlung or other emissions and
>> collision products that objects with relativistic speed normally have.
>
>First of all, this is wrong. What do you think the source of X-rays from
>an accretion disk is from?

Albeit distant galaxies contain accretion disks, typical distant
galaxies do not contain the radiation pattern of a galaxies traveling
at relativistic speeds.

>> Therefore, distant galaxies have non-relativistic speed and redshift
>> occurs during light propagation.
>>>>
>>>> Direct emission of a photon containing the extra momentum is more
>>>> likely than redshift from the Doppler effect.
>>>>
>>>> 1. The Doppler effect on a relativistic photon may violate the
>>>> conservation of energy law.
>>>
>>> No, it doesn't, as I've pointed out. Conservation of energy does NOT
>>> mean that energy is the same in different reference frames.
>>
>> Consider the relativistic Doppler effect
>> As a photon emitter and detector move apart, the time between
>> wavefronts at the detector increases. The difference in TIME between
>> wavefronts is the basis of the relativistic Doppler effect.
>> http://en.wikipedia.org/wiki/Relativistic_Doppler_effect#Motion_along_the_line_of_sight
>
>I think you've got a wrong picture of what's going on here. It's not
>like the wavetrain is being stretched like a rubber band between emitter
>and detector.

The Doppler effect is the influence of relative motion between source
and detector on the time between wavefronts that occur at regular
intervals.
http://en.wikipedia.org/wiki/Doppler_effect

Wavefronts from distant galaxies do not occur in regular intervals.
The emitters are not from a point source, (each photon from a point
source emitter propagates in one wavelength spatial increments). The
galaxy's emissions occur at random times. Random emission times
result in random temporal detection times.

As shown by the Scientific American bomb versus galactic expansion
example, distant galaxies are part of expanding space, not fragments
of a bomb moving through space. In other words, typical distant
galaxies are not moving at relativistic speeds. Therefore, distant
galaxies should not produce a relativistic Doppler effect.

>>
>> Time between wavefronts is not necessarily a measure of photon energy
>> or wavelength. Glass can have a refractive index = 1.5, 3 wavefronts
>> exist in the glass for each 2 in a vacuum. A detector moving through
>> glass would detect the same wavelength as a detector moving through a
>> vacuum
>
>Uh, no.

By Uh, no, are you saying:

Time between wavefronts is a measure of photon energy? (A single
photon does not have two wavefronts.)

A detector moving through glass would not detect the same wavelength
as a detector moving through a vacuum? (The passage of light through
transparent molecular media does not change wavelength.)

>>
>>>> 2. Molecular energy may convert to a photon instantly. because
>>>> quantized molecular and atomic energy changes occur instantly. W/o a
>>>> molecule dragging a photon during emission, the Doppler effect should
>>>> not occur.
>>>
>>> Doppler shift has nothing to do with a source "dragging" the emitted light.
>>
>> If motion of a mirror (car) indeed changed wavelength of reflected
>> radiation, w/o changing KE of the mirror (Doppler effect,) I think
>> there would be a violation of the conservation of energy law. Does
>> Doppler radar actually use the Doppler effect?
>>>>
>>>> I am not advocating that photons have momentum.
>>>
>>> The fact that they have momentum is supported by DIRECT experimental
>>> confirmation that has nothing to do with Doppler effect.
>>
>> By implication, you are saying that light has momentum.
>
>Yes, of course.
>
>> If light had
>> momentum, then light must have mass.
>
>No. Why do you think bearing momentum requires bearing mass?

1. MOMENTUM = MASS X VELOCITY
http://en.wikipedia.org/wiki/Momentum

2. Mirrors reflect photons w/o changing photon energy or mirror energy
KE.

"For photons however, the momentum imparted to a solar sail would be
p=E/c when the photon is totally absorbed (black body) and p=2E/c when
the photon is totally reflected (perfect mirror)."
http://www.asterism.org/tutorials/tut21-1.htm

If photon reflection indeed added energy to a solar sail, then
bouncing photons between two solar sails could power both sails. Each
bounce could add twice the photon's energy to the KE of the mirror.
For example, a photon that bounces 20 times before veering off could
provide about 40x the energy of the photon. Unlike gas molecules that
lose energy as the molecules expand (increased distance between
molecules), photons have the same energy before and after each bounce.

[PD]

On 12/12/2011 3:35 PM, richard....@comcast.net wrote:

Again a lot of things and I will only respond to a few.

>>
>> I think you've got a wrong picture of what's going on here. It's not
>> like the wavetrain is being stretched like a rubber band between emitter
>> and detector.
>
> The Doppler effect is the influence of relative motion between source
> and detector on the time between wavefronts that occur at regular
> intervals.
> http://en.wikipedia.org/wiki/Doppler_effect

Yes. This does not mean that the wavefronts are getting stretched
between source and receiver.

>
> Wavefronts from distant galaxies do not occur in regular intervals.
> The emitters are not from a point source, (each photon from a point
> source emitter propagates in one wavelength spatial increments). The
> galaxy's emissions occur at random times. Random emission times
> result in random temporal detection times.
>
> As shown by the Scientific American bomb versus galactic expansion
> example, distant galaxies are part of expanding space,

Yes.

> not fragments
> of a bomb moving through space. In other words, typical distant
> galaxies are not moving at relativistic speeds.

Expanding space is measurable by us as a recession velocity. It does NOT
mean that it is the same as bomb fragments flying through space.

> Therefore, distant
> galaxies should not produce a relativistic Doppler effect.
>>>
>>> Time between wavefronts is not necessarily a measure of photon energy
>>> or wavelength. Glass can have a refractive index = 1.5, 3 wavefronts
>>> exist in the glass for each 2 in a vacuum. A detector moving through
>>> glass would detect the same wavelength as a detector moving through a
>>> vacuum
>>
>> Uh, no.
>
> By Uh, no, are you saying:
>
> Time between wavefronts is a measure of photon energy? (A single
> photon does not have two wavefronts.)
>
> A detector moving through glass would not detect the same wavelength
> as a detector moving through a vacuum? (The passage of light through
> transparent molecular media does not change wavelength.)

The change in wavelength is *measurable*. The thickness of an AR coating
is determined by the wavelength IN THE COATING, which is different than
it is in air.

>>>
>>>>> 2. Molecular energy may convert to a photon instantly. because
>>>>> quantized molecular and atomic energy changes occur instantly. W/o a
>>>>> molecule dragging a photon during emission, the Doppler effect should
>>>>> not occur.
>>>>
>>>> Doppler shift has nothing to do with a source "dragging" the emitted light.
>>>
>>> If motion of a mirror (car) indeed changed wavelength of reflected
>>> radiation, w/o changing KE of the mirror (Doppler effect,) I think
>>> there would be a violation of the conservation of energy law. Does
>>> Doppler radar actually use the Doppler effect?
>>>>>
>>>>> I am not advocating that photons have momentum.
>>>>
>>>> The fact that they have momentum is supported by DIRECT experimental
>>>> confirmation that has nothing to do with Doppler effect.
>>>
>>> By implication, you are saying that light has momentum.
>>
>> Yes, of course.
>>
>>> If light had
>>> momentum, then light must have mass.
>>
>> No. Why do you think bearing momentum requires bearing mass?
>
> 1. MOMENTUM = MASS X VELOCITY
> http://en.wikipedia.org/wiki/Momentum

That formula is not a definition of momentum. It is a method of finding
momentum from other properties of an object, PROVIDED that the object is
slowing moving and massive. It does NOT work for other objects for which
those conditions do not hold.

[richard....@comcast.net]

>Again a lot of things and I will only respond to a few.

OKay

>> As shown by the Scientific American bomb versus galactic expansion
>> example, distant galaxies are part of expanding space,
>
>Yes.
>
>> not fragments
>> of a bomb moving through space. In other words, typical distant
>> galaxies are not moving at relativistic speeds.
>
>Expanding space is measurable by us as a recession velocity. It does NOT
>mean that it is the same as bomb fragments flying through space.

How do typical (relatively stationary) distant galaxies cause galactic
redshift or 'recession velocity'?

Typical distant galaxies are not traveling at relativistic speeds.
Therefore, the mass of the universe is not expanding at relativistic
speeds.

[PD]

On 12/14/2011 10:13 AM, richard....@comcast.net wrote:
>> Again a lot of things and I will only respond to a few.
>
> OKay
>
>>> As shown by the Scientific American bomb versus galactic expansion
>>> example, distant galaxies are part of expanding space,
>>
>> Yes.
>>
>>> not fragments
>>> of a bomb moving through space. In other words, typical distant
>>> galaxies are not moving at relativistic speeds.
>>
>> Expanding space is measurable by us as a recession velocity. It does NOT
>> mean that it is the same as bomb fragments flying through space.
>
> How do typical (relatively stationary) distant galaxies cause galactic
> redshift or 'recession velocity'?

What do you mean "relatively stationary"? They are not stationary
relative to us. They are receding from us at relativistic speeds.

Your claim that mass at relativistic speeds would radiate is simply
wrong. They do so in a *medium* with which they have high relative
motion, or as a result of the application of an electromagnetic field.
But no, relativistically high speed particles do not radiate.

[richard....@comcast.net]

On Wed, 14 Dec 2011 14:14:14 -0600, PD <thedrap...@gmail.com>
wrote:

>On 12/14/2011 10:13 AM, richard....@comcast.net wrote:
>>> Again a lot of things and I will only respond to a few.
>>
>> OKay
>>
>>>> As shown by the Scientific American bomb versus galactic expansion
>>>> example, distant galaxies are part of expanding space,
>>>
>>> Yes.
>>>
>>>> not fragments
>>>> of a bomb moving through space. In other words, typical distant
>>>> galaxies are not moving at relativistic speeds.
>>>
>>> Expanding space is measurable by us as a recession velocity. It does NOT
>>> mean that it is the same as bomb fragments flying through space.
>>
>> How do typical (relatively stationary) distant galaxies cause galactic
>> redshift or 'recession velocity'?
>
>What do you mean "relatively stationary"? They are not stationary
>relative to us. They are receding from us at relativistic speeds.

"As space expands, light waves get stretched. If the universe doubles
in size during the waves' journey, their wavelengths double and their
energy is halved."
http://space.mit.edu/~kcooksey/teaching/AY5/MisconceptionsabouttheBigBang_ScientificAmerican.pdf

A reflected photon that originated from Earth can return to Earth
doubly redshifted (during outward and inward journey.) Similarly,
photons from distant galaxies will redshift during their journey,
regardless of their originating distant galaxy speed.

"The cosmological redshift is not a normal Doppler shift. Astronomers
frequently refer to it as such, and in doing so they have done their
students a serious disservice."
http://space.mit.edu/~kcooksey/teaching/AY5/MisconceptionsabouttheBigBang_ScientificAmerican.pdf

In other words, the speed of a distant galaxy does not cause redshift
its photons.

Note: As shown by FEL and IFEL, relativistic emitter speed does not
red or blue shift photons.

>Your claim that mass at relativistic speeds would radiate is simply
>wrong. They do so in a *medium* with which they have high relative
>motion, or as a result of the application of an electromagnetic field.
>But no, relativistically high speed particles do not radiate.

During expansion of the universe that includes accelerating the outer
galaxies to relativistic speeds, Bremsstrahlung radiation should
exist.

If the outer galaxies have relativistic speed, but are no longer
accelerating, then the distant galaxies would still have radiation
corresponding to extreme high temperatures.

[PD]

On 12/15/2011 5:40 PM, richard....@comcast.net wrote:

>
>> Your claim that mass at relativistic speeds would radiate is simply
>> wrong. They do so in a *medium* with which they have high relative
>> motion, or as a result of the application of an electromagnetic field.
>> But no, relativistically high speed particles do not radiate.
>
> During expansion of the universe that includes accelerating the outer
> galaxies to relativistic speeds, Bremsstrahlung radiation should
> exist.

Galaxies didn't exist when the initial expansion happened. Might want to
check out big bang cosmology so you can see what happened when.

>
> If the outer galaxies have relativistic speed, but are no longer
> accelerating, then the distant galaxies would still have radiation
> corresponding to extreme high temperatures.

Sorry, but traveling fast does not translate to high temperatures. I
don't know why you would think so.

If you're thinking about jet planes in atmospheres, that's due to
friction in a medium.

The Earth's speed in orbit around the Sun is Mach 87.5. You'll notice
that the Earth does not burn up as a result of this speed.

[Sam Wormley]

On 12/8/11 6:20 PM, richard....@comcast.net wrote:
> Momentum has frame dependence. Galactic redshift (Scientific
> American) is another story.

Galactic redshift is observer dependent just like any other
relative motion observations. Look at these in order:

No Center
http://www.astro.ucla.edu/~wright/nocenter.html
http://www.astro.ucla.edu/~wright/infpoint.html

Also see Ned Wright's Cosmology Tutorial
http://www.astro.ucla.edu/~wright/cosmolog.htm
http://www.astro.ucla.edu/~wright/cosmology_faq.html
http://www.astro.ucla.edu/~wright/CosmoCalc.html

WMAP: Foundations of the Big Bang theory
http://map.gsfc.nasa.gov/m_uni.html

WMAP: Tests of Big Bang Cosmology
http://map.gsfc.nasa.gov/m_uni/uni_101bbtest.html

[be...@iwaynet.net]

On 12/15/2011 6:46 PM, PD wrote:

> If you're thinking about jet planes in atmospheres, that's due to
> friction in a medium.
>
> The Earth's speed in orbit around the Sun is Mach 87.5. You'll notice
> that the Earth does not burn up as a result of this speed.

But obviously Global warming occurs as a result of the friction of the
passage of the Earth through the aether!

[richard....@comcast.net]

On Thu, 15 Dec 2011 17:46:08 -0600, PD <thedrap...@gmail.com>
wrote:

>On 12/15/2011 5:40 PM, richard....@comcast.net wrote:
>
>>
>>> Your claim that mass at relativistic speeds would radiate is simply
>>> wrong. They do so in a *medium* with which they have high relative
>>> motion, or as a result of the application of an electromagnetic field.
>>> But no, relativistically high speed particles do not radiate.
>>
>> During expansion of the universe that includes accelerating the outer
>> galaxies to relativistic speeds, Bremsstrahlung radiation should
>> exist.
>
>Galaxies didn't exist when the initial expansion happened. Might want to
>check out big bang cosmology so you can see what happened when.

Are you saying that expansion has quit accelerating mass, but
continues to redshift photons?

>
>>
>> If the outer galaxies have relativistic speed, but are no longer
>> accelerating, then the distant galaxies would still have radiation
>> corresponding to extreme high temperatures.
>
>Sorry, but traveling fast does not translate to high temperatures. I
>don't know why you would think so.
>
>If you're thinking about jet planes in atmospheres, that's due to
>friction in a medium.
>
>The Earth's speed in orbit around the Sun is Mach 87.5. You'll notice
>that the Earth does not burn up as a result of this speed.

Is this what happened?
1. Initial big bang material flipped from being material with high
temperature, to material with high speed.
2. While at high speed, the material cooled to near absolute zero.
(Cooling within a vacuum requires radiation).
3. The cold material collapsed, Hydrogen fusion within the stars in
the galaxy heated the material to its current temperature.

[PD]

On 12/15/2011 11:00 PM, richard....@comcast.net wrote:
> On Thu, 15 Dec 2011 17:46:08 -0600, PD<thedrap...@gmail.com>
> wrote:
>
>> On 12/15/2011 5:40 PM, richard....@comcast.net wrote:
>>
>>>
>>>> Your claim that mass at relativistic speeds would radiate is simply
>>>> wrong. They do so in a *medium* with which they have high relative
>>>> motion, or as a result of the application of an electromagnetic field.
>>>> But no, relativistically high speed particles do not radiate.
>>>
>>> During expansion of the universe that includes accelerating the outer
>>> galaxies to relativistic speeds, Bremsstrahlung radiation should
>>> exist.
>>
>> Galaxies didn't exist when the initial expansion happened. Might want to
>> check out big bang cosmology so you can see what happened when.
>
> Are you saying that expansion has quit accelerating mass, but
> continues to redshift photons?

Redshift occurs for things moving at constant *velocity* with respect to
an observer. This is confirmable in a *desktop* experiment, and it
dovetails nicely with what we see in astronomical objects moving
relative to us.

>>
>>>
>>> If the outer galaxies have relativistic speed, but are no longer
>>> accelerating, then the distant galaxies would still have radiation
>>> corresponding to extreme high temperatures.
>>
>> Sorry, but traveling fast does not translate to high temperatures. I
>> don't know why you would think so.
>>
>> If you're thinking about jet planes in atmospheres, that's due to
>> friction in a medium.
>>
>> The Earth's speed in orbit around the Sun is Mach 87.5. You'll notice
>> that the Earth does not burn up as a result of this speed.
>
> Is this what happened?
> 1. Initial big bang material flipped from being material with high
> temperature, to material with high speed.
> 2. While at high speed, the material cooled to near absolute zero.
> (Cooling within a vacuum requires radiation).
> 3. The cold material collapsed, Hydrogen fusion within the stars in
> the galaxy heated the material to its current temperature.

No, that's not really what happened. Would you like a suggestion for a
reading reference for cosmology?

[richard....@comcast.net]

On Fri, 16 Dec 2011 13:46:09 -0600, PD <thedrap...@gmail.com>

wrote:

>On 12/15/2011 11:00 PM, richard....@comcast.net wrote:
>> On Thu, 15 Dec 2011 17:46:08 -0600, PD<thedrap...@gmail.com>
>> wrote:
>>
>>> On 12/15/2011 5:40 PM, richard....@comcast.net wrote:
>>>
>>>>
>>>>> Your claim that mass at relativistic speeds would radiate is simply
>>>>> wrong. They do so in a *medium* with which they have high relative
>>>>> motion, or as a result of the application of an electromagnetic field.
>>>>> But no, relativistically high speed particles do not radiate.
>>>>
>>>> During expansion of the universe that includes accelerating the outer
>>>> galaxies to relativistic speeds, Bremsstrahlung radiation should
>>>> exist.
>>>
>>> Galaxies didn't exist when the initial expansion happened. Might want to
>>> check out big bang cosmology so you can see what happened when.
>>
>> Are you saying that expansion has quit accelerating mass, but
>> continues to redshift photons?
>
>Redshift occurs for things moving at constant *velocity* with respect to
>an observer. This is confirmable in a *desktop* experiment, and it
>dovetails nicely with what we see in astronomical objects moving
>relative to us.

"In the current standard model of cosmology, galaxies with a redshift
of about 1.5--that is, whose light has a wavelength 150 percent longer
than the laboratory reference value--are receding at the speed of
light. Astronomers have observed about 1,000 galaxies with redshifts
larger than 1.5. That is, they have observed about 1,000 objects
receding from us faster than the speed of light. Equivalently, we are
receding from those galaxies faster than the speed of light."
http://space.mit.edu/~kcooksey/teaching/AY5/MisconceptionsabouttheBigBang_ScientificAmerican.pdf

The above refutes the myth that redshift is a function of the speed an
emitter toward or away from an observer. The speed between emitter
and observer cannot exceed the speed of light, yet 1,000 galaxies have
redshifts larger than 1.5. Redshift occurs during light's journey,
not during emission.

>>>
>>>>
>>>> If the outer galaxies have relativistic speed, but are no longer
>>>> accelerating, then the distant galaxies would still have radiation
>>>> corresponding to extreme high temperatures.
>>>
>>> Sorry, but traveling fast does not translate to high temperatures. I
>>> don't know why you would think so.
>>>
>>> If you're thinking about jet planes in atmospheres, that's due to
>>> friction in a medium.
>>>
>>> The Earth's speed in orbit around the Sun is Mach 87.5. You'll notice
>>> that the Earth does not burn up as a result of this speed.
>>
>> Is this what happened?
>> 1. Initial big bang material flipped from being material with high
>> temperature, to material with high speed.
>> 2. While at high speed, the material cooled to near absolute zero.
>> (Cooling within a vacuum requires radiation).
>> 3. The cold material collapsed, Hydrogen fusion within the stars in
>> the galaxy heated the material to its current temperature.
>
>No, that's not really what happened. Would you like a suggestion for a
>reading reference for cosmology?

I think that showing the difference between energy due to temperature
versus KE may be useful in explaining big bang cooling. I should
introduce a few concepts

1. Thermal energy of a gas is a function of compression.
The following equations show the relation between v (velocity), T
(temperature, P (pressure) and KE.

KE = 1/2mv^2 = 3/2kT
http://hyperphysics.phy-astr.gsu.edu/hbase/kinetic/kintem.html

Gay-Lussac's law

P = k'T

Based on Gay-Lussac's law, gas temperature, will decrease to ZERO when
releasing air from a pressurized tank, into a vacuum. Logically, the
big bang cooled by expansion into a vacuum. After expansion, some
people might think that P --> 0, T --> 0, and v --> 0. They would be
wrong, because both 1/2mv^2 and 3/2kT would be zero, a violation of

the conservation of energy law.

To avoid violation, if T --> 0, the (compression) energy due to
temperature of a gas could convert to KE (1/2mv^2) of the expanding
gas molecules.

2. Temperature is not a property of initial big bang material.
PV=nRT is a property of molecular gases. For example, adding
electrons does not necessarily compress a volume of molecular gas.

Oops, I am starting to post 'a lot of things'. Sorry.

[PD]

I'm sorry, but that's an error. There is no such statement in
relativity, and that's precisely the misconception that the authors of
the article in Scientific American are working hard to dispel.

> yet 1,000 galaxies have
> redshifts larger than 1.5. Redshift occurs during light's journey,
> not during emission.
>>>>
>>>>>
>>>>> If the outer galaxies have relativistic speed, but are no longer
>>>>> accelerating, then the distant galaxies would still have radiation
>>>>> corresponding to extreme high temperatures.
>>>>
>>>> Sorry, but traveling fast does not translate to high temperatures. I
>>>> don't know why you would think so.
>>>>
>>>> If you're thinking about jet planes in atmospheres, that's due to
>>>> friction in a medium.
>>>>
>>>> The Earth's speed in orbit around the Sun is Mach 87.5. You'll notice
>>>> that the Earth does not burn up as a result of this speed.
>>>
>>> Is this what happened?
>>> 1. Initial big bang material flipped from being material with high
>>> temperature, to material with high speed.
>>> 2. While at high speed, the material cooled to near absolute zero.
>>> (Cooling within a vacuum requires radiation).
>>> 3. The cold material collapsed, Hydrogen fusion within the stars in
>>> the galaxy heated the material to its current temperature.
>>
>> No, that's not really what happened. Would you like a suggestion for a
>> reading reference for cosmology?
>
>
> I think that showing the difference between energy due to temperature
> versus KE may be useful in explaining big bang cooling.

It's going to help if you first learn what the current model says before
proposing an alternative.

It's really not a sound approach to propose an alternative BECAUSE you
don't understand the current model.

> I should
> introduce a few concepts
>
> 1. Thermal energy of a gas is a function of compression.
> The following equations show the relation between v (velocity), T
> (temperature, P (pressure) and KE.
>
> KE = 1/2mv^2 = 3/2kT
> http://hyperphysics.phy-astr.gsu.edu/hbase/kinetic/kintem.html
>
> Gay-Lussac's law
>
> P = k'T
>
> Based on Gay-Lussac's law, gas temperature, will decrease to ZERO when
> releasing air from a pressurized tank, into a vacuum.

But the big bang is not an expanding ideal gas.

> Logically, the
> big bang cooled by expansion into a vacuum.

No sir. Just because something sounds *plausible* to you does not mean
it is "logical". It is in fact an error, no matter how plausible it
sounds to you. That's not what the big bang is.

[G=EMC^2]

On Dec 12, 4:35 pm, richard.desan...@comcast.net wrote:
> >> "Astronomers say that the galactic light has been redshifted. The
> >> explanation is straightforward: As space expands, light waves get
> >> stretched.  If the universe doubles in size during the waves' journey,
> >> their wavelengths double and their energy is halved."
>
> >> In other words, DURING the light's entire travel between emitter and
> >> detector, expansion stretches (redshifts) the light.  Do you think the
> >> energy change due to expansion during travel violates the conservation
> >> of energy law?
>
> >No. Again, it's important to understand what conservation of energy
> >actually says, and it is confined to a particular reference frame.
>
> The Scientific American article made a distinction between galactic
> expansion and a bomb.
>
> "a bomb … gets bigger by expanding into the space around it."

> "Galaxies are not traveling through space away from us."  http://space.mit.edu/~kcooksey/teaching/AY5/MisconceptionsabouttheBig...

>
> A bomb converts chemical energy into KE.  Galactic expansion converts
> compression energy into redshifting photons, w/o changing KE of
> galaxies.  In other words, most of the redshift of distant galaxies
> occurs after the photons leave the distant galaxy.
>
> "As space expands, light waves get stretched. If the universe doubles
> in size during the waves' journey, their wavelengths double and their

> energy is halved."http://space.mit.edu/~kcooksey/teaching/AY5/MisconceptionsabouttheBig...

> >>http://en.wikipedia.org/wiki/Relativistic_Doppler_effect#Motion_along...

> 1. MOMENTUM = MASS X VELOCITYhttp://en.wikipedia.org/wiki/Momentum

>
> 2. Mirrors reflect photons w/o changing photon energy or mirror energy
> KE.
>
> "For photons however, the momentum imparted to a solar sail would be
> p=E/c when the photon is totally absorbed (black body) and p=2E/c when
> the photon is totally reflected (perfect mirror)."http://www.asterism.org/tutorials/tut21-1.htm
>
> If photon reflection indeed added energy to a solar sail, then
> bouncing photons between two solar sails could power both sails.  Each
> bounce could add twice the photon's energy to the KE of the mirror.
> For example, a photon that bounces 20 times before veering off could
> provide about 40x the energy of the photon. Unlike gas molecules that
> lose energy as the molecules expand (increased distance between
> molecules), photons have the same energy before and after each bounce.

Photons do not bounce. Referring to them as rubber balls is foolish.
TreBert

[richard....@comcast.net]

>> "In the current standard model of cosmology, galaxies with a redshift
>> of about 1.5--that is, whose light has a wavelength 150 percent longer
>> than the laboratory reference value--are receding at the speed of
>> light. Astronomers have observed about 1,000 galaxies with redshifts
>> larger than 1.5. That is, they have observed about 1,000 objects
>> receding from us faster than the speed of light. Equivalently, we are
>> receding from those galaxies faster than the speed of light."
>> http://space.mit.edu/~kcooksey/teaching/AY5/MisconceptionsabouttheBigBang_ScientificAmerican.pdf
>>
>> The above refutes the myth that redshift is a function of the speed an
>> emitter toward or away from an observer. The speed between emitter
>> and observer cannot exceed the speed of light,
>
>I'm sorry, but that's an error. There is no such statement in
>relativity, and that's precisely the misconception that the authors of
>the article in Scientific American are working hard to dispel.

The authors of the Scientific American article listed two redshift
methods, w/o making a distinction between the two. 'REDSHIFT DURING
LIGHT'S JOURNEY' and 'REDSHIFT FROM RELATIVISTIC SPEEDS'.

1. Redshift during light's journey:

"As space expands, light waves get stretched. If the universe doubles

in size DURING the waves' JOURNEY, their wavelengths double and their
energy is halved."

"Although the light beam is traveling toward us at the maximum speed
possible, it cannot keep up with the stretching of space."
http://space.mit.edu/~kcooksey/teaching/AY5/MisconceptionsabouttheBigBang_ScientificAmerican.pdf

Note: Redshift during light's journey within a vacuum violates the
conservation of energy law, because energy disappears during redshift.

2. Redshift from relativistic speeds:

"the recession velocity of a galaxy away from us (v) is directly
proportional to its distance from us (d), or v = Hd."

"galaxies beyond a certain distance, known as the Hubble distance,
recede faster than the speed of light."

Note: Zero photon momentum would cause the relativistic emitter method
to violate the conservation of energy law.

Photons do not have momentum. Otherwise, a mirror that reflects a
photon must receive KE. KE receipt would be a violation of the
conservation of energy law, because photon energy does not change
during reflection.

>> I think that showing the difference between energy due to temperature
>> versus KE may be useful in explaining big bang cooling.
>
>It's going to help if you first learn what the current model says before
>proposing an alternative.
>
>It's really not a sound approach to propose an alternative BECAUSE you
>don't understand the current model.
>

No photon momentum means zero redshift from relativistic motion of an
emitter, which means average distant galaxy speeds are not
relativistic, which means galaxies do not have radial movement away
from a black hole, which means no black hole.

Note: If your reading reference for cosmology explains how galaxies
got their velocity distribution (zero to relativistic, versus all
having the same velocity), then please include it in your response.

[PD]

And in fact, the law of conservation of energy is a statement that has
some qualifications on it, as we have discussed. It is only conserved in
a particular reference frame, and is not conserved if you change
reference frames. In expanding spacetime, this qualification is important.

Please do not try to reject relativity by making the erroneous statement
that energy is conserved, as a flat and absolute statement without
qualifications about reference frame. That is not what the law of
conservation of energy actually says. Relativity and the *correct*
statement of conservation of energy are *completely* consistent.

>
> Photons do not have momentum.

Oh but they do. This is measured. Photons can knock electrons right out
of atoms, and the electrons fly in the direction of the photon's motion.

> Otherwise, a mirror that reflects a
> photon must receive KE.

And it does. This is the principle behind a solar sail.

> KE receipt would be a violation of the
> conservation of energy law, because photon energy does not change
> during reflection.

It does if the mirror recoils. Measurable effect.

There's an interesting classical physics exercise you can do. Bounce a 1
kg rubber ball off a 5000 kg brick wall mounted on a hoverpad. You will
find that the brick wall ends up with almost TWICE the momentum the ball
had originally, but practically no kinetic energy. This is all
*completely* consistent with conservation of energy and momentum, if you
work through the calculations.

>
>>> I think that showing the difference between energy due to temperature
>>> versus KE may be useful in explaining big bang cooling.
>>
>> It's going to help if you first learn what the current model says before
>> proposing an alternative.
>>
>> It's really not a sound approach to propose an alternative BECAUSE you
>> don't understand the current model.
>>
> No photon momentum means zero redshift from relativistic motion of an
> emitter, which means average distant galaxy speeds are not
> relativistic, which means galaxies do not have radial movement away
> from a black hole, which means no black hole.
>
> Note: If your reading reference for cosmology explains how galaxies
> got their velocity distribution (zero to relativistic, versus all
> having the same velocity), then please include it in your response.

It does. There are lots of good readings here.
I'd start with Peter Coles' very short book on Cosmology, and then I
would move up to Weinberg's book on the same subject.

[richard....@comcast.net]

On Fri, 23 Dec 2011 10:13:02 -0600, PD <thedrap...@gmail.com>
wrote:

>On 12/23/2011 7:31 AM, richard....@comcast.net wrote:
>>>> "In the current standard model of cosmology, galaxies with a redshift
>>>> of about 1.5--that is, whose light has a wavelength 150 percent longer
>>>> than the laboratory reference value--are receding at the speed of
>>>> light. Astronomers have observed about 1,000 galaxies with redshifts
>>>> larger than 1.5. That is, they have observed about 1,000 objects
>>>> receding from us faster than the speed of light. Equivalently, we are
>>>> receding from those galaxies faster than the speed of light."
>>>> http://space.mit.edu/~kcooksey/teaching/AY5/MisconceptionsabouttheBigBang_ScientificAmerican.pdf

Frames apply to KE of things with momentum.

>>
>> Photons do not have momentum.
>
>Oh but they do. This is measured. Photons can knock electrons right out
>of atoms, and the electrons fly in the direction of the photon's motion.
>
>> Otherwise, a mirror that reflects a
>> photon must receive KE.
>
>And it does. This is the principle behind a solar sail.

Imagine a photon zigzagging between two solar sails. If each photon
reflection added 2x energy to a solar sail, then the sails could
produce more power out than power in. For example, a photon that
reflects 20 times before veering off could provide about 40x the
energy of the photon.

>> KE receipt would be a violation of the
>> conservation of energy law, because photon energy does not change
>> during reflection.
>
>It does if the mirror recoils. Measurable effect.
>
>There's an interesting classical physics exercise you can do. Bounce a 1
>kg rubber ball off a 5000 kg brick wall mounted on a hoverpad. You will
>find that the brick wall ends up with almost TWICE the momentum the ball
>had originally, but practically no kinetic energy. This is all
>*completely* consistent with conservation of energy and momentum, if you
>work through the calculations.
>

Billiard balls are a better example. If the ball hits a pool table's
rim, compression and expansion returns the ball's KE. If the ball
hits another ball, then the incoming ball can stop, while the struck
ball gains the KE. In other words, during those collisions with low
compression / expansion, KE transfer can be high.

A photon must disappear (complete conversion of energy) if the photon
adds KE or compression energy to a mirror. The compression energy or
KE cannot return the energy to a photon that is not there anymore. A
mirror's reflection would be absorption (conversion to compression or
electron KE) and then emission in the direction of reflection.

Photon propagation through changes in refractive index can cause
reflection in a dielectric mirror, not absorption / emission from
individual molecules within the mirror. In other words, photon
reflection is a propagation phenomenon, not a function of momentum.
http://en.wikipedia.org/wiki/Dielectric_mirror

[richard....@comcast.net]

>> Photons do not have momentum.
>
>Oh but they do. This is measured. Photons can knock electrons right out
>of atoms, and the electrons fly in the direction of the photon's motion.

Are you talking about ESCA? Knocking out an electron consumes the
photon (versus reversibly exchanging momentum.) Stimulated emissions
(laser) also preserve photon direction. ESCA and stimulated emissions
preserve direction as part of a quantized molecular energy change
function, not as a collision function. Otherwise, electron direction
after a collision with a photon would be a function of the electron's
contact point on the photon's wavefront.

Perhaps the future existence of big bang theory depends on the
existence of momentum within a photon. Light energy is not a function
of speed. Therefore, photons do not contain momentum.

For example: Photons slow w/o changing energy within transparent
molecular material. Photons speed up again when they leave the
material. A photon w/ momentum would not regain original (vacuum)
speed.

[Sam Wormley]

On 12/25/11 9:56 PM, richard....@comcast.net wrote:
> For example: Photons slow w/o changing energy within transparent
> molecular material. Photons speed up again when they leave the
> material. A photon w/ momentum would not regain original (vacuum)
> speed.

Transparent molecular material doesn't absorb photons, per se -
however, the photons are absorbed by the inter atomic bonds (phonons)
and re-emitted, this is essentially why the speed of light in glass
appears slower.

From the quantum mechanical perspective, all photons travel at c.

1. photons are emitted (by charged particles including phonons)
2. photons propagate at c
3. photons are absorbed (by charged particles including phonons)

Photon momentum
p = hν/c = h/λ

Photon Energy
E = hν

[richard....@comcast.net]

On Sun, 25 Dec 2011 22:26:06 -0600, Sam Wormley <swor...@gmail.com>
wrote:

>On 12/25/11 9:56 PM, richard....@comcast.net wrote:
>> For example: Photons slow w/o changing energy within transparent
>> molecular material. Photons speed up again when they leave the
>> material. A photon w/ momentum would not regain original (vacuum)
>> speed.
>
>
> Transparent molecular material doesn't absorb photons, per se -
> however, the photons are absorbed by the inter atomic bonds (phonons)
> and re-emitted, this is essentially why the speed of light in glass
> appears slower.

The electromagnetic spectrum includes ELF (100 thousand miles
wavelength, 3Hz). Do you really claim that (all) the molecules within
the photon's volume absorb and emit the photon as it travels at the
speed of light?
http://en.wikipedia.org/wiki/Electromagnetic_spectrum

>
> From the quantum mechanical perspective, all photons travel at c.
>
> 1. photons are emitted (by charged particles including phonons)
> 2. photons propagate at c
> 3. photons are absorbed (by charged particles including phonons)

Almost by definition, wavelength specific interactions, such as
phonons can only apply to a specific electromagnetic range.
>
> Photon momentum
> p = h?/c = h/?
>
> Photon Energy
> E = h?

[Sam Wormley]

That was the assumption when talking about transparent molecular
materials.

[PD]

Frames apply to ALL physical systems.
Please don't make up a crap sentence if you're not even sure what the
words mean.

>
>>>
>>> Photons do not have momentum.
>>
>> Oh but they do. This is measured. Photons can knock electrons right out
>> of atoms, and the electrons fly in the direction of the photon's motion.
>>
>>> Otherwise, a mirror that reflects a
>>> photon must receive KE.
>>
>> And it does. This is the principle behind a solar sail.
>
> Imagine a photon zigzagging between two solar sails. If each photon
> reflection added 2x energy to a solar sail,

Hold on. Who said anything about 2x *energy* of the photon?
Read the below about the ball bouncing off a wall.

> then the sails could
> produce more power out than power in. For example, a photon that
> reflects 20 times before veering off could provide about 40x the
> energy of the photon.
>
>>> KE receipt would be a violation of the
>>> conservation of energy law, because photon energy does not change
>>> during reflection.
>>
>> It does if the mirror recoils. Measurable effect.
>>
>> There's an interesting classical physics exercise you can do. Bounce a 1
>> kg rubber ball off a 5000 kg brick wall mounted on a hoverpad. You will
>> find that the brick wall ends up with almost TWICE the momentum the ball
>> had originally, but practically no kinetic energy. This is all
>> *completely* consistent with conservation of energy and momentum, if you
>> work through the calculations.
>>
> Billiard balls are a better example. If the ball hits a pool table's
> rim, compression and expansion returns the ball's KE.
> If the ball
> hits another ball, then the incoming ball can stop, while the struck
> ball gains the KE. In other words, during those collisions with low
> compression / expansion, KE transfer can be high.

Correct, but that's in a case where you have two objects with equal
mass. That's why I gave you the example of the rubber ball bouncing off
a wall, to show what happens there. Just so that you won't make some
idiotic mistake that photons bouncing off solar sails should behave the
way billiard balls do.

Please spend a little more time learning some basics. It will save you
lots of time wasted on silly mistakes.

>
> A photon must disappear (complete conversion of energy) if the photon
> adds KE or compression energy to a mirror.

That's simply flat wrong.

[PD]

On 12/25/2011 9:56 PM, richard....@comcast.net wrote:
>>> Photons do not have momentum.
>>
>> Oh but they do. This is measured. Photons can knock electrons right out
>> of atoms, and the electrons fly in the direction of the photon's motion.
>
> Are you talking about ESCA? Knocking out an electron consumes the
> photon (versus reversibly exchanging momentum.)

Nope.

[richard....@comcast.net]

On Mon, 26 Dec 2011 08:53:29 -0600, Sam Wormley <swor...@gmail.com>

I had questioned photon propagation by means of absorption / emission.
In other words, even if phonons or individual molecules ABSORB / EMIT
photons, those phonons or individual molecules are not 'transparent'
at those particular wavelengths.

1. Even at the wavelengths corresponding to phonons, a large number of
molecules can be within the volume of a single photon. For example,
the number of STP air molecules within a 600nm photon is about 3
million (photon volume versus Avogadro's number.)

2. Within a dielectric mirror, each photon simultaneously detects two
locations of a change in refractive index (top and bottom of a layer)
and the thickness of the layer (relative to wavelength).

A million molecules within the photon are not likely absorbing and
emitting the photon, at the time photon is detecting layer thickness?
http://en.wikipedia.org/wiki/Dielectric_mirror

3. Metamaterials can produce a negative refractive index, which can
reverse the Doppler effect.
"The Doppler shift is reversed: that is, a light source moving toward
an observer appears to reduce its frequency."
http://en.wikipedia.org/wiki/Metamaterial

[Sam Wormley]

Who knows what reality is, but this model of absorption and emission
works.

>
> 1. Even at the wavelengths corresponding to phonons, a large number of
> molecules can be within the volume of a single photon. For example,
> the number of STP air molecules within a 600nm photon is about 3
> million (photon volume versus Avogadro's number.)

Size is not a property of photons.

[richard....@comcast.net]

On Mon, 26 Dec 2011 10:39:10 -0600, PD <thedrap...@gmail.com>

"You will find that the brick wall ENDS UP WITH almost TWICE the

momentum the ball had originally, but practically no kinetic energy."

In other words, the first bounce gives the wall twice the momentum of
the ball, w/o the wall gaining practically no KE. I think you are
wrong about the zero KE. The wall cannot gain momentum w/o gaining
KE. A brick that gains of 2x momentum of the ball, must also gain
four times the KE of the ball. Momentum=mv. 2x momentum = 2mv. KE =
m(2v)^2 = 4mv^2.


Throughout this thread, you have been saying that redshift caused by
relativistic emitter speed does not violate the conservation of energy
law. I still think you are wrong.

Regardless of frame, SR or GR, E=hv defines a photon's energy. If you
think otherwise, provide reference. By 'regardless of frame', I mean
that if a galaxy (moving at relativistic speed) emits a yellow photon,
that photon will be yellow within the galaxy and will be yellow on
Earth (assuming propagation through a vacuum). If redshift causes the
energy used to create the yellow photon to be higher than the energy
of the photon, a violation of the conservation of energy law occurred.

[PD]

Nope. You've not done the math correctly. This is a simple exercise in a
freshman textbook.

The ball has a mass 1 kg and the wall has a mass of 5000 kg. We'll throw
the ball at the wall at a speed of 15 m/s.

We know momentum is conserved and in this case kinetic energy is
conserved. We'll call the final speed of the ball v and the final speed
of the wall V.

Conservation of momentum tells you:
(1 kg)(15 m/s) + (5000 kg)(0 m/s) = (1 kg)(-v) + (5000 kg)(V)
Conservation of kinetic energy tells you:
(1/2)(1 kg)(15 m/s)^2 + (1/2)(5000 kg)(0 m/s)^2 = (1/2)(1 kg)v^2 +
(1/2)(5000 kg)V^2.

You now have two equations and two unknowns, so you can solve for v and
V. This is 8th grade algebra. I presume you can do this.

If you do the simple arithmetic, here are the answers you get, rounded
to two digits:
V = 0.0060 m/s (that's the final speed of the wall)
v = 15 m/s (that's the final speed of the ball)
You'll notice that if you plug these numbers in the equations above they
do end up honoring the equal sign.

Now that we have these, we can find some numbers about momentum and energy.
Initial momentum ball: (1 kg)(15 m/s) = 15 kg*m/s.
Final momentum wall: (5000 kg)(0.0060 m/s) = 30 kg*m/s.
Initial KE ball = (1/2)(1 kg)(15 m/s)^2 = 112.5 J.
Final KE wall: (1/2)(5000 kg)(0.0060 m/s)^2 = 0.090 J.

As you can see, conservation of momentum and kinetic energy produces the
results I told you:
- The momentum of the wall is twice the initial momentum of the ball.
- The final kinetic energy of the wall is small compared to the initial
kinetic energy of the ball.

>
>
> Throughout this thread, you have been saying that redshift caused by
> relativistic emitter speed does not violate the conservation of energy
> law. I still think you are wrong.

I can't help that. I can give you information that you can use to find
your mistake. I can't guarantee that any of that information would
convince you, though it does seem to convince practically everybody else.

>
> Regardless of frame, SR or GR, E=hv defines a photon's energy.

Yes, and E and v are both frame dependent.

> If you
> think otherwise, provide reference. By 'regardless of frame', I mean
> that if a galaxy (moving at relativistic speed) emits a yellow photon,
> that photon will be yellow within the galaxy and will be yellow on
> Earth (assuming propagation through a vacuum).

Hold on. What do you think a frame is? It's not a region around the
source or a region around the observer. Is that what you think a frame is?

[Darwin123]

On Dec 30 2011, 7:19 am, richard.desan...@comcast.net wrote:
It appears that you are trying to analyze this situation by
classical physics, by which I mean the physics before Einstein.
However, you are making mistakes with regards to the classical
physics. You have made some statements that aren't true in either
classical physics or relativistic physics.
You need some lessons in classical physics, wmeaning the
mechanics of Newton and the optics of classical Doppler. I will ignore
SR, GR and QM for now. I will answer your questions in terms of
classical physics.

> In other words, the first bounce gives the wall twice the momentum of
> the ball, w/o the wall gaining practically no KE.  I think you are
> wrong about the zero KE.  The wall cannot gain momentum w/o gaining
> KE.

This doesn't make sense even in Newtonian physics. Momentum is a
vector, not a scalar. Therefore, there is an issue with the direction
of the momentum. Momentum can change direction without changing the
kinetic energy.

>A brick that gains of 2x momentum of the ball, must also gain
> four times the KE of the ball.

The kinetic energy in any direction depends on the speed of the
ball. Therefore, if the ball merely reverses direction it has not
changed its kinetic energy. This is basically what happens when a ball
bounces against the wall.
There is a tiny change in kinetic energy of the ball. You can
hear some of the kinetic energy that leaves the ball. However, it is
very small.

>Momentum=mv.  2x momentum = 2mv.  KE =
> m(2v)^2 = 4mv^2.

Here, v is a speed, not a direction. Direction in this case is
characterized by the sign. The square of a number is equal the square
of the negative of that number. So,
(-v)^2=v^2
m(-2v)^2 = 4m(v)^2.
Reversal of sign changes the component of momentum in the
direction of the wall. It does not change the kinetic energy.

>
> Throughout this thread, you have been saying that redshift caused by
> relativistic emitter speed does not violate the conservation of energy
> law.

Kinetic energy is observer dependent in classical physics. One
doesn't need relativity to see that the kinetic energy changes with
the observer. The way that the kinetic energy changes in different in
classical and relativistic physics.
Conservation of energy is not violated just because an observer
sees a different kinetic energy than another observer. The kinetic
energy seen by two observers, one moving toward a source and one not
moving relative to the source, will be different even in classical
physics.
Conservation of energy is violated if the same observer traveling
at a constant speed sees a change in energy. However, that is not the
Doppler shift. It is not the relativistic Doppler shift and it is not
the classical Doppler shift. I will stick with classical Doppler for
now.

>I still think you are wrong.

The energy density of a light wave depends on the amplitude of
the electric field. The amplitude of the light wave is not changed by
the classical Doppler shift. Therefore, the energy density of the wave
is not changed by the Doppler shift.
The apparent speed of light changes for the observer headed toward
the light source. The flux of energy is the density of energy times
the speed of the light wave. The flux of energy passing a certain
point increases for the observer heading toward the source.
Therefore, observed energy absorbed by the observer increases for an
observer headed toward the source relative to an observer standing
still with respect to the source.

>
> Regardless of frame, SR or GR, E=hv defines a photon's energy.  If you
> think otherwise, provide reference.

That formula is quantum mechanical. I said I will ignore quantum
mechanics for now.
The formula defines the photons energy, not the energy of of the
light wave absorbed by the observer. The energy density of the light
wave depends on the amplitude of the wave, meaning the magnitudes of
the electric field and magnetic field. Doppler predict that the
apparent frequency of the wave is changed by the observers motion. The
number of peaks hitting the observer changes. Peaks are not photons,
but they move at the same speed as the photons.
  The total amount of energy headed toward the observer per unit
time is given by the flux of energy. It doesn't matter if the energy
is in discrete chunks or continuously distributed.

> that if a galaxy (moving at relativistic speed) emits a yellow photon,
> that photon will be yellow within the galaxy and will be yellow on
> Earth (assuming propagation through a vacuum).

Not if the instrument on earth is moving at high speed relative to
the center of the earth, or the other galaxy.

>  If redshift causes the
> energy used to create the yellow photon to be higher than the energy
> of the photon, a violation of the conservation of energy law occurred.

Two observers on earth, moving at different speeds toward or away
from the other galaxy, would measure different frequencies and
different photon energies.
This is with or without relativity. Kinetic energy is always
observer dependent. Relativity doesn't qualitatively matter in this
regard. Kinetic energy measured by two observers moving at high
velocity relative to each other will always be different.

[richard....@comcast.net]

On Wed, 4 Jan 2012 18:16:57 -0800 (PST), Darwin123
<drose...@yahoo.com> wrote:

>On Dec 30 2011, 7:19 am, richard.desan...@comcast.net wrote:
> It appears that you are trying to analyze this situation by
>classical physics, by which I mean the physics before Einstein.
>However, you are making mistakes with regards to the classical
>physics. You have made some statements that aren't true in either
>classical physics or relativistic physics.
> You need some lessons in classical physics, wmeaning the
>mechanics of Newton and the optics of classical Doppler. I will ignore
>SR, GR and QM for now. I will answer your questions in terms of
>classical physics.
>> In other words, the first bounce gives the wall twice the momentum of
>> the ball, w/o the wall gaining practically no KE.  I think you are
>> wrong about the zero KE.  The wall cannot gain momentum w/o gaining
>> KE.
> This doesn't make sense even in Newtonian physics. Momentum is a
>vector, not a scalar. Therefore, there is an issue with the direction
>of the momentum. Momentum can change direction without changing the
>kinetic energy.

I referred PD's statement that "the first bounce gives the wall twice
the momentum of the ball, w/o the wall gaining practically no KE." If
this is that statement that bothers you, why not respond directly to
PD (instead of me).

The above is primarily about the wall's change from zero momentum
(thus no direction) to acquiring twice the ball's momentum. The above
makes no mention of the wall's or the ball's direction of momentum.
Please be more explicit about why you think the above statement
indicates that either of us either of us treated momentum as a scalar,
instead of a vector.

>>A brick that gains of 2x momentum of the ball, must also gain
>> four times the KE of the ball.
> The kinetic energy in any direction depends on the speed of the
>ball. Therefore, if the ball merely reverses direction it has not
>changed its kinetic energy. This is basically what happens when a ball
>bounces against the wall.
> There is a tiny change in kinetic energy of the ball. You can
>hear some of the kinetic energy that leaves the ball. However, it is
>very small.

I assumed that ball remained virtually unchanged in magnitude of
momentum. If you think otherwise, please be explicit.

>>Momentum=mv.  2x momentum = 2mv.  KE =
>> m(2v)^2 = 4mv^2.
> Here, v is a speed, not a direction. Direction in this case is
>characterized by the sign. The square of a number is equal the square
>of the negative of that number. So,
> (-v)^2=v^2
> m(-2v)^2 = 4m(v)^2.
> Reversal of sign changes the component of momentum in the
>direction of the wall. It does not change the kinetic energy.
>

Are you claiming that the wall's KE remained zero, (instead of
becoming 4mv^2), because of the direction of the wall's velocity?

m(-2v)^2 = m(2v)^2 = 4m(v)^2 (=0?)

I claim that a violation of the conservation of momentum law occurs if
the wall's magnitude of momentum does not equal the magnitude of the
amount of momentum that the ball lost. PD claimed, "The brick wall

ends up with almost TWICE the momentum the ball had originally"
>>

>> Throughout this thread, you have been saying that redshift caused by
>> relativistic emitter speed does not violate the conservation of energy
>> law.
> Kinetic energy is observer dependent in classical physics. One
>doesn't need relativity to see that the kinetic energy changes with
>the observer. The way that the kinetic energy changes in different in
>classical and relativistic physics.

KE has frame dependence, not observer dependence.

> Conservation of energy is not violated just because an observer
>sees a different kinetic energy than another observer. The kinetic
>energy seen by two observers, one moving toward a source and one not
>moving relative to the source, will be different even in classical
>physics.

I thought both observers could have similar (or the same)
observations, because KE estimates are based on multiple measurements.
Are you referring to a special circumstance?

For example, we might know the mass and velocity of a distant galaxy,
based on galaxy characteristics and light received from the galaxy. An
observer on that galaxy can similarly estimate the mass and velocity
of his and our galaxy. Unless neither of us knew how to handle
redshift, why would the estimates of galaxy mass and velocity differ?

> Conservation of energy is violated if the same observer traveling
>at a constant speed sees a change in energy. However, that is not the
>Doppler shift. It is not the relativistic Doppler shift and it is not
>the classical Doppler shift. I will stick with classical Doppler for
>now.
>
>>I still think you are wrong.
>
> The energy density of a light wave depends on the amplitude of
>the electric field. The amplitude of the light wave is not changed by
>the classical Doppler shift. Therefore, the energy density of the wave
>is not changed by the Doppler shift.
> The apparent speed of light changes for the observer headed toward
>the light source. The flux of energy is the density of energy times
>the speed of the light wave. The flux of energy passing a certain
>point increases for the observer heading toward the source.
>Therefore, observed energy absorbed by the observer increases for an
>observer headed toward the source relative to an observer standing
>still with respect to the source.
>

RF antennae are orientated to the broad side of RF waves. To detect
the time between wavefronts, antenna orientation would instead be 90
degrees away from facing the wavefront. In other words, time between
wavefronts does not directly influence wavelength detection.

>
>>
>> Regardless of frame, SR or GR, E=hv defines a photon's energy.  If you
>> think otherwise, provide reference.
> That formula is quantum mechanical. I said I will ignore quantum
>mechanics for now.
> The formula defines the photons energy, not the energy of of the
>light wave absorbed by the observer. The energy density of the light
>wave depends on the amplitude of the wave, meaning the magnitudes of
>the electric field and magnetic field. Doppler predict that the
>apparent frequency of the wave is changed by the observers motion. The
>number of peaks hitting the observer changes. Peaks are not photons,
>but they move at the same speed as the photons.
>   The total amount of energy headed toward the observer per unit
>time is given by the flux of energy. It doesn't matter if the energy
>is in discrete chunks or continuously distributed.

RF antennae detect wavelength, not energy density. Detection is just
as efficient for RF photons that arrive at the same time, as it is for
those photons that arrive singly. The ones that arrive simultaneously
have double energy density.

>> that if a galaxy (moving at relativistic speed) emits a yellow photon,
>> that photon will be yellow within the galaxy and will be yellow on
>> Earth (assuming propagation through a vacuum).
> Not if the instrument on earth is moving at high speed relative to
>the center of the earth, or the other galaxy.

Perhaps, your photon detector detects time between wavefronts or
energy density. My money is on detection of photon diameter.

>>  If redshift causes the
>> energy used to create the yellow photon to be higher than the energy
>> of the photon, a violation of the conservation of energy law occurred.
> Two observers on earth, moving at different speeds toward or away
>from the other galaxy, would measure different frequencies and
>different photon energies.
> This is with or without relativity. Kinetic energy is always
>observer dependent. Relativity doesn't qualitatively matter in this
>regard. Kinetic energy measured by two observers moving at high
>velocity relative to each other will always be different.

I think the two observers on Earth would detect the same energy per
photon and wavelength. The detected radiation can be RF. The speed
of the detector (antenna) does not change photon wavelength. As
mentioned earlier, an RF antenna detects wavelength, not energy
density.