Re: Redshift without expansion





George Dishman wrote:
sean wrote:
George Dishman wrote:

... a lot more than is here but Sean snipped
most again. I will replace context where needed.

That is correct, the effect is still to stretch any
signal which lasts for a finite time as if the
waveform were drawn on a rubber band. The
point here is that the waveform preserves its
shape.

But you can keep things simpler than pretending EM waves are made of
stretchy rubber bands . If you think about it, a freely propogating
dipole as I suggest can also give a doppler effect.

But it does not represent a signal that lasts for
a finite time. If I stand at some point and look at
what is received on an aerial from a broadcast
station, it is an ongoing signal. Your dipole would
only give a momentary value as it passed and then
it is gone. You need something like a train of dipoles.

Yes .

And the train of dipoles gets stretched just like
a rubber band.

Right, that is basic Doppler, they have the same flip
rate measured in the frame of the source but a lower
rate measured by the moving observer.
Think of Escher's "Waterfall". You can illustrate
concepts that are physically impossible.

You have neatly edited what I said, here is the
actual context:

Then no amount of pretty pictures will prove anything.

Dont forget the saying a picture is worth a thousand words.

Think of Escher's "Waterfall". You can illustrate
concepts that are physically impossible.

True, but one can also illustrate concepts that *are* possible.

My point remains, the fact that you can illustrate
something does not prove it is possible.
Conversely..if you can illustrate something it does not prove that it
is impossible. As you seem to be suggesting.
Again restoring context:

Im not sure what you do for a living,

I am a design engineer in a communications firm
though mainly on the digital side. I am currently
working in a team developing a 1kW radio transmitter.

Interesting. Im not an transmitter expert so the importance of 1kw
eludes me. Is that unique in that it is a small power source? And
therefore portable?

No, it is built into a rack about 6 feet high and
permanently installed. Some are on ships while
others are used for air traffic control.

No, a 'photon' is the name we give to a particle
and it is the total energy carried by the particle
that is measured.

E=hv has a time t component.

No E is the amount of energy carried by the
photon.
I wasnt disputing that E is the amount of energy in each `photon`. But
how is E described? By hv. And how is v described?.. As so many per
*second*. If you say no time component is involved in measuring E then
show me how to define 200nm in Hz without refering to seconds (time).
You cant.
H is measured in joules /second. And v the
frequency is meausured in seconds. So when you measure 2 photons you
measure how much energy is in the two over the same time period.

No, you measure how much energy they carry.
What is found is that the amount is proportional
to the frequency but it is a discrete amount of
energy.
Same answer as above. Frequency is defined as " the number of complete
oscillations in unit time"And Hz is in seconds. You have to know how
many cycles were observed each second. How could you calculate E
without knowing v ? And how can you define v without knowing how many
cycles *per second*?
Thats
why a 200nm `photon` appears to have twice the energy of a 400nm
photon. You just measured them over the same time frame. But if you
measured the 200nm photon for half the amount of time you measured the
400nm photon you would find the two `photons` have the same energy

Rubbish. I can set up a detector to measure two
photons arriving a minute apart and the total
energy is just the sum of the individual energies.
It makes no difference how long you measure for.
Describe in detail how you measured those two photons and youll find
your measurements all include time.
And it does matter how long you measure for. To `observe` a photon ,
you have to measure for a mininum set time in the sense that at any
observed wavelength you will get so many photons per second.You get
more photons the longer the time measured. And you get less photons the
shorter the time measured. Both you and Eric suggest that it takes no
time at all to measure a photon. Come on , thats impossible.
But the plate in the photoelectric effect doesn't have
a specific frequency, it has a lower cutoff. What you
are ignoring is that the evidence is based not only on
the existence of the cutoff but importantly on the
linear relationship between the electron energy and
the incident frequency.

I always thought till now that pmd sensitivities to narrow bandwidths
represented the photolectric effect but going back now and reading up
on it and photovoltaic effect I realize that my model would have to
have the classical atom resonating at many frequencies to allow the
continuous effect above the cutoff. Kind of like a chord rather than a
note. (does make me ask the question why all pmds dont respond to al
frequencies above the cutoff rather than a narrow bandwidth)

Well take to the extreme, eventually gamma rays
will go right through the detctor, but over a reasonable
range there is no evidence of any sort of resonance at
all.
What a strange thing to say. If resonance in atoms is manifested to us
the observor as the photoelectric effect (as explained by the classical
model) then the fact is that the photoelectric effect is the evidence.

As a matter
of interest though .. In your model why does the cutoff level exist?
Why is it that `photons` from above a certain frequency level eject
electrons and those below dont. I can only imagine the explanation
maybe might be.. that electrons are deemed to have the basic energy
level of x and that is equivelent to the energy x of a `photon` of such
and such a wavelength, above which all shorter wavelength `photons` can
eject electrons and below they cant?

Almost right. Think of a pool table with dents. A ball
sitting in a hollow takes some energy to liberate it.
If the cue ball is moving too slowly, the target ball
isn't displaced. At higher speeds the target ball is
knocked out of the hollow and the faster the cue ball
the higher the speed of the target ball after it is ejected.
But If this is the answer then
why is it that the cutoff isnt at the same frequency for all solids?

The electrons on the surface are attracted to the positive
charge on the layer of nuclei of the metal. Different metals
have different binding strengths hence different amounts of
energy are needed to liberate the electrons.

I have to admit I find it hard to believe that all solids accept
radiation above the cutoff level
homogenously. For instance if one were to draw the response curve of an
element to light above the cutoff; is it a flat featureless standard
profile the same for all other elements?

Pretty much. There will always be slight variations
due to other effects that happen at the same time
but for the photoelectric effect it is featureless.

Or are there distinctive
profiles for each element. After all my reference says that the cutoff
level varies for each element (and doesnt explain why).The reason I ask
is also that the emission profiles of elements have distinctive spikes
for each element. Why not the same for the photoelectric effect?

Because the electrons are not being ejected from the
atoms themselves but from what is essentially a freely
moving layer of electrons on the surface. Did you look
up conduction in metals as I suggested?

We also see absorbtion lines for elements not being flat continuum but
rather spikes. That to me also suggests that different solids must
respond differently to light.

Yes but those are atomic spectra.
And those spikes are the frequencies at which the atoms must oscillate
at in my model. However Im not familar enough with experimental setups
of the photoelectric effect. So I have to find out more about the
details of photoemmision and photoelectric effect before I can supply
an explanation as to how the wave model can explain the observation
that below the cutoff no frequencies are absorbed. For instance in the
setup with light falling onto a zinc plate connected to an electroscope
...what happens if the electroscope has neither positive or negative
charge? And in millikans experiment can I assume that if no stopping
potential is applied a current flows when light falls on the metal
plate? Taken together the two seem to contradict each other. For
instance if I assume when light falls on the plate in millikans
experiment and no stopping potential is applied then it follows that a
current will flow. Two things. THis means that in the electroscope
setup the electroscope if started off with no charge should build up a
charge from the light falling on the zinc plate. Yet no mention is made
of this so I assume incident light on metal plates on its own does not
a current make. Yet millikans experiment seems to suggest just this. If
I follow my interpretation of millikans it implies that any incident
light on any metal should create a current or a charge buildup.
So for the moment I cant explain the why the cutoff level at that
partricular level exists in my model. Although If I follow the route
your model takes and explain the electron as a base unit of energy
without explaining why its that amount then its a bit easier... My
model shows that a wave atom acts as a resonating node like a macro
resonating system. In the macro system we observe it accepts incident
vibration to a certain limit where the system then `breaks`, releases
the energy catastrophically and falls back to default energy level. THe
same happens in atoms but importantly all atoms respond to any
incident radiation by accepting a standard amount before breaking and
resetting. Its common between all elements.That standard amount is the
energy level observed as the electron. And above that level the
increased energy of higher frequency radiation to stopping function
gives a corresponding straight line response graph as millikan shows.
Of course the problem is still how does my atom discriminate between
the energy in different frequencies? And why does that vary between
elements? For the moment Im not sure but one possibility is that if
atomic spectra spikes represent the natural frequencies of each element
then it may be that atoms do not oscillate naturally at frequencies
below those seen in their corresponding work function frequency
levels. So for instance maybe if zinc which has a cutoff in
ultraviolet, how does that correspond to its emmision spectra or its
lowest frequency emmision spike? Thats something I have to look up ,
although Im not sure yet where I can get an atomic spectra of zinc.
You are completely confused. What I said was that
the _voltage_ applied across the carbon on the paper
produced an electric field while the magnets produced
the magnetic field. To illustrate the concept of polarisation
we then had the paper either lying on a table (horizontal)
or glued to the wall (vertical). In both, the magnets are
glued with their _ends_ on the paper so the magnetic
field is perpendicular to the electric.

Im not confused.

You were, you said this:

Take a magnet(s) and lay it on a horizontal paper on a table apply a
ac current to it so it flips back and forth 220 times a second. You say
the field thats vertical is the magnetic field and the field thats
horizontal is electric?

That is untrue. What I said was that if you fixed the
magnets _perpendicular_ to the paper and the
paper was horizontal then the magnetic field would
be vertical. You turned the magnets around to make
them horizontal.

Im aware of your model and how it describes EM as
traverse waves etc hence I know you think there are two fields. But
please, dont keep on saying *my* model should have two fields.

I was not saying anything about your model, I pointed
out that MY analogy had the magnets vertical and you
said I was saying the were horizontal.
OK lets say your magnets are vertical on your paper. We still have
magnets on paper having their poles flipped. Why do you need a
seperatee electrical field to explian why the magnetic poles of your
magnets flip? And why if your magnets have fields that surround the
magnet in 360 degrees do you haveto describe the part of the magnetic
field that is at right angles to the n-s axis as electrical? If you
look at a magnet on its own thats not flipping. The part of the
magnetic field that is at right angles to the magnet n-s axis is part
of the magnetic field. When you rotate the magnet why does that part of
the magnetic field suddenly become an electric field? Its still part of
the magnetic field. You just pretend its electric because you need two
fields to describe polarization. I dont. Because the rotating dipole
can use one rotating mag field to describe both flux and polarization
Im just
trying to show you how the `magnets` in your paper can have their
flipping fields explained much more simply and more accurately by just
a single rotating magnetic field. I can take a line of magnets, flip
the poles of the first by hand and if the rest are secured in the right
way they will follow and flip their fields. The same thing is happening
to the point magnetic fields on your paper experiment.

No it isn't. When you flip a vertical magnet it passes
through each angle over 180 degrees and at some point
in that process it is horizontal. In my version that never
happens. What you are decribing is an good analogy
for circular polarisation but doen't work for vertical which
is what I described.
You are confusing many things together here. In your analogy initially
there werent any magnets. There was only graphite on paper with an ac
current and two imaginary fields. I say that ac current if it flows
across the paper flips the magnetic fields back and forth at any
observed point. You say the magnetic fields dont flip? Or are you
saying they do flip but not rotate? Im not sure what you are arguing
now. But if you think they dont flip and/or rotate then prove they
dont. Otherwise its speculation or theoretical premise. It seems to me
you only think there are two fields there. But you cant see them and
any measurement you make is one that only sees a flipping magnetic
fileds at any point. So where you think you find proof for an imaginary
electric field who knows. Its not observed. And if it is tell me how
its observed and describe the mechanism you imagine does this, Ill show
you that observation is actually only of magnetic fields flipping

.

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