Re: Redshift without expansion


sean wrote:
George Dishman wrote:
Ive sent a new simulation to your full name @briar..Is that the right
address?

I haven't seen it yet, the address should start
george@briar and so on. Google will show you
my address if you look at my profile after a simple
check to stop the spammers.

also wikipedia wont open for me I need an account.

Nonsense, it's freely available. You would need an
account if you wanted to edit it

Though it isnt a
simulation made from your equations.

In that case it isn't really a simulation, rather an
illustration. You have written something that shows
what you already think would happen. By using the
actual equations, the computer will tell you what
will happen and if you are lucky it is different from
what you expected and you learn something.

I set up the parameters of an oscillating dipole gave it a log
decrease in rotation and a constant speed of propogation and let it
run. I wasnt too sure what Id get and was half expecting it not to work
fortunately it did. But this is no illustration this is a simulation
that you dont want to believe works.

You are illustrating something you assume,
you are not calculating the equations.

If you emit a modulated signal the whole waveform just
gets stretched. Draw a wiggly line on a rubber band and
then pull.

This analogy doesnt work. Mainly because EMR is not joined up like a
rubber band.

That's exactly how it is. Doppler affects both the carrier
and the modulation so the entire waveform just gets
stretched.

If anything EMR is not a string as some of these equations
pretend. But thats why I beleive they fail to explain the cosmological
redshift

And with the pioneer data as its a 2v
calculation one multiplies the recession of pioneer 2 times to
calculate this additional blue shifting. And thats the exact numbers
one gets if one multiplies this out it confirms this
theory...anomaly*c=2*(pioneer recession velocity)=25!

The speed is 12km/s, the anomaly is an acceleration of
8.74*10^-10 m/s^2. They don't even have the same units
so your claim is obviously wrong.

Look at e=fv. Thats 3 unrelated units. Look at e=mc2. Thats 3 unrelated
units.

Read this and see if you can learn the technique.

http://www.physics.uoguelph.ca/tutorials/dimanaly/

Once you understand that, you'll see why I'm going to

<snip the rest>


The differential form is what I was referring to when I talked
of the 2-D array of people doing a mexican wave.

If you want to choose one, the differential form of
Gauss's Law. It should be one of the easiest for you
to understand and then see the implications of a change.

Ill have to pass on these equations . Youve got me there. Its too
complex for me to decipher what theyre up to and then correct the
equation to allow for cosmological redshift.

Then no amount of pretty pictures will prove anything.

But
your model cant explain polarization or redshift.

Of course it can, you have just never learned how
Maxwell's Equations work.

True, but If you cant explain in words how macwells equations or the
derivitives can explain polarization then no amount of maths will help
either.

But if you look later you'll see I _have_ successfully
explained it to you.

You cant have maths without a theoretical model first.

Wrong, the equations were taken from observations
and propagation was then predicted by the equations.

And it
seem syou cant supply a theoretical model. At least not one that works.
Which makes me believe that the equations are based on an incorrect
model and therefore cannot explain polarization any more than the
incorrect analogy can
No, it only applies to individual photons, it tells you
how much energy it carries. Read this (you should already
be familiar) but now consider if some of those wiggly red
lines are from a nearby static red lamp and others are a
from a blue lamp that is receding such that the Doppler
shift makes them the same colour as those from the red
lamp. The surface doesn't know which they came from and
responds identically.

You forget joules are defined with t, time.

No, you forgot the equation applies to photons.
E=hv is an equation written when it was discovered
that the energy carried by a single photon is
proportional to its frequency.

In other words you arent
measuring how much energy one wavelength of 200nm loses or gains when
it stretches to 400nm. You are measuring how much energy 200nm is, then
measuring how much energy 1/2 of the 400nm wavelength is and then
trying to compare the two . Well thats comparing oranges to apples. You
deliberately dropped out 1/2 of the 400nm wavelngths energy just to
make it look like its 1/2 that of the 200nm energy.

No, that would mean a photon emited at high energy
could dislodge an electron in the photo-electric
effect regardless of the frequency at the plate.

Youll find your reasoning doesnt hold up here. I went over this in
another thread a couple of years ago and showed how the grangier effect
was even better explained as a classical effect.

We are talking about the photoelectric effect.

Look at the proof that
you use to make this assumption. How do you know that only a high
energy photon will knock out the electron?

Because when we shine light of different frequencies onto
a metal surface, the energy of the released electrons is
equal to hv - W and if hv < W then we observe that no
electrons are emitted. W is the amount of energy needed
to liberate the electron from the surface, sometimes called
the "work function".

http://hyperphysics.phy-astr.gsu.edu/hbase/tables/photoelec.html

If I remember correctly the
only way to know this is something like the plate is only sensitive to
certain frequencies.

Wrong, the plate emits electrons regardless of the
frequency provided it is above the threshold at hv = W.

So you havent proved that a 200nm wavelength E
isnt the same as a 400nm wavelength E . You only proved that certain
elements accept only certain wavelengths in the photoelectric effect.
Its been a while since I did this argument but If I remember correctly
the japanese pmd site had different metals and compounds making
different frequency sensitive pmds.

Entirely possible, the value of W depends on the material.

And I think my argument was that as
the classical atom is a oscillating node then it only vibrates
simpathetically to certain input freqencies.Just as observed classicaly
in some closed systems like bells strings etc. And make the atom
vibrate at different frequencies and you make it sensitive to different
input frequencies. So one makes plates out of elements that suit
accepting specific frequencies.

That would create a spectral line, we don't see that,
we see electrons liberated over a wide range of
frequencies with an energy that depends on the
frequency but not the intensity.

And I have pointed out that you just need to listen
on a receiver to hear the hiss, by observation
therefore it is modulated, but you go on ....

I dont follow your logic here.

That's because you snipped the part above to
which I was responding. It isn't important.

If I argue that the reciever modulates
the unmodulated signal then listening to a reciever one must be
listening to a unmodulated signal thats just been modulated. And isnt
this what you say above? So whats your point?

That fact you you tell me something doesn't mean
I have to agree with. Go back and read my reply in
context instead of snipping.

There's something wrong there, antennas DO need to be
at a special angle to receive polarised signals.

Funny you say that. Jan seemed to be suggesting that one *doesnt* need
to orient ones antenae at any specific angle to recieve a polarized
signal?

I don't believe he said that, I think you must have
misread his post.

Who do I believe.

As I said, use Google and find out about TV aerials. Try
the term "Yagi" or "Yagi-uda".

But that's the wrong way round. If you use a vertically
polarised antenna, you don't receive a horizontally
polarised signal. That's why it is useful, if you are
in a region served by two transmitters, one vertical
and one horizontal, you use install the antenna so that
the weaker signal isn't received so doesn't cause
interference.
Different designs of antennas receive different
fields. We have to test all our equipment for
both magnetic field emissions and electric field
emissions to pass the regulations.

I can only assume Jans right and you mean that some extra electrical
feature is added to charge the antennae to effects its atoms and that
polarizes it horizontally r vertically and not that the antennae design
needs to be placed either horizontally or vertically outside your house
to work.

Wrong, I receive from a transmitter with horizontal
polarisation so my aerial is mounted with the director
and dipole elements horizontal. Do you have an aerial
or are you on a satellite dish?

Imagine the magnets are glued perpendicluar to the paper
and the graph above is that end. That is the magnetic
field component. The electric field is provided by the
battery and runs through the paper. Do you see how the
fields are perpendicular? Nothing moves, only field
strengths change.

OK lets say I understand this picture correctly. THis means that at any
point the signal on your paper is measured by the voltmeter and this is
the equivelent of what you say is what happens when one places a ccd at
an equivelnt point in space and measures the EM wave? So the ccd
measures essentially the flipping magnetic pole of the EM wave as it
passes,.. in the same way as your voltmeter measures the to and fro-ing
of the ac current on the paper?

Stick with the TV aerial for the moment, it is easier
to understand when considering the classical wave
version.

Suppose you have the paper lying on a table. the electric
field is horizontal while the magnetic field is vertical.

Now glue the paper to the wall and the elecric field is
vertical while the magnetic field is horizontal.

That is what polarisation means. Horizontal polarisation
has the electric field horizontal.

Assuming Ive got you right here, then
it seems this also fits my analogy of EM waves because my moving
oscillating dipoles would give the same results wouldnt they?

You have to combine pairs of dipoles because the field
doesn't rotate, it remains in one plane and varies in
strength. Also you need to fill space with dipoles since
a signal is extended.

But aside from that I still dont see how if the anaolgy we both seem to
be agreeing on of a flipping magnet applies then all you have is a
field that flips from n to s. Thats one field that oscillates. Not two,
as you suggest with the mag and flux fields at right angles.
I mean lets be really literal and say if you have a magnet and rotate
it so it flips back and forth. What in that mechanism is flux ...

You tell me, I have been careful to talk of the magnetic and
electric fields but you keep using the term 'flux' and you
haven't said what you mean by it.

... and what
is magnetic field? To me all I see is a mag field oscillating.

Right, but to get a propagating wave you need both electric and
magnetic fields. If you create magnetic only, it gradually changes
into a combination. This is what is known as "near field" versus
"far field". This is an extensive and complex part of the subject
so I'll leave you to dig around for yourself.

Theres
no flux unless you ask what is turning the magnetic field. In which
case I would say that to get a constant rotation ..

Note nothing is rotating in what I described.

.. in the literal
magnetic field one would have to apply a constant force to turn it. And
that measn that in a mechanistic picture flux doesnt oscillate. It only
has an amplitude. The oscillation is in the magnetic field. And in fact
this is how electricity is generated isnt it? THe ac current comes
quite literraly from the constant force of the water turning the magnet
so that in consequence the atoms in the wires have their atoms fields
physically flipped back and forth. So here the flux is the constant
strength or amplitude of the force of the water and the magnetic field
is the n-s flipping. To me thats the same thing as EM waves and it only
works if the flux has no fluctuation.

We are not talking of motors.

What about the one in between, say B:

000A00000C00000*
00A00B00C00*00*00*

It should apear on the line above but vanishes.

Ive tried to make clear that where they are there they are continuous ,
but where they disappear it is always at percentages that correspond to
e=hv.

That is nonsense. Suppose each asterisk represents
the news headlines on the hour. What you show above
is that the 1pm and 3pm news travels but the 2pm news
vanishes. The same applies to individual cycles of the
carrier.

So the graphs always 1/2 in frequency each level up and notice
that the lines always disappear by 1/2 each level up. Ie each level 1/2
the lines disappear. This I equate to the energy at each subsequent
stretched level as being spread out over twice the length as the last
level. And that ties into the conservation of energy as a wavelength
of 200nm is equivelent to the energy of a wavelength of 400nm

Then show that a wave with decreasing frequency is a
solution to your revised version.

I cant decipher the maths to work with the equations , that I admit.
But one can show in other ways how a oscillating `object` be it a wave
or a propellor , can decrease in rotation frequency and move at a
constant speed. This is indisputable and for you to deney this is to
have you ignore basic physics.

We know red shift occurs and you can produce lots of
pictures or analogies to illustrate it but they remain
purely illustrative. Nice for teaching but not much else.

There are so many macro examples in
every day life that do exactly this and its shocking that you are
unable to accept this. Take a circle of cardboard, spin it on a axis
and move it forward at a constant speed. Then as its moving slow down
the rotation of the cardboard circle. According to you this is
impossible.

No, according to me, it is irrelevant because it
doesn't illustrate a solution to Maxwell's equations.

Its only impossible if you are opposed to the concept
because it ciontradicts what GR predicts. But GR is a theory and the
effect I describe happeens in nature. And fact always comes before
theory.

What I am saying is that I don't think you can
modify the set of equations in differential
form and produce a set that has all the same
solutions over short ranges (so that they still
model antennas etc.) but which produces a
decreasing frequency for a plane wave propagating
over cosmological distances.

The maths isnt my specialty so its hard to do.But my impression is
these equations all deal with how the wave propogates at any one point

Nope, they deal with things like conservation of
charge. Find out instead of guessing - more at
the bottom.

and its only extrapolated that this occurs the same over distance so at
no time do the equations ever look for what happens over distance even
if the frequency were unchanged. So maybe I havent the skill to show a
modification but at the same time you havent actually tested what
happens over distance even if f is unchanged. So you havent any proof
that it cant decrease with distance. You just pretend you do.

The proof is pure maths, a constant frequency sine wave
is a solution to the equations while a distance-dependent
frequency is not.

OK, I see what you are doing, but I can produce that
with a variable frequecy oscillator. Plot it on time
versus frequency for a propagating wave and you'll
find you only have a variable frequency source. It's
not obvious from your drawing.

Im not familiar with a variable oscillator but isnt mine a bit
different from that?
If you look at the simulation the source isnt variable its the
propogating wavefront that is.

Draw it on a 2D graph and you'll see that the source
frequency varies.

Sure, that's easy to draw, it even represents the
conventional GR solution in a way. Now tell me how
you propose to modify Maxwell's Equations to make
that a solution. Then lab experiments can be done
to test for your change.

Like I said already thats something Ive looked at and feel it needs a
better mathematician then me. Maybe Ill hire an acountant.

ROFL! No, he wouldn't be of any use to you.

But by the
same token I can see from the equations that there is no testing that
its not possible either. It seems to me that its assumed the wave
propogates at constants over distance so teh equations just deal with
point observations. What happens at one point in space. Isnt this
correct? I cant see any distance/wavelength functions in any of the
formulas.

Remember I told you that a few days back?

They seem to deal with just one `wavelength` or phase
velocity travelling through one point in space. Im sure youll correct
me if Im wrong but please giv an example this time and specify what
part deals with multiples over distance. Because I cant see any.

Nothing deals with "multiples over distance" as I told
you, that's why what you claimed you could do is
very difficult, perhaps even impossible.

I've given you this link several times, it lists the equations:

http://en.wikipedia.org/wiki/Maxwell_equation#General_case

Take the example I choose of Gauss's Law:

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/gaulaw.html

Read that page carefully and see if you can work out what
that law is saying.

Then look at the others

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/maxeq2.html#c2

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/maxeq2.html#c3

http://en.wikipedia.org/wiki/Amp%C3%A8re%27s_law

There are many pages on these all over the net. Do some
reading and find out what they are telling you. You cannot
expect to discuss something like Maxwell's Equations
without first having studied them a bit.

George

.

Relevant Pages

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