Re: The Emission Theory of Androcles

"Androcles" <Headmaster@xxxxxxxxxxxxxxxxxx> wrote:
"Jonah Thomas" <jethomas5@xxxxxxxxx> wrote
"Androcles" <Headmaster@xxxxxxxxxxxxxxxxxx> wrote:
"Jonah Thomas" <jethomas5@xxxxxxxxx> wrote
"Androcles" <Headmaster@xxxxxxxxxxxxxxxxxx> wrote:
"Jonah Thomas" <jethomas5@xxxxxxxxx> wrote

http://yfrog.com/5ystartg
The star is seen behind where it actually is.

If the movement of the particle I highlighted in red is what's
seen,> > then the star is seen behind where it actually is.

But if the movement of the waves are what's seen, then the star
appears to be exactly in the direction that it actually is.

Nope. You see the star at 1:01 o'clock where it was at 12:00
o'clock,> it took an hour and a minute for the light to reach you.

Construct a table.
Time of emission. theta. Time of arrival:
12:00 89 12:00 + 1-cos
(89) 1:00 90 1:00 +
1-> cos(90) 2:00 91
2:00 +> 1- cos (91) 3:00 92
3:00 + 1- cos (92)

If the next
"wave" or circle is seen at a moment later it is seen in a
different> position.

Sure, but it's passing through the right spot. At any moment the
waves are pointing at the source, unless the source accelerates.
Wherever you are, whatever your own velocity, the source's waves
always face you. A result that's different from other theories
except maybe SR.

You have to "face" where the source was when the signal was emitted,
then it's the shortest distance.

That's what I'm running into. If you face where the source was when the
signal was emitted, that's the direction I'd expect particles emitted
from the source to be coming from.

But the wave front is coming from a different direction, because it
travels at c+v' and v' varies with direction of travel.

So any individual particle of light is traveling in a straight line from
the source, and it doesn't matter to the particle that there are other
particles that travel at other speeds nearby. But the wave -- a Mexican
wave, maybe -- has an existence that's different from that of the
particles, just like a mexican wave is different from the people who
make it, and doesn't travel at the same speed they do. And if you
measure interference or redshift you aren't measuring that on individual
particles, you're measuring it on the wave. It's the speed of the wave
that matters then. Unless the motion of the particles somehow gets in
the way.

Sound and light signals have different
speeds, a deaf man sees the source where it was, a blind man hears
the source where it was, you see and hear both but they are not
simultaneous.
This prompts you to say you hear the sound "behind" the source and
me to say I hear the source from where it came from.

Yes. But that's when the air you'r listening through isn't moving
relative to you. When you're listening for that sound in a wind that's
going 170 meters per second -- when you listen to somebody who's
screaming while they're blown away -- it will sound different. It's like
the case when you're traveling at half the speed of sound and the source
is stationary. You get a doppler effect because you run into the waves
at a different speed, but you run into each of them at the same angle
you would if you were sitting still at that spot. (Or is there a
calculus trick I should be applying?)

http://yfrog.com/02starmg
The star is seen behind where it actually is.

If the movement of the particle I highlighted in red is what you
see, then the star is seen behind where it actually is.

But if the movement of the waves are what's seen, then the star
appears to be in exactly the direction that it actually is.

Either way.

The blue observer can only see each circle when it arrives, the
time of> arrival is constantly changing and so is the duration of
arrival> between circles. You are looking at the diagram and see the
circles> equidistant because you are a god looking down, but the blue
observer> does NOT see them arrive at equal time intervals, only sees
one circle> at a time, only sees a point on the circle.

Yes, he only sees one at a time.

In http://yfrog.com/02starmg he only sees the first one at one time.
The trail of red squares doesn't exist, they are all history.

Yes. they are the history of the movement of one particle of light. Is
that irrelevant? I feel like we're actually looking at the wave and not
the movement of the particle through time. The movement of the particle
seems completely irrelevant to me until you need it to go down a
telescope.

I drew a history trail like yours in
http://www.androcles01.pwp.blueyonder.co.uk/Wave/Aberration.gif
but when I animated it I deleted all the cubes except one in each
frame. So one image shows the light hits the edge, but it really hits
the centre and travels down the axis. This is because the one cube
cannot be in two places at once.

I may have missed your point. I saw the cylinder moving while the dot
traveled, so that the dot could go down the center because of that
movement. It doesn't matter that the wall of the cylinder hits places
where the dot used to be, if the dot isn't there any more.

I drew the history trails because I thought it mattered where the
particles came from. But if I ignore the particles and just look at the
circles then everything seems to work out. It's only when I pay
attention to the particles that I start having objections that need more
theory to deal with them.

But he *could* post pickets who would see them at other times and
places.

The pickets would need to be on Mars, Venus, a Jovian Moon.
Perhaps you'd better work out how to get a picket to Mars before
you say "could" so glibly.

Ah, the argument from practicality. I want to look at what the theory
says will happen. In a second line of advance I want to look at what we
can observe if the theory is true. You are quibbling about what is
practical for us to observe today, when we are mostly stuck on one
planet immersed in air. The distances are so short that it's mostly
practical to measure speed with interferometry....

They could compare notes afterward. What I see from above is
that the waves are equidistant and they look equidistant no matter
what inertial frame you're in. The actual wavelength never changes,
but if you measure wavelength by seeing how fast waves roll over you
at one point then your measurements can change.

The historical wavelength never changes wrt the source, the measured
wavelength is impossible because the photon can't be in two places at
once. Your trail of red squares is a wiggle in timespace.

If you can't actually see a wave crest then you're stuck with indirect
methods.

For example -- make a standing wave. Say, with masers. Put something in
the way that absorbs light of that frequency and look for the bands
where it does absorb some.

Or is it just impossible?

Here are two different cases. The yellow source is traveling
at> >> > about 0.5c compared to the stationary observer, who is 1
distance> >> > unit away at the closest approach. It takes 1 time
unit for light> >to> > get from a stationary source at that closest
spot to the> >observer.> >
In one case the light leaves at a 60 degree angle when the
source> >is> > still 0.5 distance unit from closest approach. Since
the light> >> > travels at c+v, in one time unit it goes 1 distance
unit sideways> >> > and 0.5 distance unit forward, and reaches the
observe after 1> >time> > unit, when the source is at the closest
approach. The point> >on the> > wavefront marked in red has traveled
at this 60 degree> >angle the> > whole distance and presumably will
keep traveling at> >that angle. The> > wavefront at this moment is at
90 degrees,> >precisely facing the> > observer.

In the other case the light leaves at a 90 degree angle when
the> >> > source is at closest approach. It is leaving at a -60
degree> >angle> > relative to the source, but because of the source's
forward> >motion> > which carries over to the light, it actually
travels at a> >90 degree> > angle and will be traveling in that
direction when it> >reaches the> > observer. The wavefront, though,
is moving at a 60> >degree angle at> > that time relative to the
observer. The light> >traveles at sqrt(3)/2> > or about 0.87 c.

So, what direction do you think the light is coming from, when
the> > individual elements are coming at one angle but the
wavefront> >comes> > from another angle? I know the answer for sound
-- it sounds> >like> > the sound is coming perpendicular to the
wavefront. Sound is> >a> > compression wave, and you get the
direction by the delay in> >> > compression for one ear compared to
the other. If there's a high> >> > wind and the whole compression
wave is going sideways compared to> >> > the direction of the
wavefront you mostly don't notice -- the new> >> > air is as
compressed as the old air.> >> >
Does that work for light too? I'm confused. At first thought
it> >> > seems like it ought to, but as you point out that would
require> >> > light that's been going in one direction to make a
sudden 30> >degree> > turn when it enters your telescope.

The other picture shows the on-the-other-hand. These photons
have> >> > been traveling straight at .87c, and the ones a little bit
to> >either> > side have been traveling faster and slower. The
wavefront> >is at 60> > degrees, not 90 degrees.

The star is seen behind where it actually is in both images, and
in the direction where it was. What other hand?

With the source traveling at 0.5c, in the one case you get light
particles traveling toward you at .877c while the wave is
traveling> > at c.

Wrong. You see 1.25c, it is blue shifted on approach. You are
still making the hidden assumption that c cannot be exceeded,
that's> what confuses you and why you should condemn Einstein for the
charlatan he was. Wilson will tell you the same and he's a used car
salesman who'll sell you a used VW camper van he wants to get rid
of.

I think you must be looking at the other case. I'm talking about
starm, where the wave appears to be traveling at c while the
particle I marked travels at sqrt(3)/2.

In the "start" case, the particle of light is going at c to the side
and at .5c forward, so the particle of light travels at sqrt(5)/2
and the wave front is traveling at c when the particle arrives.

The particle is traveling faster than c, but the wave is only
traveling straight toward the observer at c. And it's waves that get
red-shifted or dopplered, which I haven't yet become certain are the
same thing.

f' = f. c+v.cos(60)/c = 1 + 0.5 * 0.5 / 1 = f* 1.25

Is that when the waves appear to be coming in at 60 degrees? The
light that left the source much earlier?

As the star leaves and passes through 60 degrees,
f' = f *0.75, red shifted.

And this is the light that left the source at the T?

No. I'm not interested in comparing where the star is now with where
it was last seen. Prophecies are for palmists, fortune tellers, tea
leaf readers, horoscope writers and relativists. I can't see where it
is now, I'll have to wait for the light to reach me.

When the light reaches you, the particles are presumably moving in
straight lines from the source. But the waves are moving in a different
direction, because every individual piece of them is moving at v in
direction x in addition to their other movement. What we can measure is
the wave.

Do you point the telescope in the direction the particles came from, or
in the direction of the wavefront?

When the star is crossing the T,
f' = f * 0.5 * cos(90) /c = 1 + 0.5 * 0.0 / 1 = f * 1

This is the light that arrives at the observer when the star is
crossing the T, that left the star when it was at -60 degrees?

No. I'm not interested in comparing where the star is now with where
it was last seen. Prophecies are for palmists, fortune tellers, tea
leaf readers, horoscope writers and relativists. I can't see where it
is now, I'll have to wait for the light to reach me.

I'm interested in which direction do you point the telescope.

** So anyway, is there any value in thinking about light particles
that travel in straight lines, or is it better to just look at how
the waves travel? **


They don't travel in straight lines. This ball doesn't travel in a
straight line.
http://ww2010.atmos.uiuc.edu/(Gh)/guides/mtr/fw/gifs/coriolis.mov

(Well, it does in the playground, but not for the observer on the
carousel).

Since you are on the moving Earth, nor does light.
http://www.androcles01.pwp.blueyonder.co.uk/Shapiro/Crapiro.htm
Being god, my frame of reference is different to you mere mortals
riding on the blue ball.

I don't see where coriolis applies to a source moving in a straight line
in direction x at velocity v, and a stationary observer.

In the other case you get light particles traveling toward you at
c> > while the wave is traveling at .877c.

I got that wrong. It should have been light at c*sqrt(5)/2 and the
wave at c.

You mean the history wave?

No, I mean the wave crest that the observer gets right now. The last
wave crest came at a slightly different direction and speed, and the
next one will be different too, but right now the wave is at 90 degrees
and traveling at c.

At least one of those cases you will get a frequency shift or a
wavelength shift or both. Is there one where that won't happen? I
say it ought to be the one I labeled "start" and not "starm"
because> > it is exactly like the sound model where the source is at
rest wrt> > the air. I don't know anybody who's actually listened
much while the> > air was rushing past him at half the speed of
sound, but the> > equation says to expect no doppler shift then.
Yesterday I was> > convinced it was correct. Now I wonder whether
that applies to> > light. And Wilson says you have to point your
telescope off at an> > angle, and the waves will be refracting
something weird that way if> > it works.

You do have to point it at an angle, that's aberration.
http://www.androcles01.pwp.blueyonder.co.uk/Wave/Aberration.gif

But that isn't what's happening in this case, right?

Yes it is.
http://www.androcles01.pwp.blueyonder.co.uk/1st/Postulates.htm
You are expected to know what relative motion is, Einstein can't tell
you.

This is important. With other theories, the direction of the waves is
always the direction of the light. That's what the direction of the
light *means* to them. With EmT you can have light that's actually
traveling in a different direction and speed from its waves. This
complication is bound to have a whole lot of implications. For one, you
will get kinds of interference etc that are forbidden to other theories,
unless it turns out somehow that they cancel out.

This is exciting!

Interference and redshifts are related problems. In the first
case> > if you point your telescope at a 60 degree angle (or
whatever> >works)> > you should get light that is traveling toward
you at> >sqrt(c^2+v^2).> > Its wavelength in the direction of the
waves'> >apparent source is the> > original wavelength -- if you look
at the> >picture at any one time> > you can't tell it from a
stationary> >source. The frequency at this> > time is the original
frequency,> >though it has been higher and is> > falling. Do you get
a blueshift?> >> >
The second picture gives you waves traveling at c at -60
degrees,> >> > but individual particles in those waves are traveling
at 90> >degrees> > at 0.87 c. Very confusing.

My intuition says to pay attention to the waves and not the
particles. But that could easily be wrong.

You don't have any waves, you have expanding circles.
Aberration of light:

http://www.androcles01.pwp.blueyonder.co.uk/Wave/Aberration.gif

I see how that would apply if you were traveling at 0.5c along
with> > the source. The problems that Einstein had to invent SR for
just> > vanish with EmT. Your picture would fit perfectly. But right
now> > we're looking at the case where the observer is not traveling
parallel to the source, and if light travels in straight lines
like> > particles who have additive velocities, and it isn't aimed
straight> > down the telescope, then that light is not going to reach
the bottom> > unreflected.

He's only got two choices, as do you. Either stand in the road and
let the oncoming traffic wave hit you or stand to the side and let
the traffic wave pass. If it passes you'll hear doppler shift, if
it> hits you, you won't hear anything except an ambulance wave and
then> only if you are lucky. It doesn't matter which way you are
facing,> either.

I don't see how that relates.
.

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