Re: SR theory is simplistic

On Mar 14, 6:17 pm, bz <bz+...@xxxxxxxxxxxxxxxxxxxx> wrote:
"G" <gehan_ameresek...@xxxxxxxxxxx> wrote innews:1173872848.421005.196720@xxxxxxxxxxxxxxxxxxxxxxxxxxxx:



On Mar 14, 7:28 am, bz <bz+...@xxxxxxxxxxxxxxxxxxxx> wrote:
"G" <gehan_ameresek...@xxxxxxxxxxx> wrote
innews:1173832374.103288.148010@xxxxxxxxxxxxxxxxxxxxxxxxxxxx:

bz

I am not for the moment adressing the experimental evidence. OK so
someone should have noticed if there were c+v effects.

And that is my point.
The evidences for c'=c+v will be there if that rule is valid.

The moons in orbit - emitting photons system can be represented by a
turntable with two light bulbs
on the end, spinning rapidly.

To further simplify, the system can be represented by, at the widest
parts of the orbit , as by the observer, two light bulbs moving
tangentially, one towards the observer one away at say velocities
v_1 and v_2 where v1=-v2

ok. Agreed.

My question is this: at any given time, measured in our frame, the
frame of the turntable

Let us NOT work in the frame of the turntable for a couple of reasons,
1) it is rotating rather than inertial.
2) once the photons are emitted, we are interested in their travel to
the observer, not their distance from the emitter.

bz, I mean the FOR of the axis of the turntable, not actually the
rotating FOR that is the same as the turntable.

Then you are just saying where the axis of the FoR go.

Be careful, that is NOT the FoR of the turntable.

In the Lab's For, we can place the '0,0' point any where we like, including
at the axis of the turntable.

From the For of the turntable, the lab (the universe also) appears to be
rotating.

When you ask questions, being precise in formulating your question will take
you most of the way toward answering your question.

OK, that I realize that...
the displacement between each emitter and the photon emitted is
different; photon 1 is closer to the
earth by s/v and so is photon 2. However the emitters are at
different distances from the earth.

They were at the same distance from earth at the moment of emission, in
your example, right?

What I mean is after a time t, the photons are at the same distance
from earth but emitters are not.

The sources move after the photons are emitted.

OK

Why should we expect them to
still be the same distance from the emitters IN OUR FoR? (I would expect each
of them to be the same distance from their emitter in their emitter's FoR.)

In our instruments FoR, we must measure distance from the point at which the
photons were emitted, NOT from where the emitters have moved to by the time
we detect the photons.

Imagine a foot race. One runner is crouched at the starting line. The other
is riding on a moving platform.
At the moment the gun sounds, both leap from their crouched positions which
are, at that moment, exactly the same distance from the finish line.
Agreed


When you time the race, you use the distance actually covered by the runners,
not what distance they would have covered if the platform one was riding on
had continued to move until they reached the finish line.

In the runners example, the one starting from the moving platform has a bit
of an advantage because of inertia. He doesn't have to overcome as much of
the inertial of his rest mass as he is already moving.

But photons have no rest mass that must be overcome. They travel at the same
velocity and take the same time to run the course.

There IS a return of the Kinetic energy boost gained by a moving start,
however, the frequency of the photon is doppler shifted by the exact right
amount to account for that moving start.

Some Ritzians believe that the photons speed changes but not its wavelength.
This is demonstrably wrong as gratings are wavelength sensitive devices and
the doppler shifted spectra of a moving source would NOT be displaced in a
grating spectrometer IF the doppler shift was due to changes in speed.

Have to look at this later





What is the mechanism whereby a photon from two identical emitters
can be a different displacement
from the emitters (source) after a given time?

The emitters moved after they emitted the photons, one away from the
observer, the other toward the observer. One emitter 'runs away from
the photon it emitted', 'the other chases after the photon it emitted',
of course they will be different distances [under the rules of Special
relativity/general relativity and the rules that also govern sound
waves and waves on water.]

Does this not violate
the principle of equivalance
when we are seeing two different sources emitting photons at
different velocities to themselves?

The sources do NOT emit the photons at different velocities with
respect to the emitters.

Just imagine looking down at the emitters and the photons along the
turntable axix.
What you willl see is two photons and two emitters, the emitters at
different distances from the photons: the photons are moving at a
different velocities relative to the emitters.

The 'closing velocity' as seen by an observer NOT in the FoR of the source or
target can be up to 2 c.

The FoR of the axis is the FOR of the target: assume the target and
turntable axis
are in the same FoR and we are viewing the arrangement from the top
and we are
in that FoR


Point two particle accelerators at each other. Each sending particles at .99
c toward a point of collision.
The 'closing velocity' is going to be 1.98 c as calculated by the observer in
the lab's FoR.

I am glad that is agreed: my point is that looking at indentical
emitters, they
appear to have ejected a photon at a different 'closing' velocity to
themselves:
two identical phenomena looking different because they are moving. We
have to conclude then
that emitters moving toward the earth appear to be sending out the
photon slower and
emitters moving away from the earth send out the photons faster when
moving away from the earth
in order to conform to the second postulate.

Of course if you assume an ether in the FOR of the earth, then it
simplifies matters
considerably as to why the photons travel at the same velocity in
ether. Without the ether?




The effective closing velocity as seen by either particle is going to be
composition(.99c,.99c) = (v+w)/(1+v*w/c^2) = 0.99994595c

As best we can tell, from all experiments with moving sources, even those
moving at high velocities, once the photons are emitted, they move at c in
the FoR of our measuring equipment. Distance must be measured from point of
emission when timing the photons, NOT from where the emitter may have moved
to.

I agree distance should be measured from the point of emission if we
are timing photons,
it does not matter what the source does: this is to measure the
photon's velocity.

However my comment is about the assemtry of the entire arrangement
when moved in the target FoR

Identical emitters, and we have photons speeding away from each at a
different velocity as
seen in the turntable and earths reference frame. Why the difference?

The source moves after the photons are emitted.



There are those (NOT I) that believe that since photons leave the
emitter at c, the photons must move at c'=c+v [contrary to
Einstein]toward the observer[Ritz et al., including some like 'Henri
Wilson', 'Ken Seto' and 'John Kennaugh' in this newsgroup] adhere to
this assertion.

It is like saying since bullets travel at s metres ber second, if a
bullet is fired at the earth
by a moving gun, it will somehow arrive at the earth at s metres per
second because bullets allways travel at that speed.

Photons, Sound waves and water waves do NOT travel like bullets, they
travel like waves. As for why they do this, I have no idea why it is
that way, it just is.

Ah just the way it is, however sound waves and water waves involve an
ether that is air and water. For EM, ether does not exist.

It may or may not exist, I make no assertions one way or the other. According
to Einstein's 1905 paper, it is NOT necessary to assume it exists. It has
proven to be 'difficult' to detect, to say the least.

Since it IS unnecessary to assume its existance, and its properties, if it
does exist, seem to be only that 'it makes light travel at c when measured by
any observer', I see no reason to use an undetectable aether.

What we do know is that for tests of various types, attempting to determine
the value for k in c'=c+kv, k has been shown to be very small.

I am surprised it exists according to SRT

If k were 1 as some insist it must be, then the effects I mentioned would be
quite visible.


Quite visible indeed.

I am looking at really AE's assumptions and they seem to be flawed:
1905 paper.
Will have to look at experimental evidence; actually will have to set
up my SRT website
and start typing away. Hopefully you can comment then.

--
bz

please pardon my infinite ignorance, the set-of-things-I-do-not-know is an
infinite set.

bz+...@xxxxxxxxxxxxxxxxxxxx remove ch100-5 to avoid spam trap


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Relevant Pages

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  • Re: SR theory is simplistic
    ... not their distance from the emitter. ... What I mean is after a time t, the photons are at the same distance ... from earth but emitters are not. ... is riding on a moving platform. ...
    (sci.physics.relativity)
  • Re: SR theory is simplistic
    ... from earth but emitters are not. ... The sources move after the photons are emitted. ... other is riding on a moving platform. ... The 'closing velocity' as seen by an observer NOT in the FoR of the ...
    (sci.physics.relativity)
  • Re: SR theory is simplistic
    ... distance from earth but emitters are not. ... The sources move after the photons are emitted. ... The other is riding on a moving platform. ... The 'closing velocity' as seen by an observer NOT in the FoR of the ...
    (sci.physics.relativity)
  • Re: SR theory is simplistic
    ... from earth but emitters are not. ... The sources move after the photons are emitted. ... other is riding on a moving platform. ... The 'closing velocity' as seen by an observer NOT in the FoR of the ...
    (sci.physics.relativity)