ha scritto nel messaggio
oups.com...
In the non relativistic doppler effect, the observed frequency of a
source depends on the speed of the source and observer.

If the source approaches the observer or of the observer approaches the
source, the frequency observed in both cases is different. Indeed, if
Observer is stationary and source approaches observer, the observed
frequency is f' = f*1/(1-b). If Source is stationary and observer
approaches source, the observed frequency is f' = f*(1+b). b = v/c.

This effect is well observed for sound waves and the similar.

Is this effect observed for EM waves?
If so, how does SR describe this effect? (taking the relativity
principle into account)
References?


There is not any difference between sound in air and light in ether (for
Doppler).

All we need is to introduce the absolute time dilatation of the source clock
and receiver clocks.

So to be quick I will use easy figures:

If we have the source S moving at Vs = .6 C (versus the air) and is
approaching the receiver.
and we have the receiver R moving at Vr = .8 C (versus the air) and
approaching the source.

The sound classical Doppler would be:

Fr = Fo* 1/ (1-Vs) = 2.5 (what the source sends in the air) (Fo =1)
Fr = 2.5 * (1+Vr) = 4.5 (what the receiver get from the air)

If we invert the speeds source S approaching at Vs = .8 and the receiver
approaching at Vr = .6 C

Fr = Fo* 1/ (1-Vs) = 5 (what the source sends in the air) (Fo =1)
Fr = 5 * (1+Vr) = 8 (what the receiver get from the air)

so we have not reciprocity.

(Note: the two ways Doppler would be 8*4.5 = 36 ) (1)

Now we calculate the "CLASSICAL" Doppler for light introducing the time
dilatation on the clocks;

a)case of source at .6 C and receiver at .8 C:

Fr = sqr(1-Vs^2) / (1-Vs) = 2 ( what the source sends in the ether )
Fr = 2 * (1+Vr) = 3.6 (what the receiver get from the ether)
Fr = 3,6 / sqr (1-Vr^2) = 6 (what is the frequency seen by the receiver
clock)

b)case of source at .8 C and receiver at .6 C:

Fr = sqr(1-Vs^2) / (1-Vs) = 3 ( what the source send in the ether)
Fr = 3 * (1+Vr) = 4.8 (what receiver get from the ether)
Fr = 4.8 / sqr (1-Vr^2) = 6 (what is the frequency seen by the receiver
clock)

as you can see there is reciprocity on the frequency received.

(Note that the two ways Doppler is 6^2 = 36) identical at (1) (2)

Now we calculate the SR RELATIVISTIC Doppler for a source and a receiver
approaching at relative speed of

Vrel. = (Vs+Vr) / (1+(Vs+Vr)) = 0.94594594594594 C

With such a speed the relativistic Doppler is:

Fr = sqr((1+Vrel.) / (1- Vrel.)) = 6

(Note: the two ways Doppler is 6^2 = 36) identical at (1) and (2)

Conclusion: light in the ether acts exactly like sound in the air (as far as
Doppler is concerned): the classical ether formulae are perfectly fit,
taken count of the time dilatation of the source and the receiver.

The behavior of the case source-observer
vs case observer-source (inverting the speeds) is asymmetrical also with
light, not any more on the frequency received, but the asymmetry
has shifted totally on the elapsed time taken for the same amount of
pulses to be received: this time is locally identical but absolutely
different.

The SR math is a way to do without the ether, while, the ether is
indispensable for the doppler phenomenon to happen.

Is like having two cars go around sending sound pulses to each other: you
can, using the relative speed of the cars, mathematically calculate the
number of sound pulses received, but would be foolish to sustain that
there is no ground on the excuse that you can do without it in your
calculations.

Or deny the function of the ground on the ridicule excuse of not knowing
what the ground is made of.

Hoping I have been helpful, if you are a math expert you may help me to
better explain the above.

Let me know.

best regards

beda pietanza

"beda pietanza" wrote in message
...

ha scritto nel messaggio
oups.com...
In the non relativistic doppler effect, the observed frequency of a
source depends on the speed of the source and observer.

If the source approaches the observer or of the observer approaches
the
source, the frequency observed in both cases is different. Indeed, if
Observer is stationary and source approaches observer, the observed
frequency is f' = f*1/(1-b). If Source is stationary and observer
approaches source, the observed frequency is f' = f*(1+b). b = v/c.

This effect is well observed for sound waves and the similar.

Is this effect observed for EM waves?
If so, how does SR describe this effect? (taking the relativity
principle into account)
References?


There is not any difference between sound in air and light in ether
(for
Doppler).

There is no aether, or the doppler shift would have shown up in MMX.


All we need is to introduce the absolute time dilatation of the source
clock
and receiver clocks.

It is zero in MMX, both clocks in MMX are at rest reative to each other.

All we need to introduce is the speed of light constant with repect to
the air,
which it is.



So to be quick I will use easy figures:

If we have the source S moving at Vs = .6 C (versus the air) and is
approaching the receiver.
and we have the receiver R moving at Vr = .8 C (versus the air) and
approaching the source.

The receiver clock is moving at the same speed as the source clock
in MMX.



The sound classical Doppler would be:

Fr = Fo* 1/ (1-Vs) = 2.5 (what the source sends in the air) (Fo =1)
Fr = 2.5 * (1+Vr) = 4.5 (what the receiver get from the air)

The air inside a plane doesn't show any doppler shift.
It moves with the plane. Same for MMX and light.
If you put the microphone and loudspeaker out on the wing,
then you'd have a point.

So put MMX on the moon and see if you get any fringe shift.
You won't, but there won't be any time dilation between source
and receiver either.


If we invert the speeds source S approaching at Vs = .8 and the
receiver
approaching at Vr = .6 C

Source and detector are not moving at different speeds in MMX.
[snip rant about two velocities}

Conclusion: light in the ether

THERE IS NO AETHER.
Androcles.

Yup. Thats what I obtain too.

If you accept ether point of view, with its time dilations, there will
be no diferences in the frequencies in this experiment.


Conclusion: light in the ether acts exactly like sound in the air (as
far as
Doppler is concerned): the classical ether formulae are perfectly fit,
taken count of the time dilatation of the source and the receiver.

The behavior of the case source-observer
vs case observer-source (inverting the speeds) is asymmetrical also
with
light, not any more on the frequency received, but the asymmetry
has shifted totally on the elapsed time taken for the same amount of
pulses to be received: this time is locally identical but absolutely
different.

[Androcles:]
There is no aether, or the doppler shift would have shown up in MMX.
[...]
THERE IS NO AETHER.
Androcles.

Nothing moves inside the MMX, thus THERE IS NO DOPPLER SHIFT.
Harald ;-)

"harry" wrote in message
oups.com...
[Androcles:]
There is no aether, or the doppler shift would have shown up in MMX.
[...]
THERE IS NO AETHER.
Androcles.

Nothing moves inside the MMX, thus THERE IS NO DOPPLER SHIFT.

The light itself moves.

Therefore its speed is constant with respect to its source.
Moving sources produce shiift.

Androcles.



Harald ;-)

In sci.physics.relativity, Androcles

wrote
on Tue, 18 Jan 2005 21:38:44 GMT
:

"harry" wrote in message
oups.com...
[Androcles:]
There is no aether, or the doppler shift would have shown up in MMX.
[...]
THERE IS NO AETHER.
Androcles.

Nothing moves inside the MMX, thus THERE IS NO DOPPLER SHIFT.

The light itself moves.

Therefore its speed is constant with respect to its source.
Moving sources produce shiift.

Androcles.

The MMX with a stationary light source is consistent with
emission theory.

I would think that someone used a moving light source (e.g.,
another planet) at some point.

[.sigsnip]

--
#191,
It's still legal to go .sigless.

"The Ghost In The Machine" wrote in
message ...
In sci.physics.relativity, Androcles

wrote
on Tue, 18 Jan 2005 21:38:44 GMT
:

"harry" wrote in message
oups.com...
[Androcles:]
There is no aether, or the doppler shift would have shown up in MMX.
[...]
THERE IS NO AETHER.
Androcles.

Nothing moves inside the MMX, thus THERE IS NO DOPPLER SHIFT.

The light itself moves.

Therefore its speed is constant with respect to its source.
Moving sources produce shiift.

Androcles.

The MMX with a stationary light source is consistent with
emission theory.

I would think that someone used a moving light source (e.g.,
another planet) at some point.

What's wrong with a star?
It's big, bright and can be seen from thousands of light years away.
If it moves, we see doppler shift. In fact we wouldn't know it was
moving
without the shift.
Androcles.



[.sigsnip]

--
#191,
It's still legal to go .sigless.

In sci.physics.relativity, Androcles

wrote
on Wed, 19 Jan 2005 18:15:30 GMT
:

"The Ghost In The Machine" wrote in
message ...
In sci.physics.relativity, Androcles

wrote
on Tue, 18 Jan 2005 21:38:44 GMT
:

"harry" wrote in message
oups.com...
[Androcles:]
There is no aether, or the doppler shift would have shown up in MMX.
[...]
THERE IS NO AETHER.
Androcles.

Nothing moves inside the MMX, thus THERE IS NO DOPPLER SHIFT.

The light itself moves.

Therefore its speed is constant with respect to its source.
Moving sources produce shiift.

Androcles.

The MMX with a stationary light source is consistent with
emission theory.

I would think that someone used a moving light source (e.g.,
another planet) at some point.

What's wrong with a star?

Not bright enough, mostly. However, feeding a star into MMX would
work, given a sufficiently sensitive interferometer. (The
requirement that no mirrors be used outside of the MMX apparatus
complicates things.)

It's big, bright and can be seen from thousands of light years away.
If it moves, we see doppler shift. In fact we wouldn't know it was
moving without the shift.

I think you're confusing the Hubble redshift with the MMX fringe shift.
Different shifts, different mechanisms.

Androcles.



[.sigsnip]

--
#191,
It's still legal to go .sigless.




--
#191,
It's still legal to go .sigless.

"The Ghost In The Machine" wrote in
message news
In sci.physics.relativity, Androcles

wrote
on Wed, 19 Jan 2005 18:15:30 GMT
:

"The Ghost In The Machine" wrote
in
message ...
In sci.physics.relativity, Androcles

wrote
on Tue, 18 Jan 2005 21:38:44 GMT
:

"harry" wrote in message
oups.com...
[Androcles:]
There is no aether, or the doppler shift would have shown up in
MMX.
[...]
THERE IS NO AETHER.
Androcles.

Nothing moves inside the MMX, thus THERE IS NO DOPPLER SHIFT.

The light itself moves.

Therefore its speed is constant with respect to its source.
Moving sources produce shiift.

Androcles.

The MMX with a stationary light source is consistent with
emission theory.

I would think that someone used a moving light source (e.g.,
another planet) at some point.

What's wrong with a star?

Not bright enough, mostly.

ROFLMAO!

You are totally impossible to discuss anything with.
Go away, mad person..
Androcles.



However, feeding a star into MMX would
work, given a sufficiently sensitive interferometer. (The
requirement that no mirrors be used outside of the MMX apparatus
complicates things.)

It's big, bright and can be seen from thousands of light years away.
If it moves, we see doppler shift. In fact we wouldn't know it was
moving without the shift.

I think you're confusing the Hubble redshift with the MMX fringe
shift.
Different shifts, different mechanisms.

Androcles.



[.sigsnip]

--
#191,
It's still legal to go .sigless.




--
#191,
It's still legal to go .sigless.

Unfortunately this discussion didn't follow any further, which shows a lake
of
clear understanding about the physical subject from all the big brains
around here.

"The Ghost In The Machine" escreveu na
mensagem news

I think you're confusing the Hubble redshift with the MMX fringe shift.

I can't see any difference.
Hubble redshift assumes the speed of light to be constant, the MMX fring
shift tries to verify if the speed of light is constant.
Hence, MMX shows the speed of light to be constant and being so the only
possible interpretation for the Hubble redshift is that a relative speed
does exist between the source and the observer. Both are one and the same
and the original consistency problem does exist.