Maybe this is something everyone knows except me (and Paul Anderson) but
here goes anyway.

In ON THE ELECTRODYNAMICS OF MOVING BODIES Sect 7 Einstein says the
frequency of the light perceived by the observer is given by the
equation

f' = f (1 - Cos(Fi)v/c)/Sqr(1-v^2/c^2) ---------------- [1]

If the source is travelling directly to or directly from an observer
then Cos(Fi) = 1. Einstein states that this gives equation [2] Which it
does.

f' = f (1 - v/c)/Sqr(1-v^2/c^2) ----------[1a]
f' = f Sqr[(1-v/c)^2 /(1-v^2/c^2)]
= f Sqr[(1-v/c)(1-v/c)/(1-v/c)(1+v/c)]
= f Sqr[(1-v/c)/(1+v/c) ---------------- [2]

I have shown the working because the equation which interests me is [1a]

We know that in acoustics you get Doppler if either the source is
moving or the observer is moving or both relative to the air, the
propagating medium. We are all familiar with the diagrams/equations in
the text book.

In Relativity we have two complications, one is the assumed absence of a
propagating medium and the other is time dilation. Time dilation is not
strictly Doppler shift but never the less represents a change in
frequency due to speed so that if you are going to produce an equation
which tells you what the frequency is when the source is moving it is
necessary to include it. Time dilation makes time intervals increase
which is the same as making frequencies reduce - frequency being the
reciprocal of time.

If we ignore the Doppler component then the frequency observed by an
inertial observer due to time dilation would be.

f' = f Sqr(1-v^2/c^2) -------------- [3]

where v is the velocity of the source relative to the observers FoR
independent of the direction of v.

If we try and derive from first principles the Doppler component then
the lack of a propagating medium is not the problem it might seem. The
second postulate says that the speed of light will be c everywhere in
the observers FoR. With or without a propagating medium this is
mathematically identical to what you would get if an observer were
stationary w.r.t a propagating medium so we know that mathematically it
is the same as Acoustic Doppler with the source and not the observer
moving and so is.

f' = f (c/(c-v)) = f/(1-v/c) -------- [4]

for a source moving towards the observer.

One can therefore combine these factors to get an equation for the
frequency observed when both affects are included so from [3] and [4]:

f' = f Sqr(1-v^2/c^2)/(1-v/c) ---------- [5]

But if we now compare it with Einstein's then clearly something is
wrong:

f' = f (1-v/c)/Sqr(1-v^2/c^2) ----------[1a]

Clearly one is the inverse of the other. One would naturally assume that
Einstein was right and me wrong except for a couple of other things
which don't fit the first is Einstein's own statement following [2].

"We see that, in contrast with the customary view, when v = -c
f'=infinity."

Time dilation when v = +/- c means that time has stopped, time intervals
are infinity and f' = 0

The second comes if we return to Einstein's original equation.

f' = f (1 - Cos(Fi)v/c)/Sqr(1-v^2/c^2) ---------------- [1]

and now make Fi = 90deg Cos(fi) = 0. This gives:

f' = f /Sqr(1-v^2/c^2) ---------------- [6]

This is the condition for what is described as transverse Doppler. But
we know that that is just time dilation. Again it is the wrong way up
and we know it should be as per equation [3]:

f' = f Sqr(1-v^2/c^2) ---------------- [3]

Have I have missed something or did Einstein have it wrong?

--
John Kennaugh
to email convert the number from hex to decimal

On Mon, 11 Oct 2004 19:47:32 +0100, John Kennaugh wrote:

Maybe this is something everyone knows except me (and Paul Anderson) but
here goes anyway.

In ON THE ELECTRODYNAMICS OF MOVING BODIES Sect 7 Einstein says the
frequency of the light perceived by the observer is given by the equation

f' = f (1 - Cos(Fi)v/c)/Sqr(1-v^2/c^2) ---------------- [1]

If the source is travelling directly to or directly from an observer then
Cos(Fi) = 1. Einstein states that this gives equation [2] Which it does.

f' = f (1 - v/c)/Sqr(1-v^2/c^2) ----------[1a] f' = f
Sqr[(1-v/c)^2 /(1-v^2/c^2)]
= f Sqr[(1-v/c)(1-v/c)/(1-v/c)(1+v/c)] = f
Sqr[(1-v/c)/(1+v/c) ---------------- [2]

I have shown the working because the equation which interests me is [1a]

That equation is correct. Einstein's got at least one mistake in the 1905
paper but it's not here; it's in section 10, which nobody cares about
anymore anyway.

I'm not going to go over your derivation in detail at this time.
However, I will say that the easiest way I know to see that E's formula
is correct for longitudinal motion is to move into the frame of the
emitter, moving in the +x direction at velocity v. Then the emitter sees
the observer moving in the -x direction at v. Let's assume the observer
is to the "left" of the emitter.

In the emitter's frame, the wavecrests travel with frequency f and
velocity c. In this frame, the receiver is running away from them, and
the crests actually catch up to the receiver at a rate of f*(c-v)/c.

In addition, the receiver's clock is running slow (in this frame!) by a
ratio of 1/gamma. So, the frequency the receiver sees will be "boosted"
by a factor of 1/(1/gamma), which is gamma. So overall, if we set c=1,
the shift will be

f' = f * gamma * (1-v) = f * (1-v)/sqrt(1-v^2)

which is exactly what's in E's paper for this case. Doing the calculation
in the emitter's frame should have made no difference to the result, of
course.

I hope this helps. If not, I'll try to find the time to look through your
derivation, maybe tonight (as Androcles says, I'm always pleading lack of
time -- but unfortunately it's true, I lack time).


We know that in acoustics you get Doppler if either the source is
moving or the observer is moving or both relative to the air, the
propagating medium. We are all familiar with the diagrams/equations in
the text book.

In Relativity we have two complications, one is the assumed absence of a
propagating medium and the other is time dilation. Time dilation is not
strictly Doppler shift but never the less represents a change in
frequency due to speed so that if you are going to produce an equation
which tells you what the frequency is when the source is moving it is
necessary to include it. Time dilation makes time intervals increase
which is the same as making frequencies reduce - frequency being the
reciprocal of time.

If we ignore the Doppler component then the frequency observed by an
inertial observer due to time dilation would be.

f' = f Sqr(1-v^2/c^2) -------------- [3]

where v is the velocity of the source relative to the observers FoR
independent of the direction of v.

If we try and derive from first principles the Doppler component then
the lack of a propagating medium is not the problem it might seem. The
second postulate says that the speed of light will be c everywhere in
the observers FoR. With or without a propagating medium this is
mathematically identical to what you would get if an observer were
stationary w.r.t a propagating medium so we know that mathematically it
is the same as Acoustic Doppler with the source and not the observer
moving and so is.

f' = f (c/(c-v)) = f/(1-v/c) -------- [4]

for a source moving towards the observer.

One can therefore combine these factors to get an equation for the
frequency observed when both affects are included so from [3] and [4]:

f' = f Sqr(1-v^2/c^2)/(1-v/c) ---------- [5]

But if we now compare it with Einstein's then clearly something is
wrong:

f' = f (1-v/c)/Sqr(1-v^2/c^2) ----------[1a]

Clearly one is the inverse of the other. One would naturally assume that
Einstein was right and me wrong except for a couple of other things
which don't fit the first is Einstein's own statement following [2].

"We see that, in contrast with the customary view, when v = -c
f'=infinity."

Time dilation when v = +/- c means that time has stopped, time intervals
are infinity and f' = 0

The second comes if we return to Einstein's original equation.

f' = f (1 - Cos(Fi)v/c)/Sqr(1-v^2/c^2) ---------------- [1]

and now make Fi = 90deg Cos(fi) = 0. This gives:

f' = f /Sqr(1-v^2/c^2) ---------------- [6]

This is the condition for what is described as transverse Doppler. But
we know that that is just time dilation. Again it is the wrong way up
and we know it should be as per equation [3]:

f' = f Sqr(1-v^2/c^2) ---------------- [3]

Have I have missed something or did Einstein have it wrong?

--
I can be contacted through http://www.physicsinsights.org

"John Kennaugh" wrote in message
news
Maybe this is something everyone knows except me (and Paul Anderson) but
here goes anyway.

In ON THE ELECTRODYNAMICS OF MOVING BODIES Sect 7 Einstein says the
frequency of the light perceived by the observer is given by the equation

f' = f (1 - Cos(Fi)v/c)/Sqr(1-v^2/c^2) ---------------- [1]

If the source is travelling directly to or directly from an observer then
Cos(Fi) = 1. Einstein states that this gives equation [2] Which it does.

f' = f (1 - v/c)/Sqr(1-v^2/c^2) ----------[1a]
f' = f Sqr[(1-v/c)^2 /(1-v^2/c^2)]
= f Sqr[(1-v/c)(1-v/c)/(1-v/c)(1+v/c)]
= f Sqr[(1-v/c)/(1+v/c) ---------------- [2]

I have shown the working because the equation which interests me is [1a]

We know that in acoustics you get Doppler if either the source is moving
or the observer is moving or both relative to the air, the propagating
medium. We are all familiar with the diagrams/equations in the text book.

In Relativity we have two complications, one is the assumed absence of a
propagating medium and the other is time dilation. Time dilation is not
strictly Doppler shift but never the less represents a change in frequency
due to speed so that if you are going to produce an equation which tells
you what the frequency is when the source is moving it is necessary to
include it. Time dilation makes time intervals increase which is the same
as making frequencies reduce - frequency being the reciprocal of time.

If we ignore the Doppler component then the frequency observed by an
inertial observer due to time dilation would be.

f' = f Sqr(1-v^2/c^2) -------------- [3]


Ok, I'll interrupt here. Let me finish before interrupting.
This has been thrashed out on this newsgroup any number of times.
Given: t' = (t-vx/c^2)/sqrt(1-v^2/c^2)
Now, v is a velocity, and is the distance x divided by the time t.
So x is vt.
Substituting for x,
t' = (t-v*vt/c^2)/sqrt(1-v^2/c^2)
t' = (t- t *v^2/c^2)/sqrt(1-v^2/c^2)
t' = t(1 - v^2/c^2)/sqrt(1-v^2/c^2)
t' = t. sqrt(1 - v^2/c^2)
and look, it agrees with the moving clock (t') running slow,
sqrt(1-v^2/c^2) is less than 1.
The amateur to this game simply ignores x, as you'll find on numerous
websites, and writes this as
t' = (t-vx/c^2)/sqrt(1-v^2/c^2), x = 0, so
t' = t/sqrt(1-v^2/c^2)
but then the moving clock (t') is running fast, he is dividing t by
something less than 1 and is confused. This is a common problem with
individuals getting their frames mixed up.

BUT... as you correctly point out, t = 1/f.
In other words,
t' = t. sqrt(1 - v^2/c^2)
and
f' = f. sqrt(1 - v^2/c^2)
So to which frame does f' belong, the "moving" or the "stationary"?
Invariably we get confused by f' meaning the observed frequency,
which is measured in the stationary frame. You have done this, you've
written above:

"If we ignore the Doppler component then the frequency observed by an
inertial observer due to time dilation would be.

f' = f Sqr(1-v^2/c^2) -------------- [3]"

To maintain the primes belonging to the correct frame, this has to be
f' = f / sqrt(1-v^2/c^2)

Remember that t' belongs to the "moving" frame, f' should as well.
This problem would be better overcome if we used greek instead
of primes, then we'd have

tau = t. sqrt(1 - v^2/c^2) (slower moving clock)
and
nu = f / sqrt(1 - v^2/c^2)

Looking at section 7 again:
"if an observer is moving with velocity v"
[he has the dilated time tau, will come home younger than
his sibling so the universal time t is racing along for him]
"the frequency nu' of the light perceived by the observer "
nu' = nu (1-cos(phi ........ etc.,
and take careful note of the confusion of frames and primes.
nu is the frequency emitted from the stationary frame, not
the frequency of the observer's clock.

Now proceed from there.
[snip]

Have I have missed something or did Einstein have it wrong?
Both. So did Koks miss something.
http://www.androc1es.pwp.blueyonder....oksDoppler.htm
Don't return to Earth too fast, it will take you longer the faster
you go, and take a 14 year supply of food or you will starve.
The moving clock runs fast on the return trip, the Andersen
Transforms apply over the Lorentz.

"That is, we can reverse the directions of the frames
which is the same as interchanging the frames,
which - as I have told you a LOT of times,
OBVIOUSLY will lead to the transform:
t = (tau-xi*v/c^2)/sqrt(1-v^2/c^2)
x = (xi - v*tau)/sqrt(1-v^2/c^2)
or:
tau = (t+xv/c^2)/sqrt(1-v^2/c^2)
xi = (x + vt)/sqrt(1-v^2/c^2)" -Paul B. Andersen

The trouble with Andersen is that he doesn't believe
his own calculations.
Dover, Page 44, line 8 (section 3 of "Electrodynamics")
"IF we place x' = x-vt..."
See that "IF"?
Well, IF we place x' = x+vt, we get the Andersen Transforms
and to get home again, we need to change the sign of v without
rotating the coordinate frame.
Androcles.



--
John Kennaugh
to email convert the number from hex to decimal

(I tried to send this earlier but my new server apparently ate it. So
I took advantage of the delay to rework it a bit, and I'm posting it
again -- maybe it'll actually go this time.)

On Mon, 11 Oct 2004 19:47:32 +0100, John Kennaugh wrote:

Maybe this is something everyone knows except me (and Paul Anderson)
but here goes anyway.

In ON THE ELECTRODYNAMICS OF MOVING BODIES Sect 7 Einstein says the
frequency of the light perceived by the observer is given by the
equation

f' = f (1 - Cos(Fi)v/c)/Sqr(1-v^2/c^2) ---------------- [1]

If the source is travelling directly to or directly from an observer
then Cos(Fi) = 1. Einstein states that this gives equation [2] Which
it does.

f' = f (1 - v/c)/Sqr(1-v^2/c^2) ----------[1a] f' = f
Sqr[(1-v/c)^2 /(1-v^2/c^2)]
= f Sqr[(1-v/c)(1-v/c)/(1-v/c)(1+v/c)] = f
Sqr[(1-v/c)/(1+v/c) ---------------- [2]

I have shown the working because the equation which interests me is [1a]

That equation is correct. Einstein's got at least one mistake in the
1905 paper but it's not here; it's in section 10, which nobody cares
about anymore anyway. But his derivation of the Doppler formula is a
little confusing, I think.

Below, I point out what I think the point of disagreement between you and
E. is. But first here's a quickie "third derivation" of the formula for
longitudinal doppler.

The easiest way I know to see that E's formula is correct for
longitudinal motion is to move into the frame of the emitter, moving
in the +x direction at velocity v. Then the emitter sees the observer
moving in the -x direction at v. Let's assume the observer is to the
"left" of the emitter.

In the emitter's frame, the wavecrests travel with frequency f and
velocity c. In this frame, the receiver is running away from them,
and the crests actually catch up to the receiver at a rate of
f*(c-v)/c.

In addition, the receiver's clock is running slow (in this frame!) by
a ratio of 1/gamma. So, the frequency the receiver sees will be
"boosted" by a factor of 1/(1/gamma), which is gamma. So overall, if
we set c=1, the shift will be

f' = f * gamma * (1-v) = f * (1-v)/sqrt(1-v^2)

which is exactly what's in E's paper for this case.

Now, on to the nit-picking mission!


We know that in acoustics you get Doppler if either the source is
moving or the observer is moving or both relative to the air, the
propagating medium. We are all familiar with the diagrams/equations
in the text book.

In Relativity we have two complications, one is the assumed absence
of a propagating medium and the other is time dilation. Time
dilation is not strictly Doppler shift but never the less represents
a change in frequency due to speed so that if you are going to
produce an equation which tells you what the frequency is when the
source is moving it is necessary to include it. Time dilation makes
time intervals increase which is the same as making frequencies
reduce - frequency being the reciprocal of time.

If we ignore the Doppler component then the frequency observed by an
inertial observer due to time dilation would be.

f' = f Sqr(1-v^2/c^2) -------------- [3]

where v is the velocity of the source relative to the observers FoR
independent of the direction of v.

You seem to be saying the source is moving, and we're viewing this
from the observer's frame. In that case, the source clock is slowed
down by time dilation, and the "wavecrests" will come out less
rapidly; the frequency would be f/gamma, which is what you've shown.
So far so good.


If we try and derive from first principles the Doppler component
then the lack of a propagating medium is not the problem it might
seem. The second postulate says that the speed of light will be c
everywhere in the observers FoR. With or without a propagating
medium this is mathematically identical to what you would get if an
observer were stationary w.r.t a propagating medium so we know that
mathematically it is the same as Acoustic Doppler with the source
and not the observer moving and so is.

f' = f (c/(c-v)) = f/(1-v/c) -------- [4]

for a source moving towards the observer.

Just for laughs, let's note that for a source moving _away_ at v,
you'd get (setting c to 1 to save typing):

f'' = f/(1 + v)

One can therefore combine these factors to get an equation for the
frequency observed when both affects are included so from [3] and
[4]:

f' = f Sqr(1-v^2/c^2)/(1-v/c) ---------- [5]

And if we look at a source moving away, rather than toward, the
observer, we see (again with C set to 1):

f'' = f * sqrt(1 - v^2)/(1+v)

Multiply top and bottom by (1-v), and we get

f'' = f * (sqrt(1-v^2) * (1-v)) / (1-v^2)

divide top and bottom by sqrt(1-v^2) and we get

f'' = f * (1-v)/sqrt(1-v^2)

which is Einstein's formula.

But if we now compare it with Einstein's then clearly something is
wrong:

f' = f (1-v/c)/Sqr(1-v^2/c^2) ----------[1a]

Yup. Einstein's got the source moving _away_ at v. You've got it
moving _toward_ the observer at v.

Set gamma to 1, and you'll see that Einstein's formula is for
redshift, yours is for blueshift. The direction of motion is clearly
different. It's a swapped sign, nothing more.

Glancing again at section 7 it appears that E. rather annoyingly
failed to say which direction his observer was moving in.


Clearly one is the inverse of the other. One would naturally assume
that Einstein was right and me wrong except for a couple of other
things which don't fit the first is Einstein's own statement
following [2].

"We see that, in contrast with the customary view, when v = -c
f'=infinity."

Time dilation when v = +/- c means that time has stopped, time intervals
are infinity and f' = 0

The second comes if we return to Einstein's original equation.

f' = f (1 - Cos(Fi)v/c)/Sqr(1-v^2/c^2) ---------------- [1]

and now make Fi = 90deg Cos(fi) = 0. This gives:

f' = f /Sqr(1-v^2/c^2) ---------------- [6]

This is the condition for what is described as transverse
Doppler. But we know that that is just time dilation. Again it is
the wrong way up and we know it should be as per equation [3]:

f' = f Sqr(1-v^2/c^2) ---------------- [3]

Have I have missed something or did Einstein have it wrong?

I'm not going to dig around in the transverse Doppler formula tonight,
but I will point out that it's _not_ just time dilation. Been there,
done that; chased my tail quite a while before I finally "got it". The
key is in the "infinitely distant" clause in his description of the setup.
If the photon is received when the line between source and observer is at
90 degrees to each, then it must have been emitted much earlier -- when
the angle between them was quite different!

Keep that in mind, work through it again, and I think you'll find it
comes out as he said.


--
I can be contacted through http://www.physicsinsights.org

"John Kennaugh" wrote in message news
Maybe this is something everyone knows except me (and Paul Anderson) but
here goes anyway.

In ON THE ELECTRODYNAMICS OF MOVING BODIES Sect 7 Einstein says the
frequency of the light perceived by the observer is given by the
equation

f' = f (1 - Cos(Fi)v/c)/Sqr(1-v^2/c^2) ---------------- [1]

If the source is travelling directly to or directly from an observer
then Cos(Fi) = 1. Einstein states that this gives equation [2] Which it
does.

f' = f (1 - v/c)/Sqr(1-v^2/c^2) ----------[1a]
f' = f Sqr[(1-v/c)^2 /(1-v^2/c^2)]
= f Sqr[(1-v/c)(1-v/c)/(1-v/c)(1+v/c)]
= f Sqr[(1-v/c)/(1+v/c) ---------------- [2]

I have shown the working because the equation which interests me is [1a]

We know that in acoustics you get Doppler if either the source is
moving or the observer is moving or both relative to the air, the
propagating medium. We are all familiar with the diagrams/equations in
the text book.

In Relativity we have two complications, one is the assumed absence of a
propagating medium and the other is time dilation. Time dilation is not
strictly Doppler shift but never the less represents a change in
frequency due to speed so that if you are going to produce an equation
which tells you what the frequency is when the source is moving it is
necessary to include it. Time dilation makes time intervals increase
which is the same as making frequencies reduce - frequency being the
reciprocal of time.

If we ignore the Doppler component then the frequency observed by an
inertial observer due to time dilation would be.

f' = f Sqr(1-v^2/c^2) -------------- [3]

where v is the velocity of the source relative to the observers FoR
independent of the direction of v.

If we try and derive from first principles the Doppler component then
the lack of a propagating medium is not the problem it might seem. The
second postulate says that the speed of light will be c everywhere in
the observers FoR. With or without a propagating medium this is
mathematically identical to what you would get if an observer were
stationary w.r.t a propagating medium so we know that mathematically it
is the same as Acoustic Doppler with the source and not the observer
moving and so is.

f' = f (c/(c-v)) = f/(1-v/c) -------- [4]

for a source moving towards the observer.

One can therefore combine these factors to get an equation for the
frequency observed when both affects are included so from [3] and [4]:

f' = f Sqr(1-v^2/c^2)/(1-v/c) ---------- [5]

But if we now compare it with Einstein's then clearly something is
wrong:

f' = f (1-v/c)/Sqr(1-v^2/c^2) ----------[1a]

Clearly one is the inverse of the other. One would naturally assume that
Einstein was right and me wrong except for a couple of other things
which don't fit the first is Einstein's own statement following [2].

"We see that, in contrast with the customary view, when v = -c
f'=infinity."

Time dilation when v = +/- c means that time has stopped, time intervals
are infinity and f' = 0

The second comes if we return to Einstein's original equation.

f' = f (1 - Cos(Fi)v/c)/Sqr(1-v^2/c^2) ---------------- [1]

and now make Fi = 90deg Cos(fi) = 0. This gives:

f' = f /Sqr(1-v^2/c^2) ---------------- [6]

This is the condition for what is described as transverse Doppler. But
we know that that is just time dilation. Again it is the wrong way up
and we know it should be as per equation [3]:

f' = f Sqr(1-v^2/c^2) ---------------- [3]

Have I have missed something or did Einstein have it wrong?

--
John Kennaugh
to email convert the number from hex to decimal
All the variations:
http://hyperphysics.phy-astr.gsu.edu...v/reldop2.html
http://hyperphysics.phy-astr.gsu.edu...eldop3.html#c1
Index:
http://hyperphysics.phy-astr.gsu.edu...reldop.html#c1
With pictures!
Kind regards,
Sue...

"John Kennaugh" skrev i melding news
Maybe this is something everyone knows except me (and Paul Anderson) but
here goes anyway.

In ON THE ELECTRODYNAMICS OF MOVING BODIES Sect 7 Einstein says the
frequency of the light perceived by the observer is given by the
equation

f' = f (1 - Cos(Fi)v/c)/Sqr(1-v^2/c^2) ---------------- [1]

If the source is travelling directly to or directly from an observer
then Cos(Fi) = 1. Einstein states that this gives equation [2] Which it
does.

f' = f (1 - v/c)/Sqr(1-v^2/c^2) ----------[1a]
f' = f Sqr[(1-v/c)^2 /(1-v^2/c^2)]
= f Sqr[(1-v/c)(1-v/c)/(1-v/c)(1+v/c)]
= f Sqr[(1-v/c)/(1+v/c) ---------------- [2]

I have shown the working because the equation which interests me is [1a]

We know that in acoustics you get Doppler if either the source is
moving or the observer is moving or both relative to the air, the
propagating medium. We are all familiar with the diagrams/equations in
the text book.

In Relativity we have two complications, one is the assumed absence of a
propagating medium and the other is time dilation. Time dilation is not
strictly Doppler shift but never the less represents a change in
frequency due to speed so that if you are going to produce an equation
which tells you what the frequency is when the source is moving it is
necessary to include it. Time dilation makes time intervals increase
which is the same as making frequencies reduce - frequency being the
reciprocal of time.

If we ignore the Doppler component then the frequency observed by an
inertial observer due to time dilation would be.

f' = f Sqr(1-v^2/c^2) -------------- [3]

where v is the velocity of the source relative to the observers FoR
independent of the direction of v.

If we try and derive from first principles the Doppler component then
the lack of a propagating medium is not the problem it might seem. The
second postulate says that the speed of light will be c everywhere in
the observers FoR. With or without a propagating medium this is
mathematically identical to what you would get if an observer were
stationary w.r.t a propagating medium so we know that mathematically it
is the same as Acoustic Doppler with the source and not the observer
moving and so is.

f' = f (c/(c-v)) = f/(1-v/c) -------- [4]

for a source moving towards the observer.

Yes.
But Einstein defined the positive velocity (phi' = 0) away from the observer.
Using the same convention, your equation becomes:
f' = f (c/(c+v)) = f/(1+v/c)

One can therefore combine these factors to get an equation for the
frequency observed when both affects are included so from [3] and [4]:

f' = f Sqr(1-v^2/c^2)/(1-v/c) ---------- [5]

Or with Einstein's convention for the positive direction:
f' = f Sqr(1-v^2/c^2)/(1+v/c) = f Sqr((1-v/c)/(1+v/c))

But if we now compare it with Einstein's then clearly something is
wrong:

f' = f (1-v/c)/Sqr(1-v^2/c^2) ----------[1a]

f' = f (1-v/c)/Sqr(1-v^2/c^2) = f Sqr((1-v/c)/(1+v/c))

Clearly one is the inverse of the other. One would naturally assume that
Einstein was right and me wrong except for a couple of other things
which don't fit the first is Einstein's own statement following [2].

The only difference is your definition of the positive direction.

"We see that, in contrast with the customary view, when v = -c
f'=infinity."

v = -c means that the source is approaching the observer at
the speed of light. So f' = infinity.


Time dilation when v = +/- c means that time has stopped, time intervals
are infinity and f' = 0

So you get 0/0 which in this case is infinity. :-)
You have to look at the limits.
Sure both sqrt(1 - v^2/c^2) and (1+v/c) - 0 when v - -c,
but the latter goes "faster", so
Sqr(1 - v^2/c^2)/(1+v/c) - infinity when v - -c
This is easy to see from the equivalent expression
f Sqr((1-v/c)/(1+v/c)) - infinity when v - -c

The second comes if we return to Einstein's original equation.

f' = f (1 - Cos(Fi)v/c)/Sqr(1-v^2/c^2) ---------------- [1]

and now make Fi = 90deg Cos(fi) = 0. This gives:

f' = f /Sqr(1-v^2/c^2) ---------------- [6]

This is the condition for what is described as transverse Doppler. But
we know that that is just time dilation. Again it is the wrong way up
and we know it should be as per equation [3]:

f' = f Sqr(1-v^2/c^2) ---------------- [3]

Have I have missed something or did Einstein have it wrong?

No, Einstein was right.
What you have missed is how the directions and angels
are defined. You have got your two equations "upside down"
for two different reasons.

I have responded to this in another thread, but due to my late
response you would not have read it when you wrote this.
You may have read it now, but I repeat it here anyway.
(I have corrected a small typo at the end and redrawn
the figures somewhat.)

John Kennaugh asked:
| Any Suggestions as to how you get from [1] to your equation?

Yes. It is quite simple.

We have:
f' = f '(1- cos(phi)*v/c)/sqr(1-v^2/c^2) [1]
where phi is the angle between the velocity vector of the observer
and the wave vector _in the source frame_.
This is clearly stated by Einstein:
"..the angle phi with the velocity of the observer referred to
a system of co-ordinates which is at rest relative to the source
of the light .."

And we have aberration:
cos(phi') = (cos(phi) - v/c)/(1 - cos(phi)*v/c) [2]
where phi' is the angle between the velocity of the source
and the wave vector in the _observer frame_.
Now [2] can be written:
cos(phi) = (cos(phi') + v/c)/(1 + cos(phi')*v/c))

Inserting this in [1] yields:
f '= f*((1 - ((cos(phi') + v/c)/(1 + cos(phi')*v/c))*v/c))/sqrt(1 -v^2/c^2)
f' =f*((1 - (v/c)^2)/(1 + cos(phi')*v/c))/sqrt(1 -v^2/c^2)
f' = f*sqrt(1-v^2/c^2)/(1 + cos(phi')*v/c)

(You may see variants of this with a minus sign in the denominator,
it depends on how the positive direction of v is defined.)
[You have demonstrated the validity of this comment. :-) ]

Note that when phi' = pi/2, the wave vector is ortogonal to
the velocity of the source in the observer's frame.
Then f' = f*sqrt(1 - v^2/c^2)
S - v
|
|
|
O


When phi = pi/2, phi' = arccos(-v/c), f' = f/sqrt(1 - v^2/c^2),
which means that the observer will see the light coming a bit from
the front (the velocity of the source has an approaching component),
and he will see the light a bit blue shifted.
S - v
\
\
\
O


When phi = 0, then phi' = 0.
There is no aberration when the wave vector is parallel
to the velocity vector.
Then we have f' = f* sqrt((1-v/c)/(1+v/c))
regardless of which of the equations you use.
(1 - v/c)/sqrt(1-v^2/c* 2) = sqrt(1-v^2/c* 2)/(1 + v/c) = sqrt((1-v/c)/(1+v/c))

Paul

John Kennaugh wrote:
Maybe this is something everyone knows except me (and Paul Anderson) but
here goes anyway.

In ON THE ELECTRODYNAMICS OF MOVING BODIES Sect 7 Einstein says the
frequency of the light perceived by the observer is given by the equation

f' = f (1 - Cos(Fi)v/c)/Sqr(1-v^2/c^2) ---------------- [1]


His equation is correct. Paul gave a detailed answer, but the
short version is that Einstein uses a (nowadays) nonstandard
convention of measuring the angle wrt the *source*, rather
than the observer.

Actually, understanding this distinction is vital to
understanding the *apparent* asymmetry in the
transverse Doppler shift.

-E

On Tue, 12 Oct 2004, EjP wrote:

John Kennaugh wrote:
Maybe this is something everyone knows except me (and Paul Anderson) but
here goes anyway.

In ON THE ELECTRODYNAMICS OF MOVING BODIES Sect 7 Einstein says the
frequency of the light perceived by the observer is given by the equation

f' = f (1 - Cos(Fi)v/c)/Sqr(1-v^2/c^2) ---------------- [1]


His equation is correct. Paul gave a detailed answer, but the
short version is that Einstein uses a (nowadays) nonstandard
convention of measuring the angle wrt the *source*, rather
than the observer.

Okay, my question here is why is someone still using Einstein's original
paper as a reference? Why not a derivation from the books used to teach
the subject now?

sal writes
(I tried to send this earlier but my new server apparently ate it. So
I took advantage of the delay to rework it a bit, and I'm posting it
again -- maybe it'll actually go this time.)

On Mon, 11 Oct 2004 19:47:32 +0100, John Kennaugh wrote:

Maybe this is something everyone knows except me (and Paul Anderson)
but here goes anyway.

In ON THE ELECTRODYNAMICS OF MOVING BODIES Sect 7 Einstein says the
frequency of the light perceived by the observer is given by the
equation

f' = f (1 - Cos(Fi)v/c)/Sqr(1-v^2/c^2) ---------------- [1]

If the source is travelling directly to or directly from an observer
then Cos(Fi) = 1. Einstein states that this gives equation [2] Which
it does.

f' = f (1 - v/c)/Sqr(1-v^2/c^2) ----------[1a]
f' = f Sqr[(1-v/c)^2 /(1-v^2/c^2)]
= f Sqr[(1-v/c)(1-v/c)/(1-v/c)(1+v/c)]
= f Sqr[(1-v/c)/(1+v/c) ---------------- [2]

I have shown the working because the equation which interests me is [1a]

That equation is correct. Einstein's got at least one mistake in the
1905 paper but it's not here; it's in section 10, which nobody cares
about anymore anyway. But his derivation of the Doppler formula is a
little confusing, I think.

Below, I point out what I think the point of disagreement between you and
E. is. But first here's a quickie "third derivation" of the formula for
longitudinal doppler.

The easiest way I know to see that E's formula is correct for
longitudinal motion is to move into the frame of the emitter, moving
in the +x direction at velocity v.

Relativity is based on a postulate which in affect says "the speed of
light is constant w.r.t the observer observing it." an observer in the
FoR of the emitter is therefore bound to see a different reality to that
of an observer moving w.r.t the source.

As far as I am concerned No 1 rule in relativity, at least for me, is to
make sure you know which FoR you are describing things from and stick to
it. I tend to distrust this kind of approach. It is the bit where having
derived something in one frame you say something similar to "..therefore
in the other FoR.....". I have seen too many people come unstuck that
way. I'm not saying you are wrong just that in taking this approach I am
not going to be sure you are right.

Then the emitter sees the observer
moving in the -x direction at v. Let's assume the observer is to the
"left" of the emitter.

O S


In the emitter's frame, the wavecrests travel with frequency f and
velocity c. In this frame, the receiver is running away from them,
and the crests actually catch up to the receiver at a rate of
f*(c-v)/c.

In addition, the receiver's clock is running slow (in this frame!) by
a ratio of 1/gamma.

Here is where my distrust surfaces because the light speed relative to
the observer is c-v in this frame, c in his, therefore I can't be sure
that anything derived in this frame will apply in the observers FoR.

So, the frequency the receiver sees will be
"boosted" by a factor of 1/(1/gamma), which is gamma. So overall, if
we set c=1, the shift will be

f' = f * gamma * (1-v) = f * (1-v)/sqrt(1-v^2)

which is exactly what's in E's paper for this case.


Now, on to the nit-picking mission!


We know that in acoustics you get Doppler if either the source is
moving or the observer is moving or both relative to the air, the
propagating medium. We are all familiar with the diagrams/equations
in the text book.

In Relativity we have two complications, one is the assumed absence
of a propagating medium and the other is time dilation. Time
dilation is not strictly Doppler shift but never the less represents
a change in frequency due to speed so that if you are going to
produce an equation which tells you what the frequency is when the
source is moving it is necessary to include it. Time dilation makes
time intervals increase which is the same as making frequencies
reduce - frequency being the reciprocal of time.

If we ignore the Doppler component then the frequency observed by an
inertial observer due to time dilation would be.

f' = f Sqr(1-v^2/c^2) -------------- [3]

where v is the velocity of the source relative to the observers FoR
independent of the direction of v.

You seem to be saying the source is moving, and we're viewing this
from the observer's frame.

Again the second postulate says the speed of light is constant w.r.t the
observer observing it so it is sensible, (for me at least) to look at
any problem in relativity from the observers PoV assuming the observer
is stationary. Speed is relative so I can legitimately do that.

In that case, the source clock is slowed
down by time dilation, and the "wavecrests" will come out less
rapidly; the frequency would be f/gamma, which is what you've shown.
So far so good.


If we try and derive from first principles the Doppler component
then the lack of a propagating medium is not the problem it might
seem. The second postulate says that the speed of light will be c
everywhere in the observers FoR. With or without a propagating
medium this is mathematically identical to what you would get if an
observer were stationary w.r.t a propagating medium so we know that
mathematically it is the same as Acoustic Doppler with the source
and not the observer moving and so is.

f' = f (c/(c-v)) = f/(1-v/c) -------- [4]

for a source moving towards the observer.

Just for laughs, let's note that for a source moving _away_ at v,
you'd get (setting c to 1 to save typing):

f'' = f/(1 + v)

One can therefore combine these factors to get an equation for the
frequency observed when both affects are included so from [3] and
[4]:

f' = f Sqr(1-v^2/c^2)/(1-v/c) ---------- [5]

And if we look at a source moving away, rather than toward, the
observer, we see (again with C set to 1):

f'' = f * sqrt(1 - v^2)/(1+v)

Multiply top and bottom by (1-v), and we get

f'' = f * (sqrt(1-v^2) * (1-v)) / (1-v^2)

divide top and bottom by sqrt(1-v^2) and we get

f'' = f * (1-v)/sqrt(1-v^2)

which is Einstein's formula.

Let me try that myself.
f' = f Sqr(1-v^2/c^2)/(1-v/c) ---------- [5]

make v = -v
f' = f Sqr(1-v^2/c^2)/(1+v/c) ---------- [5]
Take it all inside the Sqr

f' = f Sqr[(1-v^2/c^2)/(1+v/c)^2]
= f Sqr[(1-v/c)(1+v/c)/(1+v/c)^2]
= f Sqr[(1-v/c)/(1+v/c)]
which is equation [2] which I know is the same as [1a] OK I agree.

Interesting. Thank you for that. I have always thought of the equation
as being made of two distinct components. Those components are obviously
related mathematically but then I recall something about Lorentz's
starting point (or one of them) being the Doppler equation so there is
perhaps an underlying relationship Doppler - LTs - Time dilation.


But if we now compare it with Einstein's then clearly something is
wrong:

f' = f (1-v/c)/Sqr(1-v^2/c^2) ----------[1a]

Yup. Einstein's got the source moving _away_ at v. You've got it
moving _toward_ the observer at v.

Set gamma to 1, and you'll see that Einstein's formula is for
redshift, yours is for blueshift. The direction of motion is clearly
different. It's a swapped sign, nothing more.

Glancing again at section 7 it appears that E. rather annoyingly
failed to say which direction his observer was moving in.

I noticed that. I would dock him a mark for that if I were you )



Clearly one is the inverse of the other. One would naturally assume
that Einstein was right and me wrong except for a couple of other
things which don't fit the first is Einstein's own statement
following [2].

"We see that, in contrast with the customary view, when v = -c
f'=infinity."

Time dilation when v = +/- c means that time has stopped, time intervals
are infinity and f' = 0

The second comes if we return to Einstein's original equation.

f' = f (1 - Cos(Fi)v/c)/Sqr(1-v^2/c^2) ---------------- [1]

and now make Fi = 90deg Cos(fi) = 0. This gives:

f' = f /Sqr(1-v^2/c^2) ---------------- [6]

This is the condition for what is described as transverse
Doppler. But we know that that is just time dilation. Again it is
the wrong way up and we know it should be as per equation [3]:

f' = f Sqr(1-v^2/c^2) ---------------- [3]

Have I have missed something or did Einstein have it wrong?


I'm not going to dig around in the transverse Doppler formula tonight,
but I will point out that it's _not_ just time dilation. Been there,
done that; chased my tail quite a while before I finally "got it". The
key is in the "infinitely distant" clause in his description of the setup.
If the photon is received when the line between source and observer is at
90 degrees to each, then it must have been emitted much earlier -- when
the angle between them was quite different!

If it is an infinite distance then the angle between them when it was
emitted is indeterminate surely.


Keep that in mind, work through it again, and I think you'll find it
comes out as he said.

He defines his co-ordinates as follows:

"the connecting line ``source-observer'' makes the angle Fi with the
velocity of the observer referred to a system of co-ordinates which is
at rest relatively to the source of light"

The infinite distance is surely to avoid complications with parallax,
light coming from a fixed direction.

"a system of co-ordinates which is at rest relatively to the source" to
avoid aberration complications. To make Fi the real angle between the
direction of the light and the direction of the observer.

I agree it is not simple. I for one would not like to argue about any
angle of Fi other than 0 and 90 because surely one of the affects of
length dilation is that an angle in one FoR changes when viewed in
another and I am not sure in which FoR Einstein has defined the angle.
Apparently in the source's yet he says it is what the observer would
measure.

Let us just say that there is an awful lot of room for confusion.

Thank you for taking the time. Very interesting. We have two equations
which look entirely different and you have shown that in reality they
only differ in which direction you call positive which is arbitrary
anyway. I am never the less gratified that my derivation from 'first
principles' did give a valid solution and I did define the direction of
v.

My head hurts )

If you have time I would be interested in why transverse Doppler is
"_not_ just time dilation".
--
John Kennaugh
to email convert the number from hex to decimal

"Creighton Hogg" wrote in message
...


On Tue, 12 Oct 2004, EjP wrote:

John Kennaugh wrote:
Maybe this is something everyone knows except me (and Paul Anderson)
but
here goes anyway.

In ON THE ELECTRODYNAMICS OF MOVING BODIES Sect 7 Einstein says the
frequency of the light perceived by the observer is given by the
equation

f' = f (1 - Cos(Fi)v/c)/Sqr(1-v^2/c^2) ---------------- [1]


His equation is correct. Paul gave a detailed answer, but the
short version is that Einstein uses a (nowadays) nonstandard
convention of measuring the angle wrt the *source*, rather
than the observer.

Okay, my question here is why is someone still using Einstein's original
paper as a reference? Why not a derivation from the books used to teach
the subject now?
If you prefer to examine the tail that wags the dog, you won't find all the
fleas the dog has hidden.
Androcles