"John Kennaugh" wrote in message
.uk...
Androcles writes

"John Kennaugh" wrote in message
. co.uk...
Androcles writes

"John Kennaugh" wrote in message
news 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.

No.

Suppose t is the time between two 1s ticks. If you were right t' is
smaller then the interval between ticks is smaller - the clock is
ticking
faster - time has speeded up. Look up 'dilate' = "become wider or
larger"
as in eyes dilated.

Put some numbers in. For v = 0.866c (always convenient),
sqrt(1-v^2/c^2) = sqrt(1- 3/4) = sqrt( 1/4) = 1/2.
5 hours = 10 hours * 1/2.
The moving clock records 5 hours (t') for every 10 hours (t) of the
stationary clock, so:
t' = t.sqrt(1-v^2/c^2)

The interval between ticks of the moving clock equals two intervals of the
stationary clock and is therefore "dilated."

0----------1----------2----------3----------4---------5 time t'
0-----1----2----3----4----5----6----7-----8----9---10 time t


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,

The 'amateur' has it right. t't. A tick is stretched so the clock
appears
to be going slower. It is actually easier to think in terms of
frequency.

t' = t/sqrt(1-v^2/c^2)

converts to

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

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

This says the frequency of the clock ticks appears less = the clock
going
slower. f' is a lower frequency than f because Sqr(1-v^2/c^2)1

Put some numbers in. I don't see why I should do it for you again.

Had you put the numbers in to this equation rather than the other you
would have spotted your own mistake. I really don't enjoy putting you
right all the time.

It is with regret that I have to continue to correct you. However, it
appears it is necessary.


You once defined a clock, quite rightly as an oscillator and a counter.
Suppose we have a 100Hz oscillator and a counter which increments every
100 cycles i.e. once per second. Suppose we now make that clock travel at
0.866c. The equation, which you claim is wrong says the frequency will
appear to a stationary observer as

f' = f Sqr(1-v^2/c^2) = 100 x 0.5 = 50 Hz

The equation which I say is right, and Doppler would confirm, is
f' = f(c+v)/c = 100 * 1.866 = 186.6 Hz. for blue shift,
and
f' = f(c-v)/c = 100 * 0.134 = 13.4 for red shift.

Your equation above doesn't belong to any theory I know of.

Einstein would claim
f' = 100. sqrt [1.866/ 0.134] = 373.1 Hz for blue shift,
and
f' = 100. sqrt [0.134/1.866] = 0.268 Hz for blue shift.

Since 0.268 13.4, it follows that the receding clock is running
SLOWER than the stationary clock.
Everybody is happy, that is SR's prediction.
Since we agree that t = 1/f, 0.01 seconds elapsed between ticks
for the stationary clock, as measured by the stationary clock.
t = 1/f' = 1/13.4 = 0.0746 seconds elapsed between ticks for the
moving clock, as measured by the stationary clock.
The interval is dilated 7.5 times.
The moving clock is running slower, and everybody is still happy.


Since 371.1 186.6, it follows that the approaching clock is running FASTER
than the stationary clock.
You do the arithmetic and spot your own mistake.

The amateur has it wrong.
There is no 50Hz in these equations, so I need not consider your agument
further, my turn to snip.
[snip]
Androcles.

Thanks for your comments. I will try and digest them over the weekend

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

Bilge writes
John Kennaugh:
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.

It's much simpler to start with the relation,

k.x - wt = k'.x' = w't',

substitute the lorentz transforms of x and t for x' and t', then solve
for k' and w'.

That would be a totally pointless exercise. The whole point of the
exercise was to try and get some insight into what the equation is
describing by deriving it from first principles.


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

On Wed, 13 Oct 2004 20:23:18 +0100, John Kennaugh wrote:

In message ,
Creighton Hogg writes


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?



While I want to be sure of those parts of relativity which interest me
(i.e. the basics) I have not great desire to become an expert in
relativity and have neither the time nor the inclination to get my head
around the modern space-time geometry approach. Perhaps I have been
unlucky in the books I have looked at in that they either insult your
intelligence or jump into space time geometry. I need something in between
and Einstein's original will do me for now. In any case Einstein's paper I
can download via the WWW free. If you know a good text book I can download
let me know.

Do you have Einstein's "Relativity, the Special and the General Theory"?
It's not terribly deep but it's very readable, and it's available on
Project Gutenberg. I wouldn't recommend the online version,
though, because ... A hard copy is about 2 bucks (used), plus shipping,
and has the illustrations and sensible-looking equations which are missing
from the Gutenberg version. It's a nice book and "reads well". Check
Barnes&Noble or Amazon for exact information if you want buy a copy.

If you wanted to strain your brain and learn the geometric version of SR
Rindler's "Intro to SR" would be a good follow-on. But it's _not_ light
reading, unlike the Einstein book.

By the way, if you favor Ritz theory, as one of your other posts seemed to
suggest, I'm curious about your take on the Hubble redshift results. Ritz
(and Fox, for that matter) wrote at a time when all spectroscopes were
immersed in air, so there was not, as yet, anything to explain. However,
these days there's a lot to explain. The Hubble spectroscope is in vacuum,
and the wavelength it measures is the wavelength of the light before it
gets to the Earth's atmosphere. What's your preferred explanation for the
fact that redshift measured in space appears identical to the values we
get from ground-based spectroscopes? FWIW Androcles says V=F*L is
incorrect for photons, and Henri Wilson says it's the Cosmic Fart -- er,
interstellar gas -- that normalizes the velocity and hence causes the
wavelength to show the same redshift as the frequency. I have some
problems with both of these explanations, and I was wondering if you agree
with one or both, or have taken a different tack.

If you don't know what I'm talking about, google "extinction length" along
with "ritz" and "fox".


The book I have which does have a reasonable chapter on relativity does
not cover the Doppler equation. I did a search on WWW for 'Doppler
Einstein relativity' and came up with nothing useful surprisingly
enough. Maybe gave up too soon )

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

John Kennaugh:
Bilge writes
John Kennaugh:
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.

It's much simpler to start with the relation,

k.x - wt = k'.x' = w't',

substitute the lorentz transforms of x and t for x' and t', then solve
for k' and w'.

That would be a totally pointless exercise. The whole point of the
exercise was to try and get some insight into what the equation is
describing by deriving it from first principles.


Given that the mist fundamental principle in relativity is invariance,
that _is_ a first principle. The phase is an invariant. How much more
fundamental can you get? Let me give you a clue. ``First principles'' is
not synonymous with ``the way which offers the least insight and the most
opportunity for arithmetic mistakes.'' On the contrary, doing a calculation
the hard way usually indicates an understanding of first principles is
lacking.

Androcles writes

"John Kennaugh" wrote in message
o.uk...
Androcles writes

"John Kennaugh" wrote in message
.co.uk...
Androcles writes

"John Kennaugh" wrote in message
news 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.

No.

Suppose t is the time between two 1s ticks. If you were right t' is
smaller then the interval between ticks is smaller - the clock is
ticking
faster - time has speeded up. Look up 'dilate' = "become wider or
larger"
as in eyes dilated.

Put some numbers in. For v = 0.866c (always convenient),
sqrt(1-v^2/c^2) = sqrt(1- 3/4) = sqrt( 1/4) = 1/2.
5 hours = 10 hours * 1/2.
The moving clock records 5 hours (t') for every 10 hours (t) of the
stationary clock, so:
t' = t.sqrt(1-v^2/c^2)

The interval between ticks of the moving clock equals two intervals of the
stationary clock and is therefore "dilated."

0----------1----------2----------3----------4---------5 time t'
0-----1----2----3----4----5----6----7-----8----9---10 time t


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,

The 'amateur' has it right. t't. A tick is stretched so the clock
appears
to be going slower. It is actually easier to think in terms of
frequency.

t' = t/sqrt(1-v^2/c^2)

converts to

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

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

This says the frequency of the clock ticks appears less = the clock
going
slower. f' is a lower frequency than f because Sqr(1-v^2/c^2)1

Put some numbers in. I don't see why I should do it for you again.

Had you put the numbers in to this equation rather than the other you
would have spotted your own mistake. I really don't enjoy putting you
right all the time.

It is with regret that I have to continue to correct you. However, it
appears it is necessary.


You once defined a clock, quite rightly as an oscillator and a counter.
Suppose we have a 100Hz oscillator and a counter which increments every
100 cycles i.e. once per second. Suppose we now make that clock travel at
0.866c. The equation, which you claim is wrong says the frequency will
appear to a stationary observer as

f' = f Sqr(1-v^2/c^2) = 100 x 0.5 = 50 Hz

The equation which I say is right, and Doppler would confirm, is
f' = f(c+v)/c = 100 * 1.866 = 186.6 Hz. for blue shift,
and
f' = f(c-v)/c = 100 * 0.134 = 13.4 for red shift.

Your equation above doesn't belong to any theory I know of.

Time dilation is independent of direction. If you want to talk about
time dilation and not confuse the issue with Doppler shift we will have
to have the space ship going round us in a circle. If time is dilated
the clock ticks we hear will be at a lower frequency indicating that the
clock is going slower. Each tick is a time interval. If the frequency is
lower then the ticks are longer because time is the reciprocal of
frequency.

If you return to our discussion on the twin paradox where I analysed a
problem set by you I showed that you get the same number of ticks using
two methods:

1/ If I ignored Doppler and assumed time dilation for the whole of the
trip

2/ If I used the Doppler equation and counted the number of ticks at one
frequency and then at another.

Time Dilation effects the actual number of ticks, Doppler shift only
affects their distribution in space. On a return trip you can ignore
Doppler because Doppler effects cancels.

You have clearly realised I am right and are shovelling in red herrings
by the bucket load rather than admit it.

I'm not playing anymore. Don't bother to reply to any more of my posts I
will ignore you. There is no point in trying to help someone who refuses
to be helped and I doubt you are in any position to constructively
contribute to my knowledge.
--
John Kennaugh
to email convert the number from hex to decimal

"sal" wrote in message
news
On Wed, 13 Oct 2004 20:23:18 +0100, John Kennaugh wrote:

In message ,
Creighton Hogg writes


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?



While I want to be sure of those parts of relativity which interest me
(i.e. the basics) I have not great desire to become an expert in
relativity and have neither the time nor the inclination to get my head
around the modern space-time geometry approach. Perhaps I have been
unlucky in the books I have looked at in that they either insult your
intelligence or jump into space time geometry. I need something in
between
and Einstein's original will do me for now. In any case Einstein's paper
I
can download via the WWW free. If you know a good text book I can
download
let me know.

Do you have Einstein's "Relativity, the Special and the General Theory"?
It's not terribly deep but it's very readable, and it's available on
Project Gutenberg. I wouldn't recommend the online version,
though, because ... A hard copy is about 2 bucks (used), plus shipping,
and has the illustrations and sensible-looking equations which are missing
from the Gutenberg version. It's a nice book and "reads well". Check
Barnes&Noble or Amazon for exact information if you want buy a copy.

If you wanted to strain your brain and learn the geometric version of SR
Rindler's "Intro to SR" would be a good follow-on. But it's _not_ light
reading, unlike the Einstein book.

By the way, if you favor Ritz theory, as one of your other posts seemed to
suggest, I'm curious about your take on the Hubble redshift results. Ritz
(and Fox, for that matter) wrote at a time when all spectroscopes were
immersed in air, so there was not, as yet, anything to explain. However,
these days there's a lot to explain. The Hubble spectroscope is in vacuum,
and the wavelength it measures is the wavelength of the light before it
gets to the Earth's atmosphere. What's your preferred explanation for the
fact that redshift measured in space appears identical to the values we
get from ground-based spectroscopes? FWIW Androcles says V=F*L is
incorrect for photons,

Huh?
V= F*L is quite correct, because F = 1/T and V = L/T by definition.
Ohh.... I see your problem. You are not familiar with the good ol' American
phrase "spinning your wheels". It won't get you anywhere.
FWIW, I'll try to educate.
Ever heard of refraction? That's a change of speed, V. Refraction occurs
when the photon enters a medium, or changes from one medium to another.
What happens when a revolving wheel, on a road, hits a patch of ice? Nothing
really.... unless you put the brakes on, or depress the accelerator.
Then what happens? Nothing really, until you get past the ice and back on
the friction. Photons don't know how to apply brakes, and Hubble doesn't
know how to measure their frequency either, and HST certainly can't measure
their velocity by timing them over a distance.
FWIW.
Androcles.



and Henri Wilson says it's the Cosmic Fart -- er,
interstellar gas -- that normalizes the velocity and hence causes the
wavelength to show the same redshift as the frequency. I have some
problems with both of these explanations, and I was wondering if you agree
with one or both, or have taken a different tack.

If you don't know what I'm talking about, google "extinction length" along
with "ritz" and "fox".


The book I have which does have a reasonable chapter on relativity does
not cover the Doppler equation. I did a search on WWW for 'Doppler
Einstein relativity' and came up with nothing useful surprisingly
enough. Maybe gave up too soon )

FWIW, this critique of Einstein's doppler is free.
http://www.androc1es.pwp.blueyonder....oksDoppler.htm
Androcles.



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

"John Kennaugh" wrote in message
.uk...
Androcles writes

"John Kennaugh" wrote in message
. co.uk...
Androcles writes

"John Kennaugh" wrote in message
e.co.uk...
Androcles writes

"John Kennaugh" wrote in
message
news 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.

No.

Suppose t is the time between two 1s ticks. If you were right t' is
smaller then the interval between ticks is smaller - the clock is
ticking
faster - time has speeded up. Look up 'dilate' = "become wider or
larger"
as in eyes dilated.

Put some numbers in. For v = 0.866c (always convenient),
sqrt(1-v^2/c^2) = sqrt(1- 3/4) = sqrt( 1/4) = 1/2.
5 hours = 10 hours * 1/2.
The moving clock records 5 hours (t') for every 10 hours (t) of the
stationary clock, so:
t' = t.sqrt(1-v^2/c^2)

The interval between ticks of the moving clock equals two intervals of
the
stationary clock and is therefore "dilated."

0----------1----------2----------3----------4---------5 time t'
0-----1----2----3----4----5----6----7-----8----9---10 time t


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,

The 'amateur' has it right. t't. A tick is stretched so the clock
appears
to be going slower. It is actually easier to think in terms of
frequency.

t' = t/sqrt(1-v^2/c^2)

converts to

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

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

This says the frequency of the clock ticks appears less = the clock
going
slower. f' is a lower frequency than f because Sqr(1-v^2/c^2)1

Put some numbers in. I don't see why I should do it for you again.

Had you put the numbers in to this equation rather than the other you
would have spotted your own mistake. I really don't enjoy putting you
right all the time.

It is with regret that I have to continue to correct you. However, it
appears it is necessary.


You once defined a clock, quite rightly as an oscillator and a counter.
Suppose we have a 100Hz oscillator and a counter which increments every
100 cycles i.e. once per second. Suppose we now make that clock travel
at
0.866c. The equation, which you claim is wrong says the frequency will
appear to a stationary observer as

f' = f Sqr(1-v^2/c^2) = 100 x 0.5 = 50 Hz

The equation which I say is right, and Doppler would confirm, is
f' = f(c+v)/c = 100 * 1.866 = 186.6 Hz. for blue shift,
and
f' = f(c-v)/c = 100 * 0.134 = 13.4 for red shift.

Your equation above doesn't belong to any theory I know of.

Time dilation is independent of direction.

WRONG!
I dont mind discussing Einstein's fairy tale with you, but let's get the
story straight.
Quote:
"IF we place x' = x-vt....." - Einstein.
Unquote
then.....
xi = (x') / sqrt(1-v^2/c^2)
tau = (t-vx/c^2) / sqrt(1-v^2/c^2)

But of course if we place x' = x+vt, then
xi = (x') / sqrt(1-v^2/c^2)
tau = (t + vx/c^2) / sqrt(1-v^2/c^2)

Go ahead, put some numbers into that.

If you want to talk about time dilation and not confuse the issue with
Doppler shift we will have to have the space ship going round us in a
circle.

You can't talk about one independently of the other, John as you seem to
be attempting.
Take a look at Koks pathetic attempt to deny the moving clock speeds up
when direction changes.

If time is dilated

Which it isn't, but let's pretend..

the clock ticks we hear will be at a lower frequency indicating that the
clock is going slower. Each tick is a time interval. If the frequency is
lower then the ticks are longer because time is the reciprocal of
frequency.

You need to determine which time frame and which frequency frame you are
referring to.
If you mean t = 1/f and tau = 1/nu, that's just fine.
If you mean t = 1/nu or tau = 1/f, it isn't.

As to going round in circles, let's send two ships in opposite directions.
They can pass each other twice each orbit and run parallel to each
other twice each orbit. Now each can determine the other's clock is
running slow and telling exactly the same time.



If you return to our discussion on the twin paradox where I analysed a
problem set by you I showed that you get the same number of ticks using
two methods:

1/ If I ignored Doppler and assumed time dilation for the whole of the
trip

2/ If I used the Doppler equation and counted the number of ticks at one
frequency and then at another.

If you get the same number of ticks, you get the same time.


Time Dilation effects the actual number of ticks, Doppler shift only
affects their distribution in space. On a return trip you can ignore
Doppler because Doppler effects cancels.

You have clearly realised I am right and are shovelling in red herrings by
the bucket load rather than admit it.

You clearly have no understanding of physics, or even relativity,
and you are not right.



I'm not playing anymore. Don't bother to reply to any more of my posts I
will ignore you.

Well **** off then. I'll still continue to criticise your garbage for the
benefit
of others.



There is no point in trying to help someone who refuses to be helped

That's too right, and you don't want to be helped. You are just another
think-you-know-it-all that doesn't.


and I doubt you are in any position to constructively
contribute to my knowledge.

Since your "knowledge" is incorrect, I have no desire to contribute to
nonsense. Pout all you want to, time dilation is directional and you are
wrong.
Androcles


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

I believe you are stuck in a false regime.



John Kennaugh wrote:

Bilge writes
John Kennaugh:
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.

It's much simpler to start with the relation,

k.x - wt = k'.x' = w't',

substitute the lorentz transforms of x and t for x' and t', then solve
for k' and w'.

That would be a totally pointless exercise. The whole point of the
exercise was to try and get some insight into what the equation is
describing by deriving it from first principles.

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

Bilge writes
John Kennaugh:
Bilge writes
John Kennaugh:
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.

It's much simpler to start with the relation,

k.x - wt = k'.x' = w't',

substitute the lorentz transforms of x and t for x' and t', then solve
for k' and w'.

That would be a totally pointless exercise. The whole point of the
exercise was to try and get some insight into what the equation is
describing by deriving it from first principles.


Given that the mist fundamental
~~~~
Freudian slip? Sub-consciously you know that the fundamentals are
immersed in fog.

principle in relativity is invariance,

The fundamental principle of relativity, that which distinguishes it
from other theories (other than LET) is source independence. Your
constant attempts to deny the true origin of relativity are perfectly
understandable considering what they were, but cut no ice with me.
Whether you like it or not Source independence was believed in for 200
years because of a belief in the ether - something few believe in now,
incorporated into SR for no logical reason and sustained as a belief for
a further 60 years by now discredited 'evidence'. That is FACT whether
you like it or not.

Your 'magical mystery diagram' based on Minkowski's pictorial
representation of Einstein's (subconsciously?) ether based theory may be
gods revelation of the mystery of the universe to you. Something you
have based your faith on but it means nothing to me. It is just an
alternative mathematical technique which I don't need to bother with
because unlike you I have no desire to impress anyone by my supposed
mastery of the subject.

that _is_ a first principle. The phase is an invariant. How much more
fundamental can you get? Let me give you a clue. ``First principles'' is
not synonymous with ``the way which offers the least insight and the most
opportunity for arithmetic mistakes.'' On the contrary, doing a calculation
the hard way usually indicates an understanding of first principles is
lacking.

The first principles of relativity are and always were two postulates.
From the second postulate I can derive quite simply the time dilation
equation. By examining the second postulate I can conclude that ether or
no ether what it states is that the observer is at least mathematically
equivalent to being stationary w.r.t the ether and from that I can
simply write the affect Doppler will have.

Go Away.

--
John Kennaugh
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