Fallacy of Relativistic Doppler Effect

On Mar 14, 4:23*pm, Koobee Wublee wrote:
As we all know, the time transformation of the Lorentz transform is

[snip]

Your analysis is flawed .. you are comparing time dilation alone with
the combination of time dilation and doppler (ie relativeitc
doppler). In terms you might understand, you've done the equivalent
of proving regular doppler shift to be wrong by comparing the
frequency of a train whistle on the train vs the one an observer hears
and saying it is wrong because they are different.

I could point out your errors in more detail if you were at all
interested in learning why you are wrong.

But I douct that your arrogance will even let you even consider that
you made a mistake ss a possibility .. obviously (to you) the rest of
the physicists and mathematicians in the world don't know what they're
doing and you only you know how to do the math 'correctly' shrug.

On Mar 14, 1:23*am, Koobee Wublee wrote:
As we all know, the time transformation of the Lorentz transform is

** *dt’ = (dt + [v] * d[s] / c^2) / sqrt(1 – v^2 / c^2)

Where

** *[v] = velocity of dt frame as observed by the dt’ frame
** *d[s] = observed displacement vector by the dt frame
** *[] * [] = dot product of two vectors

It was attributed to Einstein the nitwit, the plagiarist, and the liar
who first wrote down the relativistic Doppler effect of light or
whoever the author of that 1905 paper was. *The above equation becomes
the following.

** *f’ / f = sqrt(1 – v^2 / c^2) / (1 + [v] * [c] / c^2)

Where

** *f’ = 1 / dt’
** *f = 1 / dt
** *d[s]/dt = [c]

Of course, Einstein the nitwit, the plagiarist, and the liar was not
bright enough to realize the above equation in general. *The nitwit
and almost all self-styled physicists can only rationalize in the very
special case where [v] and [c] are in parallel to each other. *If so,
the above equation can be simplified according to the following.

** *f’ / f = sqrt(1 – v^2 / c^2) / (1 - v / c)

Or

** *f’ / f = sqrt(1 + v / c) / sqrt(1 - v / c)

Where

** *[c] is always propagating from dt frame to dt’
** *v 0 means dt is moving away from dt’

In this case, the relativistic Doppler effect according to the Lorentz
transform would always predict an opposite to the classical one.
Oops! *How can the self-styled physicists miss this blatant math error
for over 100 years?

Interestingly, there is another way of deriving the relativistic
Doppler effect. *All the infinite non-ballistic-theory-of-light
transforms that satisfy the null results of the MMX share the same
equation of energy transform derive from the geodesic equations. *In
doing so, the energy transform can be written as follows.

** *E’ = (E + [v] * [p]) / sqrt(1 – v^2 / c^2)

Where

** *E’, E = observed energies
** *[p] = observed momentum by the dt frame

Using the same, previous criteria where [v] and [p] are in parallel
for the intellect-deficient self-styled physicists, the above
equations simplifies into the following.

** *f’ / f = (1 – v / c) / sqrt(1 – v^2 / c^2)

Or

** *f’ / f = sqrt(1 – v / c) / sqrt(1 + v / c)

Where

** *[p] is always going from dt frame to dt’
** *v 0 means dt is moving away from dt’
** *E’ = h f’
** *E = h f
** *[p] = h f [c] / c^2

This version of the relativistic Doppler effect is the exact opposite
of the one derived earlier in this post. *Thus, yours truly demands to
know why the self-styled physicists have allowed this blatant math
error to go through to justify the validity of SR in the past 100
years.

Oh, would any wise Dingleberry suggest that [v] is the velocity of dt’
frame as observed by dt frame instead? *If so, you can count on the
Guillotine is coming down hard in the reply post. *Einstein the
nitwit, the plagiarist, and the liar was a fudger of mathematics. *The
nitwit understood nothing about SR and GR. *The nitwit could not have
analyzed anything rationally and correctly to save his life. *shrug

I have studied your derivation No. 1 of the relativistic Doppler
effect,
and I believe your error is in claiming that [ds]/dt = [c]. I believe
that [ds]/dt = -[v].

I find it helpful to describe the frames of reference as the rest
frame of a star (x,y,z,t) and our observatory (x',y',z',t'). Thus the
direction of light is from star to observatory. In your presentation
[v] is the velocity of the star with respect to the laboratory, which
points from the observatory to the star. (This confirms your equation
of [v]*[c] = -vc.)

Now the laboratory is moving away from the star. The coordinates in
the LT concern events at the star and the laboratory. The laboratory
is moving in the +x direction with respect to the star. Thus [ds] is
the displacement of the laboratory w.r.t. the star in star-time dt.
[ds]/dt = the velocity of the observatory with respect to the star.
This is -[v].

Uncle Ben

On Mar 16, 8:21 pm, Uncle Ben wrote:
On Mar 14, 1:23 am, Koobee Wublee wrote:

As we all know, the time transformation of the Lorentz transform is

** dt’ = (dt + [v] * d[s] / c^2) / sqrt(1 – v^2 / c^2)

Where

** [v] = velocity of dt frame as observed by the dt’ frame
** d[s] = observed displacement vector by the dt frame
** [] * [] = dot product of two vectors

It was attributed to Einstein the nitwit, the plagiarist, and the liar
who first wrote down the relativistic Doppler effect of light or
whoever the author of that 1905 paper was. The above equation becomes
the following.

** f’ / f = sqrt(1 – v^2 / c^2) / (1 + [v] * [c] / c^2)

Where

** f’ = 1 / dt’
** f = 1 / dt
** d[s]/dt = [c]

Of course, Einstein the nitwit, the plagiarist, and the liar was not
bright enough to realize the above equation in general. The nitwit
and almost all self-styled physicists can only rationalize in the very
special case where [v] and [c] are in parallel to each other. If so,
the above equation can be simplified according to the following.

** f’ / f = sqrt(1 – v^2 / c^2) / (1 - v / c)

Or

** f’ / f = sqrt(1 + v / c) / sqrt(1 - v / c)

Where

** [c] is always propagating from dt frame to dt’
** v 0 means dt is moving away from dt’

In this case, the relativistic Doppler effect according to the Lorentz
transform would always predict an opposite to the classical one.
Oops! How can the self-styled physicists miss this blatant math error
for over 100 years?

Interestingly, there is another way of deriving the relativistic
Doppler effect. All the infinite non-ballistic-theory-of-light
transforms that satisfy the null results of the MMX share the same
equation of energy transform derive from the geodesic equations. In
doing so, the energy transform can be written as follows.

** E’ = (E + [v] * [p]) / sqrt(1 – v^2 / c^2)

Where

** E’, E = observed energies
** [p] = observed momentum by the dt frame

Using the same, previous criteria where [v] and [p] are in parallel
for the intellect-deficient self-styled physicists, the above
equations simplifies into the following.

** f’ / f = (1 – v / c) / sqrt(1 – v^2 / c^2)

Or

** f’ / f = sqrt(1 – v / c) / sqrt(1 + v / c)

Where

** [p] is always going from dt frame to dt’
** v 0 means dt is moving away from dt’
** E’ = h f’
** E = h f
** [p] = h f [c] / c^2

This version of the relativistic Doppler effect is the exact opposite
of the one derived earlier in this post. Thus, yours truly demands to
know why the self-styled physicists have allowed this blatant math
error to go through to justify the validity of SR in the past 100
years.

Oh, would any wise Dingleberry suggest that [v] is the velocity of dt’
frame as observed by dt frame instead? If so, you can count on the
Guillotine is coming down hard in the reply post. Einstein the
nitwit, the plagiarist, and the liar was a fudger of mathematics. The
nitwit understood nothing about SR and GR. The nitwit could not have
analyzed anything rationally and correctly to save his life. shrug

I have studied your derivation No. 1 of the relativistic Doppler
effect, and I believe your error is in claiming that [ds]/dt = [c].
I believe that [ds]/dt = -[v].

No, this is wrong. d[s] is the displacement vector of an event as
observed by the unprimed frame, and d[s’] is the displacement vector
of the same event as observed by the primed frame. The event is the
photon leaving the unprimed frame, and [v] is the velocity between the
two frames that has nothing to do with the event. shrug

You are not alone to make this mistake. Daryl McCullough tried to
fudge the answer into his wishing as explained by the following post.
Since then, Daryl has not been a man to confront his own mistake.
Because in doing so would invalidate SR since in actuality SR is not
accountable for and does not degenerate (at low speeds) to the
observed classical Doppler effect. shrug

http://groups.google.com/group/sci.p...e96cb951?hl=en

Notice this is a very simple algebra problem. If yours truly is
wrong, all these self-styled physicists would be all over my ass.
shrug

I find it helpful to describe the frames of reference as the rest
frame of a star (x,y,z,t) and our observatory (x',y',z',t'). Thus the
direction of light is from star to observatory. In your presentation
[v] is the velocity of the star with respect to the laboratory, which
points from the observatory to the star. (This confirms your equation
of [v]*[c] = -vc.)

Yes, that is correct. shrug

Now the laboratory is moving away from the star. The coordinates in
the LT concern events at the star and the laboratory. The laboratory
is moving in the +x direction with respect to the star. Thus [ds] is
the displacement of the laboratory w.r.t. the star in star-time dt.
[ds]/dt = the velocity of the observatory with respect to the star.
This is -[v].

Two days ago, you did not understand the Lorentz transform. After two
days of staring at a textbook, you are still confused with the Lorentz
transform. It is time to convert the pages of that textbook into
toilet papers unless the place you bought it from would give you a
complete refund in return. shrug

Is there anyone else who agrees with Uncle Ben? Tom? PD? Paul
Andersen?

On Mar 17, 4:05*pm, Koobee Wublee wrote:
On Mar 16, 8:21 pm, Uncle Ben wrote:
I have studied your derivation No. 1 of the relativistic Doppler
effect, and I believe your error is in claiming that [ds]/dt = [c].
I believe that [ds]/dt = -[v].

That is one of the errors .. yes

No, this is wrong.

You are wrong

*d[s] is the displacement vector of an event

d[s] means a difference in displacement (ie how far something moves
for the layman such as yourself) .. an event doesn't have a difference
in displacement. An event doesn't move.

as
observed by the unprimed frame, and d[s’] is the displacement vector
of the same event as observed by the primed frame.

*The event is the
photon leaving the unprimed frame,

That's makes no sense

and [v] is the velocity between the
two frames that has nothing to do with the event. *shrug

Your 'explanation' above is nonsense

You are not alone to make this mistake.

He made no mistake .. you have, however. It would help if you knew
enough of physics to know what the terms in the equations your are
transcribing actually mean. I suggest a basic course in physics

On Mar 16, 11:55*pm, artful wrote:
On Mar 17, 4:05*pm, Koobee Wublee wrote:

On Mar 16, 8:21 pm, Uncle Ben wrote:
I have studied your derivation No. 1 of the relativistic Doppler
effect, and I believe your error is in claiming that [ds]/dt = [c].
I believe that [ds]/dt = -[v].

That is one of the errors .. yes

No, this is wrong.

You are wrong

*d[s] is the displacement vector of an event

d[s] means a difference in displacement (ie how far something moves
for the layman such as yourself) .. an event doesn't have a difference
in displacement. *An event doesn't move.

The thing is, nobody ever writes out 'ds'. He's the only one who does
it.


as
observed by the unprimed frame, and d[s’] is the displacement vector
of the same event as observed by the primed frame.

*The event is the
photon leaving the unprimed frame,

That's makes no sense

That's because he does not know what he is talking about. He is
generally close with words and concepts, but things like this
indicates he doesn't actually understand.

He does not know what the symbols mean, or how to compute anything
with them. You can go back years and find dozens of examples of him
being unable to compute the surface area of a sphere using the metric.


and [v] is the velocity between the
two frames that has nothing to do with the event. *shrug

Your 'explanation' above is nonsense

You are not alone to make this mistake.

He made no mistake .. you have, however. *It would help if you knew
enough of physics to know what the terms in the equations your are
transcribing actually mean. *I suggest a basic course in physics

On Mar 14, 1:23 am, Koobee Wublee wrote:
As we all know, the time transformation of the Lorentz transform is

** dt’ = (dt + [v] * d[s] / c^2) / sqrt(1 – v^2 / c^2)

Where

** [v] = velocity of dt frame as observed by the dt’ frame
** d[s] = observed displacement vector by the dt frame
** [] * [] = dot product of two vectors

It was attributed to Einstein the nitwit, the plagiarist, and the liar
who first wrote down the relativistic Doppler effect of light or
whoever the author of that 1905 paper was. The above equation becomes
the following.

** f’ / f = sqrt(1 – v^2 / c^2) / (1 + [v] * [c] / c^2)

Oh, my gosh. Koobee is completely confused. That isn't a correct
derivation, at all. What he seems to be deriving is something
that has nothing to do with the Doppler shift. It's the correct
derivation for the following situation:

Suppose we have an observer at rest in the F' frame.
This observer is traveling at speed v in the +x direction,
as measured in the F frame. (I'm going to assume that
v is parallel to the line between the two observers,
so that I can dispense with the dot product business;
that's a complication that doesn't really introduce
any new insight).

Let the F' observer send a light
signal to an observer at rest in frame F.

Let delta-x be the displacement in the x-direction between
the sending event and the receiving event (that is, the change
in the x-coordinate), as measured in frame F.
Let delta-t be the time between these events, as measured
in frame F. Let delta-x' and delta-t' be the displacement and
time between the events, as measured in frame F'. Then
we have:

delta-x = - c delta-t

(delta-x is negative, since the light signal is traveling
in the -x direction).

We can use the Lorentz transformations to compute
delta-t' in terms of delta-t:

delta-t' = gamma (delta-t - v/c^2 delta-x)
= gamma (delta-t + v/c^2 c delta-t)
= (1+v/c) gamma delta-t
= (1+v/c)/square-root(1-(v/c)^2) delta-t
= square-root((1+v/c)/(1-v/c)) delta-t

This delta-t' and delta-t has NOTHING to do with any
frequency. Note: there is nothing in anything that
was said here that has anything to do with the *frequency*
of the light used. delta-t has nothing to do with
the frequency of the light, and this derivation has
NOTHING to do with Doppler shift.

What Koobee apparently has done is to reason:

delta-t is a time. If you take the reciprocal,
you can call that a frequency, f. Similarly,
we can let f' = 1/delta-t'. Then we calculate
the ratio of the frequencies as:

f'/f = (1/delta-t')/(1/delta-t)
= delta-t/delta-t'
= square-root((1+v/c)/(1-v/c))

This ratio looks like the relativistic Doppler shift
formula, but it has NOTHING to
do with any Doppler shift, because f and f' have
NOTHING to do with the light frequency. The real
frequency of the light didn't even enter into the
problem.

This is C-student level work. It is an indication
that the student really has not taken the time to
think through what it is he is deriving, and is
instead doing purely formal manipulations without
understanding.

If you want to derive the Doppler shift for
the frequency of light, you have to have some
place where the frequency of light goes into
the derivation.

Koobee's derivation didn't come close to being
correct. At some point, he blames this on *EINSTEIN*,
but Koobee's bogus derivation has nothing to do with
Einstein. It's Koobee's own incompetence at work.

--
Daryl McCullough
Ithaca, NY

On Mar 17, 11:29*am, (Daryl McCullough)
wrote:
On Mar 14, 1:23 am, Koobee Wublee wrote:
As we all know, the time transformation of the Lorentz transform is

** *dt = (dt + [v] * d[s] / c^2) / sqrt(1 v^2 / c^2)

Where

** *[v] = velocity of dt frame as observed by the dt frame
** *d[s] = observed displacement vector by the dt frame
** *[] * [] = dot product of two vectors

It was attributed to Einstein the nitwit, the plagiarist, and the liar
who first wrote down the relativistic Doppler effect of light or
whoever the author of that 1905 paper was. *The above equation becomes
the following.

** *f / f = sqrt(1 v^2 / c^2) / (1 + [v] * [c] / c^2)

Oh, my gosh. Koobee is completely confused. That isn't a correct
derivation, at all. What he seems to be deriving is something
that has nothing to do with the Doppler shift. It's the correct
derivation for the following situation:

Suppose we have an observer at rest in the F' frame.
This observer is traveling at speed v in the +x direction,
as measured in the F frame. (I'm going to assume that
v is parallel to the line between the two observers,
so that I can dispense with the dot product business;
that's a complication that doesn't really introduce
any new insight).

Let the F' observer send a light
signal to an observer at rest in frame F.

Let delta-x be the displacement in the x-direction between
the sending event and the receiving event (that is, the change
in the x-coordinate), as measured in frame F.
Let delta-t be the time between these events, as measured
in frame F. Let delta-x' and delta-t' be the displacement and
time between the events, as measured in frame F'. Then
we have:

delta-x = - c delta-t

(delta-x is negative, since the light signal is traveling
in the -x direction).

We can use the Lorentz transformations to compute
delta-t' in terms of delta-t:

delta-t' = gamma (delta-t - v/c^2 delta-x)
= gamma (delta-t + v/c^2 c delta-t)
= (1+v/c) gamma delta-t
= (1+v/c)/square-root(1-(v/c)^2) delta-t
= square-root((1+v/c)/(1-v/c)) delta-t

This delta-t' and delta-t has NOTHING to do with any
frequency. Note: there is nothing in anything that
was said here that has anything to do with the *frequency*
of the light used. delta-t has nothing to do with
the frequency of the light, and this derivation has
NOTHING to do with Doppler shift.

What Koobee apparently has done is to reason:

delta-t is a time. If you take the reciprocal,
you can call that a frequency, f. Similarly,
we can let f' = 1/delta-t'. Then we calculate
the ratio of the frequencies as:

f'/f = (1/delta-t')/(1/delta-t)
= delta-t/delta-t'
= square-root((1+v/c)/(1-v/c))

This ratio looks like the relativistic Doppler shift
formula, but it has NOTHING to
do with any Doppler shift, because f and f' have
NOTHING to do with the light frequency. The real
frequency of the light didn't even enter into the
problem.

This is C-student level work. It is an indication
that the student really has not taken the time to
think through what it is he is deriving, and is
instead doing purely formal manipulations without
understanding.

If you want to derive the Doppler shift for
the frequency of light, you have to have some
place where the frequency of light goes into
the derivation.

Koobee's derivation didn't come close to being
correct. At some point, he blames this on *EINSTEIN*,
but Koobee's bogus derivation has nothing to do with
Einstein. It's Koobee's own incompetence at work.

--
Daryl McCullough
Ithaca, NY- Hide quoted text -

- Show quoted text -

alas, poor yoricK.

It's Koobee's own work.

On Mar 17, 8:29 am, Daryl McCullough wrote:
On Mar 16, 10:05 pm, Koobee Wublee wrote:

As we all know, the time transformation of the Lorentz transform is

** dt’ = (dt + [v] * d[s] / c^2) / sqrt(1 – v^2 / c^2)

Where

** [v] = velocity of dt frame as observed by the dt’ frame
** d[s] = observed displacement vector by the dt frame
** [] * [] = dot product of two vectors

It was attributed to Einstein the nitwit, the plagiarist, and the liar
who first wrote down the relativistic Doppler effect of light or
whoever the author of that 1905 paper was. The above equation becomes
the following.

** f’ / f = sqrt(1 – v^2 / c^2) / (1 + [v] * [c] / c^2)

Where

** f’ = 1 / dt’
** f = 1 / dt
** d[s]/dt = [c]

Of course, Einstein the nitwit, the plagiarist, and the liar was not
bright enough to realize the above equation in general. The nitwit
and almost all self-styled physicists can only rationalize in the very
special case where [v] and [c] are in parallel to each other. If so,
the above equation can be simplified according to the following.

** f’ / f = sqrt(1 – v^2 / c^2) / (1 - v / c)

Or

** f’ / f = sqrt(1 + v / c) / sqrt(1 - v / c)

Where

** [c] is always propagating from dt frame to dt’
** v 0 means dt is moving away from dt’

Oh, my gosh. Koobee is completely confused. That isn't a correct
derivation, at all. What he seems to be deriving is something
that has nothing to do with the Doppler shift. It's the correct
derivation for the following situation:

[mumblings with no substance and no points mercifully snipped]

Yours truly has been telling you that is not the correct derivation
for Doppler effects for years since with the same derivation, the
Galilean transform does not yield any Doppler shift which is just
wrong. shrug

In this case, the relativistic Doppler effect according to the Lorentz
transform would always predict an opposite to the classical one.
Oops! How can the self-styled physicists miss this blatant math error
for over 100 years?

Interestingly, there is another way of deriving the relativistic
Doppler effect. All the infinite non-ballistic-theory-of-light
transforms that satisfy the null results of the MMX share the same
equation of energy transform derive from the geodesic equations. In
doing so, the energy transform can be written as follows.

** E’ = (E + [v] * [p]) / sqrt(1 – v^2 / c^2)

Where

** E’, E = observed energies
** [p] = observed momentum by the dt frame

Using the same, previous criteria where [v] and [p] are in parallel
for the intellect-deficient self-styled physicists, the above
equations simplifies into the following.

** f’ / f = (1 – v / c) / sqrt(1 – v^2 / c^2)

Or

** f’ / f = sqrt(1 – v / c) / sqrt(1 + v / c)

Where

** [p] is always going from dt frame to dt’
** v 0 means dt is moving away from dt’
** E’ = h f’
** E = h f
** [p] = h f [c] / c^2

This version of the relativistic Doppler effect is the exact opposite
of the one derived earlier in this post. Thus, yours truly demands to
know why the self-styled physicists have allowed this blatant math
error to go through to justify the validity of SR in the past 100
years.

Oh, would any wise Dingleberry suggest that [v] is the velocity of dt’
frame as observed by dt frame instead? If so, you can count on the
Guillotine is coming down hard in the reply post. Einstein the
nitwit, the plagiarist, and the liar was a fudger of mathematics. The
nitwit understood nothing about SR and GR. The nitwit could not have
analyzed anything rationally and correctly to save his life. shrug

What Koobee apparently has done is to reason:

delta-t is a time. If you take the reciprocal,
you can call that a frequency, f. Similarly,
we can let f' = 1/delta-t'. Then we calculate
the ratio of the frequencies as:

f'/f = (1/delta-t')/(1/delta-t)
= delta-t/delta-t'
= square-root((1+v/c)/(1-v/c))

However, this was exactly how your god Einstein the nitwit, the
plagiarist, and the liar derived the relativistic Doppler shift from
after some mathemaGical inversion of v. shrug

This is C-student level work. It is an indication
that the student really has not taken the time to
think through what it is he is deriving, and is
instead doing purely formal manipulations without
understanding.

Right, your god Einstein the nitwit, the plagiarist, and the liar was
nothing but a nitwit, a plagiarist, and a liar. shrug

If you want to derive the Doppler shift for
the frequency of light, you have to have some
place where the frequency of light goes into
the derivation.

Gee! Yours truly also has been telling you the following for years.
shrug

** f = (observed speed) / wavelength

Koobee's derivation didn't come close to being
correct. At some point, he blames this on *EINSTEIN*,
but Koobee's bogus derivation has nothing to do with
Einstein.

In Maxwell’s Aether, the observed speed is dependent on the velocity
of the observer while the wavelength is invariant embedded in the
medium itself, and that explains the classical Doppler effect.
shrug

Under SR, the observed speed is the invariant one. So, the wavelength
must be dependent on the observer. The FitzGerald-Lorentz length
contraction does not bode very well for SR’s explanation of Doppler
effect. shrug

It's Koobee's own incompetence at work.

You are the one who is incompetent. Remember that you made the same
stupid mistake as Uncle Ben the phd (what else is new)? MathemaGics
rules in you and Uncle Ben. You have no analytic skills but merely
are a fudger of mathematics like your good Einstein the nitwit, the
plagiarist, and the liar was. shrug

On Mar 18, 12:59*am, Koobee Wublee wrote:
On Mar 17, 7:42 am, Uncle Ben wrote:





On Mar 17, 1:05 am, Koobee Wublee wrote:
On Mar 16, 8:21 pm, Uncle Ben wrote:

On Mar 14, 1:23 am, Koobee Wublee wrote:
As we all know, the time transformation of the Lorentz transform is

** *dt’ = (dt + [v] * d[s] / c^2) / sqrt(1 – v^2 / c^2)

Where

** *[v] = velocity of dt frame as observed by the dt’ frame
** *d[s] = observed displacement vector by the dt frame
** *[] * [] = dot product of two vectors

It was attributed to Einstein the nitwit, the plagiarist, and the liar
who first wrote down the relativistic Doppler effect of light or
whoever the author of that 1905 paper was. *The above equation becomes
the following.

** *f’ / f = sqrt(1 – v^2 / c^2) / (1 + [v] * [c] / c^2)

Where

** *f’ = 1 / dt’
** *f = 1 / dt
** *d[s]/dt = [c]

Of course, Einstein the nitwit, the plagiarist, and the liar was not
bright enough to realize the above equation in general. *The nitwit
and almost all self-styled physicists can only rationalize in the very
special case where [v] and [c] are in parallel to each other. *If so,
the above equation can be simplified according to the following.

** *f’ / f = sqrt(1 – v^2 / c^2) / (1 - v / c)

Or

** *f’ / f = sqrt(1 + v / c) / sqrt(1 - v / c)

Where

** *[c] is always propagating from dt frame to dt’
** *v 0 means dt is moving away from dt’

In this case, the relativistic Doppler effect according to the Lorentz
transform would always predict an opposite to the classical one.
Oops! *How can the self-styled physicists miss this blatant math error
for over 100 years?

Interestingly, there is another way of deriving the relativistic
Doppler effect. *All the infinite non-ballistic-theory-of-light
transforms that satisfy the null results of the MMX share the same
equation of energy transform derive from the geodesic equations. *In
doing so, the energy transform can be written as follows.

** *E’ = (E + [v] * [p]) / sqrt(1 – v^2 / c^2)

Where

** *E’, E = observed energies
** *[p] = observed momentum by the dt frame

Using the same, previous criteria where [v] and [p] are in parallel
for the intellect-deficient self-styled physicists, the above
equations simplifies into the following.

** *f’ / f = (1 – v / c) / sqrt(1 – v^2 / c^2)

Or

** *f’ / f = sqrt(1 – v / c) / sqrt(1 + v / c)

Where

** *[p] is always going from dt frame to dt’
** *v 0 means dt is moving away from dt’
** *E’ = h f’
** *E = h f
** *[p] = h f [c] / c^2

This version of the relativistic Doppler effect is the exact opposite
of the one derived earlier in this post. *Thus, yours truly demands to
know why the self-styled physicists have allowed this blatant math
error to go through to justify the validity of SR in the past 100
years.

Oh, would any wise Dingleberry suggest that [v] is the velocity of dt’
frame as observed by dt frame instead? *If so, you can count on the
Guillotine is coming down hard in the reply post. *Einstein the
nitwit, the plagiarist, and the liar was a fudger of mathematics. *The
nitwit understood nothing about SR and GR. *The nitwit could not have
analyzed anything rationally and correctly to save his life. *shrug

I have studied your derivation No. 1 of the relativistic Doppler
effect, and I believe your error is in claiming that [ds]/dt = [c].
I believe that [ds]/dt = -[v].

No, this is wrong. *d[s] is the displacement vector of an event as
observed by the unprimed frame, and d[s’] is the displacement vector
of the same event as observed by the primed frame. *The event is the
photon leaving the unprimed frame, and [v] is the velocity between the
two frames that has nothing to do with the event. *shrug

You are not alone to make this mistake. *Daryl McCullough tried to
fudge the answer into his wishing as explained by the following post.
Since then, Daryl has not been a man to confront his own mistake.
Because in doing so would invalidate SR since in actuality SR is not
accountable for and does not degenerate (at low speeds) to the
observed classical Doppler effect. *shrug

http://groups.google.com/group/sci.p...sg/7aaf065fe96....

Notice this is a very simple algebra problem. *If yours truly is
wrong, all these self-styled physicists would be all over my ass.
shrug

I find it helpful to describe the frames of reference as the rest
frame of a star (x,y,z,t) and our observatory (x',y',z',t'). Thus the
direction of light is from star to observatory. *In your presentation
[v] is the velocity of the star with respect to the laboratory, which
points from the observatory to the star. (This confirms your equation
of [v]*[c] = -vc.)

Yes, that is correct. *shrug

Now the laboratory is moving away from the star. The coordinates in
the LT concern events at the star and the laboratory. The laboratory
is moving in the +x direction with respect to the star. Thus [ds] is
the displacement of the laboratory w.r.t. the star in star-time dt.
[ds]/dt = the velocity of the observatory with respect to the star.