Fallacy of Relativistic Doppler Effect

On Mar 18, 12:56*am, Koobee Wublee wrote:
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- Hide quoted text -

- Show quoted text -- Hide quoted text -

- Show quoted text -

Your "derivation" of the relativistic Doppler shift (in the wrong
direction) and blaming it on Einstein puts you in a class with John
Parker, who "derived" the Einstein Expansion a few years back,
claiming that Einstein erred in driving the Lorentz Contraction.

John, at least did the LT correctly and flubbed only in the last step.
He showed that the moving rod measured w.r.t. its rest frame is longer
than the same rod w.r.t. the stationary frame. His idiocy was that he
assigned the length w.r.t. the stationary frame as L and concluded
that the proper length must have expanded to L*gamma. He could not see
that the moving frame IS the rest frame of the moving rod.

Your presentation is in some kind of code. (What does it mean that a
vector points to "dt" or to "dt'", as if they were points in space?) I
guessed your meaning in spite of the poor description, and you did mis-
identify the vector d[s]. That is why you could not even derive the
classical Doppler shift by means of the Galilean transformation.

Koobee Wooblee and Androcles, both believing in isolation that the
emperor has no clothes.

Uncle Ben

Uncle Ben

Uncle Ben says...

On Mar 18, 12:56=A0am, Koobee Wublee wrote:
On Mar 17, 8:29 am, Daryl McCullough wrote:

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.

And *NOBODY* derives the Doppler shift that way. Koobee makes
up a derivation that is completely wrong, and then complains
about his *OWN* derivation.

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.

Koobee is unable to follow any derivation, and he blames
his stupidity on Einstein, for some reason.

Your "derivation" of the relativistic Doppler shift (in the wrong
direction) and blaming it on Einstein puts you in a class with John
Parker, who "derived" the Einstein Expansion a few years back,
claiming that Einstein erred in driving the Lorentz Contraction.

Right. When Koobee makes a mistake, he blames it on Einstein.
Einstein had nothing to do with Koobee's errors.

--
Daryl McCullough
Ithaca, NY

In article , Daryl McCullough says...

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.

Here's an interesting question: Since this ratio
has nothing (apparently) to do with Doppler shifts,
*why* does it have the same form as a Doppler shift?

It occurred to me why. Let's identify three events:
e_0 = the two observers depart from each other
e_1 = the "traveling" observer sends a light signal
e_2 = the "stay-at-home" observer receives the light
signal.

Now, imagine that we make a movie of these events and
play it *backwards*. In the time-reversed movie, things
happen like this:

First (event e_2) a light signal is "sent" by the stay-at-home
observer. (The time reversal of receiving a signal is sending
a signal).

Second (event e_1) the signal is "received" by the traveling
observer. (The time reversal of sending is receiving).

Third (event e_0) the two observers are together. If
a second light signal were sent at this point, it would
arrive instantaneously, since the two are side-by-side.

So we can interpret the time-reversed sequence of events
like this:

The stay-at-home observer is sending signals at a rate
of one signal every delta-t seconds (where delta-t is the
absolute value of the time between e_2 and e_1, as measured by
the stay-at-home observer). The first signal is
sent at event e_2 and the second is sent at e_0.

The traveling observer is traveling *toward* the
stay-at-home observer (because the time-reverse of
a positive velocity is a negative velocity). He
receives events at a rate of one signal every delta-t'
seconds (where delta-t' is the absolute value of the
time between e_1 and e_0).

Since the traveling observer is receiving signals
from an *approaching* sender (in this time-reversed
sequence of events), he uses the relativistic
Doppler shift for an approaching sender:

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

which is exactly the ratio computed in my previous
post, using the Lorentz transformations. This explains
why the ratio looks like a Doppler shift, and why
the sign of v seems the opposite what it should.

--
Daryl McCullough
Ithaca, NY

Uncle Ben says...

Your "derivation" of the relativistic Doppler shift (in the wrong
direction) and blaming it on Einstein puts you in a class with John
Parker, who "derived" the Einstein Expansion a few years back,
claiming that Einstein erred in driving the Lorentz Contraction.

We can see here http://www.fourmilab.ch/etexts/einstein/specrel/www/
in section 7, Einstein derives the relativistic Doppler shift in a
way that has nothing to do with Koobee's derivation, nor with my
derivation. Instead, he directly uses the transformational properties
of the electromagnetic field, which is the hard way, but is more
directly relevant, I suppose.

So Koobee is lying (or mistaken, to be charitable) in calling the
erroneous derivation Einstein's.

--
Daryl McCullough
Ithaca, NY

On Mar 18, 8:44 am, Daryl McCullough wrote:

Here's an interesting question: Since this ratio
has nothing (apparently) to do with Doppler shifts,
*why* does it have the same form as a Doppler shift?

[bull**** snipped]

which is exactly the ratio computed in my previous
post, using the Lorentz transformations. This explains
why the ratio looks like a Doppler shift, and why
the sign of v seems the opposite what it should.

You have no idea what you are talking about which is emphasized with
your line of “why the sign of v seems the opposite what it should”.
You are just fudging the mathematics to justify your personal belief.
shrug

On Mar 18, 8:48 am, Daryl McCullough wrote:

We can see here
http://www.fourmilab.ch/etexts/einstein/specrel/www/
in section 7, Einstein derives the relativistic Doppler shift in a
way that has nothing to do with Koobee's derivation, nor with my
derivation.

In section 7, Einstein the nitwit, the plagiarist, and the liar used
energy transformation which was not yet derived by anyone including
Einstein the nitwit, the plagiarist, and the liar. In section, 8,
Einstein the nitwit, the plagiarist, and the liar used time
transformation with a flipping v to arrive at an answer in which he
knew should be correct and accepted. shrug

Instead, he directly uses the transformational properties
of the electromagnetic field, which is the hard way, but is more
directly relevant, I suppose.

Bull****! shrug

So Koobee is lying (or mistaken, to be charitable) in calling the
erroneous derivation Einstein's.

shrug

On Mar 18, 12:04*pm, Koobee Wublee wrote:
On Mar 18, 8:48 am, Daryl McCullough wrote:

We can see here
http://www.fourmilab.ch/etexts/einstein/specrel/www/
in section 7, Einstein derives the relativistic Doppler shift in a
way that has nothing to do with Koobee's derivation, nor with my
derivation.

In section 7, Einstein the nitwit, the plagiarist, and the liar used
energy transformation which was not yet derived by anyone including
Einstein the nitwit, the plagiarist, and the liar.

You apparently cannot read. Section 7 is entitled (English
translation) Theory of Doppler's Principle and of Aberration. And
there what is applied is are the electric and magnetic field
transformations, and coordinate transformations. Energy
transformations do not appear in this section.

*In section, 8,
Einstein the nitwit, the plagiarist, and the liar used time
transformation with a flipping v to arrive at an answer in which he
knew should be correct and accepted. *shrug

Instead, he directly uses the transformational properties
of the electromagnetic field, which is the hard way, but is more
directly relevant, I suppose.

Bull****! *shrug

So Koobee is lying (or mistaken, to be charitable) in calling the
erroneous derivation Einstein's.

shrug

.... ahahahah... AHAHAHAHA... ahahahaha...

"Uncle Ben" wrote:
On Mar 18, 12:56 am, Koobee Wublee wrote:
On Mar 17, 8:29 am, Daryl McCullough wrote:
On Mar 16, 10:05 pm, Koobee Wublee wrote:

KW 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’

ED Daryl wrote:
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]

KW wrote:
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

KW wrote:
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

ED Daryl wrote:
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))

KW wrote:
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

ED Daryl wrote:
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.

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

ED Daryl wrote:
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.

KW wrote:
Gee! Yours truly also has been telling you the following for years.
shrug
** f = (observed speed) / wavelength

ED Daryl wrote:
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.

KW wrote:
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

ED Daryl wrote:
It's Koobee's own incompetence at work.

KW wrote:
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

U-Ben made a ballroom dance & worte:
Your "derivation" of the relativistic Doppler shift (in the wrong
direction) and blaming it on Einstein puts you in a class with John
Parker, who "derived" the Einstein Expansion a few years back,
claiming that Einstein erred in driving the Lorentz Contraction.
John, at least did the LT correctly and flubbed only in the last step.
He showed that the moving rod measured w.r.t. its rest frame is longer
than the same rod w.r.t. the stationary frame. His idiocy was that he
assigned the length w.r.t. the stationary frame as L and concluded
that the proper length must have expanded to L*gamma. He could not see
that the moving frame IS the rest frame of the moving rod.
Your presentation is in some kind of code. (What does it mean that a
vector points to "dt" or to "dt'", as if they were points in space?) I
guessed your meaning in spite of the poor description, and you did mis-
identify the vector d[s]. That is why you could not even derive the
classical Doppler shift by means of the Galilean transformation.

Koobee Wooblee and Androcles, both believing in isolation that the
emperor has no clothes. --- Uncle Ben Green, Ballroom Dancer.

hanson wrote:
ahahahaha... "in isolation"??... aka.. "in a solution"
... aka... "insulation"... aka "in an illusion"... etc...

Ballroom man, listen. You will never find out nor know.
The reason is simple. You, Ben AND Androcles &
Koobee Wooblee and Daryl are ALL correct and right.
But each one of you is ONLY right from his own
perspective, that is from your own [1] place where
you sit and explain what you [1] perceives...
.... and that makes all the other guys' [2,3 & 4] view
automatically wrong in the eyes of [1]... & naturally
also when they exchange places.... ahahaha...

This is the wonder of relativity!! It's a beautiful thing
to justify your very own superiority, in the Gedanken
world... (of Jewish origin)... BUT in the real world
where you have to feed yourself and your family...
-------- SR is short for SILLY RANT ------- and
----- GR stands for GULLIBLE RECITAL -----
Relativity is as useless as are its trains, whose 100
milion mph speed "faded into Minkowski's non-existent
locations & time" of his mental turbidity and intellectual
subduction... ahahahahaha...

So, keep on dancing and entertain each other.
The dancing may come to an end the day when
the first one of you explains and DEFINES in
no uncertain terms, in street lingo, with examples
and besides his personal formalism what he means
by "frame"... "w.r.t. ... "dt" or to "dt'", "observer" etc...

But that is not gonna happen, because when done
so, a winner will emerge... and they all know that.
ahaha... So, the dancing, the foxtrot & especially
the side-step-tango will continue... ahahahaha...
Thanks for the laughs, guys... ahahahanson

In article ,
Koobee Wublee says...

On Mar 18, 8:44 am, Daryl McCullough wrote:

Here's an interesting question: Since this ratio
has nothing (apparently) to do with Doppler shifts,
*why* does it have the same form as a Doppler shift?

[bull**** snipped]

which is exactly the ratio computed in my previous
post, using the Lorentz transformations. This explains
why the ratio looks like a Doppler shift, and why
the sign of v seems the opposite what it should.

You have no idea what you are talking about which is emphasized with
your line of why the sign of v seems the opposite what it should.

The derivation is *YOURS*. That's not the way that Einstein
derived the Doppler shift, it's not the way I derive the Doppler shift.
It's the way *YOU* derived it, and blamed it on Einstein. I noticed
that the result looked like a Doppler shift (with the wrong sign of
v), even though there was no apparent reason for it to.

I'm not justifying *MY* derivation, because it's *YOUR* derivation,
wrong sign and all.

You posted it. You didn't get it from Einstein. You didn't get it
from me.

--
Daryl McCullough
Ithaca, NY

Koobee Wublee says...

On Mar 18, 8:48 am, Daryl McCullough wrote:

We can see here
http://www.fourmilab.ch/etexts/einstein/specrel/www/
in section 7, Einstein derives the relativistic Doppler shift in a
way that has nothing to do with Koobee's derivation, nor with my
derivation.

In section 7, Einstein the nitwit, the plagiarist, and the liar used
energy transformation

He was not using the energy transformation. He was using the
transformation properties of the electromagnetic field. You
are deeply, deeply confused. That confusion is not Einstein's
fault.

--
Daryl McCullough
Ithaca, NY