Fallacy of Relativistic Doppler Effect

Eric Gisse says...

On Mar 26, 2:42=A0pm, Koobee Wublee wrote:

What is the transverse Doppler effect under relativity? =A0According to
the energy transformation and also your derivation, it should predict
a blue shift while experiments time after time all have indicated
red. oops! oshrug

In what way are your arguments credible? It has already been
established that you were COMPLETELY WRONG when discussing your
strawman derivation of the relativistic Doppler effect. What are the
odds you are correct about the transverse Doppler effect?

Koobee's problem is that physics is hard, and Koobee is lazy.
Whenever he runs into something that he doesn't understand, he
gives up, and declares it to be nonsense.

But Koobee has noticed an ambiguity in the interpretation of
the phrase "the transverse Doppler effect". That's actually
to his credit. But rather than trying to *resolve* the ambiguity,
he's taking at as yet another argument that relativity is nonsense.
I could explain the situation to him, but Koobee is incapable of
following arguments that require thought, and dismisses them as
"fudging".

The question is: What is the formula for the transverse relativistic
Doppler effect? Let's add some background information to make this
question more precise:

Suppose we have two observers, A and B, traveling inertially. Let
F be the frame in which A is at rest, and let F' be the frame in
which B is at rest. Assume that, according to the coordinate system
of frame F, B is traveling in the +x direction at speed v.
Rather than assuming that the separation between A and B is
in the x-direction, we will assume that they are at different
y locations. For definiteness, we will assume that, as measured
in frame F, B is traveling in the x-direction along the line y=L,
and A is sitting at x=0, y=0.

Assume that A is transmitting a periodic electromagnetic
wave in the +y direction. Let e_1 be the
event at A corresponding to the start of a cycle, and let
e_2 be the event at A corresponding to the end of that
cycle. Let T be the time between e_1 and e_2, in frame F,
which is also the period of the wave.

Now, about this set up, we can ask two *different* questions
about what things look like in frame F':

(1) Let T'' be the period of the electromagnetic wave produced
by A, as measured by frame F'. What is the ratio
T''/T?

(2) Let e_3 be the event at which the signal from event e_1
reaches the line y=L. Let e_4 be the event at which the signal
from event e_2 reaches the line y=L. Let T' be the time between
e_3 and e_4, as measured in frame F'. What is the ratio
T'/T?

If B were traveling in the same direction as the electromagnetic
wave, straight away from A, then there would be no difference
between T' and T''. But in the transverse case, they are not
the same. This is not an inconsistency; the two quantities
T' and T'' have different definitions, and there is no
logical reason for them to be equal, and they are not equal
according to SR.

Solution to (1).

The simplest to derive is T''. The phase phi of the
electromagnetic wave is given in frame F by
phi = k y - w t,
where k = w/c and where w = 2pi/T.

Phase is an invariant. So when we switch to frame F',
we have:

phi' = phi = k y - w t

We want to re-express this in terms of F' coordinates,
so we use the inverse Lorentz transform:

y = y'
t = gamma (t' + v/c^2 x')

to get

phi' = k y' - gamma w t' - gamma vw/c^2 x'

We can write this in the form: phi' = k_x' x' + k_y' y' - w' t'
with the definitions:

k_x' = - gamma vw/c^2
k_y' = k
w' = gamma w

Since w' = 2pi/period, we defined T'' to be the period in F',
we have:

T'' = 2pi/w' = 2pi/(gamma w) = 1/gamma (2pi/w) = T/gamma

So T''/T = 1/gamma.

So T'' is less than T, by a factor of 1/gamma.

Solution to (2).

To derive T', we need to compute the coordinates of
the events e_1, e_2, e_3 and e_4 in both frames.
Once again, e_1 is the event at sender A at rest
in the F frame at the start of a cycle. e_2 is
the event at sender A at the end of the same cycle
(time T later, according to frame F). e_3 is the
event at which the light from e_1 crosses the line
y=L. e_4 is the event at which the light from e_2
crosses y=L.

e_1 and e_2 take place at A, which we can assume
is the origin of the F coordinate system. We may
as well assume that e_1 takes place at t=0.
So we have:

x_1 = 0
y_1 = 0
t_1 = 0

x_2 = 0
y_2 = 0
t_2 = T

Light propagating in the y-direction will reach
the line y=L after a time period of L/c. So we
have:

x_3 = 0
y_3 = L
t_3 = L/c

x_4 = 0
y_4 = L
t_4 = T + L/c

Letting delta-x be x_4 - x_3, delta-y be y_4 - y_3,
and delta-t be t_4 - t_3, we have:

delta-x = 0
delta-y = 0
delta-t = T

Now, transform to frame F' to get:

delta-x' = - gamma vT
delta-y' = 0
delta-t' = gamma T

delta-t' is just the T' introduced earlier. So we have:

T'/T = gamma

So T' is greater than T, by a factor of gamma.

Reconciliation of (1) and (2).

As we saw, T' and T'' are not the same: T' T, but T'' T.
But the two results are completely compatible. Let's look
at the change in phase between e_3 and e_4 in both frames.
In frame F, the phase is given by: phi = ky - wt, so the
change in phase between e_3 and e_4 is
delta-phi = k delta-y - w delta-t
= 0 - wT
= - 2pi
(because w = 2pi/T)

Now, look at the same change in phase from the point of
view of frame F':

delta-phi = k_x' delta-x' + k_y' delta-y' - w' delta-t'

We've already calculated
k_x' = - gamma vw/c^2
k_y' = k
w' = gamma w
delta-x' = - gamma vT
delta-t' = gamma T

So

delta-phi = (- gamma vw/c^2)(- gamma vT) + 0 - (gamma w) (gamma T)
= gamma^2 v^2/c^2 wT - gamma^2 wT
= -gamma^2 wT (1-v^2/c^2)
= - wT
(since gamma^2 = 1/(1-v^2/c^2))
= -2pi

So the two equations T' = gamma T and T'' = 1/gamma T are
perfectly consistent, once you realize that T' is *NOT*
the period of the electromagnetic wave in frame F'. Why
not? It's because e_3 and e_4 are not at the same location
in frame F'; they don't have the same value for x'. To
directly compute the period of a wave, you have to have
two events such that the first event is at the start of
one cycle (which is the case with e_3), and the second
event is at the end of that cycle (which is the case
with e_4), and the two events are at the *SAME* location
(which is not the case with e_3 and e_4). So the time
between e_3 and e_4 is NOT the period in frame F'.

--
Daryl McCullough
Ithaca, NY

On Mar 27, 9:01 am, Daryl McCullough wrote:
On Mar 26, Koobee Wublee wrote:

What is the transverse Doppler effect under relativity?
According to the energy transformation and also your derivation,
it should predict a blue shift while experiments time after time
all have indicated red. Oops! shrug

This disagreement of SR with experiments is serious and fatal, no?

[So] checkmate

Koobee's problem...

It should be very clear at this stage the Doppler shift no matter how
you fudge it to be should agree with the energy transform as described
below.

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

Where

** [v] = Velocity vector between frames of f and f’
** [c] = Velocity vector of light
** [] * [] = dot product of two vectors

Under the transverse case,

** f’ / f = 1 / sqrt(1 – v^2 / c^2)

Where

** [v] * [c] = 0

This means SR predicts a blue Doppler shift in the transverse
direction, and that is totally wrong. shrug

Anyone with half a brain would attempt to execute a graceful retreat
from that. shrug

It smells like a bunch of sour ass losers with unsportsmanlike
conducts. shrug

The bottom line is that SR does not produce what is observed in
experiments. Thus, SR is merely garbage. Just how difficult can that
be? shrug

On Mar 27, 9:28*pm, Koobee Wublee wrote:
On Mar 27, 9:01 am, Daryl McCullough wrote:

On Mar 26, Koobee Wublee wrote:
What is the transverse Doppler effect under relativity?
According to the energy transformation and also your derivation,
it should predict a blue shift while experiments time after time
all have indicated red. *Oops! *shrug

This disagreement of SR with experiments is serious and fatal, no?

[So] checkmate

Koobee's problem...

It should be very clear at this stage the Doppler shift no matter how
you fudge it to be should agree with the energy transform as described
below.

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


Its' already been established that you don't know how to derive
anything in relativity. What's the point in going over old ground?

[snip rest, unread]

Koobee Wublee says...

On Mar 27, 9:01 am, Daryl McCullough wrote:
On Mar 26, Koobee Wublee wrote:

What is the transverse Doppler effect under relativity?
According to the energy transformation and also your derivation,
it should predict a blue shift while experiments time after time
all have indicated red. Oops! shrug

This disagreement of SR with experiments is serious and fatal, no?

[So] checkmate

You are deeply confused. You haven't pointed out any
disagreements.

It should be very clear at this stage the Doppler shift no matter how
you fudge it to be should agree with the energy transform as described
below.

You never derive any of your results, which makes them useless
for the purposes of discussion. You have made claims, without
deriving those claims, and every single claim has proved to be
false.

As I explained, there are two different questions:
(1) Given an electromagnetic wave described by attributes:
k_x, k_y, k_z, w
in one frame F, what are the attributes in the frame F' of
an observer moving at speed v in the x-direction relative to
F?

This is easily answered by realizing that phase is an invariant.
So

k_x x + k_y y + k_z z - w t
=
k_x gamma (x' + v t') + k_y y' + k_z z' - w gamma (t' + v/c^2 x')
= gamma (k_x - wv/c^2) x' + k_y y' + k_z z' - gamma (w - k_x v) t'

So it is clear that in frame F', the wave attributes are given by:
k_x' = gamma (k_x - wv/c^2)
k_y' = k_y
k_z' = k_z
w' = gamma (w - k_x v)

This is exactly the same as the transformation of the energy-momentum
four-vector:

p_x' = gamma (p_x - Ev/c^2)
p_y' = p_y
p_z' = p_z
E' = gamma (E - p_x v)

I don't know why you think that there is some kind of incompatibility.

This means SR predicts a blue Doppler shift in the transverse
direction, and that is totally wrong

According to what, or who?

The bottom line is that SR does not produce what is observed in
experiments.

The bottom line is that you have made many false claims and
that's another.

--
Daryl McCullough
Ithaca, NY

Koobee Wublee says...

On Mar 27, 9:01 am, Daryl McCullough wrote:
On Mar 26, Koobee Wublee wrote:

What is the transverse Doppler effect under relativity?
According to the energy transformation and also your derivation,
it should predict a blue shift while experiments time after time
all have indicated red. Oops! shrug

This disagreement of SR with experiments is serious and fatal, no?

[So] checkmate

Koobee's problem...

It should be very clear at this stage the Doppler shift no matter how
you fudge it to be should agree with the energy transform as described
below.

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

Under the transverse case,

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

It's kind of funny that you should quote these as evidence that
I'm wrong, when I derived exactly those results. But the transverse
result you give is *NOT* describing how the frequency of an
electromagnetic wave changes under a change of coordinate systems.
Instead, it is a description of the ratio of frequencies of sending
and receiving signals. Why are these not the same thing? Well,
I explained why not, in another post.

The scenario is this: We have two inertial observers, A and B.
According to A's frame, A is at rest at location x=0, y=0.
B is traveling in the +x direction along the line y=L.
A is transmitting signals in the +y direction.

Now, think about it: B is moving perpendicularly to the path
of the signals sent by A. That means that *if* B intercepts
one signal, then the *next* signal will miss him. So if A
is transmitting signals in the y direction, then B is not
going to receive more than one signal, and so will be unable
to measure any kind of frequency at all.

Two ways to rectify this:

(1) You can imagine that instead of just
one sender, there is an *array* of senders, all sending in the
y-direction, all sending in phase. Then if B catches the signal
from one sender, he will catch a different signal from a *different*
sender. In that case, B will see the signals *BLUESHIFTED*.

(2) Instead of A sending in the y-direction, he continually
adjusts his signals so that they always reach B. That means
he has to aim his signals *ahead* of where B is now.

These are *NOT* the same case. In case (2), A is *NOT* transmitting
in the y-direction, the angle is changing with time.

--
Daryl McCullough
Ithaca, NY

On Mar 27, 11:28*pm, Koobee Wublee wrote:
On Mar 27, 9:01 am, Daryl McCullough wrote:

On Mar 26, Koobee Wublee wrote:
What is the transverse Doppler effect under relativity?
According to the energy transformation and also your derivation,
it should predict a blue shift while experiments time after time
all have indicated red. *Oops! *shrug

This disagreement of SR with experiments is serious and fatal, no?

[So] checkmate

Koobee's problem...

It should be very clear at this stage the Doppler shift no matter how
you fudge it to be should agree with the energy transform as described
below.

So, just to recap, what you are saying is that, in your view, the
Doppler shift should agree with the derivation that YOU produced and
which does NOT agree with experiment, and that on the basis of this
result, relativity should be found fault with.

Aha.

Seto has a similar agenda -- to try to show that relativity says
something which it does not say, so that relativity can be
discredited.

Seto, however, is notoriously stupid, and you have adopted his
tactics.


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

Where

** *[v] = Velocity vector between frames of f and f’
** *[c] = Velocity vector of light
** *[] * [] = dot product of two vectors

Under the transverse case,

** *f’ / f = 1 / sqrt(1 – v^2 / c^2)

Where

** *[v] * [c] = 0

This means SR predicts a blue Doppler shift in the transverse
direction, and that is totally wrong. *shrug

Anyone with half a brain would attempt to execute a graceful retreat
from that. *shrug

It smells like a bunch of sour ass losers with unsportsmanlike
conducts. *shrug

The bottom line is that SR does not produce what is observed in
experiments. *Thus, SR is merely garbage. *Just how difficult can that
be? *shrug

On Mar 28, 7:50 am, PD wrote:
On Mar 27, 11:28 pm, Koobee Wublee wrote:

It should be very clear at this stage the Doppler shift no matter how
you fudge it to be should agree with the energy transform as described
below.

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

Where

** [v] = Velocity vector between frames of f and f’
** [c] = Velocity vector of light
** [] * [] = dot product of two vectors

Under the transverse case,

** f’ / f = 1 / sqrt(1 – v^2 / c^2)

Where

** [v] * [c] = 0

This means SR predicts a blue Doppler shift in the transverse
direction, and that is totally wrong. shrug

Anyone with half a brain would attempt to execute a graceful
retreat from that. shrug

It smells like a bunch of sour ass losers with unsportsmanlike
conducts. shrug

The bottom line is that SR does not produce what is observed in
experiments. Thus, SR is merely garbage. Just how difficult can
that be? shrug

So, just to recap, what you are saying is that, in your view, the
Doppler shift should agree with the derivation that YOU produced and
which does NOT agree with experiment, and that on the basis of this
result, relativity should be found fault with.

OK, let’s do a recap. Are the following equations valid under the
Lorentz transform and SR?

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

Where

** [v] = Velocity vector between frames of f and f’
** [c] = Velocity vector of light
** [] * [] = dot product of two vectors

Under the transverse case,

** f’ / f = 1 / sqrt(1 – v^2 / c^2)

Where

** [v] * [c] = 0

If the answer is yes, WTF is the problem? shrug

If the answer is no, what do you think is the correct equation
describing the transverse Doppler shift under SR?

On Mar 28, 3:53 am, Daryl McCullough wrote:
Koobee Wublee says...

It should be very clear at this stage the Doppler shift no matter how
you fudge it to be should agree with the energy transform as described
below.

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

Where

** [v] = Velocity vector between frames of f and f’
** [c] = Velocity vector of light
** [] * [] = dot product of two vectors

Under the transverse case,

** f’ / f = 1 / sqrt(1 – v^2 / c^2)

Where

** [v] * [c] = 0

This means SR predicts a blue Doppler shift in the transverse
direction, and that is totally wrong. shrug

Anyone with half a brain would attempt to execute a graceful
retreat from that. shrug

It smells like a bunch of sour ass losers with unsportsmanlike
conducts. shrug

The bottom line is that SR does not produce what is observed in
experiments. Thus, SR is merely garbage. Just how difficult can
that be? shrug

p_x' = gamma (p_x - Ev/c^2)
p_y' = p_y
p_z' = p_z
E' = gamma (E - p_x v)

Is the transverse Doppler equation not the following?

** E' = gamma (E - p_x v)

Where

** p_x = 0

If yes, WTF is the problem? The transverse Doppler shift under SR is
blue. shrug?

If no, what should the correct equation for transverse Doppler shift
look like according to you?

I don't know why you think that there is some kind of incompatibility.

Is the last equation the same as the following?

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

Where

** [v] = Velocity vector between frames of f and f’
** [c] = Velocity vector of light
** [] * [] = dot product of two vectors

As yours truly has said, you guys are bunch of sour ass losers.
shrug

The bottom line is that SR is just garbage. shrug

On Mar 28, 9:24*am, Koobee Wublee wrote:
On Mar 28, 7:50 am, PD wrote:









On Mar 27, 11:28 pm, Koobee Wublee wrote:
It should be very clear at this stage the Doppler shift no matter how
you fudge it to be should agree with the energy transform as described
below.

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

Where

** *[v] = Velocity vector between frames of f and f’
** *[c] = Velocity vector of light
** *[] * [] = dot product of two vectors

Under the transverse case,

** *f’ / f = 1 / sqrt(1 – v^2 / c^2)

Where

** *[v] * [c] = 0

This means SR predicts a blue Doppler shift in the transverse
*direction, and that is totally wrong. *shrug

Anyone with half a brain would attempt to execute a graceful
*retreat from that. *shrug

It smells like a bunch of sour ass losers with unsportsmanlike
*conducts. *shrug

The bottom line is that SR does not produce what is observed in
experiments. *Thus, SR is merely garbage. *Just how difficult can
that be? *shrug

So, just to recap, what you are saying is that, in your view, the
Doppler shift should agree with the derivation that YOU produced and
which does NOT agree with experiment, and that on the basis of this
result, relativity should be found fault with.

OK, let’s do a recap. *Are the following equations valid under the
Lorentz transform and SR?

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

You were just given the derivation of the Doppler effect and you just
agreed that your derivation is wrong. Yet here we are again.

If you don't know what the symbols mean, just ask. Don't just start
swapping them around and call it a day.

[snip rest, unread]

On Mar 27, 10:57 pm, Eric Gisse wrote:
Koobee Wublee wrote:

It should be very clear at this stage the Doppler shift no matter how
you fudge it to be should agree with the energy transform as described
below.

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

Where

** [v] = Velocity vector between frames of f and f’
** [c] = Velocity vector of light
** [] * [] = dot product of two vectors

Its' already been established that you don't know how to derive
anything in relativity. What's the point in going over old ground?

So, the college dropout thinks the equation above is wrong under SR.
shrug

You were just given the derivation of the Doppler effect and you just
agreed that your derivation is wrong. Yet here we are again.

What derivation? Well, the college dropout cannot weasel its way out
of this one. Under the transverse case,

** f’ / f = 1 / sqrt(1 – v^2 / c^2)

Where

** [v] * [c] = 0

If you don't know what the symbols mean, just ask. Don't just start
swapping them around and call it a day.

[snip rest, unread]

Let’s get this straight. The college dropout is bitching about
something it has not read. What else is new and expected from a
college dropout? shrug