Randy Poe wrote in message . ..
[EL]
Hi Bilge.

Why is it so difficult for people to imagine group velocities as wave
envelopes!

All decent empirical wave-group-velocity measurements show that they
are much lower than phase velocity in the same dispersive medium.

Except in negatively-dispersive media, where both theory and
experiment show the opposite.

- Randy

[EL]
Bring back the grudge and let us grind.
What is the merit in being a beautifully coloured parrot?
The merit is to admire the colourful feathers (mostly in a mirror).
However, a parrot is a sound mimicking bird with a bird's brain.

Dispersion is another word for scattering.
Scattering if you did not know is like what happened to the Jews all
over history.
Scattering is like holding a handful of seeds and then tossing them
for random distribution during planting processes.
A dispersive medium is a medium as media are defined in being either
homogenous or not and isotropic or not.
The quality of a medium indicates if it was able to induce scattering
or not.

Now try to imagine pinball the game and look for Pachinko.
The many steal marbles are supposed to be scattered randomly by design
to induce a factor of luck.

The dispersion of light and this means its scattering among air
molecules or any transparent or semitransparent medium should make you
understand that the waves are being physically scattered by deflection
and reflection on the particles of that medium.

Of course by now you should have realised that negative-dispersion is
an expression coined by an idiot.

The phenomenon being the heart of this debate is definition dependant.
In optics dispersion is also defined as the Rate of Change of the
Refractive Index over wavelength scale at a specific wavelength.

Therefore, that definition implies that a wavelength scale must be
constructed by arbitrating a periodical interval indicating wavelength
increments against which we plot the refractive index to extract a
rate of change of that RI with respect to the change in wavelength
about the wavelength in question to illustrate a dispersion figure.

If my sentence was too complex for you to understand, here it is again
in different words.

We have a specific frequency of a wave of light.
We put a point on a graph's x-axis to represent that frequency and we
extend our scale to the left and to the right.
The graph's y-axis then represents the refractive index.
This means that in that specific material the refractive index is
frequency dependant.
By plotting all the different refractive index values we illustrate
the scattering of the refractive index about (before and after) the
frequency at question.

What does the idiotic negative-dispersion supposed to mean!

Pick up any respectable reference that tabulates the refractive index
of materials and try to find any negative value.
The overwhelming majority of indexes have a value between 1 and 2 and
they are POSITIVE VALUES.

Now the rate of change of positive values over positive intervals is
quite unlikely to make sense being negative.
A faster rate and a slower rate are simply seen in the aggregation/
dispersion of the plotted points of the refractive index against the
wavelengths.

So please educate yourself before defending idiots because it only
makes an idiot out of you too.

EL

On 25 Oct 2003 14:01:38 -0700, (EL) wrote:

Randy Poe wrote in message . ..
[EL]
Hi Bilge.

Why is it so difficult for people to imagine group velocities as wave
envelopes!

All decent empirical wave-group-velocity measurements show that they
are much lower than phase velocity in the same dispersive medium.

Except in negatively-dispersive media, where both theory and
experiment show the opposite.

[EL]
Bring back the grudge and let us grind.
What is the merit in being a beautifully coloured parrot?
The merit is to admire the colourful feathers (mostly in a mirror).
However, a parrot is a sound mimicking bird with a bird's brain.

Nice poetry, if unfathomable. Shall we discuss physics?


Dispersion is another word for scattering.

Incorrect.

Dispersion in wave physics means wavelength-dependence of the speed of
propagation. A medium can be dispersive but, in theory, lossless.
Scattering is inherently lossy. It refers to energy being sent in many
other directions away from the line of propagation.

Normal (positive) dispersion does have the tendency to spread signals
out in time, but this is a different phenomenon than dispersion.

Scattering if you did not know is like what happened to the Jews all
over history.
Scattering is like holding a handful of seeds and then tossing them
for random distribution during planting processes.

Again, nice poetry, but not very closely related to either dispersion
or scattering as the terms are used in physics.

A dispersive medium is a medium as media are defined in being either
homogenous or not and isotropic or not.

A dispersive medium is one in which speed depends on frequency. As we
all know from prisms, glass is a dispersive medium. Most real media
area, though the slope depends on the details of the interaction
between the energy and the molecules of the medium. Most of the time,
the slope of a speed vs. wavelength curve is positive, i.e., the speed
increases with wavelength, or decreases with frequency. Low frequency
waves propagate faster. You can see this in water with storm systems,
with low frequency components traveling reaching shore far ahead of
the rest of the system.

But there are materials which exhibit an opposite slope over some
wavelength regimes, with higher frequencies having higher speeds. This
is not magic, but it leads to a different effect on the shape of a
multiple-frequency pulse as it propagates.

What does the idiotic negative-dispersion supposed to mean!

Indicated above.


Pick up any respectable reference that tabulates the refractive index
of materials and try to find any negative value.
The overwhelming majority of indexes have a value between 1 and 2 and
they are POSITIVE VALUES.

Dispersion refers to slope. How low frequencies move compared to high
frequencies is what leads to the effect.


Now the rate of change of positive values over positive intervals is
quite unlikely to make sense being negative.

Huh? You're saying it doesn't make sense for a positive value to
decrease? Why the hell not?

http://en.wikipedia.org/wiki/Dispersion_(optics)
http://www.corning.com/opticalfiber/...ture_page2.asp
http://adsabs.harvard.edu/cgi-bin/np...ptCo.179..107W

- Randy

Randy Poe wrote in message . ..

Huh? You're saying it doesn't make sense for a positive value to
decrease? Why the hell not?

[EL]
Because less positive and more positive is still positive and I am
talking about refractive indexes.
The refractive index may increase or decrease away from the tested
frequency (wavelength) about which dispersion is being measured but in
all those cases dispersion is positive, absolute or simply unsigned.
What you are talking about is not negative dispersion at all but we
may describe it better as the rate of dispersion.
Since dispersion itself is the rate of change in the refractive index
then what you are talking about is the rate of the rate of change,
which is irrelevant to group velocity.



http://en.wikipedia.org/wiki/Dispersion_(optics)

[EL
Just as an example to demonstrate how silly you can be Randy I copied
the contents of that link and here it is.
{{{
Dispersion (optics
From Wikipedia, the free encyclopedia.
Find out how you can help support Wikipedia's phenomenal growth.

(There is currently no text in this page)
}}}

So you are referring me to a page that has no text in it which means
that you did not even read the content of the links you supplied.

Stop fabricating responses and get serious please.
I do not even know why you are being so enthusiastic defending those
idiots while you are much better a parrot than that.

Please invest your precious time in mathematics where you know better.

Regards.

EL

"Bilge" wrote in message
...
Robert Clark:
wrote in message news:
...
I'll add here as a comment that the issue of group velocity is
generally misunderstood, perhaps due to the fact that lower level
textbooks don't explain it well. Group velocity *is not* signal
velocity. Under some circumstances, when the dependence of phase
velocity on frequency over the bandwidth of the signal is weak, group
velocity is a good approximation to signal velocity over distances
short enough so that the pulse shape does not change appreciably in
propagation. That's all. The conditions listed above are reasonably
well satisfied in most practical situations, but they totally fail
under anomalous dispersion situation.

Mati Meron | "When you argue with a fool,
| chances are he is doing just the
same"

You're aware of the discussions on sci.physics.relativity that to
determine if a signal travelled superluminally what would be required
is a round-trip measurement.

That is not the case.

This is because of the uncertainty of synchronizing clocks in
two different locations.

It's completely unnecessary to synchronize anything.

Besides, as long as two clocks are synchronized from the frame in which the
experiment is being carried out, the measurement they give for the speed of
light should be C.

On 25 Oct 2003 23:07:46 -0700, (EL) wrote:

Randy Poe wrote in message . ..

Huh? You're saying it doesn't make sense for a positive value to
decrease? Why the hell not?

[EL]
Because less positive and more positive is still positive and I am
talking about refractive indexes.

Yes. But why does that mean a positive function can't decrease?

Consider exponential decay: a function which is positive everywhere,
but always decreasing.

The refractive index may increase or decrease away from the tested
frequency (wavelength) about which dispersion is being measured but in
all those cases dispersion is positive, absolute or simply unsigned.

What? Dispersion is the slope of the curve. If the slope is negative,
it's not "positive, absolute, or unsigned", it's negative.

What you are talking about is not negative dispersion at all but we
may describe it better as the rate of dispersion.

Instead of making up definitions, why don't you do a search for
"negative dispersion" and read a definition? Speed is positive. Index
of refraction is generally greater than 1. We can all agree on that.
Now imagine a curve which is always above 1. Can you picture such a
curve sometimes increasing and sometimes decreasing? Where it
decreases, the grownups say "this has a negative slope".

As the name for this slope is "dispersion", then where the slope is
negative we call that "negative dispersion". It has nothing to do with
negative speeds, with speeds greater than c, or with any other
crackpot strawman you seem to have latched onto.

Since dispersion itself is the rate of change in the refractive index

Very good. The slope of the curve vs. frequency.

then what you are talking about is the rate of the rate of change,
which is irrelevant to group velocity.

No, what I'm talking about is the rate of change. In a negatively
dispersive regime, shorter waves move faster than longer waves. They
have a higher speed. A lower index of refraction.

Draw a line representing index of refraction vs. frequency. Draw the
line so it slopes downward. Can we agree we're talking about the slope
of the index of refraction curve? I hope so. Can we agree that a
downward slope on this plot means higher frequencies have a lower
index, i.e. a higher speed? I hope so.

Negative dispersion: higher frequencies have higher speed. Is that
really so difficult?

And you're incorrect, the dn/d(lambda) value has everything to do with
group velocity.

- Randy

(Randy Poe) wrote in message . com...

[EL]
I think that you are only trying to make me angry by telling lies and
putting words in my mouth.

With unaltered quotes from your own post?
[EL]
No, it is about the dumb interpretation of what is being written.


I know very well that dn/d(lambda) is the rate of change in the
refractive index with respect to the change in frequency.

That's the part where I said "correct". Earlier you said this:

[EL]
Amen.


The refractive index may increase or decrease away from the tested
frequency (wavelength) about which dispersion is being measured but in
all those cases dispersion is positive, absolute or simply unsigned.

Which is incorrect. The slope can be positive or it can be
negative. As it happens, interaction theory predicts that
the anomalous case of negative dispersion happens only
under very specific circumstances. Hence it is called
"anomalous".

[EL]
I shall try to make it very simple and in plain English.
Let us put some marks at regular intervals on the Asphalt with chalk.
We mark one of those marks as a reference and we place one single
stone at the mark on the left of that mark.
We then skip the right mark and place a stone on the next and skip
even two and place one stone and we say that our stone distribution is
dispersed.
If we put ten stones on the first mark and 15 on the next and 3 on the
next and 100 on the next and 1 on the next the stones are still
dispersed but they are less dispersed.
If you cannot understand English forget about the mathematics and all
semantics.

If you dumbly see my analogy as irrelevant then you are dumber than I
thought.

Now let us get back to frequency dependant refractive indexes.
Let us use now more civil techniques like paper and pencils.
Take a mean wavelength of 100 nano-meters and label your intervals
regularly with 10 nm increments and decrements to the right and left
respectively.
Now you have an x-axis with a frequency scale all of which are
positive frequencies to the right side of the origin labelled zero.
Your vertical axis the y-axis shall be used for refractive indexes as
a dependant variable.

The origin is also zero and your maximum may be assumed at 2 and your
effective plotting zone is trapped between y = 1 and y = 2 while the
points are scattered to the left and right of the 100 nm line extended
vertically at x =100 nm.
This means that any discreet frequency must have a positive refractive
index that is more than one and less than 2 for air as an example.

Now let us take a real and practical and emptirical example and work
it out to cut all the crap.

Phosphate crown glass has a refractive index set that correspond a set
of frequencies as follows.

1060.0 nm - 1.51519
546.1 nm - 1.52736
365.0 nm - 1.54503
312.6 nm - 1.5574

As you can see, as the wavelength decreases (the frequency increases)
the RI increases.

Now the rate of increase can be calculated from the ABSOLUTE deltas
and we may build a new set of |dn|.

{0.01217, 0.01767, 0.01237}

These values are not even a refractive index value category because
they are deltas.
As you can OBVIOUSLY SEE (at least I hope) from this empirical example
that there is no ****ing slope.
The first delta is smaller than the second and the second is greater
than the third.
The variance of those deltas is an expression of how much the
refractive-indexes are dispersed or scattered apart.
If the frequency dependant refractive indexes were linearly
proportional over regular wavelength intervals, The variance would be
Zero.
The only slope there is that of the refractive indexes themselves and
the meaningful slope is that of represented as a mean refractive index
for a mean wavelength but delta refractive indexes is not a refractive
index and it is totally irrelevant here.

Delta refractive index is a good measure for scattering or dispersion
and you may wish to convert it into a standard deviation concept
during analysis such that if the group refractive index was 1.5 (±
0.09) the consequential group velocity should is more or less vg = c /
1.5 (± 0.09).
There is no such anomaly in which the refractive index dives below [1]
and fabricating equations that effects otherwise is pure bull****.

Stop fighting it.

Given that the wave number k = 2pi/lambda and omega = 2pi.f
The group velocity
vg = dw/dk

Now you are a mathematician so work it out for yourself as you did
here before to save bandwidth.
This means that the group velocity itself could be negative and could
be positive envelope-wise but deltas do not reflect absolute speeds.

In other words, even if the envelope's relative location seems to be
advancing over identical parts of the carrier wave as a superposition
the past waves that had advanced in propagation are totally unaffected
and the propagation speed is medium dependent constant when that
medium is homogenous and isotropic.

EL



If you understand that dispersion refers to the sign of
dn/d(lambda) how can you simultaneously say that no matter
what the sign of dn/d(lambda), dispersion is never
negative?

I did say up here and there that that is what dispersion is.

Then since the slope can be negative, and you claim to agree
dispersion is the slope, why claim dispersion can't be
negative? You don't see a contradiction there?

This is analogous to speed being a rate of change in distance with
respect to change in time (mathematically speaking) and we know that
such a scalar may be represented as a slope and that speed may
increase or decrease while stile being a positive mathematical entity.

To decrease speed you need NEGATIVE acceleration (deceleration), which
is the rate of change in the rate of change in distance with respect
to time.

dn/d(lambda) is not an acceleration. Time is not involved.

The same applies here and that is precisely what I said.
We do not care about the rate of dispersion here

Are you now saying dn/d(lambda) is the "rate of dispersion"
and "not dispersion itself"?

Do you realize that a few sentences above you said that
dispersion and dn/d(lambda) are the same, and now you appear
to be saying something different?

because dispersion
itself is the rate we need for group velocity not group acceleration.

What we need is dn/d(lambda). I'm calling that "dispersion".
Can we agree that what we need is dn/d(lambda) so we don't
have a moving target here? Can we agree that this can be
either positive or negative?

A negative velocity is an absolute-quantity-velocity in the opposite
direction of a positive velocity reference direction.

There are no negative velocities here.

However, the speed scalar is an absolute quantity that cannot be less
than zero.

What does dn/d(lambda) have to do with this? n is a positive
number greater than or equal to 1. What does dn/d(lambda)
have to do with velocities being negative or speed being
negative? Didn't you just agree dn/d(lambda) can be
negative?

[I keep confusing myself on the sign. I think of this in terms
of dv/df, the rate of change of speed with frequency. In
anomalous dispersion, v increases with frequency. That means
n decreases with frequency, or increases with wavelength.
So "negative dispersion" = dn/d(lambda) is POSITIVE, while
the more common case is that dn/d(lambda) is NEGATIVE]

That is why I insist that the terminology of {negative-dispersion} is
Oxymoronic because an increasing index with an increasing frequency or
a decreasing index with an increasing frequency does not imply that
rate of change itself is negative

What? The rate of change with frequency IS the slope. You're
now saying that a negative derivative does not imply a
negative slope?

"Negative derivative" and "downward" are the same thing.
A decreasing index with wavelength does indeed imply that
dn/d(lambda) is negative. Let me repeat your words again:

An increasing index with an increasing frequency
[dn/df being positive...]
or a decreasing index with an increasing frequency
[dn/df being negative...]
does not imply that the rate of change itself is
negative
[does not imply that dn/df is negative...]

I'm sorry, I really can't make sense of that remark. But I
don't really care if you can't get your mind around the
concept of first derivatives and their signs. Let's work
through a really simple example. I'm going to try to derive
the group velocity for this example.

Suppose we have a really simple wave composed of two
frequencies, f1 and f2. Let us suppose that f2 = 2f1.
Let us also suppose that the speed of propagation for
wave 1 is v, and for wave 2 is 1.5v. So the higher
frequency wave (wave 2) has 50% higher velocity.

The wave is described by
S(x,t) = cos(2*pi*f1*(x/v1 - t)) + cos(2*pi*f2*(x/v2 - t))

You should first make sure you believe that. Each term
describes a wave which is:
- constant for x - v*t = constant
- for fixed t, has spatial period (wavelength)
v/f = lambda
- for fixed x, has temporal period 1/f1
- has phase 0 at x = 0, t = 0.

The first wave has a wavelength of v1/f1 = lambda1, the
second has a wavelength of v2/f2 = (1.5*v1)/(2*f2)
= (3/4)*lambda1

So every 4 wavelengths of wave 2 corresponds to
3 wavelengths of wave 1, and the packet is in phase
every 4*lambda2 = 3*lambda1. If you fix time t, you
will see the waves add up constructively at x=0
and every 4*lambda1 afterward. These peaks will move
as the wave moves. The rate of advance of the peaks
will be the group velocity, as it is the motion of our
coherent pulse.

So let's analyze what happens as time evolves. Suppose
it is no longer time 0, but a little later, time T.
The phase of wave 1 is f1*(x/v1-T) and of wave 2 is
f2*(x/v2-T) = 2*f1*(x/(1.5*v1)-T)

These phases are equal where
f1*x/v1 - f1*T = 1.33*f1*x/v1 - 2*f1*T

or f1*T = 0.33*f1*x/v1

or x = 3*v1*T

The place where the waves are in phase has moved by a
distance 3*v1*T in time T. Thus, the peak appears to
be moving forward at 3*v1, despite the fact that
one wave is moving at v1 and the other at 1.5*v1.
As the whole thing has a spatial periodicity of 3*lambda1,
you will find that all of the peaks, spaced 3*lambda1
apart, are similarly marching forward at 3*v1.

The fact that their point of coincidence can move
forward faster than either wave alone is a consequence
of the shorter wave moving faster than the longer one.
In other words, of the longer wave having higher index
of refraction, or of dn/df being negative.

If the longer wave moved faster (had lower index of
refraction), you would not get this "superluminal"
effect.

You can do everything I just did in the more general
case of a complex, multi-frequency wave packet and
an arbitrary slope. You can further work out that
there is no information being propagated at speed v1.

This has nothing to do with relativity. It's pure
classical wave physics, analyzing sines and cosines.

- Randy

Robert Clark:
(Bilge) wrote in message
e-al.net...
Robert Clark:
You're aware of the discussions on sci.physics.relativity that to
determine if a signal travelled superluminally what would be required
is a round-trip measurement.

That is not the case.

This is because of the uncertainty of synchronizing clocks in
two different locations.

It's completely unnecessary to synchronize anything.

It is well known among researchers in the foundations of relativity
the need to synchronize clocks at two locations for comparing times at
those locations.

Why is it necessary to compare times at two locations? That would
seem to be the hard way to determine whether or not a signal is
superluminal and it would be less accurate in making the determination.
It would seem to me that the simplest way to make this determination
is for the source to arrange that the pulse be split, with one part
of te pulse propagating in vacuum and the other through the apparatus.
You then compare the two pulses and see which leads which.

Mark Palenik:

"Bilge" wrote in message
ue-al.net...
Robert Clark:
wrote in message news:
...
I'll add here as a comment that the issue of group velocity is
generally misunderstood, perhaps due to the fact that lower level
textbooks don't explain it well. Group velocity *is not* signal
velocity. Under some circumstances, when the dependence of phase
velocity on frequency over the bandwidth of the signal is weak, group
velocity is a good approximation to signal velocity over distances
short enough so that the pulse shape does not change appreciably in
propagation. That's all. The conditions listed above are reasonably
well satisfied in most practical situations, but they totally fail
under anomalous dispersion situation.

Mati Meron | "When you argue with a fool,
| chances are he is doing just the
same"

You're aware of the discussions on sci.physics.relativity that to
determine if a signal travelled superluminally what would be required
is a round-trip measurement.

That is not the case.

This is because of the uncertainty of synchronizing clocks in
two different locations.

It's completely unnecessary to synchronize anything.

Besides, as long as two clocks are synchronized from the frame in which the
experiment is being carried out, the measurement they give for the speed of
light should be C.

Huh?

(EL) wrote in message . com...
(Randy Poe) wrote in message . com...

[EL]
I think that you are only trying to make me angry by telling lies and
putting words in my mouth.

With unaltered quotes from your own post?
[EL]
No, it is about the dumb interpretation of what is being written.


I know very well that dn/d(lambda) is the rate of change in the
refractive index with respect to the change in frequency.

That's the part where I said "correct". Earlier you said this:

[EL]
Amen.


The refractive index may increase or decrease away from the tested
frequency (wavelength) about which dispersion is being measured but in
all those cases dispersion is positive, absolute or simply unsigned.

Which is incorrect. The slope can be positive or it can be
negative. As it happens, interaction theory predicts that
the anomalous case of negative dispersion happens only
under very specific circumstances. Hence it is called
"anomalous".

[EL]
I shall try to make it very simple and in plain English.
Let us put some marks at regular intervals on the Asphalt with chalk.
We mark one of those marks as a reference and we place one single
stone at the mark on the left of that mark.
We then skip the right mark and place a stone on the next and skip
even two and place one stone and we say that our stone distribution is
dispersed.

If I throw flour up in the air in a high wind, I say that
the flour is dispersed as well. But that is not what is
meant by "dispersion" in wave propagation. And you know
it, since you admitted (but then backed away from agreeing
with) that dispersion is another name for dn/df.

If we put ten stones on the first mark and 15 on the next and 3 on the
next and 100 on the next and 1 on the next the stones are still
dispersed but they are less dispersed.
If you cannot understand English forget about the mathematics and all
semantics.

The thing about the English language is that words have
multiple meanings. I can not guess why you choose this one
in a discussion of wave physics.

I agree the semantics is getting us into ridiculous arguments
about everything but the physics. Let me skip all the
semantic part and see how you reacted to the physics
part: The demonstration that if dn/df is negative (that
if higher frequencies propagate with higher speeds) you
can get wave packets moving faster than the propagation
velocity.

[snip]
Now let us take a real and practical and emptirical example and work
it out to cut all the crap.

Phosphate crown glass has a refractive index set that correspond a set
of frequencies as follows.

1060.0 nm - 1.51519
546.1 nm - 1.52736
365.0 nm - 1.54503
312.6 nm - 1.5574

As you can see, as the wavelength decreases (the frequency increases)
the RI increases.

Right. Therefore this would be a case of positive dispersion,
the usual case, not the anomalous case that gives rise to
superluminal group velocities.


Now the rate of increase can be calculated from the ABSOLUTE deltas
and we may build a new set of |dn|.

{0.01217, 0.01767, 0.01237}

These values are not even a refractive index value category because
they are deltas.
As you can OBVIOUSLY SEE (at least I hope) from this empirical example
that there is no ****ing slope.

There is no slope? What?

The first delta is smaller than the second and the second is greater
than the third.

Oh, you mean the trend is not linear over a factor of
3 in wavelength. Do you know what a derivative is?
Apparently not.

Are you aware that the curve y = x^2 has a slope
everywhere, a nonzero derivative everywhere but at
x = 0?

The variance of those deltas is an expression of how much the
refractive-indexes are dispersed or scattered apart.

Apparently not only do you not know what a derivative is,
you don't know what a slope is. That explains the
misunderstanding. Remember "rise over run"?

You didn't calculate the slope between points,
but just the delta between y values. The slope dn/d(lambda)
is APPROXIMATED by delta-n divided by delta-lambda.
(rise over run).

In your table I get the values:
-0.0000237 -0.0000976 -0.0002361

The slope dn/d(lambda) is clearly getting more negative
as wavelength decreases.

However, this is only an approximation of slope. A better
one is found by interpolating a smooth curve that passes
through these points, and taking its derivative at
each point. Using that procedure I find these estimates
(second column below):

lambda dn/d(lambda) linear estimate
1060.0 nm -0.0
546.1 nm -0.0000422 -0.0000237
365.0 nm -0.0001495 -0.0000976
312.6 nm -0.0002671 -0.0002361

The estimate of 0 at 1060 was an artifact of the
interpolation procedure I used. It estimated that the
curve was bottoming out at that point, but obviously
if you included more data beyond 1060 nm it would do
a better job.

The last column shows the estimate from "rise over
run". You can see it's the right order of magnitude
but not really a good estimate.

Again, let me skip this obvious misconception and see what
you did with my example of an actual propagating wave
packet.

[snip]

Nothing. You ignored it entirely to focus on the semantic
argument again.

I'm getting off this part of the train. I'm sorry that you
don't understand what a derivative or the slope of a curve
means, or what the sign of a derivative means. That's
irrelevant. I'm going to ask you again, politely, to please
look at my derivation, in which I took:

- one wave moving at speed v1
- another wave of higher frequency moving at speed 1.5*v1
- summed together they give a wave packet with peaks
wherever they are in phase (wavelength 3*lambda1 where
lambda1 = wavelength of first wave)

and showed
- the peaks of that wave packet propagate at 3*v1

This demonstrates that by taking a sum of waves with the
property that higher frequency wave move faster, I get
a wave packet that moves much faster than any individual
wave.

- Randy

[EL]
Randy wrote
{{{
[Randy]
I'm going to ask you again, politely, to please
look at my derivation, in which I took:

- one wave moving at speed v1
- another wave of higher frequency moving at speed 1.5*v1
- summed together they give a wave packet with peaks
wherever they are in phase (wavelength 3*lambda1 where
lambda1 = wavelength of first wave)

and showed
- the peaks of that wave packet propagate at 3*v1

This demonstrates that by taking a sum of waves with the
property that higher frequency wave move faster, I get
a wave packet that moves much faster than any individual
wave.

- Randy
}}}

And per his request I decided to include his example here to test its
physical validity while having confidence in Randy's mathematical
knowledge.
His example is included complete with his comments and my comments
would be interleaved.

{{{
[Randy]
Suppose we have a really simple wave composed of two
frequencies, f1 and f2. Let us suppose that f2 = 2f1.
Let us also suppose that the speed of propagation for
wave 1 is v, and for wave 2 is 1.5v. So the higher
frequency wave (wave 2) has 50% higher velocity.

The wave is described by
S(x,t) = cos(2*pi*f1*(x/v1 - t)) + cos(2*pi*f2*(x/v2 - t))

You should first make sure you believe that. Each term
describes a wave which is:
- constant for x - v*t = constant
- for fixed t, has spatial period (wavelength)
v/f = lambda
- for fixed x, has temporal period 1/f1
- has phase 0 at x = 0, t = 0.

The first wave has a wavelength of v1/f1 = lambda1, the
second has a wavelength of v2/f2 = (1.5*v1)/(2*f2)
= (3/4)*lambda1
}}}

[EL]
This is a very small glitch and here is the correction.
v2/f2 = (1.5*v1)/(2*f1)

{{{
[Randy]
So every 4 wavelengths of wave 2 corresponds to
3 wavelengths of wave 1, and the packet is in phase
every 4*lambda2 = 3*lambda1. If you fix time t, you
will see the waves add up constructively at x=0
and every 4*lambda1 afterward. These peaks will move
as the wave moves. The rate of advance of the peaks
will be the group velocity, as it is the motion of our
coherent pulse.

So let's analyze what happens as time evolves. Suppose
it is no longer time 0, but a little later, time T.
}}}

[EL]
Here I would like to emphasize on the word "later".
In fact, relativity did not screw anything else than the conception of
time and certainly every thing else consequently.
The classical calculation of wave modulation represented by Randy is
quite legitimate but the problem is in the interpretation of time
"what happens where".
Bare with me because this example should be a perfect example to see
what relativity destroyed, it destroyed the ability of quality minds
to distinguish between reckless sequencing and the impeccable
precision of logical event sequencing.

{{{
[Randy]
The phase of wave 1 is f1*(x/v1-T) and of wave 2 is
f2*(x/v2-T) = 2*f1*(x/(1.5*v1)-T)

These phases are equal where
f1*x/v1 - f1*T = 1.33*f1*x/v1 - 2*f1*T

or f1*T = 0.33*f1*x/v1

or x = 3*v1*T
}}}

[EL]
As you have noticed that I emphasized on the word "later", I would
like to drive the attention of the reader to some facts. When we draw
a graph representing a wave with time on the x-axis and amplitude on
the y-axis, we should pay attention to the meaning of zero time and
positive time more than zero where zero time comes first and more than
zero time comes "later".
This means that looking at the wave graph we should imagine the wave
evolving towards the left side and not to the right side as in
oscilloscopes where raster scanning begins at the left side of the
screen.
The implication of this fact is so great but quite overlooked by many
physicists and almost all mathematicians who take the hype O thesis
from physicists for granted.

Randy demonstrated that x = 3.v1.T, where the product of velocity and
time is a distance of course and that distance is where identifiable
group-wave-peaks may appear in LATER.

I shall not discuss the out-of-synchronisation artefacts of
oscilloscopes here again as you can read it up in this thread.
Now I shall focus on the paper graph and what the physical meaning of
wave modulation means.

It is quite easy to confuse the Time versus Amplitude chart with
velocity chart where Time is versus distance.
If you can avoid that confusion then that is precisely what we need
from the reader here.

{{{
[Randy]
The place where the waves are in phase has moved by a
distance 3*v1*T in time T.
}}}

As you can read in Randy's own statement, he used the expression "a
distance in time T".
We know that the velocity v1 is the distance per unit time traversed
by the wave W1.
We know that the velocity v2 is the distance per unit time traversed
by the wave W2.
To understand how the modulation proceeds we need a spatial reference
GATE through which the two waves propagate and modulate. That gate is
an infinite plane placed orthogonal to the waves' propagational
direction axis, assuming that they are coincident in direction.
T must be a multiple of time units in which a finite portion of each
wavelength is propagating through the medium and across our
referential gate.
Now that portion of the wave is what advances in one time unit.
So what does that distance [3.v1.T] mean?
Randy said that there is a periodically repeating event at which the
two waves become in phase once more and then go out of phase for some
time.
Here are Randy's own words again.
"So every 4 wavelengths of wave 2 corresponds to 3 wavelengths of wave
1"
Naturally W2 is given to be faster than W1 such that every 4 cycles of
W2 correspond to 3 cycles of W1.
The resulting modulation IS a consequence of those two physical
velocities of wave propagation in Length over time dimensions.

{{{
[Randy]
Thus, the peak appears to
be moving forward at 3*v1, despite the fact that
one wave is moving at v1 and the other at 1.5*v1.
As the whole thing has a spatial periodicity of 3*lambda1,
you will find that all of the peaks, spaced 3*lambda1
apart, are similarly marching forward at 3*v1.
}}}

[EL]
This is the crux of the confusion.
Here we ask; what is it that is moving forward and relative to what?
In this particular case, as time advances, the difference in the two
velocities causes the in-phase event to show up at our referential
gate at regular time intervals when 3 cycles of W1 have passed through
the gate or 4 cycles of W2 have passed through the gate.

This means that the frequency of the in-phase event is a Third of the
frequency of W1 or a Quarter of the frequency of W2 if we assumed
proper time to dominate such frequencies.

Let us call this in-phase frequency F, then;

F = f1/3 = f2/4
Hence f2 = 4/3 f1, which contradicts our premise where we assumed that
f2 = 2 f1.

Something hidden must be screwed up here; can you guess what is it?

TIME.

Good guess!

If wave number one was 1000 Hz and wave number two was 2000 Hz then
our referential gate must be THE OBSERVER through which relative
velocities cause a frequency shift, such that 1000 complete cycles are
introduced less frequently than 2000 complete cycles being introduced
more frequently.

Here we propose a standard time window of one proper second in which
1000 complete cycles of wave number one happens and 2000 complete
cycles of wave number two happens concurrently.

By taking the proper time window during which W1 passes through the
gate of observation at its own velocity as our standard time frame we
realise that W2 introduces 3000 complete cycles rather than 2000
because the velocity of observation is 1.5 times faster. This means
that from the Observational gate's point of view the number of
complete cycles of W2 within one time window is 3 times greater than
the complete cycles encountered from W1 and it is not the velocity of
anything at all. When those two waves interfere the less frequent wave
becomes an envelope for the more frequent wave Such that each
composite wave is W1 subdivided into 3 cycles of W2.

This modulated wave becomes a moving reference envelope inside which
the other wave is moving and we end up with (v2/v1).(f2/f1) being a
dimensionless product of ratios that would yield the number 3 that
means nothing in physical essence.

You can see clearly that we could have taken the faster wave as the
reference to which modulation is happening less frequent rather than
more frequent.

So we have the observational freedom to see the faster wave slipping
forward inside the slower wave or to see the slower wave as a
peristaltic motion moving backwards over the faster wave as a
modulation of some constant peak amplitude.

Here I would like to make a historical declaration.
The observational gate that I have proposed does not observe any
velocities once both wavefronts have arrived and being observed, and
all that that gate could observe is the frequency of events because
the gate is stationary in the space of both wave velocities.

You could repeat this exercise by assuming a gate on a railway and let
one train be 1.5 times faster than the other train but both arrive at
the gate simultaneously. Let the size of the cars be twice longer on
the slow train and record your observations regarding the coincidence
of car-joints on both trains.
The gate-observer may have a clock but all he may observe is a
frequency of coincidences and no velocities are perceived at all. The
true velocities are what the trains' wheels make on the iron railway.

This in-phase frequency (car-joints on both trains coinciding) may be
(as in our example) the frequency of 3 cars on the slow train or 4
cars on the fast train.
Now we have two lambdas and one frequency so how can you decide on a
single velocity?

3 cars take 3 times as much as one to pass on the slow train and 4
cars take 4 times as much as one to pass on the fast train. The static
length of 3 slow cars is equivalent to 6 fast cars not 4 and that is
what Lorentz Fitzgerald contraction is all about. Time dilation
follows if we compare times and fix the lengths.

So this idiotic game of juggling numbers makes no physics and makes no
science.

We can hybridize mathematics and physics and come up with negative
dispersion, time dilation, length contraction, black holes and big
bangs, but for what end is this clownish path taking us Randy?

What is so funny and pleasing in ****ing with innocent minds?
There are so many people out there that believe that such fiction is
true and real.
I am seriously asking why.
Why the academic establishment allows this comedy?
Are all scientists becoming incompetent to realise and figure out how
this modern wave of fiction is screwing with their sanity and sound
logic?

I see Lemmings, plenty of them, all the time and everywhere.
Rarely do I see like those men that made the solid foundations on
which we stand today.
Lemmings and clones are what the academic institutions are producing
today.

Perhaps it is time for humans to go extinct and rid nature from the
asshole species.

EL

{{{
[Randy]
The fact that their point of coincidence can move
forward faster than either wave alone is a consequence
of the shorter wave moving faster than the longer one.
In other words, of the longer wave having higher index
of refraction, or of dn/df being negative.

If the longer wave moved faster (had lower index of
refraction), you would not get this "superluminal"
effect.

You can do everything I just did in the more general
case of a complex, multi-frequency wave packet and
an arbitrary slope. You can further work out that
there is no information being propagated at speed v1.

This has nothing to do with relativity. It's pure
classical wave physics, analyzing sines and cosines.

- Randy
}}}