"Porky" wrote in message
...

Here's my original statement: If Doppler distortion actually exists at
the levels postulated (I believe the amount 26% was posted), then the
human
hearing must have a mechanism to compensate for it because we don't hear
it.

You're lying again. Your original statement was, "From a practical
standpoint, the question isn't whether Doppler shift exists in a
loudspeaker, the real issue is whether or not it's audible, and if it is
audible, whether or not our hearing has a mechanism to compensate for it."
http://tinyurl.com/4b3xg

There was no discussion for 26% at the time. In fact in that same post you
wrote, "I would suggest that until someone builds a computer model of a real
loudspeaker reproducing real music in a real room, comes up with the
necessary algorythms and runs the simulation, no accurate, or even
approximate solution for the real world issue is going to be found."

My personal take on it is that if Doppler distortion does exist it is at
inaudible levels.
If you have another alternative, you're welcome to post it and argue in
its favor, but don't start taking things I say out of context and trying
to
twist them to mean something I obviously didn't intend.

Explain to me how I twisted your actual statement. YOU twisted it. There's
no way the brain automatically compensates for it. Many of us explained why.
It's not a "real" issue. Instead of agreeing, you started with that
either/or crap so as not to take responsibility for what you said.

Once again if you'd just admit you were wrong instead of arguing endlessly
about what you said, we wouldn't be in this mess.

"Jim Carr" wrote in message
news:gcL9d.10577$_a3.4635@fed1read05...
"Porky" wrote in message
. ..

"Jim Carr" wrote in message
news:jJp9d.3557$_a3.1033@fed1read05...
"Porky" wrote in message
t...

It was Doppler's train experiment, not Einstein's.

You've obviously never read "Relativity - The Special and General
Theory"
by
some old guy named Albert Einstein. He only mentions the train and
embankment about a zillion times.

But since this thread is about Doppler shift in a speaker, Doppler's
train
experiment is the more relevant one, is it not?

The person to whom you were replying talked about the Lorentz
Transformation
and referred to "relatavistic mumbo-jumbo" so he was certainly entitled to
refer to Einstein. Since you were a smart-ass to him, I was a smart-ass
back
to you.

Fair enough, but did you understand anything he said? Specifically, "The
brain
has nothing to do with light or sound. It's the geometric relationship
bewteen eye sockets, ear sockets, and wind that create the sound." or "Most
mediums, including the Earth's crust, the Ocean Bottom, clouds, and the
sun's photosphere don't have a fixed speed of sound. So the "wavelength" in
those media doesn't really exist in the Classical Fourier sense. Which is
why you need complex frequencies in those cases.
And the imaginary part of the frequency is going to be traveling through a
different state of matter than the real part of the frequency. And it's
going to travel infinitely faster
than the real part. Which is where matrix mechanics and the Uncertainty
Prinicple came from. And what the wavelength is in a matrix, nobody knows."
Can you tell us what any of that means and, more importantly, what it has
to do with recording or home studio?

"Jim Carr" wrote in message
news:PCL9d.10898$_a3.1977@fed1read05...
"Porky" wrote in message
...

Wavelength and frequency are directly related and in any medium where
the
velocity of propagation is constant, as with sound, if one changes, the
other has to change as well and the change will be directly
proportional,
there are no exceptions.

Are you being bullheaded or do you really not understand? I've given you
mind experiments and mathematics to prove you wrong, and all you can do is
repeat your same assertation with no proof. You never admit you're wrong,
do
you? Show me one post in the hundreds where people have disagreed with you
where you admit you're wrong.

One thing you're overlooking is that if the
receiver is moving, in order to make a valid observation about what he
observes, your measuring device has to be moving with him, so the
apparent
wavelength

Here we go again. Go back to the stationary source. The first wave strikes
the receiver. The receiver moves towards the source then STOPS BEFORE THE
SECOND WAVE ARRIVES. According to you since the receiver is not moving I
don't have to worry about the measuring stick moving. At the moment the
second wave strikes the receiver I take my measurement between it and the
first wave.

If the receiver stops, he will hear the sound at it's actual frequency, it
is ONLY when there is relative motion between the source and the receiver
that a pitch shift is heard, if you stop the motion you stop the shift.

What's that distance? If you said the same as at the source, you would be
correct.

Yes, because the receiver has stopped and there is NO motion between the
source and the receiver, and thus, no Doppler shift! You can't stop to take
measurements because that changes the relative motion between sourece and
receiver on which Doppler shift is based.

What's the frequency? Obviously the frequency is greater than at the
source
because the second wave traveled for less time/distance before striking
the
receiver.

But the instant the receiver stopped, the frequency shift stopped too and
the receiver hears the actual frequency of the source, As I said, the
relationship between frequency and the wavelength is fixed, it cannot be
changed without changing the propagation speed of the medium.

Thus it is proved again.

No, it isn't! Will someone please explain it to Jim? He obviously isn't
listening to me.
To make it even more confusing, if you are traveling at a different
velocity and direction relative to the source than is the receiver,
you'll
observations will be totally different than his.

My formula will correctly determine the wavelength even if the source and
receiver are moving. Plug in the numbers and try it yourself.

Frequency is time dependent. It tells how often some event occurs such as
the wave passing by us. Wavelength is not time dependent. It's the
distance
between identical points between waves. If the speed is constant, there is
a
mathematical relationship - on that we agree.

If we introduce other movement into the picture, we had better account for
it. You can't pretend it didn't happen, can we. You have no problems
adjusting the frequency based on the movement (Doppler Shift), so you had
better adjust the wavelength based on the movement as well.

This is my last post on the topic, Porky. If you don't understand by now,
you never will.

Look, the apparent frequency the listener hears is directly related to the
wavelength he sees, in a relationship that is locked together. If C is the
speed of sound, W is the wavelength and F the frequency, then the equations
for determining wavelength from frequency or frequency from wavelength,
W=C/F and F=C/W apply to both sound waves and electromagnetic waves, the
only difference being the relative velocities of C for each. If C is a
constant, and it is in this case, then the relationship between F and W is
absolutely fixed. If you change one by any given percentage, you change the
other as well by exactly the same percentage. See:
http://cnx.rice.edu/content/m11060/latest/
http://www.installer.com/tech/freqandwave.html - Look at the table, the
relationship between frequency and wavelength is fixed, if you change one,
you change the other.
http://hyperphysics.phy-astr.gsu.edu...ound/dopp.html This one gives
graphic representation of the apparent shift in wavelength and thus in
frequency as well.
http://www.safetyline.wa.gov.au/inst...e53/l53_03.asp
http://rockpile.phys.virginia.edu/arc00/arch32.pdf
http://www.isvr.soton.ac.uk/SPCG/Tut...-frequency.htm
These links cover electromagnetic waves, but if you make C equal the
speed of sound, it works for sound too.
http://www.1728.com/freqwave.htm
http://www.qrg.northwestern.edu/proj...h-related.html
http://hubblesite.org/reference_desk...d=72&cat=light
http://www.zyra.org.uk/freqwav.htm

Jim, you don't seem to understand that the receiver cannot stop to take
measurements without the Doppler shift stopping as well. I'll try one last
time:
It doesn't matter which is moving, as long as there is relative motion there
between the source and the receiver there will be Doppler shift in the
apparent frequency picked up by the receiver BECAUSE the motion results in
the wavelength's apparent shortening or lengthening (depending on direction
of the relative motion) as seen by the receiver. If the motion stops, the
Doppler shift will stop too, and the receiver will pick up the actual
frequency of the source.
If I'm being stubborn and bullheaded, it's because I'm trying to help you,
because this time I'm right!

Porky wrote:
"Jim Carr" wrote in message
news:PCL9d.10898$_a3.1977@fed1read05...

"Porky" wrote in message
et...


Wavelength and frequency are directly related and in any medium where

the

velocity of propagation is constant, as with sound, if one changes, the
other has to change as well and the change will be directly

proportional,

there are no exceptions.

Are you being bullheaded or do you really not understand? I've given you
mind experiments and mathematics to prove you wrong, and all you can do is
repeat your same assertation with no proof. You never admit you're wrong,

do

you? Show me one post in the hundreds where people have disagreed with you
where you admit you're wrong.


One thing you're overlooking is that if the
receiver is moving, in order to make a valid observation about what he
observes, your measuring device has to be moving with him, so the

apparent

wavelength

Here we go again. Go back to the stationary source. The first wave strikes
the receiver. The receiver moves towards the source then STOPS BEFORE THE
SECOND WAVE ARRIVES. According to you since the receiver is not moving I
don't have to worry about the measuring stick moving. At the moment the
second wave strikes the receiver I take my measurement between it and the
first wave.


If the receiver stops, he will hear the sound at it's actual frequency, it
is ONLY when there is relative motion between the source and the receiver
that a pitch shift is heard, if you stop the motion you stop the shift.


What's that distance? If you said the same as at the source, you would be
correct.


Yes, because the receiver has stopped and there is NO motion between the
source and the receiver, and thus, no Doppler shift! You can't stop to take
measurements because that changes the relative motion between sourece and
receiver on which Doppler shift is based.


What's the frequency? Obviously the frequency is greater than at the

source

because the second wave traveled for less time/distance before striking

the

receiver.


But the instant the receiver stopped, the frequency shift stopped too and
the receiver hears the actual frequency of the source, As I said, the
relationship between frequency and the wavelength is fixed, it cannot be
changed without changing the propagation speed of the medium.


Thus it is proved again.


No, it isn't! Will someone please explain it to Jim? He obviously isn't
listening to me.

To make it even more confusing, if you are traveling at a different
velocity and direction relative to the source than is the receiver,

you'll

observations will be totally different than his.

My formula will correctly determine the wavelength even if the source and
receiver are moving. Plug in the numbers and try it yourself.

Frequency is time dependent. It tells how often some event occurs such as
the wave passing by us. Wavelength is not time dependent. It's the

distance

between identical points between waves. If the speed is constant, there is

a

mathematical relationship - on that we agree.

If we introduce other movement into the picture, we had better account for
it. You can't pretend it didn't happen, can we. You have no problems
adjusting the frequency based on the movement (Doppler Shift), so you had
better adjust the wavelength based on the movement as well.

This is my last post on the topic, Porky. If you don't understand by now,
you never will.


Look, the apparent frequency the listener hears is directly related to the
wavelength he sees, in a relationship that is locked together. If C is the
speed of sound, W is the wavelength and F the frequency, then the equations
for determining wavelength from frequency or frequency from wavelength,
W=C/F and F=C/W apply to both sound waves and electromagnetic waves, the
only difference being the relative velocities of C for each. If C is a
constant, and it is in this case, then the relationship between F and W is
absolutely fixed. If you change one by any given percentage, you change the
other as well by exactly the same percentage. See:
http://cnx.rice.edu/content/m11060/latest/
http://www.installer.com/tech/freqandwave.html - Look at the table, the
relationship between frequency and wavelength is fixed, if you change one,
you change the other.
http://hyperphysics.phy-astr.gsu.edu...ound/dopp.html This one gives
graphic representation of the apparent shift in wavelength and thus in
frequency as well.
http://www.safetyline.wa.gov.au/inst...e53/l53_03.asp
http://rockpile.phys.virginia.edu/arc00/arch32.pdf
http://www.isvr.soton.ac.uk/SPCG/Tut...-frequency.htm
These links cover electromagnetic waves, but if you make C equal the
speed of sound, it works for sound too.
http://www.1728.com/freqwave.htm
http://www.qrg.northwestern.edu/proj...h-related.html
http://hubblesite.org/reference_desk...d=72&cat=light
http://www.zyra.org.uk/freqwav.htm

Jim, you don't seem to understand that the receiver cannot stop to take
measurements without the Doppler shift stopping as well. I'll try one last
time:
It doesn't matter which is moving, as long as there is relative motion there
between the source and the receiver there will be Doppler shift in the
apparent frequency picked up by the receiver BECAUSE the motion results in
the wavelength's apparent shortening or lengthening (depending on direction
of the relative motion) as seen by the receiver. If the motion stops, the
Doppler shift will stop too, and the receiver will pick up the actual
frequency of the source.
If I'm being stubborn and bullheaded, it's because I'm trying to help you,
because this time I'm right!

Both of you have completely omitted the role of the medium in
determining wavelength. The wavelength is given by

wavelength = (u - cos v)/freq

where u is the velocity of the wave wrt the medium, and v is the
velocity of the source wrt the medium. Take the cosine of the angle
formed between a specific "ray" and the direction of motion of the
source wrt the that ray.

freq is the source frequency, i.e. wrt the source frame. Obviously,
according to this equation, when the source is in motion wrt the medium,
the wavelength varies with direction of wave propagation from the point
of origin in the medium. Thus the detected frequency will differ for
various observers that are perfectly at rest wrt each other, i.e. when
those observers are positioned along the circumference of a circle that
surrounds the source. The frequency detected by any detector will be
given by

freq' = (u - cos' v')/wavelength

where freq' is the detected frequency, and v' is the velocity of the
detector wrt the medium.

Since the wavelength is invariant, then by substitution:

freq' = freq(u - cos' v')/ (u - cos v)

When the source and observer are moving wrt each other in a head on
path, then

cos = cos' = 1 and the equation reduces to

freq' = freq(u - v')/ (u - v)

or

freq' = freq(1 - v'/u)/ (1 - v/u)

The doppler shift is thus not proportional to the velocity of the source
wrt the detector.
IOW, the equation

wavelength = u/freq

is only valid when the frequency is that measured by a detector or by a
source that is at rest wrt the medium.

Since in the context of special relativity the source and detector are
always at rest wrt the medium in their own frames respectively, then the
equation

wavelength = c/freq

is adopted as universally true.

IOW, this form is invalid for sound waves, or any type of matter wave,
where the terms
(cos v) and (cos' v') don't drop out of the doppler equations if and
when they are non zero. Because they are zero in the case of light
propagation and special relativity, they do in fact drop out, leaving
the equation above. Hope this helps. Pick up "Schaum's outlines:
Physics for Science and Engineering". The section on doppler theory is
very clearly written.

I don't know what doppler distortion might be. I think someone has too
much time on their hands.

Richard Perry

"RP" wrote in message
...


Porky wrote:
"Jim Carr" wrote in message
news:PCL9d.10898$_a3.1977@fed1read05...

"Porky" wrote in message
et...


Wavelength and frequency are directly related and in any medium
where

the

velocity of propagation is constant, as with sound, if one changes, the
other has to change as well and the change will be directly

proportional,

there are no exceptions.

Are you being bullheaded or do you really not understand? I've given you
mind experiments and mathematics to prove you wrong, and all you can do
is
repeat your same assertation with no proof. You never admit you're
wrong,

do

you? Show me one post in the hundreds where people have disagreed with
you
where you admit you're wrong.


One thing you're overlooking is that if the
receiver is moving, in order to make a valid observation about what he
observes, your measuring device has to be moving with him, so the

apparent

wavelength

Here we go again. Go back to the stationary source. The first wave
strikes
the receiver. The receiver moves towards the source then STOPS BEFORE
THE
SECOND WAVE ARRIVES. According to you since the receiver is not moving I
don't have to worry about the measuring stick moving. At the moment the
second wave strikes the receiver I take my measurement between it and
the
first wave.


If the receiver stops, he will hear the sound at it's actual frequency,
it
is ONLY when there is relative motion between the source and the
receiver
that a pitch shift is heard, if you stop the motion you stop the shift.


What's that distance? If you said the same as at the source, you would
be
correct.


Yes, because the receiver has stopped and there is NO motion between the
source and the receiver, and thus, no Doppler shift! You can't stop to
take
measurements because that changes the relative motion between sourece
and
receiver on which Doppler shift is based.


What's the frequency? Obviously the frequency is greater than at the

source

because the second wave traveled for less time/distance before striking

the

receiver.


But the instant the receiver stopped, the frequency shift stopped too
and
the receiver hears the actual frequency of the source, As I said, the
relationship between frequency and the wavelength is fixed, it cannot be
changed without changing the propagation speed of the medium.


Thus it is proved again.


No, it isn't! Will someone please explain it to Jim? He obviously
isn't
listening to me.

To make it even more confusing, if you are traveling at a different
velocity and direction relative to the source than is the receiver,

you'll

observations will be totally different than his.

My formula will correctly determine the wavelength even if the source
and
receiver are moving. Plug in the numbers and try it yourself.

Frequency is time dependent. It tells how often some event occurs such
as
the wave passing by us. Wavelength is not time dependent. It's the

distance

between identical points between waves. If the speed is constant, there
is

a

mathematical relationship - on that we agree.

If we introduce other movement into the picture, we had better account
for
it. You can't pretend it didn't happen, can we. You have no problems
adjusting the frequency based on the movement (Doppler Shift), so you
had
better adjust the wavelength based on the movement as well.

This is my last post on the topic, Porky. If you don't understand by
now,
you never will.


Look, the apparent frequency the listener hears is directly related to
the
wavelength he sees, in a relationship that is locked together. If C is
the
speed of sound, W is the wavelength and F the frequency, then the
equations
for determining wavelength from frequency or frequency from wavelength,
W=C/F and F=C/W apply to both sound waves and electromagnetic waves, the
only difference being the relative velocities of C for each. If C is a
constant, and it is in this case, then the relationship between F and W
is
absolutely fixed. If you change one by any given percentage, you change
the
other as well by exactly the same percentage. See:
http://cnx.rice.edu/content/m11060/latest/
http://www.installer.com/tech/freqandwave.html - Look at the table, the
relationship between frequency and wavelength is fixed, if you change
one,
you change the other.
http://hyperphysics.phy-astr.gsu.edu...ound/dopp.html This one
gives
graphic representation of the apparent shift in wavelength and thus in
frequency as well.

http://www.safetyline.wa.gov.au/inst...e53/l53_03.asp
http://rockpile.phys.virginia.edu/arc00/arch32.pdf

http://www.isvr.soton.ac.uk/SPCG/Tut...-frequency.htm
These links cover electromagnetic waves, but if you make C equal the
speed of sound, it works for sound too.
http://www.1728.com/freqwave.htm

http://www.qrg.northwestern.edu/proj...h-related.html
http://hubblesite.org/reference_desk...d=72&cat=light
http://www.zyra.org.uk/freqwav.htm

Jim, you don't seem to understand that the receiver cannot stop to
take
measurements without the Doppler shift stopping as well. I'll try one
last
time:
It doesn't matter which is moving, as long as there is relative motion
there
between the source and the receiver there will be Doppler shift in the
apparent frequency picked up by the receiver BECAUSE the motion results
in
the wavelength's apparent shortening or lengthening (depending on
direction
of the relative motion) as seen by the receiver. If the motion stops,
the
Doppler shift will stop too, and the receiver will pick up the actual
frequency of the source.
If I'm being stubborn and bullheaded, it's because I'm trying to help
you,
because this time I'm right!

Both of you have completely omitted the role of the medium in
determining wavelength. The wavelength is given by

wavelength = (u - cos v)/freq

where u is the velocity of the wave wrt the medium, and v is the
velocity of the source wrt the medium. Take the cosine of the angle
formed between a specific "ray" and the direction of motion of the
source wrt the that ray.

freq is the source frequency, i.e. wrt the source frame. Obviously,
according to this equation, when the source is in motion wrt the medium,
the wavelength varies with direction of wave propagation from the point
of origin in the medium. Thus the detected frequency will differ for
various observers that are perfectly at rest wrt each other, i.e. when
those observers are positioned along the circumference of a circle that
surrounds the source. The frequency detected by any detector will be
given by

freq' = (u - cos' v')/wavelength

where freq' is the detected frequency, and v' is the velocity of the
detector wrt the medium.

I believe I covered that in a general sense under the relative motion and
direction of travel of the source and receiver(s), and with this quote from
my previous post, " To make it even more confusing, if you are traveling at
a different velocity and direction relative to the source than is the
receiver, you'll observations will be totally different than his.".


Since the wavelength is invariant, then by substitution:

freq' = freq(u - cos' v')/ (u - cos v)

When the source and observer are moving wrt each other in a head on
path, then

cos = cos' = 1 and the equation reduces to

freq' = freq(u - v')/ (u - v)

or

freq' = freq(1 - v'/u)/ (1 - v/u)

The doppler shift is thus not proportional to the velocity of the source
wrt the detector.
IOW, the equation

wavelength = u/freq

is only valid when the frequency is that measured by a detector or by a
source that is at rest wrt the medium.

True but "apparent wavelength = u/ apparent frequency" is valid when the
measrements are done by the receiver. Regardless, the relationship of the
receiver's perceived frequency to perceived wavelength is a fixed one, if
one changes the other will too. To figure Doppler shift you have to take
into account both relative velocity and direction, otherwise you have no
idea which way the freuqnecy is shifting.


Since in the context of special relativity the source and detector are
always at rest wrt the medium in their own frames respectively, then the
equation

wavelength = c/freq

is adopted as universally true.

IOW, this form is invalid for sound waves, or any type of matter wave,
where the terms
(cos v) and (cos' v') don't drop out of the doppler equations if and
when they are non zero. Because they are zero in the case of light
propagation and special relativity, they do in fact drop out, leaving
the equation above. Hope this helps. Pick up "Schaum's outlines:
Physics for Science and Engineering". The section on doppler theory is
very clearly written.

I was trying to keep it simple, but your observations are correct and valid,
thanks. :-)

I don't know what doppler distortion might be. I think someone has too
much time on their hands.

That's what we're trying to figure out, and you might be right. :-)

Porky wrote:

"RP" wrote in message
...


Porky wrote:

"Jim Carr" wrote in message
news:PCL9d.10898$_a3.1977@fed1read05...


"Porky" wrote in message
.net...



Wavelength and frequency are directly related and in any medium

where

the


velocity of propagation is constant, as with sound, if one changes, the
other has to change as well and the change will be directly

proportional,


there are no exceptions.

Are you being bullheaded or do you really not understand? I've given you
mind experiments and mathematics to prove you wrong, and all you can do

is

repeat your same assertation with no proof. You never admit you're

wrong,

do


you? Show me one post in the hundreds where people have disagreed with

you

where you admit you're wrong.



One thing you're overlooking is that if the
receiver is moving, in order to make a valid observation about what he
observes, your measuring device has to be moving with him, so the

apparent


wavelength

Here we go again. Go back to the stationary source. The first wave

strikes

the receiver. The receiver moves towards the source then STOPS BEFORE

THE

SECOND WAVE ARRIVES. According to you since the receiver is not moving I
don't have to worry about the measuring stick moving. At the moment the
second wave strikes the receiver I take my measurement between it and

the

first wave.


If the receiver stops, he will hear the sound at it's actual frequency,

it

is ONLY when there is relative motion between the source and the

receiver

that a pitch shift is heard, if you stop the motion you stop the shift.



What's that distance? If you said the same as at the source, you would

be

correct.


Yes, because the receiver has stopped and there is NO motion between the
source and the receiver, and thus, no Doppler shift! You can't stop to

take

measurements because that changes the relative motion between sourece

and

receiver on which Doppler shift is based.



What's the frequency? Obviously the frequency is greater than at the

source


because the second wave traveled for less time/distance before striking

the


receiver.


But the instant the receiver stopped, the frequency shift stopped too

and

the receiver hears the actual frequency of the source, As I said, the
relationship between frequency and the wavelength is fixed, it cannot be
changed without changing the propagation speed of the medium.



Thus it is proved again.


No, it isn't! Will someone please explain it to Jim? He obviously

isn't

listening to me.


To make it even more confusing, if you are traveling at a different
velocity and direction relative to the source than is the receiver,

you'll


observations will be totally different than his.

My formula will correctly determine the wavelength even if the source

and

receiver are moving. Plug in the numbers and try it yourself.

Frequency is time dependent. It tells how often some event occurs such

as

the wave passing by us. Wavelength is not time dependent. It's the

distance


between identical points between waves. If the speed is constant, there

is

a


mathematical relationship - on that we agree.

If we introduce other movement into the picture, we had better account

for

it. You can't pretend it didn't happen, can we. You have no problems
adjusting the frequency based on the movement (Doppler Shift), so you

had

better adjust the wavelength based on the movement as well.

This is my last post on the topic, Porky. If you don't understand by

now,

you never will.


Look, the apparent frequency the listener hears is directly related to

the

wavelength he sees, in a relationship that is locked together. If C is

the

speed of sound, W is the wavelength and F the frequency, then the

equations

for determining wavelength from frequency or frequency from wavelength,
W=C/F and F=C/W apply to both sound waves and electromagnetic waves, the
only difference being the relative velocities of C for each. If C is a
constant, and it is in this case, then the relationship between F and W

is

absolutely fixed. If you change one by any given percentage, you change

the

other as well by exactly the same percentage. See:
http://cnx.rice.edu/content/m11060/latest/
http://www.installer.com/tech/freqandwave.html - Look at the table, the
relationship between frequency and wavelength is fixed, if you change

one,

you change the other.
http://hyperphysics.phy-astr.gsu.edu...ound/dopp.html This one

gives

graphic representation of the apparent shift in wavelength and thus in
frequency as well.


http://www.safetyline.wa.gov.au/inst...e53/l53_03.asp

http://rockpile.phys.virginia.edu/arc00/arch32.pdf


http://www.isvr.soton.ac.uk/SPCG/Tut...-frequency.htm

These links cover electromagnetic waves, but if you make C equal the
speed of sound, it works for sound too.
http://www.1728.com/freqwave.htm


http://www.qrg.northwestern.edu/proj...h-related.html

http://hubblesite.org/reference_desk...d=72&cat=light
http://www.zyra.org.uk/freqwav.htm

Jim, you don't seem to understand that the receiver cannot stop to

take

measurements without the Doppler shift stopping as well. I'll try one

last

time:
It doesn't matter which is moving, as long as there is relative motion

there

between the source and the receiver there will be Doppler shift in the
apparent frequency picked up by the receiver BECAUSE the motion results

in

the wavelength's apparent shortening or lengthening (depending on

direction

of the relative motion) as seen by the receiver. If the motion stops,

the

Doppler shift will stop too, and the receiver will pick up the actual
frequency of the source.
If I'm being stubborn and bullheaded, it's because I'm trying to help

you,

because this time I'm right!

Both of you have completely omitted the role of the medium in
determining wavelength. The wavelength is given by

wavelength = (u - cos v)/freq

where u is the velocity of the wave wrt the medium, and v is the
velocity of the source wrt the medium. Take the cosine of the angle
formed between a specific "ray" and the direction of motion of the
source wrt the that ray.

freq is the source frequency, i.e. wrt the source frame. Obviously,
according to this equation, when the source is in motion wrt the medium,
the wavelength varies with direction of wave propagation from the point
of origin in the medium. Thus the detected frequency will differ for
various observers that are perfectly at rest wrt each other, i.e. when
those observers are positioned along the circumference of a circle that
surrounds the source. The frequency detected by any detector will be
given by

freq' = (u - cos' v')/wavelength

where freq' is the detected frequency, and v' is the velocity of the
detector wrt the medium.


I believe I covered that in a general sense under the relative motion and
direction of travel of the source and receiver(s), and with this quote from
my previous post, " To make it even more confusing, if you are traveling at
a different velocity and direction relative to the source than is the
receiver, you'll observations will be totally different than his.".



Since the wavelength is invariant, then by substitution:

freq' = freq(u - cos' v')/ (u - cos v)

When the source and observer are moving wrt each other in a head on
path, then

cos = cos' = 1 and the equation reduces to

freq' = freq(u - v')/ (u - v)

or

freq' = freq(1 - v'/u)/ (1 - v/u)

The doppler shift is thus not proportional to the velocity of the source
wrt the detector.
IOW, the equation

wavelength = u/freq

is only valid when the frequency is that measured by a detector or by a
source that is at rest wrt the medium.


True but "apparent wavelength = u/ apparent frequency" is valid when the
measrements are done by the receiver. Regardless, the relationship of the
receiver's perceived frequency to perceived wavelength is a fixed one, if
one changes the other will too. To figure Doppler shift you have to take
into account both relative velocity and direction, otherwise you have no
idea which way the freuqnecy is shifting.

Of course direction cannot be discounted, which is why the equations
that I posted refer to the velocities rather than to the scalar speeds.
In hindsight, my last statement above should have read:

wavelength = u/freq

is only valid when the frequency is that measured by a detector or
source that is at rest wrt the medium, or that is in motion at 90 or 270 deg
wrt the projected ray (in which case cos_theta = 0).
Or IOW, when (cos v) = 0.

But I have no idea what "apparent" wavelength and frequency is supposed
to mean, except maybe that the observer was "apparently" wrong about one
or the other

BTW, for those reading this in the sci.physics NG, note that in the
following argument we had to assume a medium in order to derive the
special relativistic wavelength-frequency relationship. OTOH the mere
existence of a "wavelength" of em radiation should offer a clue

Richard Perry




Since in the context of special relativity the source and detector are
always at rest wrt the medium in their own frames respectively, then the
equation

wavelength = c/freq

is adopted as universally true.

IOW, this form is invalid for sound waves, or any type of matter wave,
where the terms
(cos v) and (cos' v') don't drop out of the doppler equations if and
when they are non zero. Because they are zero in the case of light
propagation and special relativity, they do in fact drop out, leaving
the equation above. Hope this helps. Pick up "Schaum's outlines:
Physics for Science and Engineering". The section on doppler theory is
very clearly written.


I was trying to keep it simple, but your observations are correct and valid,
thanks. :-)


I don't know what doppler distortion might be. I think someone has too
much time on their hands.


That's what we're trying to figure out, and you might be right. :-)

"RP" wrote in message
...

Of course direction cannot be discounted, which is why the equations
that I posted refer to the velocities rather than to the scalar speeds.
In hindsight, my last statement above should have read:

wavelength = u/freq

is only valid when the frequency is that measured by a detector or
source that is at rest wrt the medium, or that is in motion at 90 or 270
deg
wrt the projected ray (in which case cos_theta = 0).
Or IOW, when (cos v) = 0.

Which agrees with what I was saying, though I admit to never seeing your
formulas with cos before. Like I said, this is not my field. If it's okay
I'm gonna assume the medium doesn't move for the purposes of this
discussion.

But I have no idea what "apparent" wavelength and frequency is supposed
to mean, except maybe that the observer was "apparently" wrong about one
or the other

Which, again, agrees with my point. I'll attempt to address "apparent
wavelength" if you'll bear with me to the end before responding. You and I
are not in disgreement. I'm ignoring Porky.

We take the simple wavelength = u/freq formula with everything at rest and
we see an inverse relationship between frequency and wavelength. I never
disputed that. It's not unique to waves. If automobiles driving at 10 feet
per second pass us at a rate of two per second, we can calculate the
distance between the cars. It's just grade school algebra that even I can
do. Of course, we'll have to measure the same point on each car such as the
front bumper (gotta be precise with you physics guys).

So, as we discussed Doppler shift, I got to thinking about what was really
happening. If my source is moving, then the physical distance between two
waves is the speed of the waves times the time interval between waves plus
or minus the speed of the source times the time interval. With a stationary
receiver and medium we can mathematically derive the frequency at which the
receiver encounters the waves. Again, I believe we agree on this part.

But as I was thinking about the receiver moving with a stationary medium and
source, it occured to me that while his movement alters the frequency at
which the waves arrive over time, the distance between the waves really
didn't change. Please keep reading as I address this.

Therefore, our formula didn't apply anymore. This agrees with your
assertation that the formula only works for everything being still. All I
did was add to the formula some compensation for the movement of the
receiver. With that new formula I ended up calculating the same physical
distance for the wavelength at the stationary source regardless of whether
the receiver was moving or not. I also used Doppler to get the correct
frequency as measured at either source.

It's easy to visualize. Drop pebbles into a still pond at a constant rate.
We watch the rings radiate from the center. From above we can watch them
stay the same distance apart. We see a bee fly over the pond on a bee-line
for the point where the pebbles are dropped. How do we calculate the
frequency at which the bee passes over each wave? If we know the speed of
the waves, the speed of the bee and the frequency of the waves, we can
calculate that frequency. This is the Doppler shift formula simply stated.

But if we use the simple wavelength formula, we arrive at a number that
disagrees with what we see from above. If we use mine, we get the same
number. If we use the simple formula (which assumes everything to be still)
in this situation with the receiver moving, it REALLY tells us the how far
Wave B traveled after the bee crossed over Wave A. Is that wavelength?

No. We we know wavelength is defined as the distance between the same points
on two waves, NOT the distance traveled. A subtle but important difference.
Only in the case of everything being still are the wavelength and the
distance traveled the same. For most discussions of audio that assumption
suffices.

Thus if we use my formula we can arrive at the correct wavelength in all
cases of the source or receiver moving in a stationary medium. Am I totally
whacked or is my theory, pardon the pun, sound?

Just to comment on the semantics: Apparent can mean either readily seen *or*
readily clear/understood *or* appearing but not necesarily so. If you say
the apparent wavelength changes using the third definition, then we can
ignore the movement of the receiver when calculating that number. But why
would we do that when we know that formula is only measuring the distance
the second wave traveled, not the distance between the same points on the
wave? As we've seen we can easily calcuate the correct wavelength.

Jim Carr wrote:
"RP" wrote in message
...


Of course direction cannot be discounted, which is why the equations
that I posted refer to the velocities rather than to the scalar speeds.
In hindsight, my last statement above should have read:

wavelength = u/freq

is only valid when the frequency is that measured by a detector or
source that is at rest wrt the medium, or that is in motion at 90 or 270

deg

wrt the projected ray (in which case cos_theta = 0).
Or IOW, when (cos v) = 0.


Which agrees with what I was saying, though I admit to never seeing your
formulas with cos before. Like I said, this is not my field. If it's okay
I'm gonna assume the medium doesn't move for the purposes of this
discussion.


But I have no idea what "apparent" wavelength and frequency is supposed
to mean, except maybe that the observer was "apparently" wrong about one
or the other


Which, again, agrees with my point. I'll attempt to address "apparent
wavelength" if you'll bear with me to the end before responding. You and I
are not in disgreement. I'm ignoring Porky.

We take the simple wavelength = u/freq formula with everything at rest and
we see an inverse relationship between frequency and wavelength. I never
disputed that. It's not unique to waves. If automobiles driving at 10 feet
per second pass us at a rate of two per second, we can calculate the
distance between the cars. It's just grade school algebra that even I can
do. Of course, we'll have to measure the same point on each car such as the
front bumper (gotta be precise with you physics guys).

So, as we discussed Doppler shift, I got to thinking about what was really
happening. If my source is moving, then the physical distance between two
waves is the speed of the waves times the time interval between waves plus
or minus the speed of the source times the time interval. With a stationary
receiver and medium we can mathematically derive the frequency at which the
receiver encounters the waves. Again, I believe we agree on this part.

But as I was thinking about the receiver moving with a stationary medium and
source, it occured to me that while his movement alters the frequency at
which the waves arrive over time, the distance between the waves really
didn't change. Please keep reading as I address this.

Therefore, our formula didn't apply anymore. This agrees with your
assertation that the formula only works for everything being still. All I
did was add to the formula some compensation for the movement of the
receiver. With that new formula I ended up calculating the same physical
distance for the wavelength at the stationary source regardless of whether
the receiver was moving or not. I also used Doppler to get the correct
frequency as measured at either source.

It's easy to visualize. Drop pebbles into a still pond at a constant rate.
We watch the rings radiate from the center. From above we can watch them
stay the same distance apart. We see a bee fly over the pond on a bee-line
for the point where the pebbles are dropped. How do we calculate the
frequency at which the bee passes over each wave? If we know the speed of
the waves, the speed of the bee and the frequency of the waves, we can
calculate that frequency. This is the Doppler shift formula simply stated.

But if we use the simple wavelength formula, we arrive at a number that
disagrees with what we see from above. If we use mine, we get the same
number. If we use the simple formula (which assumes everything to be still)
in this situation with the receiver moving, it REALLY tells us the how far
Wave B traveled after the bee crossed over Wave A. Is that wavelength?

No. We we know wavelength is defined as the distance between the same points
on two waves, NOT the distance traveled. A subtle but important difference.
Only in the case of everything being still are the wavelength and the
distance traveled the same. For most discussions of audio that assumption
suffices.

Thus if we use my formula we can arrive at the correct wavelength in all
cases of the source or receiver moving in a stationary medium. Am I totally
whacked or is my theory, pardon the pun, sound?

Just to comment on the semantics: Apparent can mean either readily seen *or*
readily clear/understood *or* appearing but not necesarily so. If you say
the apparent wavelength changes using the third definition, then we can
ignore the movement of the receiver when calculating that number. But why
would we do that when we know that formula is only measuring the distance
the second wave traveled, not the distance between the same points on the
wave? As we've seen we can easily calcuate the correct wavelength.

I see that you provided:

" W=(C+R)/F where R is the speed of the Receiver"

This equation is ok as long as you know what the special case is that
you are applying it to. Since you didn't define the system, the
equation may or may not get the correct answer when applied by others.

Richard Perry

"RP" wrote in message
...

Of course direction cannot be discounted, which is why the equations
that I posted refer to the velocities rather than to the scalar speeds.
In hindsight, my last statement above should have read:

wavelength = u/freq

is only valid when the frequency is that measured by a detector or
source that is at rest wrt the medium, or that is in motion at 90 or 270
deg
wrt the projected ray (in which case cos_theta = 0).
Or IOW, when (cos v) = 0.

But I have no idea what "apparent" wavelength and frequency is supposed
to mean, except maybe that the observer was "apparently" wrong about one
or the other

The "apparent" wavelength and frequency are what the receiver observes, as
opposed to the actual frequency and wavelength as it appears at the source.
"Apparent" in this case meaning "as it appears" to the receiver. If there is
relative motion between the source and receiver, the apparent wavelength
observed by the reseiver will be different from the actual wavelength and
frequency produced by the source, this difference is Doppler shift.
Therefore, you have the actual frequency and wavelength emitted by the
source and the apparent frequency and wavelength as it is observed by the
receiver. We are worried only about Doppler shift in sound as it applies to
recording and home studio.

BTW, for those reading this in the sci.physics NG, note that in the
following argument we had to assume a medium in order to derive the
special relativistic wavelength-frequency relationship. OTOH the mere
existence of a "wavelength" of em radiation should offer a clue

I don't think we're really worried about relativistic aspects, and since
we live in the earth's atmosphere and of necessity use air as our primary
medium of acoustic propogation when listening to music, there is no reason
to assume a medium, we are not concerned with EM radiation, just sound in
air. For that reason, I'm taking sci.physics off the cross posting list,
just to keep it simple and on topic.

"Jim Carr" wrote in message
news:HHZ9d.14497$_a3.4819@fed1read05...
"RP" wrote in message
...

Of course direction cannot be discounted, which is why the equations
that I posted refer to the velocities rather than to the scalar speeds.
In hindsight, my last statement above should have read:

wavelength = u/freq

is only valid when the frequency is that measured by a detector or
source that is at rest wrt the medium, or that is in motion at 90 or 270
deg
wrt the projected ray (in which case cos_theta = 0).
Or IOW, when (cos v) = 0.

Which agrees with what I was saying, though I admit to never seeing your
formulas with cos before. Like I said, this is not my field. If it's okay
I'm gonna assume the medium doesn't move for the purposes of this
discussion.

But I have no idea what "apparent" wavelength and frequency is supposed
to mean, except maybe that the observer was "apparently" wrong about one
or the other

Which, again, agrees with my point. I'll attempt to address "apparent
wavelength" if you'll bear with me to the end before responding. You and I
are not in disgreement. I'm ignoring Porky.

We take the simple wavelength = u/freq formula with everything at rest and
we see an inverse relationship between frequency and wavelength. I never
disputed that. It's not unique to waves. If automobiles driving at 10 feet
per second pass us at a rate of two per second, we can calculate the
distance between the cars. It's just grade school algebra that even I can
do. Of course, we'll have to measure the same point on each car such as
the
front bumper (gotta be precise with you physics guys).

So, as we discussed Doppler shift, I got to thinking about what was really
happening. If my source is moving, then the physical distance between two
waves is the speed of the waves times the time interval between waves plus
or minus the speed of the source times the time interval. With a
stationary
receiver and medium we can mathematically derive the frequency at which
the
receiver encounters the waves. Again, I believe we agree on this part.

But as I was thinking about the receiver moving with a stationary medium
and
source, it occured to me that while his movement alters the frequency at
which the waves arrive over time, the distance between the waves really
didn't change. Please keep reading as I address this.

Therefore, our formula didn't apply anymore. This agrees with your
assertation that the formula only works for everything being still. All I
did was add to the formula some compensation for the movement of the
receiver. With that new formula I ended up calculating the same physical
distance for the wavelength at the stationary source regardless of whether
the receiver was moving or not. I also used Doppler to get the correct
frequency as measured at either source.

It's easy to visualize. Drop pebbles into a still pond at a constant rate.
We watch the rings radiate from the center. From above we can watch them
stay the same distance apart. We see a bee fly over the pond on a bee-line
for the point where the pebbles are dropped. How do we calculate the
frequency at which the bee passes over each wave? If we know the speed of
the waves, the speed of the bee and the frequency of the waves, we can
calculate that frequency. This is the Doppler shift formula simply stated.

But if we use the simple wavelength formula, we arrive at a number that
disagrees with what we see from above. If we use mine, we get the same
number. If we use the simple formula (which assumes everything to be
still)
in this situation with the receiver moving, it REALLY tells us the how far
Wave B traveled after the bee crossed over Wave A. Is that wavelength?

No. We we know wavelength is defined as the distance between the same
points
on two waves, NOT the distance traveled. A subtle but important
difference.
Only in the case of everything being still are the wavelength and the
distance traveled the same. For most discussions of audio that assumption
suffices.

Thus if we use my formula we can arrive at the correct wavelength in all
cases of the source or receiver moving in a stationary medium. Am I
totally
whacked or is my theory, pardon the pun, sound?

Just to comment on the semantics: Apparent can mean either readily seen
*or*
readily clear/understood *or* appearing but not necesarily so. If you say
the apparent wavelength changes using the third definition, then we can
ignore the movement of the receiver when calculating that number. But why
would we do that when we know that formula is only measuring the distance
the second wave traveled, not the distance between the same points on the
wave? As we've seen we can easily calcuate the correct wavelength.


My whole point was that the frequency and wavelength of a sound have a
fixed relationship and if one change, the other must change as well. All
other things being equal, you simply cannot have a sound in air that changes
in frequency without changing in wavelength, period! If you think otherwise,
you think wrong!

Porky wrote:

My whole point was that the frequency and wavelength of a sound have a
fixed relationship and if one change, the other must change as well. All
other things being equal, you simply cannot have a sound in air that changes
in frequency without changing in wavelength, period! If you think otherwise,
you think wrong!

As Richard points out, these statements must be carefully
qualified. The relationship depends on the speed of sound.
The speed of sound in your frame of reference depends on its
velocity with respect to the medium and the direction you
chose to measure it in with respect to your frame's
velocity. That's what Michaelson and Morely were trying to
measure to see how fast and in what direction we were moving
with respect to the ether. That it never varies with
direction is a property of light, not sound because sound
actually has a medium whereas light doesn't, as Einstein
figured out.

If you measure W from within one frame of reference you'd
better be using the C that applies to the direction of
measurement in that frame of reference to calculate F in
that same frame of reference.

When you start mixing frames of reference and not
considering angles of measurement as y'all have been doing
in this sub-thread it gets very confusing, as you've seen.


Bob
--

"Things should be described as simply as possible, but no
simpler."

A. Einstein

"Bob Cain" wrote in message
...


Porky wrote:

My whole point was that the frequency and wavelength of a sound have
a
fixed relationship and if one change, the other must change as well. All
other things being equal, you simply cannot have a sound in air that
changes
in frequency without changing in wavelength, period! If you think
otherwise,
you think wrong!

As Richard points out, these statements must be carefully
qualified. The relationship depends on the speed of sound.
The speed of sound in your frame of reference depends on its
velocity with respect to the medium and the direction you
chose to measure it in with respect to your frame's
velocity. That's what Michaelson and Morely were trying to
measure to see how fast and in what direction we were moving
with respect to the ether. That it never varies with
direction is a property of light, not sound because sound
actually has a medium whereas light doesn't, as Einstein
figured out.

If you measure W from within one frame of reference you'd
better be using the C that applies to the direction of
measurement in that frame of reference to calculate F in
that same frame of reference.

When you start mixing frames of reference and not
considering angles of measurement as y'all have been doing
in this sub-thread it gets very confusing, as you've seen.

Is it not true that for a specific value of the speed of propagation, C,
the relationship of wavelength to frequency is a fixed one, and in order for
that relationship of wavelength to frequency to change, C must change as
well?
The whole point I was trying to make is that if one assumes that the
velocity and direction of motion between them is constant, the receiver will
see a fixed relationship between the frequency and wavelength of the sound
he hears. I'm talking about the typical Doppler shift one is likely to
encounter in everyday life, like a train whistle, a siren on an emergency
vehicle, or a car engine as it approaches and passes. In other words,
Doppler shift as one is likely to encounter it in a recording or home studio
enviornment. This assumes that the air will retain roughly the same
characteristics during the time one observes a particular instance of
Doppler shift. I think most of those interested in the phenomena as it
affects the topics of the various recording groups have little interest in
anything other than the basics.
Jim made statements about observed frequency being shifted while observed
wavelength stayed the same, and everything I've ever seen about Doppler
shift, as it applies to sound in air, indicated that the apparent shift in
frequency the receiver observed is due to the apparent shift in wavelength
caused by the relative motion between the source and the receiver. Does the
value of C actually change because of the relative motion between the source
and the receiver? If so, how does that happen since that motion is relative
and not absolute? And if so, how does the absolute velocity of either of or
both the source and receiver affect the change in C?

"RP" wrote in message
...

" W=(C+R)/F where R is the speed of the Receiver"

This equation is ok as long as you know what the special case is that
you are applying it to. Since you didn't define the system, the
equation may or may not get the correct answer when applied by others.

The above will work whether the receiver is moving or the source is moving
and we use Doppler to calculate F at the source *or* we skip Doppler and
simply measure F at the receiver. It does not take into account the medium
moving. We're pretty much in agreement that I'm talking about specific
scenarios.

Doppler was explained to me as a kid. It made perfect sense. I know you
fully understand it better than I. Just for kicks I decided to look at a
dozen sites all explaining Doppler. In most of them they *only* mentioned
the source moving. Some included a formula that handled the receiver moving
as well.

However, in *every* site I saw that used a picture, they used some form of
the concentric circles similar to what I described regarding the pond. One
picture was a stationary source with perfectly concentric circles. The
distances between the lines we are told is the wavelength.

With the source moving in a straight line we see a different picture. You
can clearly see that on one side the lines are closer together and on the
opposite side they are farther apart than in the first picture. Since the
distance between the lines is the wavelength, we can see how the wavelength
changes. This link is a great example:
http://www.fearofphysics.com/Sound/dopwhy2.html

Seeing it this way makes it easy to understand.

But nobody showed a picture of the receiver moving. What would the lines
look like if the receiver was moving and the source was stationary? The
lines would look exactly the same as the first case when the source was
stationary. But we're told that the receiver sees a different frequency. It
doesn't just "appear" to be a different frequency. Every tool we use to
measure the frequency will arrive at the same value.

How can the distance between the lines be the same when we acknowledge the
frequency at the point of the observer is different than that at the emitter
when Porky says the wavelength/frequency relationship is fixed - period?

It's simple, and I know you know. After you calculate the wavelength
mathematically using Porky's formula, you gotta add/subtract the distance
the receiver traveled. If you're moving, Porky's formula tells you how far
the wave traveled, not the distance between two equal points on a wave
(wavelength).

Why he refuses to acknowledge this escapes me. If he wants to argue that
there's not much practical use for it, I wouldn't mind. Here's an example
using a horn. My limited understanding is that there is a mathematical
relationship between the length and flare of the horn and the lowest note it
can reproduce properly/well/at all. If I am misstating the above, humor me
and pretend I'm right for the sake of argument.

If someone is using a horn and says "this is the lowest note it can play
properly/well/at all," you could measure the frequency and calculate the
length of the horn. If the horn is moving, you're moving (or both) you would
still get the right number so long as you know the speed of the movement and
apply the Doppler formula. If you didn't account for Doppler, you'd never
know the length of the horn.

The application of Doppler above implicitly removes the effect of the
movement. Intuitively we say we do it because we don't know the right
frequency. Once we know the right frequency we can use the fixed
relationship between frequency and wavelength to perform our calculations.

Another person might argue that there are no wrong frequencies. Either
something happens at a certain time interval or it doesn't. There's no
ambiguity about how many times per second sound waves strike my microphone.
The problem is that when you're moving, you can no longer apply the
frequency and speed of the waves to determine the wavelength. You have to
account for the movement somehow.

In both cases you use Doppler to convert the frequency back to that of a
stationary source, then do the calculation. Or, if you're oddball like me,
you simply modify your wavelength formula to account for the movement of the
source and skip the Doppler formula. Either way the end result is always the
same.

It's no big deal.

Porky wrote:


Is it not true that for a specific value of the speed of propagation, C,
the relationship of wavelength to frequency is a fixed one, and in order for
that relationship of wavelength to frequency to change, C must change as
well?

Yes, for whatever specific value of C pertains to your frame
of reference.

Youse guys are using too many words and not enough algebra
for me to figure out what you're saying.

And if so, how does the absolute velocity of either of or
both the source and receiver affect the change in C?

Therein lies the crux of both your arguments. What do you
think the answer is? Hint: your question is without meaning
as stated. You should be asking what is the speed of sound
in the frame of reference of the TX, the RX, and the fixed
frame of reference of the quiescent air (that's the easy
one) and how do you transform the variables among them.

Richard spelled that out.


Bob
--

"Things should be described as simply as possible, but no
simpler."

A. Einstein

Bob Cain wrote:


Porky wrote:

My whole point was that the frequency and wavelength of a sound have a
fixed relationship and if one change, the other must change as well. All
other things being equal, you simply cannot have a sound in air that
changes
in frequency without changing in wavelength, period! If you think
otherwise,
you think wrong!


As Richard points out, these statements must be carefully qualified.
The relationship depends on the speed of sound. The speed of sound in
your frame of reference depends on its velocity with respect to the
medium and the direction you chose to measure it in with respect to your
frame's velocity. That's what Michaelson and Morely were trying to
measure to see how fast and in what direction we were moving with
respect to the ether. That it never varies with direction is a property
of light, not sound because sound actually has a medium whereas light
doesn't, as Einstein figured out.

No, Einstein figured no so thing, only that a particular state of motion
couldn't be ascribed to it, i.e. it "seems" to move along with every
observer in his own frame. Not so, however, when light speed isn't
defined to be constant. The statement that the traveling twin aged less
is an admission that his clock ticked slower, i.e. that his standard of
measure differed from the stay at home twin's standard. Though both call
their unit of time the second, these seconds are not equivalent, thus
they should have agreed that the traveling twin call his time unit by
some other name. IOW, though both calculated the velocity of some beam
of light to be c, it must follow that the velocity actually differed wrt
these two, and in fact wrt some third observer the velocity of the first
two wrt the beam did in fact differ, since the latter has no difficulty
seeing a beam move at 2c wrt another observer, even though that other
observer only clocks it at 1c. Thus, if each relies upon a universal
standard, i.e. one that remains in equilibrium throughout, then the
anisotropy becomes apparent.

Richard Perry



If you measure W from within one frame of reference you'd better be
using the C that applies to the direction of measurement in that frame
of reference to calculate F in that same frame of reference.

When you start mixing frames of reference and not considering angles of
measurement as y'all have been doing in this sub-thread it gets very
confusing, as you've seen.


Bob

RP wrote:

That it never varies with direction is a
property of light, not sound because sound actually has a medium
whereas light doesn't, as Einstein figured out.


No, Einstein figured no so thing, only that a particular state of motion
couldn't be ascribed to it, i.e. it "seems" to move along with every
observer in his own frame.

So it was others then that have interpreted his conclusions
as making the ether an unnecessary idea? I thought he
actually stated that in his paper but it's been quite a
while since I read it.

Not so, however, when light speed isn't
defined to be constant. The statement that the traveling twin aged less
is an admission that his clock ticked slower, i.e. that his standard of
measure differed from the stay at home twin's standard. Though both call
their unit of time the second, these seconds are not equivalent, thus
they should have agreed that the traveling twin call his time unit by
some other name. IOW, though both calculated the velocity of some beam
of light to be c, it must follow that the velocity actually differed wrt
these two, and in fact wrt some third observer the velocity of the first
two wrt the beam did in fact differ, since the latter has no difficulty
seeing a beam move at 2c wrt another observer, even though that other
observer only clocks it at 1c.

They both see each other's clock's ticking slower but they
also see a contraction of distance in the direction of
motion of the other. Don't these effects combine to leave
the observed c constant for all observers regardless of the
frame of reference from which it originated?


Bob
--

"Things should be described as simply as possible, but no
simpler."

A. Einstein

Bob Cain wrote:


RP wrote:

That it never varies with direction is a property of light, not sound
because sound actually has a medium whereas light doesn't, as
Einstein figured out.



No, Einstein figured no so thing, only that a particular state of
motion couldn't be ascribed to it, i.e. it "seems" to move along with
every observer in his own frame.


So it was others then that have interpreted his conclusions as making
the ether an unnecessary idea?

Yes indeed. Einstein saw the necessity of a medium, on logical grounds,
though he discounted the common interpretations of it, i.e. as being
something that objects have a definite velocity wrt.


I thought he actually stated that in his
paper but it's been quite a while since I read it.

Not so, however, when light speed isn't defined to be constant. The
statement that the traveling twin aged less is an admission that his
clock ticked slower, i.e. that his standard of measure differed from
the stay at home twin's standard. Though both call their unit of time
the second, these seconds are not equivalent, thus they should have
agreed that the traveling twin call his time unit by some other name.
IOW, though both calculated the velocity of some beam of light to be
c, it must follow that the velocity actually differed wrt these two,
and in fact wrt some third observer the velocity of the first two wrt
the beam did in fact differ, since the latter has no difficulty
seeing a beam move at 2c wrt another observer, even though that other
observer only clocks it at 1c.


They both see each other's clock's ticking slower but they also see a
contraction of distance in the direction of motion of the other. Don't
these effects combine to leave the observed c constant for all observers
regardless of the frame of reference from which it originated?

Yes indeed again, but keep in mind that these effects occur by sheer
fiat, i.e. they are taken as logical premises, and the universe and its
interactions are afterward adjusted accordingly to accommodate those
presuppositions. The same can be done with virtually any other set of
initial premises. OTOH, again, without a universally agreed upon
standard, special relativity is automatically disjoint from Newtonian
Mechanics without so much as one pen laid to a paper. They are mutually
exclusive logical systems, and as such the conclusions of one bear no
effect whatsoever on the conclusions of the other. They can both be
correct, and moreover one can be correct and the incorrect given exactly
the same mathematical conclusion. As B. Russell noted "the resolution
to any philosophical argument lies in the definitions of the terms."

Richard Perry



Bob

RP wrote:


They are mutually
exclusive logical systems, and as such the conclusions of one bear no
effect whatsoever on the conclusions of the other.

Only because they have a mutually exclusive premise.
Newton's theory requires that light veloctities be additive.
Einstein's requires the opposite, that it be constant for
all observers. From only this difference, the first part of
his paper develops his theory purely kinematically and
arrives at the Lorentz transformation.

They can both be
correct, and moreover one can be correct and the incorrect given exactly
the same mathematical conclusion.

Not sure what you mean here. There is an incorrect premise
in Newton's theory that leads to incorrect prediction.


Bob
--

"Things should be described as simply as possible, but no
simpler."

A. Einstein

Bob Cain wrote:


RP wrote:


They are mutually exclusive logical systems, and as such the
conclusions of one bear no effect whatsoever on the conclusions of the
other.


Only because they have a mutually exclusive premise. Newton's theory
requires that light veloctities be additive. Einstein's requires the
opposite, that it be constant for all observers. From only this
difference, the first part of his paper develops his theory purely
kinematically and arrives at the Lorentz transformation.

They can both be correct, and moreover one can be correct and the
incorrect given exactly the same mathematical conclusion.


Not sure what you mean here. There is an incorrect premise in Newton's
theory that leads to incorrect prediction.

Which prediction would that be?

Richard Perry

On Fri, 08 Oct 2004 11:42:55 -0700, Bob Cain
wrote:

If the the motion of a test particle normally at rest at a
point is purely sinusoidal and containing two tones then

First, to expirience doppler effect, we don't need 2 tones. One is
enough, but source and listener have to travell (change the distance)
relative to each other.

Second, molecules of air do not really travell from the speaker towards
the listener. If that was the case we'd have wind, not sound. In sound,
molecules move in such way, to form areas of higher or lower air
pressure (Speaker gives nescesserry energy by means of alternate forward
and backward movements). Because of this, pressure wave travell - not
matery, length of cones excursion does not influence the freq (or
wavelength).

Third, the distance btw source and listener is always the same. Fact
that cone is sometimes closer, sometimes further changes nothing,
because it started "compressing" the air from the neutral position Now,
you can say the pressure wave started to travell from he point of max
excursion. Yes, but it always does. Therefore the distance does not
change. Also, from that same point (moment in time) speaker started
"decompressing" the air. All the speaker has to do is to keep pushing
and pulling in propper interval so to keep pressure zones equally
spaced. Therefore, we can say the distance is constant. If it is
constant source dos not travell. If it does not travell there can not
possibly be Doppler effect.

I think "doppler in speaker" advocates are analyzing the problem, as if
there would be two sources, one behind another, appart by the distance
of max cone excursion, emmiting sound of same freq. Well, if that was
the case, what would be the result heard by listener (being third dot on
the same line with two sources)? Can that be called doppler effect?

there is a cross phase modulation of each by the other and
by itself that appears as IM distortion in the fluid
velocity at that same point. It can be called Doppler
because standard Doppler shift can be drived from a constant
modulation of frequency. Or it can be called Billy. :-)

Experiment to prove Doppler effect due to speaker movement, would be to
feed the amp with constant tone put a mic infront of a speaker, than
measure captured frequency. If ther's fluctuation coinciding with
movement, you have doppler. Say we have tone of 100Hz. That's 100 pushes
(and 100 pulls). Doppler effect would be present if captured freq would
constantly dance arround 100Hz, rise and fall 200 times per second (100
rises and 100 falls) being exactly 100Hz only momentary, while "passing
through" neutral position. Obviously, that's not the case.

"Porky" wrote in message et...
"Paul Draper" wrote in message
om...
We're talking at crossed purposes here, and I suppose it depends on
what you are measuring when you look at the scope and see a square
wave. There are several possibilities:
1. The voltage applied across the terminals of the speaker coil.
2. The current driven through the speaker coil
3. The position of the speaker coil
4. The velocity of the speaker coil
5. The acceleration of the speaker coil
6. The pressure in the air at a point in front of the speaker
diaphragm

I've been arguing that there is a connection between 4 and 6. Those
two, I think, follow the same profile. The others are related through
derivatives or integrals, possibly with a phase shift.

PD

A pure square wave has zero rise and fall times, the "slopes" are
vertical. If you apply a square wave to a speaker's terminals, when the
positive voltage rise happens, the cone will jump forward and will push the
air in front of it. Because of the inertia of the mass of the cone and the
air in front of it, the cone will not instantly spring to it's maximum
forward position, it will require some period of time, thus the slope of the
soundwave being generated will not have a vertical rise time. The same
applies to the fall time, again, the slope will not be vertical. Also if the
applied square wave is of a low enough frequency, since the average
listening room is not air tight, the "top" and "bottom" of the generated
soundwave will not be level, they will contain a slope even though the
applied electrical square wave has a level "top" and "bottom". As for item
(4), in order to follow the square wave's zero rise time, the cone's
velocity would have to be infinite, and that can't happen. There is a direct
relationship between the velocity of the cone and the rate of change in
pressure, so there is a connection between (4) and (6). If one were to apply
say, a 20 Hz square wave to a speaker, and record the resulting sound wave,
one would see a positive sawtooth wave corresponding to the rise of the wave
with a rather abrupt (but not vertical) positive slope and a more gradual
negative slope, and then a negative sawtooth wave corresponding to the fall
of the wave with a rather abrupt (but not vertical) negative slope and a
more gradual positive slope. This is what happens in a real world speaker
placed in a real world typical listening room. At higher frequencies, it is
possible to reproduce a fair approximation of a square wave, but not a true
pure square wave.

OK, and now I think I've narrowed down the point of our debate. You
think that a voltage applied to the speaker's terminals corresponds to
the position of the speaker cone. In the low frequency limit, that may
be right. Certainly, if I appy a DC current to the electromagnet, the
speaker cone will move to a position where the force applied by the
electromagnet balances the elastic restoring force of the cone. But at
higher frequencies, I think the situation gets more complicated.
First of all, the electromagnet is an inductor, so
V = L dI/dt
Since the electromagnet is a solenoid, there is something like the
relationship
B = (mu)nI
and so B is proportional to the integral of V with respect to time.
And the *force* on the speaker diaphragm is proportional to B, and
proportional to the acceleration, not the position of the diaphragm
(constant)B = F = ma = m d2(x)/dt2

Thus we have Int(V)dt = (constant) d2(x)/dt2

Thus, the voltage across the terminal is not proportional to the
position of the speaker diaphragm, but to the third derivative of the
position with respect to time.

PD

Vladan wrote:

First, to expirience doppler effect, we don't need 2 tones. One is
enough, but source and listener have to travell (change the distance)
relative to each other.

Correct, but the effect for a single frequency is very small
in comparison to two or more.

Second, molecules of air do not really travell from the speaker towards
the listener. If that was the case we'd have wind, not sound. In sound,
molecules move in such way, to form areas of higher or lower air
pressure (Speaker gives nescesserry energy by means of alternate forward
and backward movements).

They oscillate about a rest position.

Because of this, pressure wave travell - not
matery, length of cones excursion does not influence the freq (or
wavelength).

That's what I thought too for a long time. I was wrong.

Third, the distance btw source and listener is always the same. Fact
that cone is sometimes closer, sometimes further changes nothing,
because it started "compressing" the air from the neutral position Now,
you can say the pressure wave started to travell from he point of max
excursion. Yes, but it always does. Therefore the distance does not
change. Also, from that same point (moment in time) speaker started
"decompressing" the air. All the speaker has to do is to keep pushing
and pulling in propper interval so to keep pressure zones equally
spaced. Therefore, we can say the distance is constant. If it is
constant source dos not travell. If it does not travell there can not
possibly be Doppler effect.

Assume that there is an oscilatory motion of particles about
a rest position that is imparted linearly by a driver doing
the same. Then figure out the fluid velocity at that rest
position as a function of the motion of the particles about
it. You will find that the relationship predicts
intermodulation at that fixed point in an amount related to
the size of the excursion of the particles about it.

The pressure at that point will be linearly related to the
fluid velocity at the point. Thus, a pressure microphone
fixed at that rest position will yield signal that contains
intermodulation products.

If you take that point to be the rest position of the
speaker, the potentially large excursions there, on the
order of centimeters, can result in a pressure signal
propegating from that rest position which contains very
signifigant amounts of this IM distortion.

The Mackie HR824 has an interesting design that minimizes
this. It is only a two way system so should be very prone
to low frequencies modulating the mid frequencies. However,
the excursion of the Low/Mid driver is small at low
frequencies compared to what it would be in a typical ported
cabinet. The HR824 cabinet has a tuned passive radiator in
the back where the large LF excursion occur separately from
the mid range. This effectively makes it a three way
system. Tuned ports somewhat accomplish the same thing but
not as effecitvely as a passively coupled mass system does.


I think "doppler in speaker" advocates are analyzing the problem, as if
there would be two sources, one behind another, appart by the distance
of max cone excursion, emmiting sound of same freq. Well, if that was
the case, what would be the result heard by listener (being third dot on
the same line with two sources)? Can that be called doppler effect?

An analyis starting from the view of having a high frequency
radiator attached and moving with respect to the a low
frequency radiator of the same size does correctly depict
the situation of the mixed signal being applied to a single
radiator. It's just linear superposition.

Trying to view this as the velocity of the LF component
modulating the frequence of the HF component, as is often
done and justified with "dynamic" Doppler shift, is not
correct because it does not take into account the fact that
the LF component is also generating a wave, not just moving
through the air. It predicts an error that is minimum at
the ends of the LF excursion where the veolocity due to it
is a minimum and a maximum error as the LF motion passes
through the rest position where its velocity is a maximum.
An analysis that proceeds just from the conditions in a
propegating wave, as defined above, predicts the opposite.
The latter analysis also applies to the situation where the
point in question is the rest position of the speaker due to
boundry condition requirements. I don't know what to call
the real effect that follows from the correct analysis but
it does not follow from thinking of it as "Doppling."


Bob
--

"Things should be described as simply as possible, but no
simpler."

A. Einstein

RP wrote:

Not sure what you mean here. There is an incorrect premise in
Newton's theory that leads to incorrect prediction.


Which prediction would that be?

How about that time and space are a fixed background?


Bob
--

"Things should be described as simply as possible, but no
simpler."

A. Einstein

"Paul Draper" wrote in message
om...
"Porky" wrote in message
et...
"Paul Draper" wrote in message
om...
We're talking at crossed purposes here, and I suppose it depends on
what you are measuring when you look at the scope and see a square
wave. There are several possibilities:
1. The voltage applied across the terminals of the speaker coil.
2. The current driven through the speaker coil
3. The position of the speaker coil
4. The velocity of the speaker coil
5. The acceleration of the speaker coil
6. The pressure in the air at a point in front of the speaker
diaphragm

I've been arguing that there is a connection between 4 and 6. Those
two, I think, follow the same profile. The others are related through
derivatives or integrals, possibly with a phase shift.

PD

A pure square wave has zero rise and fall times, the "slopes" are
vertical. If you apply a square wave to a speaker's terminals, when the
positive voltage rise happens, the cone will jump forward and will push
the
air in front of it. Because of the inertia of the mass of the cone and
the
air in front of it, the cone will not instantly spring to it's maximum
forward position, it will require some period of time, thus the slope of
the
soundwave being generated will not have a vertical rise time. The same
applies to the fall time, again, the slope will not be vertical. Also if
the
applied square wave is of a low enough frequency, since the average
listening room is not air tight, the "top" and "bottom" of the generated
soundwave will not be level, they will contain a slope even though the
applied electrical square wave has a level "top" and "bottom". As for
item
(4), in order to follow the square wave's zero rise time, the cone's
velocity would have to be infinite, and that can't happen. There is a
direct
relationship between the velocity of the cone and the rate of change in
pressure, so there is a connection between (4) and (6). If one were to
apply
say, a 20 Hz square wave to a speaker, and record the resulting sound
wave,
one would see a positive sawtooth wave corresponding to the rise of the
wave
with a rather abrupt (but not vertical) positive slope and a more
gradual
negative slope, and then a negative sawtooth wave corresponding to the
fall
of the wave with a rather abrupt (but not vertical) negative slope and
a
more gradual positive slope. This is what happens in a real world
speaker
placed in a real world typical listening room. At higher frequencies, it
is
possible to reproduce a fair approximation of a square wave, but not a
true
pure square wave.

OK, and now I think I've narrowed down the point of our debate. You
think that a voltage applied to the speaker's terminals corresponds to
the position of the speaker cone.

Old magnetic meter style voltmeters work on the same principle, but it is
the current across a known resistance, not specifically the voltage that
makes them work. However, that wasn't what I was trying to get across. With
a zero rise time on the squarewave, the voltage rise will be instantaneous
from negative value to positive value, However, the cone is not capable of
moving instantanoeusly, so there will be some lag in motion both ways. This
alone makes the generation of a pure square wave impossible.

In the low frequency limit, that may
be right. Certainly, if I appy a DC current to the electromagnet, the
speaker cone will move to a position where the force applied by the
electromagnet balances the elastic restoring force of the cone. But at
higher frequencies, I think the situation gets more complicated.

You are quite correct, so it will always be impossible to reproduce a pure
squarewave with zero rise and fall times, but at higher frequencies the
recovery slope of the air in the room will allow fairly (but not perfectly)
flat tops and bottoms, so one can get a fair approximation (but only an
approximation) of a square wave at higher frequencies.

First of all, the electromagnet is an inductor, so
V = L dI/dt
Since the electromagnet is a solenoid, there is something like the
relationship
B = (mu)nI
and so B is proportional to the integral of V with respect to time.
And the *force* on the speaker diaphragm is proportional to B, and
proportional to the acceleration, not the position of the diaphragm
(constant)B = F = ma = m d2(x)/dt2

Thus we have Int(V)dt = (constant) d2(x)/dt2

Thus, the voltage across the terminal is not proportional to the
position of the speaker diaphragm, but to the third derivative of the
position with respect to time.

I'm not disagreeing with you, I'm just pointing out that a real speaker
in a real listening room cannot even approximate a square wave at low
frequencies, but can reproduce a fair approximation at higher frequencies.
Note that an approximation is not the real thing, and a fair approximation
isn't all that accurate an approximation, but it does look fairly close on
an oscilloscope. :-)
With all due respect to the sci.physics folks, the recording news groups
are not particularly concerned with the fine points of theory, we're
primarily concerned with good studio techniques and practices. We want
practical applications, and rule of thumb is often more practical for us
than are complicated exact equations. This is my opinion and others in the
recording groups may disagree with me, but for the most part I think many of
those in the recording groups simply don't have the physics and math
background, they're just trying to get good sounding recordings. They're
interested in basic engineering solutions, not complex background theory. I
hope no offense is taken, because there certainly is none intended. :-)

"Bob Cain" wrote in message
...


Vladan wrote:

First, to expirience doppler effect, we don't need 2 tones. One is
enough, but source and listener have to travell (change the distance)
relative to each other.

Correct, but the effect for a single frequency is very small
in comparison to two or more.

Second, molecules of air do not really travell from the speaker towards
the listener. If that was the case we'd have wind, not sound. In sound,
molecules move in such way, to form areas of higher or lower air
pressure (Speaker gives nescesserry energy by means of alternate forward
and backward movements).

They oscillate about a rest position.

Because of this, pressure wave travell - not
matery, length of cones excursion does not influence the freq (or
wavelength).

That's what I thought too for a long time. I was wrong.

Third, the distance btw source and listener is always the same. Fact
that cone is sometimes closer, sometimes further changes nothing,
because it started "compressing" the air from the neutral position Now,
you can say the pressure wave started to travell from he point of max
excursion. Yes, but it always does. Therefore the distance does not
change. Also, from that same point (moment in time) speaker started
"decompressing" the air. All the speaker has to do is to keep pushing
and pulling in propper interval so to keep pressure zones equally
spaced. Therefore, we can say the distance is constant. If it is
constant source dos not travell. If it does not travell there can not
possibly be Doppler effect.

Assume that there is an oscilatory motion of particles about
a rest position that is imparted linearly by a driver doing
the same. Then figure out the fluid velocity at that rest
position as a function of the motion of the particles about
it. You will find that the relationship predicts
intermodulation at that fixed point in an amount related to
the size of the excursion of the particles about it.

The pressure at that point will be linearly related to the
fluid velocity at the point. Thus, a pressure microphone
fixed at that rest position will yield signal that contains
intermodulation products.

If you take that point to be the rest position of the
speaker, the potentially large excursions there, on the
order of centimeters, can result in a pressure signal
propegating from that rest position which contains very
signifigant amounts of this IM distortion.

The Mackie HR824 has an interesting design that minimizes
this. It is only a two way system so should be very prone
to low frequencies modulating the mid frequencies. However,
the excursion of the Low/Mid driver is small at low
frequencies compared to what it would be in a typical ported
cabinet. The HR824 cabinet has a tuned passive radiator in
the back where the large LF excursion occur separately from
the mid range. This effectively makes it a three way
system. Tuned ports somewhat accomplish the same thing but
not as effecitvely as a passively coupled mass system does.


I think "doppler in speaker" advocates are analyzing the problem, as if
there would be two sources, one behind another, appart by the distance
of max cone excursion, emmiting sound of same freq. Well, if that was
the case, what would be the result heard by listener (being third dot on
the same line with two sources)? Can that be called doppler effect?

An analyis starting from the view of having a high frequency
radiator attached and moving with respect to the a low
frequency radiator of the same size does correctly depict
the situation of the mixed signal being applied to a single
radiator. It's just linear superposition.

Trying to view this as the velocity of the LF component
modulating the frequence of the HF component, as is often
done and justified with "dynamic" Doppler shift, is not
correct because it does not take into account the fact that
the LF component is also generating a wave, not just moving
through the air. It predicts an error that is minimum at
the ends of the LF excursion where the veolocity due to it
is a minimum and a maximum error as the LF motion passes
through the rest position where its velocity is a maximum.
An analysis that proceeds just from the conditions in a
propegating wave, as defined above, predicts the opposite.
The latter analysis also applies to the situation where the
point in question is the rest position of the speaker due to
boundry condition requirements. I don't know what to call
the real effect that follows from the correct analysis but
it does not follow from thinking of it as "Doppling."



Vladan made the same point that I was trying to make. it isn't Doppler
distortion, however Bob is correct that there are IM distortion products,
therfore we should try to come up with a different label for the IM
distortion Bob is describing, calling it Doppler distortion is just too
confusing.

Bob Cain wrote:


RP wrote:

Not sure what you mean here. There is an incorrect premise in
Newton's theory that leads to incorrect prediction.



Which prediction would that be?


How about that time and space are a fixed background?

Time, when a universal standard is adopted for use by all observers,
whether or not on board clocks perpetually agree with that standard,
must necessarily be universal time. Obvious why, eh?

Now as for space, acceleration most certainly is absolute, evidenced
especially by angular acceleration. I doubt that anyone disputes this.
If however, you are referring implicitly to Mach's arguments, then I
must concur, i.e. that space is indeed not some static and fixed
Euclidean metric in which masses are embedded in. OTOH, whether Newton
personally perceived space as such an entity or not, it is immaterial,
because his sentiments have no part nor parcel with his math. Thus, show
me in a quantitative fashion that Newtonian Mechanics makes an incorrect
prediction within its domain of applicability, i.e. the kinematics of
neutral masses. If you can do so without utilizing a premise that he
did not actually pen, then good luck. Don't try Fizeau's measurements of
light speed in moving media, because I've embarrassed enough Ph D's
already in that argument by showing them how to correctly compose
"average speeds" with the Galilean transform

Richard Perry

"Jim Greenfield" wrote in message
om...

Porky should realise at once that you haven't given him sufficient
information (your speed ref the room). As he doesn't understand the
implications of mistakenly identifying the _cause_ of Doppler, he
might fall for the trap! :-)

Jim G
c'=c+v

Aw, you're no fun!! To recap this entire discussion, it started like this.
Porky described Doppler for sound as a shift in the wavelength whether the
source or receiver is moving (one should describe it using frequency of the
receiver IMHO). I replied that the wavelength certainly does change when the
source moves, but it does NOT change when only the receiver moves. He said I
was wrong because wavelength and frequency are always in a direct
relationship to one another.

He cited a few references that did not explicitly discuss the movement of
the receiver *and* wavelength specifically. I didn't care because it seemed
readily obvious to my uneducated mind. I decided to find a link that
discusses wavelength: http://physics.nad.ru/Physics/English/dopp_txt.htm .
This site writes, "Next, we shall consider the case when observer moves and
the source of the wave is still. In this case the wavelength is NOT changed
and Doppler frequency shift appears because the velocity w of the wave
relatively the observer is changed."

That is certainly a good way of explaining it (which is how you did as
well). I never said it that way. My little formula certainly worked that
way, but I should have used the phrase "the velocity of the wave relative to
the observer is changed."

No matter how you slice it if you observe a sound frequency while you're
moving, you need to account for that movement when calculating wavelength.
You could use Doppler to try to compute the wavelength at the source first
and use *that* number, but that requires you know the speed of the source.
My method doesn't require it.

So as to avoid arguments of relativity I've been talking about sound waves
here where the frequency change is different depending on whether the source
moves towards the receiver or vice versa.

On Mon, 11 Oct 2004 14:40:17 -0700, Bob Cain
wrote:

Correct, but the effect for a single frequency is very small
in comparison to two or more.

I think this is not an argument in the case of wether there is Doppler
effect present, or not. I think we agreed there is not.

Because of this, pressure wave travell - not
matery, length of cones excursion does not influence the freq (or
wavelength).

That's what I thought too for a long time. I was wrong.

I dissagree. I think you have wrong premisse

Assume that there is an oscilatory motion of particles about
a rest position that is imparted linearly by a driver doing
the same....

I think "doppler in speaker" advocates are analyzing the problem, as if
there would be two sources, one behind another...

An analyis starting from the view of having a high frequency
radiator attached and moving with respect to the a low
frequency radiator of the same size does correctly depict
the situation of the mixed signal being applied to a single
radiator. It's just linear superposition....

O.K. I think you say if cone is to produce 2 waves of different freqs at
same time, and since excursion it makes to produce one is unrelated to
excursion it would make to produce another, that other wave is shifted
in a manner similar to doppler effect.

I say, speaker does not produce two, or more, pressure waves, for each
freq separately, at one time, but rather one result of all the freqs it
has to reproduce at that time. Only because ear and mics work analogue
to that, we are abele to distinguish two or more tones from that
resultant.
Also, for producing certain freq, it doesn't matter how far cone goes,
but how often. If cone make X full cycles in second, freq is always X,
regardles if cone moved 0.1cm, or 1cm. If ther's no shift in freq -
ther's no doppler effect.

I agree there can be some phase shift, due to various reasons, like
inertia, viscosity, displacement....., but none of them produces Doppler
effect.

Vladan wrote:
On Mon, 11 Oct 2004 14:40:17 -0700, Bob Cain
wrote:


Correct, but the effect for a single frequency is very small
in comparison to two or more.


I think this is not an argument in the case of wether there is Doppler
effect present, or not. I think we agreed there is not.

I agree that what's there doesn't follow from a dynamic
Doppler analysis where it is the LF velocity shifting the HF
frequency. That analysis fails because it assumes the LF
motion is not generating any signifigant wave.

There is an effect, however, that is very similar in the
sense of producing IM distortion that can be very signifigant.

Because of this, pressure wave travell - not
matery, length of cones excursion does not influence the freq (or
wavelength).


That's what I thought too for a long time. I was wrong.


I dissagree. I think you have wrong premisse

An analysis that derives the fluid velocity at a point as a
function of the motion of particles normally at rest at that
point gives, to a close aproximation,

Vf(d,t) = Vd(t - (d + Sd(t - d/c))/c)

where Vd(t) and Sd(t) are functions describing the velocity
and position of the driver and d is the distance from the
driver's rest position to the point. This is a plane wave
solution and is a modulation of effective distance between
the rest position of the driver and a point a fixed distance
away from it.

If Vd(t) and Sd(t) are a two frequency mix, the resulting
fluid velocity Vf(t) and the proportional pressure at a
point a fixed distance from the rest position of the driver
will show IM products.

O.K. I think you say if cone is to produce 2 waves of different freqs at
same time, and since excursion it makes to produce one is unrelated to
excursion it would make to produce another, that other wave is shifted
in a manner similar to doppler effect.

No, I was only trying to say that the compound motion of two
drivers receiving separate signals and one driver recieving
a compound signal are the same.


I say, speaker does not produce two, or more, pressure waves, for each
freq separately, at one time, but rather one result of all the freqs it
has to reproduce at that time. Only because ear and mics work analogue
to that, we are abele to distinguish two or more tones from that
resultant.

Yes.

Also, for producing certain freq, it doesn't matter how far cone goes,
but how often. If cone make X full cycles in second, freq is always X,
regardles if cone moved 0.1cm, or 1cm. If ther's no shift in freq -
ther's no doppler effect.

That's what I also thought for too long. The flaw in my
view was that the fluid velocity at a point was equal to the
velocity of the particles normally at rest there about the
point. This produces

Vf(d,t) = Vd(t - d/c)

instead of the more correct form above and does indicate
that there is no distortion at all.

This simplifying approximation is made early on in the
acoustics texts I've seen without an analyisis the
signifigance of that approximation.

I don't know if the effect should be called "Doppler"
distortion but it's real and in many ways similar.


Bob
--

"Things should be described as simply as possible, but no
simpler."

A. Einstein

RP wrote:


Time, when a universal standard is adopted for use by all observers,
whether or not on board clocks perpetually agree with that standard,
must necessarily be universal time. Obvious why, eh?

Nope. That implies a preferred reference frame and there
isn't one.

Thus, show
me in a quantitative fashion that Newtonian Mechanics makes an incorrect
prediction within its domain of applicability, i.e. the kinematics of
neutral masses. If you can do so without utilizing a premise that he
did not actually pen, then good luck. Don't try Fizeau's measurements of
light speed in moving media, because I've embarrassed enough Ph D's
already in that argument by showing them how to correctly compose
"average speeds" with the Galilean transform

I can't. Don't have the Principia at hand. :-)


Bob
--

"Things should be described as simply as possible, but no
simpler."

A. Einstein

"Jim Carr" wrote in message
news:g%Kad.29368$_a3.1912@fed1read05...
"Jim Greenfield" wrote in message
om...

Porky should realise at once that you haven't given him sufficient
information (your speed ref the room). As he doesn't understand the
implications of mistakenly identifying the _cause_ of Doppler, he
might fall for the trap! :-)

Jim G
c'=c+v

Aw, you're no fun!! To recap this entire discussion, it started like this.
Porky described Doppler for sound as a shift in the wavelength whether the
source or receiver is moving (one should describe it using frequency of
the
receiver IMHO). I replied that the wavelength certainly does change when
the
source moves, but it does NOT change when only the receiver moves. He said
I
was wrong because wavelength and frequency are always in a direct
relationship to one another.

He cited a few references that did not explicitly discuss the movement of
the receiver *and* wavelength specifically. I didn't care because it
seemed
readily obvious to my uneducated mind. I decided to find a link that
discusses wavelength: http://physics.nad.ru/Physics/English/dopp_txt.htm .
This site writes, "Next, we shall consider the case when observer moves
and
the source of the wave is still. In this case the wavelength is NOT
changed
and Doppler frequency shift appears because the velocity w of the wave
relatively the observer is changed."

That is certainly a good way of explaining it (which is how you did as
well). I never said it that way. My little formula certainly worked that
way, but I should have used the phrase "the velocity of the wave relative
to
the observer is changed."

No matter how you slice it if you observe a sound frequency while you're
moving, you need to account for that movement when calculating wavelength.
You could use Doppler to try to compute the wavelength at the source first
and use *that* number, but that requires you know the speed of the source.
My method doesn't require it.

So as to avoid arguments of relativity I've been talking about sound waves
here where the frequency change is different depending on whether the
source
moves towards the receiver or vice versa.

Well, I did qualify it by saying for a given value of C, the relationship
of frequency to wavelength is fixed, and that is correct. :-) It would
appear that either I was mistakenly arguing EM Doppler shift where relative
velocity is what matters, or acoustic Doppler shift works a bit differently
in Russia. *LOL* All the references I could find where the source was moving
stated that it was relative velocity that mattered, not which was in motion,
but now it's obvious that they were refering to EM Doppler.
Simply stated, your reference says that, with acoustic Doppler shift,
when the source is moving, the wavelength observed by the receiver changes,
but when the receiver is moving, C, the speed of propagation in the medium,
as it is observed by the receiver changes. Sounds logical to me, and it
certainly isn't the first time I've made a mistake. I suppose that means
that, when both are moving, both wavelength and C change, but all changes
are relative to the receiver's perception. Relative to the source, neither
changes. :-)

Bob Cain wrote:


RP wrote:


Time, when a universal standard is adopted for use by all observers,
whether or not on board clocks perpetually agree with that standard,
must necessarily be universal time. Obvious why, eh?


Nope. That implies a preferred reference frame and there isn't one.

LOL. Well it isn't a mysterious undertaking; the preferred frame would
be exactly the one that we preferred, i.e. upon which we agreed. The
point is, that any clock in equilibrium will do, as these, regardless of
actual ticking rate, will tick at proportional rates to each other.
Certainly you can find no preferred frame, in the sense that you
implied, if all observers were to use the Earth/Sun orbital system as
reference clock. The ambiguities that you have in mind are relevant
only in the special relativistic view of simultaneity.

It isn't difficult to see that even within the Galilean system, that the
redefinition of time to: What an observers clock reads, also leads to a
transform other then the Galilean transform, even though it will
necessarily derive the same outcomes, that is, since we've not altered
the physics of space time with our redefinition of measures.


Thus, show
me in a quantitative fashion that Newtonian Mechanics makes an
incorrect prediction within its domain of applicability, i.e. the
kinematics of neutral masses. If you can do so without utilizing a
premise that he did not actually pen, then good luck. Don't try
Fizeau's measurements of light speed in moving media, because I've
embarrassed enough Ph D's already in that argument by showing them how
to correctly compose "average speeds" with the Galilean transform


I can't. Don't have the Principia at hand. :-)

This will do:

F = dp/dt
E = 0.5mv^2
Galilean transform
Energy is conserved.
Angular momentum is conserved.

That's about all there is to Newton, barring his gravitation equation,
which one could argue is simply incomplete.
Even so, within experimental error, his equation hasn't been disproved.
Any deviations can be corrected for within the context of the premises
listed above, or by the inclusion of extraneous electromagnetic forces,
which latter are not necessarily contradictory to Newton, as I've proved
also by formulating just such an electromagnetic synthesis.

This argument has a high prospect of degrading into philosophical
banter, and with good reason, it began that way
I'll leave it with this: They are disjoint logical systems, bottom line.

Richard Perry
http://www.cswnet.com/~rper/Electromagnetism.html



Bob

"Porky" wrote in message
.. .

Simply stated, your reference says that, with acoustic Doppler shift,
when the source is moving, the wavelength observed by the receiver
changes,

Not quite. I think I see where your practical experience, which I lack, is
throwing you off. You think of wavelength solely as W=C/F. In practical
terms such as figuring out horn length, loudspeaker dimensions, room
acoustics, and so forth, that works fine because everything is stationary.
In effect you're saying that wavelength is the distance the second wave
travels after the receiver encounters the first wave.

However, wavelength by definition is the distance between the crest or peak
of one wave and the peak/crest of the next wave. Wavelength is a distance.
We can stop time and motion and figure out the wavelength with a ruler.
Distance is time independent (Los Angeles and New York are 3,000 miles apart
last night, tonight and tomorrow). W=C/F is a quick way of calculating
wavelength if everything is stationary. But it's not the *definition*.
Again, since the above practical examples require things to be stationary,
it's easy to fall into the trap of thinking of it that way.

When the source emits a wave, moves, then emits the second wave, the
distance is different than if the source had just stood still. If the
receiver is still, it can use W=C/F to figure out the wavelength. If the
receiver is moving, W=C/F has to include an adjustment to C to calculate the
distance between the crest/peak of one wave and the crest/peak of the next
wave. Simply stated it's the distance the second wave traveled plus or minus
the distance the receiver traveled as opposed to just the distance traveled
by itself.

Saying the apparent or "observed" wavelength changes when the source moves
is incorrect. It really is different. When the receiver moves you might say
that the "observed" wavelength is different, but that again implies that
wavelength is the distance the second wave travels after the receiver
encounters the first wave, which we know it's not. We can't use W=C/F if the
receiver is moving - we use W=(C+/-V)/F where V is the velocity of the
receiver relative to the source.

"RP" wrote in message
...

LOL. Well it isn't a mysterious undertaking; the preferred frame would
be exactly the one that we preferred, i.e. upon which we agreed. The
point is, that any clock in equilibrium will do, as these, regardless of
actual ticking rate, will tick at proportional rates to each other.
Certainly you can find no preferred frame, in the sense that you
implied, if all observers were to use the Earth/Sun orbital system as
reference clock. The ambiguities that you have in mind are relevant
only in the special relativistic view of simultaneity.

Is that why somebody can post a message in this newsgroup from New York at
midnight and I see it here in Phoenix at 9:00 pm??

"Jim Carr" wrote in message news:g%Kad.29368$_a3.1912@fed1read05...
"Jim Greenfield" wrote in message
om...

Porky should realise at once that you haven't given him sufficient
information (your speed ref the room). As he doesn't understand the
implications of mistakenly identifying the _cause_ of Doppler, he
might fall for the trap! :-)

Jim G
c'=c+v

Aw, you're no fun!! To recap this entire discussion, it started like this.
Porky described Doppler for sound as a shift in the wavelength whether the
source or receiver is moving (one should describe it using frequency of the
receiver IMHO). I replied that the wavelength certainly does change when the
source moves, but it does NOT change when only the receiver moves.

It seems to me my bike sounds the same at 5000rpm in the shed, or when
I am on it at 120 kliks! Though to someone I am approaching, they
would think I am pulling (roughly) 6000rpm

He said I
was wrong because wavelength and frequency are always in a direct
relationship to one another.

Nope. Frequency is "actions at a POINT per time". I despair at the
inability of people to grasp this seemingly simple concept. If the
point is moving, things alter (relative speed/frequency/sine waves
passed per second)

He cited a few references that did not explicitly discuss the movement of
the receiver *and* wavelength specifically. I didn't care because it seemed
readily obvious to my uneducated mind. I decided to find a link that
discusses wavelength: http://physics.nad.ru/Physics/English/dopp_txt.htm .
This site writes, "Next, we shall consider the case when observer moves and
the source of the wave is still. In this case the wavelength is NOT changed
and Doppler frequency shift appears because the velocity w of the wave
relatively the observer is changed."

That is certainly a good way of explaining it (which is how you did as
well). I never said it that way. My little formula certainly worked that
way, but I should have used the phrase "the velocity of the wave relative to
the observer is changed."

No matter how you slice it if you observe a sound frequency while you're
moving, you need to account for that movement when calculating wavelength.
You could use Doppler to try to compute the wavelength at the source first
and use *that* number, but that requires you know the speed of the source.
My method doesn't require it.

So as to avoid arguments of relativity I've been talking about sound waves
here where the frequency change is different depending on whether the source
moves towards the receiver or vice versa.

Drat! I was coming to the Rel bit. It always comes up in discussion of
Doppler, because DHR's like to compare sound waves with light. I don't
think there is any correllation between light (photons through a
vacuum) and sound (pressure fronts through a medium, but there you go!
YOU realise that those waves in the hall aren't actually changing
their wavelength when you move towards them, but try explaining to a
DHR that a stream of photons will have a changed 'frequency' of
hitting a receiver, WITHOUT having an altered wavelength (distance
between them), if the source or observer are moving rel each other.

Sorry to spoil your fun :-)
There are plenty more who think that Doppler is a cause, instead of an
effect.

Jim G
c'=c+v

"Jim Greenfield" wrote in message
om...

Drat! I was coming to the Rel bit. It always comes up in discussion of
Doppler, because DHR's like to compare sound waves with light. I don't
think there is any correllation between light (photons through a
vacuum) and sound (pressure fronts through a medium, but there you go!
YOU realise that those waves in the hall aren't actually changing
their wavelength when you move towards them, but try explaining to a
DHR that a stream of photons will have a changed 'frequency' of
hitting a receiver, WITHOUT having an altered wavelength (distance
between them), if the source or observer are moving rel each other.

In case you missed it before, I am entirely untrained in this field. It's
not even a hobby. I play and record music as a hobby, but it doesn't require
knowing any of this stuff. With that said...

Looking at a formula I found for determing Doppler in electromagnetic waves
I see it is different than for sound. Looking at *that* formula I would have
to say that it doesn't matter if the source or observer is moving, you'll
see the same apparent shift in frequency. With Doppler in sound, if the
source is moving at X you get a different frequency shift than if you do if
the observer is moving at X (assuming a head-on movement).

With electromagnetic waves I would hazard to guess that the "actual"
wavelength changes in either case (source or observer moving) since the
speed of light is my speed limit. Or maybe I'm falling into a layman's trap.
If so, I'm sure someone will correct me.

Sorry to spoil your fun :-)
There are plenty more who think that Doppler is a cause, instead of an
effect.

Very well put.

"Jim Carr" wrote in message
news:Fh2bd.29584$_a3.17034@fed1read05...
"Porky" wrote in message
.. .

Simply stated, your reference says that, with acoustic Doppler shift,
when the source is moving, the wavelength observed by the receiver
changes,

Not quite. I think I see where your practical experience, which I lack, is
throwing you off. You think of wavelength solely as W=C/F. In practical
terms such as figuring out horn length, loudspeaker dimensions, room
acoustics, and so forth, that works fine because everything is stationary.
In effect you're saying that wavelength is the distance the second wave
travels after the receiver encounters the first wave.

Actually, I was referring to the apparent wavelength that the receiver
"sees", the actual wavelength does not change, but the wavelength as it
appears to the receiver isn't the same as the actual wavelength.

However, wavelength by definition is the distance between the crest or
peak
of one wave and the peak/crest of the next wave. Wavelength is a distance.
We can stop time and motion and figure out the wavelength with a ruler.
Distance is time independent (Los Angeles and New York are 3,000 miles
apart
last night, tonight and tomorrow). W=C/F is a quick way of calculating
wavelength if everything is stationary. But it's not the *definition*.
Again, since the above practical examples require things to be stationary,
it's easy to fall into the trap of thinking of it that way.

When the source emits a wave, moves, then emits the second wave, the
distance is different than if the source had just stood still. If the
receiver is still, it can use W=C/F to figure out the wavelength. If the
receiver is moving, W=C/F has to include an adjustment to C to calculate
the
distance between the crest/peak of one wave and the crest/peak of the next
wave. Simply stated it's the distance the second wave traveled plus or
minus
the distance the receiver traveled as opposed to just the distance
traveled
by itself.

But as it appears to the receiver, the apparent wavelength will be longer or
shorter, depending on direction of motion. This doesn't change the actual
wavelength of course.

Saying the apparent or "observed" wavelength changes when the source moves
is incorrect. It really is different. When the receiver moves you might
say
that the "observed" wavelength is different, but that again implies that
wavelength is the distance the second wave travels after the receiver
encounters the first wave, which we know it's not. We can't use W=C/F if
the
receiver is moving - we use W=(C+/-V)/F where V is the velocity of the
receiver relative to the source.

According to the link you posted, if the source is moving, the apparent
wavelength changes, but if the receiver is moving C, the apparent speed of
sound changes. Therefore, F=C/W always applies in the general sense, the
difference is that if the source is moving, W is the variable, and if the
receiver is moving, C is the variable. Your equation above is correct and it
calculates the change in apparent C if the receiver is moving. Correct me if
I'm wrong, but won't
Frc = Ftx ((C + V)/C) work for motion of either Tx or Rc? V, of course, will
be a positive number if the motion is toward the other and a negative number
if the motion is away from the other. The trick is that whether W or C is
the variable depends on which is moving, the source or the receiver. This
doesn't take the variation of velocity/time into account, but for the home
studioist, that can safely be ignored.

"Jim Carr" wrote in message
news:rZ3bd.29612$_a3.12325@fed1read05...
"Jim Greenfield" wrote in message
om...

Drat! I was coming to the Rel bit. It always comes up in discussion of
Doppler, because DHR's like to compare sound waves with light. I don't
think there is any correllation between light (photons through a
vacuum) and sound (pressure fronts through a medium, but there you go!
YOU realise that those waves in the hall aren't actually changing
their wavelength when you move towards them, but try explaining to a
DHR that a stream of photons will have a changed 'frequency' of
hitting a receiver, WITHOUT having an altered wavelength (distance
between them), if the source or observer are moving rel each other.

In case you missed it before, I am entirely untrained in this field. It's
not even a hobby. I play and record music as a hobby, but it doesn't
require
knowing any of this stuff. With that said...

Looking at a formula I found for determing Doppler in electromagnetic
waves
I see it is different than for sound. Looking at *that* formula I would
have
to say that it doesn't matter if the source or observer is moving, you'll
see the same apparent shift in frequency. With Doppler in sound, if the
source is moving at X you get a different frequency shift than if you do
if
the observer is moving at X (assuming a head-on movement).

With electromagnetic waves I would hazard to guess that the "actual"
wavelength changes in either case (source or observer moving) since the
speed of light is my speed limit. Or maybe I'm falling into a layman's
trap.
If so, I'm sure someone will correct me.

When dealing with relativistic speeds you not only have C, W and F to
contend with, the time contraction factor Tau comes into play and at any
velocity which is a respectable percentage of the speed of light, Tau must
be taken into account to get accurate results. For the home recordist who
just wants to figure acoustic Doppler shift, Tau is insignificant.

"Jim Carr" wrote in message news:rZ3bd.29612$_a3.12325@fed1read05...
"Jim Greenfield" wrote in message
om...

Drat! I was coming to the Rel bit. It always comes up in discussion of
Doppler, because DHR's like to compare sound waves with light. I don't
think there is any correllation between light (photons through a
vacuum) and sound (pressure fronts through a medium, but there you go!
YOU realise that those waves in the hall aren't actually changing
their wavelength when you move towards them, but try explaining to a
DHR that a stream of photons will have a changed 'frequency' of
hitting a receiver, WITHOUT having an altered wavelength (distance
between them), if the source or observer are moving rel each other.

In case you missed it before, I am entirely untrained in this field. It's
not even a hobby. I play and record music as a hobby, but it doesn't require
knowing any of this stuff. With that said...

Me too! until I had my left fingers chopped!!!!!!!!!!!!!
With that said.........

Looking at a formula I found for determing Doppler in electromagnetic waves
I see it is different than for sound. Looking at *that* formula I would have
to say that it doesn't matter if the source or observer is moving, you'll
see the same apparent shift in frequency. With Doppler in sound, if the
source is moving at X you get a different frequency shift than if you do if
the observer is moving at X (assuming a head-on movement).

Are you sure?
If I am on my bike at 5000rpm, and my mate behind is pulling the same,
it sounds just the same as if we are both stationary; his motion
(source) has not effected the frequency to me, but we will both sound
to be pulling 6000 to someone standing by the road ahead.
You must measure the wavelength from where it is produced relative to
the medium, and not to the source, in the case of sound in air. If the
source were moving at 1/2 speed of sound (same direction), you should
still measure the wavelength from where it was produced (coordinate in
the air), to where it is after one cycle. This is confusing for
longituudinal waves, but just think in terms of cycles (points on a
sine)
So I don't think that wavelength alters in either case relative to the
medium against which I am hearing(measuring)

With electromagnetic waves I would hazard to guess that the "actual"
wavelength changes in either case (source or observer moving) since the
speed of light is my speed limit. Or maybe I'm falling into a layman's trap.
If so, I'm sure someone will correct me.

Sorry to spoil your fun :-)
There are plenty more who think that Doppler is a cause, instead of an
effect.

Very well put.

Thanks

Jim G
c'=c+v

"Porky" wrote in message ...
"Jim Carr" wrote in message
news:rZ3bd.29612$_a3.12325@fed1read05...
"Jim Greenfield" wrote in message
om...

Drat! I was coming to the Rel bit. It always comes up in discussion of
Doppler, because DHR's like to compare sound waves with light. I don't
think there is any correllation between light (photons through a
vacuum) and sound (pressure fronts through a medium, but there you go!
YOU realise that those waves in the hall aren't actually changing
their wavelength when you move towards them, but try explaining to a
DHR that a stream of photons will have a changed 'frequency' of
hitting a receiver, WITHOUT having an altered wavelength (distance
between them), if the source or observer are moving rel each other.

In case you missed it before, I am entirely untrained in this field. It's
not even a hobby. I play and record music as a hobby, but it doesn't
require
knowing any of this stuff. With that said...

Looking at a formula I found for determing Doppler in electromagnetic
waves
I see it is different than for sound. Looking at *that* formula I would
have
to say that it doesn't matter if the source or observer is moving, you'll
see the same apparent shift in frequency. With Doppler in sound, if the
source is moving at X you get a different frequency shift than if you do
if
the observer is moving at X (assuming a head-on movement).

With electromagnetic waves I would hazard to guess that the "actual"
wavelength changes in either case (source or observer moving) since the
speed of light is my speed limit. Or maybe I'm falling into a layman's
trap.
If so, I'm sure someone will correct me.

When dealing with relativistic speeds you not only have C, W and F to
contend with, the time contraction factor Tau comes into play and at any
velocity which is a respectable percentage of the speed of light, Tau must
be taken into account to get accurate results. For the home recordist who
just wants to figure acoustic Doppler shift, Tau is insignificant.

Jim Carr:
As I intimated, Porky, as a believing DHR, has stepped up to the plate
equipped with the standard SR opening play.
Just because you are a layman (as am I), don't be coerced into
accepting something (Relativity), which is patently impossible/wrong,
just because it is in fashion (they can get their stuff "peer
reviewed" by OTHER true believers).

As you realise, there are three things involved; velocity, wavelength,
frequency.
Notice immediately Relativity is introduced, Porky introduces a
fourth! Tau
What Porky fails to mention, is that this is the result of circular
logic which runs like this:
c (velocity of light in vacuum) is a constant, independent of the
velocity of the source of the light.
Due to this magic, length contracts as c is approached
time dilates (changes) as c is approached
Because time and length both change, c remains the same!

Do you think at school you could get away with that?
Given a simple 3 part algebra, with 2 unknowns, provide an accepted
answer by throwing in an extra (arbitrary and imaginary) factor??????

Relativity will go down in history as the greatest con of the 20th
century.
(It surely wont see this one out)

Jim G
c'=c+v