| Topic: |
Science > Physics |
| User: |
"Old Man" |
| Date: |
11 Apr 2004 12:04:06 AM |
| Object: |
Transmitting Sound Through a Vacuum at c/3 |
Imagine a long evacuated tube. The interior hollow of the
tube is free of any gas, but is filled with thermal black body
radiation in equilibrium with the interior wall of the tube
which is maintained at some finite temperature above 0K.
Each end of the tube is capped with a thin light reflecting
membrane which is strong enough to withstand external
atmospheric pressure.
We can cause one of the membranes to vibrate in response
to external atmospheric sound waves. On the evacuated side,
the membrane's longitudinal vibrations create photon density
and pressure fluctuations which travel along the tube axis,
without loss, and at a third of the speed of light, to the
membrane at the other end of the tube,, causing it to move
in true fidelity with the arriving photon pressure fluctuations.
It's simultaneously bi-directional too. Old Man calls it his
"light-speed vacuum-conducting sound-conductor". Probably
should copyright that. Gadget names are everything.
[Old Man]
.
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| User: "Franz Heymann" |
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| Title: Re: Transmitting Sound Through a Vacuum at c/3 |
13 Apr 2004 03:03:36 AM |
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"Old Man" <nomail@nomail.net> wrote in message
news:RpGdnY9dXqJMXuXdRVn-sA@prairiewave.com...
Imagine a long evacuated tube. The interior hollow of the
tube is free of any gas, but is filled with thermal black body
radiation in equilibrium with the interior wall of the tube
which is maintained at some finite temperature above 0K.
Each end of the tube is capped with a thin light reflecting
membrane which is strong enough to withstand external
atmospheric pressure.
We can cause one of the membranes to vibrate in response
to external atmospheric sound waves. On the evacuated side,
the membrane's longitudinal vibrations create photon density
and pressure fluctuations which travel along the tube axis,
without loss,
So there is no attenuation?
Why, then, does a doorbell hanging inside a bell jar become inaudible
when the jar is evacuated?
[snip]
Franz
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| User: "Old Man" |
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| Title: Re: Transmitting Sound Through a Vacuum at c/3 |
13 Apr 2004 11:24:25 PM |
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"Franz Heymann" <notfranz.heymann@btopenworld.com> wrote in message
news:c5g6sn$laq$2@titan.btinternet.com...
"Old Man" <nomail@nomail.net> wrote in message
news:RpGdnY9dXqJMXuXdRVn-sA@prairiewave.com...
Imagine a long evacuated tube. The interior hollow of the
tube is free of any gas, but is filled with thermal black body
radiation in equilibrium with the interior wall of the tube
which is maintained at some finite temperature above 0K.
Each end of the tube is capped with a thin light reflecting
membrane which is strong enough to withstand external
atmospheric pressure.
We can cause one of the membranes to vibrate in response
to external atmospheric sound waves. On the evacuated side,
the membrane's longitudinal vibrations create photon density
and pressure fluctuations which travel along the tube axis,
without loss,
So there is no attenuation?
Why, then, does a doorbell hanging inside a bell jar become inaudible
when the jar is evacuated?
[snip]
Franz
Old Man can see no reason for the sound wave in the gas to be
attenuated, but the transfer of energy from membrane to photon
gas and from photon gas to membrane is very small.
[Old Man]
.
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| User: "TimR" |
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| Title: Re: Transmitting Sound Through a Vacuum at c/3 |
15 Apr 2004 09:38:51 AM |
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I apologize for meandering slightly from the direction this is going
but I'm really interested in sound propagation in the tube of a
musical instrument.
In a musical instrument a sound wave (I assume slightly spherical)
propagates down the the tube and partially or completely reflects at
the end. Length of the tube is chosen so that the reflection arrives
at the starting point at the correct point in time to do something
useful. (like resonate, etc.)
I had always assumed that mean free path was short relative to the
tube length, and the pressure pulse propagated by means of multiple
collisions. Is that in fact not true? Your discussion would seem to
suggest propagation without collisions being important, and that makes
it a little unclear why the impedance mismatch at the end causes a
reflection.
Sorry if this is a long way from what you are talking about. I have
debated points close to this with fellow trombone players and not
achieved consensus. A trombone not extended is 9 feet long. Mean free
path must be on the order of micrometers? nanometers? yet you say it
is unimportant? I do not understand.
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| User: "Old Man" |
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| Title: Re: Transmitting Sound Through a Vacuum at c/3 |
15 Apr 2004 06:08:01 PM |
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"TimR" <timothy42b@aol.com> wrote in message
news:87af0be7.0404150638.412feac9@posting.google.com...
I apologize for meandering slightly from the direction this is going
but I'm really interested in sound propagation in the tube of a
musical instrument.
In a musical instrument a sound wave (I assume slightly spherical)
propagates down the the tube and partially or completely reflects at
the end. Length of the tube is chosen so that the reflection arrives
at the starting point at the correct point in time to do something
useful. (like resonate, etc.)
I had always assumed that mean free path was short relative to the
tube length, and the pressure pulse propagated by means of multiple
collisions. Is that in fact not true? Your discussion would seem to
suggest propagation without collisions being important, and that makes
it a little unclear why the impedance mismatch at the end causes a
reflection.
Sorry if this is a long way from what you are talking about. I have
debated points close to this with fellow trombone players and not
achieved consensus. A trombone not extended is 9 feet long. Mean free
path must be on the order of micrometers? nanometers? yet you say it
is unimportant? I do not understand.
Mean free path is not decisive for the ideal gas law, PV = NRT.
The V specifies a volume in which no collisions are observed.
All inter-gas collisions occur outside of that volume. Alternatvely,
the boundry for V can be taken as the walls of a container
whereat gas-wall interactions occur.
Real gases do things that aren't predicted by the ideal gas law.
A mean free path of less than the dimensions of V doesn't
yield PV = NRT. Sound propagation is supported by an ideal
gas, but some aspects of sound propagation that are observed
in real gases aren't predicted by the ideal gas law.
[Old Man]
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| User: "Richard Herring" |
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| Title: Re: Transmitting Sound Through a Vacuum at c/3 |
16 Apr 2004 03:33:43 AM |
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In message <lKSdnZJvL7JIiOLdRVn_iw@prairiewave.com>, Old Man
<nomail@nomail.net> writes
"TimR" <timothy42b@aol.com> wrote in message
news:87af0be7.0404150638.412feac9@posting.google.com...
I apologize for meandering slightly from the direction this is going
but I'm really interested in sound propagation in the tube of a
musical instrument.
In a musical instrument a sound wave (I assume slightly spherical)
propagates down the the tube and partially or completely reflects at
the end. Length of the tube is chosen so that the reflection arrives
at the starting point at the correct point in time to do something
useful. (like resonate, etc.)
I had always assumed that mean free path was short relative to the
tube length, and the pressure pulse propagated by means of multiple
collisions. Is that in fact not true? Your discussion would seem to
suggest propagation without collisions being important, and that makes
it a little unclear why the impedance mismatch at the end causes a
reflection.
Sorry if this is a long way from what you are talking about. I have
debated points close to this with fellow trombone players and not
achieved consensus. A trombone not extended is 9 feet long. Mean free
path must be on the order of micrometers? nanometers? yet you say it
is unimportant? I do not understand.
Mean free path is not decisive for the ideal gas law, PV = NRT.
Assumptions about it are hidden deep in the derivation of the law.
The V specifies a volume in which no collisions are observed.
"No collisions" would mean no equipartition, no equilibrium, and P and T
would be meaningless.
All inter-gas collisions occur outside of that volume. Alternatvely,
the boundry for V can be taken as the walls of a container
whereat gas-wall interactions occur.
Real gases do things that aren't predicted by the ideal gas law.
A mean free path of less than the dimensions of V doesn't
yield PV = NRT. Sound propagation is supported by an ideal
gas,
Not by your definition of it. Consider a tube filled with your
ultra-ideal gas of totally non-interacting particles, with a membrane at
each end. Apply an impulse to one membrane. It imparts excess momentum
to the nearby gas particles. When they reach the other membrane, they
give up that momentum to it, as one would expect. But *when* do they
reach the far end? Their velocities are all different, so the impulse
will be smeared out into a Maxwell-Boltzmann shaped mess. If that's
sound propagation, it's not as we know it.
(It's not a problem with the "photon gas" because the photons all travel
at the same speed.)
but some aspects of sound propagation that are observed
in real gases aren't predicted by the ideal gas law.
--
Richard Herring
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| User: "Richard Herring" |
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| Title: Re: Transmitting Sound Through a Vacuum at c/3 |
15 Apr 2004 10:28:46 AM |
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In message <87af0be7.0404150638.412feac9@posting.google.com>, TimR
<timothy42b@aol.com> writes
I apologize for meandering slightly from the direction this is going
but I'm really interested in sound propagation in the tube of a
musical instrument.
In a musical instrument a sound wave (I assume slightly spherical)
propagates down the the tube and partially or completely reflects at
the end. Length of the tube is chosen so that the reflection arrives
at the starting point at the correct point in time to do something
useful. (like resonate, etc.)
I had always assumed that mean free path was short relative to the
tube length, and the pressure pulse propagated by means of multiple
collisions. Is that in fact not true?
It is true. Mean free path in air at standard pressure is of the order
of 10^{-7} metres.
Your discussion would seem to
suggest propagation without collisions being important,
Not at all. Momentum transfer is essential to the propagation of normal
non-dispersive pressure waves.
and that makes
it a little unclear why the impedance mismatch at the end causes a
reflection.
Sorry if this is a long way from what you are talking about. I have
debated points close to this with fellow trombone players and not
achieved consensus. A trombone not extended is 9 feet long. Mean free
path must be on the order of micrometers? nanometers?
Something between the two, yes.
yet you say it
is unimportant? I do not understand.
I think _your_ understanding is just fine.
--
Richard Herring
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| User: "Franz Heymann" |
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| Title: Re: Transmitting Sound Through a Vacuum at c/3 |
11 Apr 2004 05:28:23 AM |
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"Old Man" <nomail@nomail.net> wrote in message
news:RpGdnY9dXqJMXuXdRVn-sA@prairiewave.com...
Imagine a long evacuated tube. The interior hollow of the
tube is free of any gas, but is filled with thermal black body
radiation in equilibrium with the interior wall of the tube
which is maintained at some finite temperature above 0K.
Each end of the tube is capped with a thin light reflecting
membrane which is strong enough to withstand external
atmospheric pressure.
We can cause one of the membranes to vibrate in response
to external atmospheric sound waves. On the evacuated side,
the membrane's longitudinal vibrations create photon density
and pressure fluctuations which travel along the tube axis,
without loss, and at a third of the speed of light, to the
membrane at the other end of the tube,, causing it to move
in true fidelity with the arriving photon pressure fluctuations.
It's simultaneously bi-directional too. Old Man calls it his
"light-speed vacuum-conducting sound-conductor". Probably
should copyright that. Gadget names are everything.
Old Man has forgotten to calculate the speed of sound in photons.
Franz
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| User: "Old Man" |
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| Title: Re: Transmitting Sound Through a Vacuum at c/3 |
11 Apr 2004 10:06:53 PM |
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"Franz Heymann" <notfranz.heymann@btopenworld.com> wrote in message
news:c5b6k7$2cm$1@sparta.btinternet.com...
"Old Man" <nomail@nomail.net> wrote in message
news:RpGdnY9dXqJMXuXdRVn-sA@prairiewave.com...
Imagine a long evacuated tube. The interior hollow of the
tube is free of any gas, but is filled with thermal black body
radiation in equilibrium with the interior wall of the tube
which is maintained at some finite temperature above 0K.
Each end of the tube is capped with a thin light reflecting
membrane which is strong enough to withstand external
atmospheric pressure.
We can cause one of the membranes to vibrate in response
to external atmospheric sound waves. On the evacuated side,
the membrane's longitudinal vibrations create photon density
and pressure fluctuations which travel along the tube axis,
without loss, and at a third of the speed of light, to the
membrane at the other end of the tube,, causing it to move
in true fidelity with the arriving photon pressure fluctuations.
It's simultaneously bi-directional too. Old Man calls it his
"light-speed vacuum-conducting sound-conductor". Probably
should copyright that. Gadget names are everything.
Old Man has forgotten to calculate the speed of sound in photons.
Franz
Franz is a purist. Old Man has faith.
As stated, the speed of sound in a gas of photons is c / 3. Old
Man doesn't have to calculate it. Seen it derived before, but got
it most recently from a book by Barbara Ryden in "Introduction
to Cosmology", wherein she references it to someone else. Old
Man believes her because he wants to believe it. He who has little
faith, doesn't Franz?
Besides, if I gave credit to Landau and Lifshitz, I'd probably be
right. They've derived practically everything that's anything!
[Old Man]
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| User: "Edward Green" |
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| Title: Re: Transmitting Sound Through a Vacuum at c/3 |
11 Apr 2004 06:45:09 PM |
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"Franz Heymann" <notfranz.heymann@btopenworld.com> wrote in message news:<c5b6k7$2cm$1@sparta.btinternet.com>...
"Old Man" <nomail@nomail.net> wrote in message
news:RpGdnY9dXqJMXuXdRVn-sA@prairiewave.com...
Imagine a long evacuated tube. The interior hollow of the
tube is free of any gas, but is filled with thermal black body
radiation in equilibrium with the interior wall of the tube
which is maintained at some finite temperature above 0K.
Each end of the tube is capped with a thin light reflecting
membrane which is strong enough to withstand external
atmospheric pressure.
We can cause one of the membranes to vibrate in response
to external atmospheric sound waves. On the evacuated side,
the membrane's longitudinal vibrations create photon density
and pressure fluctuations which travel along the tube axis,
without loss, and at a third of the speed of light, to the
membrane at the other end of the tube,, causing it to move
in true fidelity with the arriving photon pressure fluctuations.
It's simultaneously bi-directional too. Old Man calls it his
"light-speed vacuum-conducting sound-conductor". Probably
should copyright that. Gadget names are everything.
Copyright, schmopyright: patent it, quick!
Old Man has forgotten to calculate the speed of sound in photons.
Ed has great confidence in Old Man, and is liable to invest screwball
schemes of his with high Bayesian confidence levels which from other
sources would have vanishing probabilities. Therefore Ed assumes Old
Man has carried out the indicate calculation, and concluded that the
speed of transverse pressure fluctuations in a themral photon bath
confined to a tube is c/3.
Ed thinks this is a very cool idea, since it presents a plausible
mechanism for a thing plausibly called "sound" in vacuum, which, as
everybody knows, is impossible. Though I suppose some wet rag might
insist that since photons are involved, this amounts to a speed of
light in a waveguide.
OTOH, wait just a gosh darn minute here: propagation of pressure
fluctuations implies something like photon/photon collisions, does it
not? But everybody knows photons don't collide. What Old Man
proposes is beginning to look very much like a waveguide mode which we
have probably known and loved since childhood, but always called
"light" before this.
Is Old Man possibly trying to pull the wool over our psi's?
BTW: isn't Old Man's scheme a version of an existing known method of
coupling light to sound, using lasers and mirrors?
.
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| User: "Old Man" |
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| Title: Re: Transmitting Sound Through a Vacuum at c/3 |
11 Apr 2004 10:29:46 PM |
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"Edward Green" <spamspamspam3@netzero.com> wrote in message
news:eca320d0.0404111545.419586ab@posting.google.com...
"Franz Heymann" <notfranz.heymann@btopenworld.com> wrote in message
news:<c5b6k7$2cm$1@sparta.btinternet.com>...
"Old Man" <nomail@nomail.net> wrote in message
news:RpGdnY9dXqJMXuXdRVn-sA@prairiewave.com...
Imagine a long evacuated tube. The interior hollow of the
tube is free of any gas, but is filled with thermal black body
radiation in equilibrium with the interior wall of the tube
which is maintained at some finite temperature above 0K.
Each end of the tube is capped with a thin light reflecting
membrane which is strong enough to withstand external
atmospheric pressure.
We can cause one of the membranes to vibrate in response
to external atmospheric sound waves. On the evacuated side,
the membrane's longitudinal vibrations create photon density
and pressure fluctuations which travel along the tube axis,
without loss, and at a third of the speed of light, to the
membrane at the other end of the tube,, causing it to move
in true fidelity with the arriving photon pressure fluctuations.
It's simultaneously bi-directional too. Old Man calls it his
"light-speed vacuum-conducting sound-conductor". Probably
should copyright that. Gadget names are everything.
Copyright, schmopyright: patent it, quick!
Old Man has forgotten to calculate the speed of sound in photons.
Ed has great confidence in Old Man, and is liable to invest screwball
schemes of his with high Bayesian confidence levels which from other
sources would have vanishing probabilities. Therefore Ed assumes Old
Man has carried out the indicate calculation, and concluded that the
speed of transverse pressure fluctuations in a themral photon bath
confined to a tube is c/3.
Ed thinks this is a very cool idea, since it presents a plausible
mechanism for a thing plausibly called "sound" in vacuum, which, as
everybody knows, is impossible. Though I suppose some wet rag might
insist that since photons are involved, this amounts to a speed of
light in a waveguide.
OTOH, wait just a gosh darn minute here: propagation of pressure
fluctuations implies something like photon/photon collisions, does it
not? But everybody knows photons don't collide. What Old Man
proposes is beginning to look very much like a waveguide mode which we
have probably known and loved since childhood, but always called
"light" before this.
Is Old Man possibly trying to pull the wool over our psi's?
BTW: isn't Old Man's scheme a version of an existing known method of
coupling light to sound, using lasers and mirrors?
Nope, but Old Man thinks that sound waves in a gas of photons
are longitudinal, not transverse. As Ed said, they travel at c / 3.
Maybe Franz assumed that Old Man was taking credit for an
original derivation. Old Man is a parrot, not a thief. Polly says
that the speed of sound for dark energy is - c (negative). Ain't
that Whacko? Where's my cracker? [Old Man]
.
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| User: "Edward Green" |
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| Title: Re: Transmitting Sound Through a Vacuum at c/3 |
12 Apr 2004 07:37:26 PM |
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"Old Man" <nomail@nomail.net> wrote in message news:<erednZg4XN2rYuTdRVn-jA@prairiewave.com>...
"Edward Green" <spamspamspam3@netzero.com> wrote in message
news:eca320d0.0404111545.419586ab@posting.google.com...
"Old Man" <nomail@nomail.net> wrote in message
news:RpGdnY9dXqJMXuXdRVn-sA@prairiewave.com...
Imagine a long evacuated tube. The interior hollow of the
tube is free of any gas, but is filled with thermal black body
radiation in equilibrium with the interior wall of the tube
which is maintained at some finite temperature above 0K.
Each end of the tube is capped with a thin light reflecting
membrane which is strong enough to withstand external
atmospheric pressure.
We can cause one of the membranes to vibrate in response
to external atmospheric sound waves. On the evacuated side,
the membrane's longitudinal vibrations create photon density
and pressure fluctuations which travel along the tube axis,
without loss, and at a third of the speed of light, to the
membrane at the other end of the tube,, causing it to move
in true fidelity with the arriving photon pressure fluctuations.
... isn't Old Man's scheme a version of an existing known method of
coupling light to sound, using lasers and mirrors?
Nope, but Old Man thinks that sound waves in a gas of photons
are longitudinal, not transverse.
Of course they are! Couldn't you tell from my post that that was the
kind of "transverse" which means "longitudinal"? :-)
Note you did side step my question about how it is possible to have
pressure waves in a medium which has no internal collisions. I'm not
saying it is impossible, just eyebrow raising -- you might consider it
a density wave.
Perhaps we (implying for a heady moment that I am on Old Man's level,
which fantasy he is sometimes want to allow me) should consider a
simpler problem first (of course). Forget the walls of the tube:
consider a space containing a spherical black body of finite radius at
the origin. Let the black body hum. Call the body Aretha.
How do we describe the resulting spherically symmetric radiation
field? What is the boundary condition at the surface of the black
body?
Presumably the moving surface of the body continues to emit black body
radiation characteristic of its temperature, but doppler shifted. So
what we see at a distant detector is a time varying spectrum. Is this
the same thing as a pressure wave in the light bath, or is this a
different effect?
Maybe we should forget the spectrum. Maybe we should concentrate on
conservation of energy. A source moving towards us will correspond to
an uptick in power at the detector, the same source receding to a
downtick: this much simply from considering the time the source
transmits corresponding to one second's reception at the sink. I
_think_ this is a seperate effect from the doppler shift, but it
points in the same direction.
Ok... there are your pressure fluctuations ... power fluctuations are
pressure fluctuations in light. It looks to me like in this case the
power fluctuations should transmit at c?
As Ed said, they travel at c / 3.
So what is the effect of the walls...
Maybe Franz assumed that Old Man was taking credit for an
original derivation. Old Man is a parrot, not a thief. Polly says
that the speed of sound for dark energy is - c (negative).
Oh ... let's work out light first. Then we can worry about this
academic frippery ;-) This second result does sound like a result for
free space, though, and not a tube; doesn't it?
.
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| User: "Old Man" |
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| Title: Re: Transmitting Sound Through a Vacuum at c/3 |
12 Apr 2004 11:27:28 PM |
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"Edward Green" <spamspamspam3@netzero.com> wrote in message
news:eca320d0.0404121637.606ae73a@posting.google.com...
"Old Man" <nomail@nomail.net> wrote in message
news:<erednZg4XN2rYuTdRVn-jA@prairiewave.com>...
"Edward Green" <spamspamspam3@netzero.com> wrote in message
news:eca320d0.0404111545.419586ab@posting.google.com...
"Old Man" <nomail@nomail.net> wrote in message
news:RpGdnY9dXqJMXuXdRVn-sA@prairiewave.com...
Imagine a long evacuated tube. The interior hollow of the
tube is free of any gas, but is filled with thermal black body
radiation in equilibrium with the interior wall of the tube
which is maintained at some finite temperature above 0K.
Each end of the tube is capped with a thin light reflecting
membrane which is strong enough to withstand external
atmospheric pressure.
We can cause one of the membranes to vibrate in response
to external atmospheric sound waves. On the evacuated side,
the membrane's longitudinal vibrations create photon density
and pressure fluctuations which travel along the tube axis,
without loss, and at a third of the speed of light, to the
membrane at the other end of the tube,, causing it to move
in true fidelity with the arriving photon pressure fluctuations.
... isn't Old Man's scheme a version of an existing known method of
coupling light to sound, using lasers and mirrors?
Nope, but Old Man thinks that sound waves in a gas of photons
are longitudinal, not transverse.
Of course they are! Couldn't you tell from my post that that was the
kind of "transverse" which means "longitudinal"? :-)
Note you did side step my question about how it is possible to have
pressure waves in a medium which has no internal collisions. I'm not
saying it is impossible, just eyebrow raising -- you might consider it
a density wave.
The calculation for the speed of sound in a baryonic gas, is
according to the ideal gas law, wherein gas molecules do
not interact with each other. The gas interacts only with the
container walls. An oscillating piston, as one of the walls, is
the source of traveling density and pressure waves through
the gas. The speed of sound in an ideal gas with molecular
mass, m, is proportional to and somewhat less than sqrt(kT / m).
Molecule size and interaction cross section are assummed zero.
Perhaps we (implying for a heady moment that I am on Old Man's level,
which fantasy he is sometimes want to allow me) should consider a
simpler problem first (of course). Forget the walls of the tube:
consider a space containing a spherical black body of finite radius at
the origin. Let the black body hum. Call the body Aretha.
How do we describe the resulting spherically symmetric radiation
field? What is the boundary condition at the surface of the black
body?
Presumably the moving surface of the body continues to emit black body
radiation characteristic of its temperature, but doppler shifted. So
what we see at a distant detector is a time varying spectrum. Is this
the same thing as a pressure wave in the light bath, or is this a
different effect?
Maybe we should forget the spectrum. Maybe we should concentrate on
conservation of energy. A source moving towards us will correspond to
an uptick in power at the detector, the same source receding to a
downtick: this much simply from considering the time the source
transmits corresponding to one second's reception at the sink. I
_think_ this is a seperate effect from the doppler shift, but it
points in the same direction.
Ok... there are your pressure fluctuations ... power fluctuations are
pressure fluctuations in light. It looks to me like in this case the
power fluctuations should transmit at c?
As Ed said, they travel at c / 3.
Old Man blundered. The speed of sound in a gas of photons
is c / sqrt(3), not c / 3.
Old man thinks that sqrt(3) is a clue. the diagonal length of a unit
cube is sqrt(3). The sound wave created by an oscillating piston
travels in but one direction, but the radiation is emitted over 2pi
steradians from every point on the cylinder face. Take the mean
path length to a point a small unit distance from the cylinder face
in the photon gas from all points on the cylinder face with
appropriate weighting for solid angle. Does Ed get sqrt(3)?
So what is the effect of the walls...
Maybe Franz assumed that Old Man was taking credit for an
original derivation. Old Man is a parrot, not a thief. Polly says
that the speed of sound for dark energy is - c (negative).
Oh ... let's work out light first. Then we can worry about this
academic frippery ;-) This second result does sound like a result for
free space, though, and not a tube; doesn't it?
.
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| User: "Edward Green" |
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| Title: Re: Old Man pulls the wool over our psis |
13 Apr 2004 05:29:32 PM |
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"Old Man" <nomail@nomail.net> wrote in message news:<OPqdnYdrVfmnw-bdRVn-vg@prairiewave.com>...
The calculation for the speed of sound in a baryonic gas, is
according to the ideal gas law, wherein gas molecules do
not interact with each other. The gas interacts only with the
container walls. An oscillating piston, as one of the walls, is
the source of traveling density and pressure waves through
the gas. The speed of sound in an ideal gas with molecular
mass, m, is proportional to and somewhat less than sqrt(kT / m).
Molecule size and interaction cross section are assummed zero.
I notice this is essentially thermal velocity.
So this says again what we could have guessed: that density waves in
gases of non-interacting particles will travel as fast as the
particles carrying them.
This result will be nuanced somewhat by velocity distributions,
reflections from the walls of a guide tube (effective path length) and
questions where on the rise of a smeared pulse we declare the signal
received.
....
Old Man blundered. The speed of sound in a gas of photons
is c / sqrt(3), not c / 3.
Old man thinks that sqrt(3) is a clue. the diagonal length of a unit
cube is sqrt(3). The sound wave created by an oscillating piston
travels in but one direction, but the radiation is emitted over 2pi
steradians from every point on the cylinder face. Take the mean
path length to a point a small unit distance from the cylinder face
in the photon gas from all points on the cylinder face with
appropriate weighting for solid angle. Does Ed get sqrt(3)?
Before I do this (hedge, evasion and delaying tactic), I note that
essentially a similar argument can be made about ordinary sound in an
ordinary gas: yet we seem to be able to settle on a speed of sound
which is equal to itself, and not some factor near unity times itself.
:-)
Anyway, I want to say something else frightfully insightful: your
gedanken is a bit of a red herring. What you are really encouraging
us to see is that "sound" and "light" are degenerate in a photon gas.
Oh, you call it a "photon gas" to encourage us to think of it more
like an ordinary material gaseous medium (even though its molecules
don't interact), and you throw in some distractions about "black body"
to hint that the photon gas being at thermal equilibrium -- just like
an ordinary gas -- is somehow important, and so forth. But really
this is all irrelevant to your model: your "sound tube" would work
with classical monochromatic radiation -- or radiation derived from
monochromatic by a simple transformation, since the moving source
creates doppler shifts.
But the essential points are (1) light does couple mechanically to
matter, exerting pressure and transferring momentum, and (2) momentum
fluctuations can be introduced into directed light.
If Plato, leading us by the nose, gets us to accept some definition of
"sound" as "transferring mechanical motion by momentum fluctuations in
an intervening medium", then we are suddenly stuck with "light" as
carrying "sound"! Oh, we may try to back out then and claim that
sound can only be carried by matter, but it's too late: we should have
consulted our lawyer before we bought into his seemingly harmless
ideas -- cloaked, as I said, to be sure, in a curtain of red herrings
about "photon gas" and "thermal radiation" to make it look as if the
light he introduces were somehow a special kind of light, qualified to
carry "sound" as a special case: but it's not -- it's damn ordinary
light.
Ok ... I buy it: a signal carried by the electromagnetic field is
really always a form of "sound", subject to a broad but plausible
definition of sound which emphasizes the transfer of signals by means
of/encoded in momentum fluctuations in a directed disturbance. Are
you happy now, Old Man? Light is sound.
.
|
|
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| User: "Old Man" |
|
| Title: Re: Old Man pulls the wool over our psis |
13 Apr 2004 11:12:33 PM |
|
|
"Edward Green" <spamspamspam3@netzero.com> wrote in message
news:eca320d0.0404131429.55507057@posting.google.com...
"Old Man" <nomail@nomail.net> wrote in message
news:<OPqdnYdrVfmnw-bdRVn-vg@prairiewave.com>...
The calculation for the speed of sound in a baryonic gas, is
according to the ideal gas law, wherein gas molecules do
not interact with each other. The gas interacts only with the
container walls. An oscillating piston, as one of the walls, is
the source of traveling density and pressure waves through
the gas. The speed of sound in an ideal gas with molecular
mass, m, is proportional to and somewhat less than sqrt(kT / m).
Molecule size and interaction cross section are assummed zero.
I notice this is essentially thermal velocity.
So this says again what we could have guessed: that density waves in
gases of non-interacting particles will travel as fast as the
particles carrying them.
This result will be nuanced somewhat by velocity distributions,
reflections from the walls of a guide tube (effective path length) and
questions where on the rise of a smeared pulse we declare the signal
received.
...
Old Man blundered. The speed of sound in a gas of photons
is c / sqrt(3), not c / 3.
Old man thinks that sqrt(3) is a clue. the diagonal length of a unit
cube is sqrt(3). The sound wave created by an oscillating piston
travels in but one direction, but the radiation is emitted over 2pi
steradians from every point on the cylinder face. Take the mean
path length to a point a small unit distance from the cylinder face
in the photon gas from all points on the cylinder face with
appropriate weighting for solid angle. Does Ed get sqrt(3)?
Before I do this (hedge, evasion and delaying tactic), I note that
essentially a similar argument can be made about ordinary sound in an
ordinary gas: yet we seem to be able to settle on a speed of sound
which is equal to itself, and not some factor near unity times itself.
:-)
Anyway, I want to say something else frightfully insightful: your
gedanken is a bit of a red herring. What you are really encouraging
us to see is that "sound" and "light" are degenerate in a photon gas.
Oh, you call it a "photon gas" to encourage us to think of it more
like an ordinary material gaseous medium (even though its molecules
don't interact), and you throw in some distractions about "black body"
to hint that the photon gas being at thermal equilibrium -- just like
an ordinary gas -- is somehow important, and so forth. But really
this is all irrelevant to your model: your "sound tube" would work
with classical monochromatic radiation -- or radiation derived from
monochromatic by a simple transformation, since the moving source
creates doppler shifts.
But the essential points are (1) light does couple mechanically to
matter, exerting pressure and transferring momentum, and (2) momentum
fluctuations can be introduced into directed light.
If Plato, leading us by the nose, gets us to accept some definition of
"sound" as "transferring mechanical motion by momentum fluctuations in
an intervening medium", then we are suddenly stuck with "light" as
carrying "sound"! Oh, we may try to back out then and claim that
sound can only be carried by matter, but it's too late: we should have
consulted our lawyer before we bought into his seemingly harmless
ideas -- cloaked, as I said, to be sure, in a curtain of red herrings
about "photon gas" and "thermal radiation" to make it look as if the
light he introduces were somehow a special kind of light, qualified to
carry "sound" as a special case: but it's not -- it's damn ordinary
light.
Ok ... I buy it: a signal carried by the electromagnetic field is
really always a form of "sound", subject to a broad but plausible
definition of sound which emphasizes the transfer of signals by means
of/encoded in momentum fluctuations in a directed disturbance. Are
you happy now, Old Man? Light is sound.
No. Ed can't sneak around it with metaphysics.
The expression for the speed of sound in an ideal gas is
v_s = sqrt[ dP / d(rho) ]
under adiabatic conditions. The speed of sound in a photon
gas follows from the ideal gas law for a photon gas
PV = E / 3
as given by Landau and Lifshitz in "Statistical Physics"
ISBN 0 7506 3372 7. Combining equations yields
v_s = c / sqrt(3)
The ideal gas law for a gas of photons is identical to that
for a gas of ultra-relativistic electrons, which obviously
supports the propagation ofsound waves independent of
gas electron-electron interaction. [Old Man]
.
|
|
|
| User: "Edward Green" |
|
| Title: Re: Old Man pulls the wool over our psis |
14 Apr 2004 09:14:07 PM |
|
|
"Old Man" <nomail@nomail.net> wrote in message news:<Eq6dnQu1ws2hMeHdRVn-vw@prairiewave.com>...
"Edward Green" <spamspamspam3@netzero.com> wrote in message
news:eca320d0.0404131429.55507057@posting.google.com...
....
Ok ... I buy it: a signal carried by the electromagnetic field is
really always a form of "sound", subject to a broad but plausible
definition of sound which emphasizes the transfer of signals by means
of/encoded in momentum fluctuations in a directed disturbance. Are
you happy now, Old Man? Light is sound.
No. Ed can't sneak around it with metaphysics.
I fear you misunderstand my purpose. That was intended as a sincere
appreciation of the points you raise, whether intentionally or not.
The expression for the speed of sound in an ideal gas is
v_s = sqrt[ dP / d(rho) ]
Ah ... sqrt: at least the units work out now.
How does this compare to the earlier expression suggesting v_s in an
ideal gas is essential equal to thermal velocity of the molecules? Is
this the same?
under adiabatic conditions.
Speaking of methaphysics though, I wonder what assumptions are
inherent in the simple expression for v_s above. Is it possible the
given equation of state (P = rho RT) is required to hold locally and
intensively, not simply globally?
This condition would not be met for a Knudsen gas -- what you are
calling an ideal gas -- which has no internal collisions: and a photon
gas seems to be an example of a gas with no internal collisions. In
which case the derivation:
The speed of sound in a photon
gas follows from the ideal gas law for a photon gas
PV = E / 3
as given by Landau and Lifshitz in "Statistical Physics"
ISBN 0 7506 3372 7. Combining equations yields
v_s = c / sqrt(3)
Would be specious.
The ideal gas law for a gas of photons is identical to that
for a gas of ultra-relativistic electrons, which obviously
supports the propagation ofsound waves independent of
gas electron-electron interaction. [Old Man]
I am convinced there is no problem for a collisionless gas to conduct
what I would be happy to call "sound". I am not so convinced now that
the derivation given for its velocity here is correct. The only room
I see for the velocity of light-sound to deviate from c is in the
geometry of the tube, which doesn't enter into the derivation, not the
equation of state of photon gas. This c/sqrt(3) _might_ turn out to be
the velocity of light-sound in a cavity foam.*
And I still think (even if I am wrong about the velocity part) that
while the derivation has been carefully massaged here to suggest we
have singled out a special sound-like phenomenon unlike other behavior
of light, that in fact all disturbances of the EM field imply
disturbances in pressure (momentum flux), and might equally well be
considered "light sound", though we don't look at it this way.
*As Old Man says, names are everything: it is left to the reader's
imagination to decode this one. ;-)
.
|
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|
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|
|
| User: "Richard Herring" |
|
| Title: Re: Transmitting Sound Through a Vacuum at c/3 |
13 Apr 2004 04:36:15 AM |
|
|
In message <OPqdnYdrVfmnw-bdRVn-vg@prairiewave.com>, Old Man
<nomail@nomail.net> writes
"Edward Green" <spamspamspam3@netzero.com> wrote in message
news:eca320d0.0404121637.606ae73a@posting.google.com...
"Old Man" <nomail@nomail.net> wrote in message
news:<erednZg4XN2rYuTdRVn-jA@prairiewave.com>...
"Edward Green" <spamspamspam3@netzero.com> wrote in message
news:eca320d0.0404111545.419586ab@posting.google.com...
"Old Man" <nomail@nomail.net> wrote in message
news:RpGdnY9dXqJMXuXdRVn-sA@prairiewave.com...
Imagine a long evacuated tube. The interior hollow of the
tube is free of any gas, but is filled with thermal black body
radiation in equilibrium with the interior wall of the tube
which is maintained at some finite temperature above 0K.
Each end of the tube is capped with a thin light reflecting
membrane which is strong enough to withstand external
atmospheric pressure.
We can cause one of the membranes to vibrate in response
to external atmospheric sound waves. On the evacuated side,
the membrane's longitudinal vibrations create photon density
and pressure fluctuations which travel along the tube axis,
without loss, and at a third of the speed of light, to the
membrane at the other end of the tube,, causing it to move
in true fidelity with the arriving photon pressure fluctuations.
... isn't Old Man's scheme a version of an existing known method of
coupling light to sound, using lasers and mirrors?
Nope, but Old Man thinks that sound waves in a gas of photons
are longitudinal, not transverse.
Of course they are! Couldn't you tell from my post that that was the
kind of "transverse" which means "longitudinal"? :-)
Note you did side step my question about how it is possible to have
pressure waves in a medium which has no internal collisions. I'm not
saying it is impossible, just eyebrow raising -- you might consider it
a density wave.
The calculation for the speed of sound in a baryonic gas, is
according to the ideal gas law, wherein gas molecules do
not interact with each other. The gas interacts only with the
container walls.
That's a Knudsen gas, not an ideal gas.
--
Richard Herring
.
|
|
|
| User: "Old Man" |
|
| Title: Re: Transmitting Sound Through a Vacuum at c/3 |
14 Apr 2004 12:06:32 AM |
|
|
"Richard Herring" <junk@[127.0.0.1]> wrote in message
news:w7Hu1mOPS7eAFw$n@baesystems.com...
In message <OPqdnYdrVfmnw-bdRVn-vg@prairiewave.com>, Old Man
<nomail@nomail.net> writes
"Edward Green" <spamspamspam3@netzero.com> wrote in message
news:eca320d0.0404121637.606ae73a@posting.google.com...
"Old Man" <nomail@nomail.net> wrote in message
news:<erednZg4XN2rYuTdRVn-jA@prairiewave.com>...
"Edward Green" <spamspamspam3@netzero.com> wrote in message
news:eca320d0.0404111545.419586ab@posting.google.com...
"Old Man" <nomail@nomail.net> wrote in message
news:RpGdnY9dXqJMXuXdRVn-sA@prairiewave.com...
Imagine a long evacuated tube. The interior hollow of the
tube is free of any gas, but is filled with thermal black body
radiation in equilibrium with the interior wall of the tube
which is maintained at some finite temperature above 0K.
Each end of the tube is capped with a thin light reflecting
membrane which is strong enough to withstand external
atmospheric pressure.
We can cause one of the membranes to vibrate in response
to external atmospheric sound waves. On the evacuated side,
the membrane's longitudinal vibrations create photon density
and pressure fluctuations which travel along the tube axis,
without loss, and at a third of the speed of light, to the
membrane at the other end of the tube,, causing it to move
in true fidelity with the arriving photon pressure
fluctuations.
... isn't Old Man's scheme a version of an existing known method of
coupling light to sound, using lasers and mirrors?
Nope, but Old Man thinks that sound waves in a gas of photons
are longitudinal, not transverse.
Of course they are! Couldn't you tell from my post that that was the
kind of "transverse" which means "longitudinal"? :-)
Note you did side step my question about how it is possible to have
pressure waves in a medium which has no internal collisions. I'm not
saying it is impossible, just eyebrow raising -- you might consider it
a density wave.
The calculation for the speed of sound in a baryonic gas, is
according to the ideal gas law, wherein gas molecules do
not interact with each other. The gas interacts only with the
container walls.
That's a Knudsen gas, not an ideal gas.
--
Richard Herring
Zero molecular volume, zero molecular interaction, except
with walls is precisely an ideal gas => PV = NRT. That's
way it's called IDEAL. It supports sound propagation.
[Old Man]
.
|
|
|
| User: "Richard Herring" |
|
| Title: Re: Transmitting Sound Through a Vacuum at c/3 |
14 Apr 2004 04:38:27 AM |
|
|
In message <hcadnR1FfsR_JeHdRVn-vw@prairiewave.com>, Old Man
<nomail@nomail.net> writes
"Richard Herring" <junk@[127.0.0.1]> wrote in message
news:w7Hu1mOPS7eAFw$n@baesystems.com...
In message <OPqdnYdrVfmnw-bdRVn-vg@prairiewave.com>, Old Man
<nomail@nomail.net> writes
"Edward Green" <spamspamspam3@netzero.com> wrote in message
news:eca320d0.0404121637.606ae73a@posting.google.com...
"Old Man" <nomail@nomail.net> wrote in message
news:<erednZg4XN2rYuTdRVn-jA@prairiewave.com>...
"Edward Green" <spamspamspam3@netzero.com> wrote in message
news:eca320d0.0404111545.419586ab@posting.google.com...
"Old Man" <nomail@nomail.net> wrote in message
news:RpGdnY9dXqJMXuXdRVn-sA@prairiewave.com...
Imagine a long evacuated tube. The interior hollow of the
tube is free of any gas, but is filled with thermal black body
radiation in equilibrium with the interior wall of the tube
which is maintained at some finite temperature above 0K.
Each end of the tube is capped with a thin light reflecting
membrane which is strong enough to withstand external
atmospheric pressure.
We can cause one of the membranes to vibrate in response
to external atmospheric sound waves. On the evacuated side,
the membrane's longitudinal vibrations create photon density
and pressure fluctuations which travel along the tube axis,
without loss, and at a third of the speed of light, to the
membrane at the other end of the tube,, causing it to move
in true fidelity with the arriving photon pressure
fluctuations.
... isn't Old Man's scheme a version of an existing known method of
coupling light to sound, using lasers and mirrors?
Nope, but Old Man thinks that sound waves in a gas of photons
are longitudinal, not transverse.
Of course they are! Couldn't you tell from my post that that was the
kind of "transverse" which means "longitudinal"? :-)
Note you did side step my question about how it is possible to have
pressure waves in a medium which has no internal collisions. I'm not
saying it is impossible, just eyebrow raising -- you might consider it
a density wave.
The calculation for the speed of sound in a baryonic gas, is
according to the ideal gas law, wherein gas molecules do
not interact with each other. The gas interacts only with the
container walls.
That's a Knudsen gas, not an ideal gas.
Zero molecular volume, zero molecular interaction, except
with walls is precisely an ideal gas => PV = NRT. That's
way it's called IDEAL.
No. That's a typical dictionary definition, and it's wrong. A lot of
texts also seem to get this wrong. There's one more condition: *elastic
collisions*.
See for example
http://www.ucdsb.on.ca/tiss/stretton/chem1/gases9.html
"1. A gas consists of molecules in constant random motion.
2. Gas molecules influence each other only by collision; they exert no
other forces on each other.
3. All collisions between gas molecules are perfectly elastic; all
kinetic energy is conserved.
4. The volume actually occupied by the molecules of a gas is
negligibly small; the vast majority of the volume of the gas is empty
space through which the gas molecules are moving."
It supports sound propagation.
Only if the mean free path is much less than the container dimensions is
pressure a meaningful quantity. You can't maintain pressure equilibrium
without a mechanism for exchanging momentum between particles.
If the mean free path is larger than the container you get a Knudsen
gas, which has many interesting properties, but is rather different
because pressure equilibrium is not one of them.
--
Richard Herring
.
|
|
|
| User: "Old Man" |
|
| Title: Re: Transmitting Sound Through a Vacuum at c/3 |
14 Apr 2004 01:44:57 PM |
|
|
"Richard Herring" <junk@[127.0.0.1]> wrote in message
news:WRhHffITaQfAFwAd@baesystems.com...
In message <hcadnR1FfsR_JeHdRVn-vw@prairiewave.com>, Old Man
<nomail@nomail.net> writes
"Richard Herring" <junk@[127.0.0.1]> wrote in message
news:w7Hu1mOPS7eAFw$n@baesystems.com...
In message <OPqdnYdrVfmnw-bdRVn-vg@prairiewave.com>, Old Man
<nomail@nomail.net> writes
"Edward Green" <spamspamspam3@netzero.com> wrote in message
news:eca320d0.0404121637.606ae73a@posting.google.com...
"Old Man" <nomail@nomail.net> wrote in message
news:<erednZg4XN2rYuTdRVn-jA@prairiewave.com>...
"Edward Green" <spamspamspam3@netzero.com> wrote in message
news:eca320d0.0404111545.419586ab@posting.google.com...
"Old Man" <nomail@nomail.net> wrote in message
news:RpGdnY9dXqJMXuXdRVn-sA@prairiewave.com...
Imagine a long evacuated tube. The interior hollow of the
tube is free of any gas, but is filled with thermal black
body
radiation in equilibrium with the interior wall of the tube
which is maintained at some finite temperature above 0K.
Each end of the tube is capped with a thin light reflecting
membrane which is strong enough to withstand external
atmospheric pressure.
We can cause one of the membranes to vibrate in response
to external atmospheric sound waves. On the evacuated side,
the membrane's longitudinal vibrations create photon density
and pressure fluctuations which travel along the tube axis,
without loss, and at a third of the speed of light, to the
membrane at the other end of the tube,, causing it to move
in true fidelity with the arriving photon pressure
fluctuations.
... isn't Old Man's scheme a version of an existing known method
of
coupling light to sound, using lasers and mirrors?
Nope, but Old Man thinks that sound waves in a gas of photons
are longitudinal, not transverse.
Of course they are! Couldn't you tell from my post that that was
the
kind of "transverse" which means "longitudinal"? :-)
Note you did side step my question about how it is possible to have
pressure waves in a medium which has no internal collisions. I'm
not
saying it is impossible, just eyebrow raising -- you might consider
it
a density wave.
The calculation for the speed of sound in a baryonic gas, is
according to the ideal gas law, wherein gas molecules do
not interact with each other. The gas interacts only with the
container walls.
That's a Knudsen gas, not an ideal gas.
Zero molecular volume, zero molecular interaction, except
with walls is precisely an ideal gas => PV = NRT. That's
way it's called IDEAL.
No. That's a typical dictionary definition, and it's wrong. A lot of
texts also seem to get this wrong. There's one more condition: *elastic
collisions*.
See for example
http://www.ucdsb.on.ca/tiss/stretton/chem1/gases9.html
"1. A gas consists of molecules in constant random motion.
2. Gas molecules influence each other only by collision; they exert no
other forces on each other.
3. All collisions between gas molecules are perfectly elastic; all
kinetic energy is conserved.
4. The volume actually occupied by the molecules of a gas is
negligibly small; the vast majority of the volume of the gas is empty
space through which the gas molecules are moving."
This is inconsistent. It's metaphysics. There can't be a finite
cross section for elastic scattering if the gas molecules occupy
zero volume. The volume derives from the scattering cross
section. If the gas molecules don't scatter, the volume is zero.
Richard would have use believe in hidden variables. If sound
propagation depended on an elastic scattering cross section,
Then it's magnitude would appear in the result for the speed
of sound. It doesn't. It doesn't appear in the ideal gas law.
It doesn't appear in a rigorous derivation. It's nothing but
words. It's a figment of Richard's metaphysical imagination.
Hidden variables and physics don't blend.
The effects of a finite scattering cross section upon the
Ideal gas law and the propagation of sound can be and have
been measured. But in that case, the variables aren't hidden.
They appear explicitly in the theoretical predictions.
[Old Man]
It supports sound propagation.
Only if the mean free path is much less than the container dimensions is
pressure a meaningful quantity. You can't maintain pressure equilibrium
without a mechanism for exchanging momentum between particles.
If the mean free path is larger than the container you get a Knudsen
gas, which has many interesting properties, but is rather different
because pressure equilibrium is not one of them.
--
Richard Herring
.
|
|
|
| User: "Richard Herring" |
|
| Title: Re: Transmitting Sound Through a Vacuum at c/3 |
15 Apr 2004 03:55:56 AM |
|
|
In message <aYSdncuA0uop5eDdRVn-sw@prairiewave.com>, Old Man
<nomail@nomail.net> writes
"Richard Herring" <junk@[127.0.0.1]> wrote in message
news:WRhHffITaQfAFwAd@baesystems.com...
In message <hcadnR1FfsR_JeHdRVn-vw@prairiewave.com>, Old Man
<nomail@nomail.net> writes
"Richard Herring" <junk@[127.0.0.1]> wrote in message
news:w7Hu1mOPS7eAFw$n@baesystems.com...
In message <OPqdnYdrVfmnw-bdRVn-vg@prairiewave.com>, Old Man
<nomail@nomail.net> writes
[...]
The calculation for the speed of sound in a baryonic gas, is
according to the ideal gas law, wherein gas molecules do
not interact with each other. The gas interacts only with the
container walls.
That's a Knudsen gas, not an ideal gas.
Zero molecular volume, zero molecular interaction, except
with walls is precisely an ideal gas => PV = NRT. That's
way it's called IDEAL.
No. That's a typical dictionary definition, and it's wrong. A lot of
texts also seem to get this wrong. There's one more condition: *elastic
collisions*.
See for example
http://www.ucdsb.on.ca/tiss/stretton/chem1/gases9.html
"1. A gas consists of molecules in constant random motion.
2. Gas molecules influence each other only by collision; they exert no
other forces on each other.
3. All collisions between gas molecules are perfectly elastic; all
kinetic energy is conserved.
4. The volume actually occupied by the molecules of a gas is
negligibly small; the vast majority of the volume of the gas is empty
space through which the gas molecules are moving."
This is inconsistent. It's metaphysics.
Take it up with the author of that web page. Or with Flowers and
Mendoza, the authors of the text I used as an undergraduate.
There can't be a finite
cross section for elastic scattering if the gas molecules occupy
zero volume. The volume derives from the scattering cross
section. If the gas molecules don't scatter, the volume is zero.
Just so. That's why point 4 above says "negligibly small", not "zero".
Richard would have use believe in hidden variables.
Don't put words into my mouth. What I said was that your definition of
"ideal gas" was incomplete.
If sound
propagation depended on an elastic scattering cross section,
Then it's magnitude would appear in the result for the speed
of sound. It doesn't.
Bad syllogism. Did you miss the section on limits in Analysis 101?
It doesn't appear in the ideal gas law.
I'll tell you what *does* appear in the ideal gas law, and that's a term
usually denoted by T. For it to be meaningful, the gas has to be in
thermal equilibrium. How do you propose to achieve that if the particles
don't interact?
It doesn't appear in a rigorous derivation.
You haven't *shown* a rigorous derivation of the speed of sound. Hint:
at low enough pressures it's dispersive.
It's nothing but
words.
It's a necessity for thermal equilibrium.
It's a figment of Richard's metaphysical imagination.
"Most of the time, of course, the molecules are far apart and their
interatomic potential energy is negligible. But when they collide with
one another, their potential energy is, for a short time, by no means
negligible. Now it is essential that there should be collisions between
gas molecules because this is the mechanism whereby a gas reaches
thermal equilibrium; if it were not for these collisions the gas could
never reach uniform temperature. "
Flowers & Mendoza, "Properties of Matter", Wiley, 1970, section 4.4
Hidden variables and physics don't blend.
All kinds of variables can vanish when you take a suitable limit. That
doesn't make them "hidden". The relevant one here is mean free path.
The effects of a finite scattering cross section upon the
Ideal gas law and the propagation of sound can be and have
been measured. But in that case, the variables aren't hidden.
They appear explicitly in the theoretical predictions.
It supports sound propagation.
Only if the mean free path is much less than the container dimensions is
pressure a meaningful quantity. You can't maintain pressure equilibrium
without a mechanism for exchanging momentum between particles.
If the mean free path is larger than the container you get a Knudsen
gas, which has many interesting properties, but is rather different
because pressure equilibrium is not one of them.
--
Richard Herring
.
|
|
|
| User: "Old Man" |
|
| Title: Re: Transmitting Sound Through a Vacuum at c/3 |
15 Apr 2004 04:52:24 AM |
|
|
"Richard Herring" <junk@[127.0.0.1]> wrote in message
news:EXCjufDc4kfAFwDw@baesystems.com...
In message <aYSdncuA0uop5eDdRVn-sw@prairiewave.com>, Old Man
<nomail@nomail.net> writes
"Richard Herring" <junk@[127.0.0.1]> wrote in message
news:WRhHffITaQfAFwAd@baesystems.com...
In message <hcadnR1FfsR_JeHdRVn-vw@prairiewave.com>, Old Man
<nomail@nomail.net> writes
"Richard Herring" <junk@[127.0.0.1]> wrote in message
news:w7Hu1mOPS7eAFw$n@baesystems.com...
In message <OPqdnYdrVfmnw-bdRVn-vg@prairiewave.com>, Old Man
<nomail@nomail.net> writes
[...]
The calculation for the speed of sound in a baryonic gas, is
according to the ideal gas law, wherein gas molecules do
not interact with each other. The gas interacts only with the
container walls.
That's a Knudsen gas, not an ideal gas.
Zero molecular volume, zero molecular interaction, except
with walls is precisely an ideal gas => PV = NRT. That's
way it's called IDEAL.
No. That's a typical dictionary definition, and it's wrong. A lot of
texts also seem to get this wrong. There's one more condition: *elastic
collisions*.
See for example
http://www.ucdsb.on.ca/tiss/stretton/chem1/gases9.html
"1. A gas consists of molecules in constant random motion.
2. Gas molecules influence each other only by collision; they exert
no
other forces on each other.
3. All collisions between gas molecules are perfectly elastic; all
kinetic energy is conserved.
4. The volume actually occupied by the molecules of a gas is
negligibly small; the vast majority of the volume of the gas is empty
space through which the gas molecules are moving."
This is inconsistent. It's metaphysics.
Take it up with the author of that web page. Or with Flowers and
Mendoza, the authors of the text I used as an undergraduate.
There can't be a finite
cross section for elastic scattering if the gas molecules occupy
zero volume. The volume derives from the scattering cross
section. If the gas molecules don't scatter, the volume is zero.
Just so. That's why point 4 above says "negligibly small", not "zero".
Richard would have use believe in hidden variables.
Don't put words into my mouth. What I said was that your definition of
"ideal gas" was incomplete.
If sound
propagation depended on an elastic scattering cross section,
Then it's magnitude would appear in the result for the speed
of sound. It doesn't.
Bad syllogism. Did you miss the section on limits in Analysis 101?
It doesn't appear in the ideal gas law.
I'll tell you what *does* appear in the ideal gas law, and that's a term
usually denoted by T. For it to be meaningful, the gas has to be in
thermal equilibrium. How do you propose to achieve that if the particles
don't interact?
It doesn't appear in a rigorous derivation.
You haven't *shown* a rigorous derivation of the speed of sound. Hint:
at low enough pressures it's dispersive.
It's nothing but
words.
It's a necessity for thermal equilibrium.
It's a figment of Richard's metaphysical imagination.
"Most of the time, of course, the molecules are far apart and their
interatomic potential energy is negligible. But when they collide with
one another, their potential energy is, for a short time, by no means
negligible. Now it is essential that there should be collisions between
gas molecules because this is the mechanism whereby a gas reaches
thermal equilibrium; if it were not for these collisions the gas could
never reach uniform temperature. "
Flowers & Mendoza, "Properties of Matter", Wiley, 1970, section 4.4
Hidden variables and physics don't blend.
All kinds of variables can vanish when you take a suitable limit. That
doesn't make them "hidden". The relevant one here is mean free path.
"mean free path" doesn't appear in PV = NRT for an ideal gas nor
can "mean free path" be obtained from it. [Old Man]
The effects of a finite scattering cross section upon the
Ideal gas law and the propagation of sound can be and have
been measured. But in that case, the variables aren't hidden.
They appear explicitly in the theoretical predictions.
It supports sound propagation.
Only if the mean free path is much less than the container dimensions
is
pressure a meaningful quantity. You can't maintain pressure equilibrium
without a mechanism for exchanging momentum between particles.
If the mean free path is larger than the container you get a Knudsen
gas, which has many interesting properties, but is rather different
because pressure equilibrium is not one of them.
--
Richard Herring
.
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| User: "Richard Herring" |
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| Title: Re: Transmitting Sound Through a Vacuum at c/3 |
15 Apr 2004 05:46:49 AM |
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In message <FIqdnSejINDGxuPdRVn-jA@prairiewave.com>, Old Man
<nomail@nomail.net> writes
"Richard Herring" <junk@[127.0.0.1]> wrote in message
news:EXCjufDc4kfAFwDw@baesystems.com...
In message <aYSdncuA0uop5eDdRVn-sw@prairiewave.com>, Old Man
<nomail@nomail.net> writes
"Richard Herring" <junk@[127.0.0.1]> wrote in message
news:WRhHffITaQfAFwAd@baesystems.com...
In message <hcadnR1FfsR_JeHdRVn-vw@prairiewave.com>, Old Man
<nomail@nomail.net> writes
"Richard Herring" <junk@[127.0.0.1]> wrote in message
news:w7Hu1mOPS7eAFw$n@baesystems.com...
In message <OPqdnYdrVfmnw-bdRVn-vg@prairiewave.com>, Old Man
<nomail@nomail.net> writes
[...]
The calculation for the speed of sound in a baryonic gas, is
according to the ideal gas law, wherein gas molecules do
not interact with each other. The gas interacts only with the
container walls.
That's a Knudsen gas, not an ideal gas.
Zero molecular volume, zero molecular interaction, except
with walls is precisely an ideal gas => PV = NRT. That's
way it's called IDEAL.
No. That's a typical dictionary definition, and it's wrong. A lot of
texts also seem to get this wrong. There's one more condition: *elastic
collisions*.
See for example
http://www.ucdsb.on.ca/tiss/stretton/chem1/gases9.html
"1. A gas consists of molecules in constant random motion.
2. Gas molecules influence each other only by collision; they exert
no
other forces on each other.
3. All collisions between gas molecules are perfectly elastic; all
kinetic energy is conserved.
4. The volume actually occupied by the molecules of a gas is
negligibly small; the vast majority of the volume of the gas is empty
space through which the gas molecules are moving."
This is inconsistent. It's metaphysics.
Take it up with the author of that web page. Or with Flowers and
Mendoza, the authors of the text I used as an undergraduate.
There can't be a finite
cross section for elastic scattering if the gas molecules occupy
zero volume. The volume derives from the scattering cross
section. If the gas molecules don't scatter, the volume is zero.
Just so. That's why point 4 above says "negligibly small", not "zero".
Richard would have use believe in hidden variables.
Don't put words into my mouth. What I said was that your definition of
"ideal gas" was incomplete.
If sound
propagation depended on an elastic scattering cross section,
Then it's magnitude would appear in the result for the speed
of sound. It doesn't.
Bad syllogism. Did you miss the section on limits in Analysis 101?
It doesn't appear in the ideal gas law.
I'll tell you what *does* appear in the ideal gas law, and that's a term
usually denoted by T. For it to be meaningful, the gas has to be in
thermal equilibrium. How do you propose to achieve that if the particles
don't interact?
It doesn't appear in a rigorous derivation.
You haven't *shown* a rigorous derivation of the speed of sound. Hint:
at low enough pressures it's dispersive.
It's nothing but
words.
It's a necessity for thermal equilibrium.
It's a figment of Richard's metaphysical imagination.
"Most of the time, of course, the molecules are far apart and their
interatomic potential energy is negligible. But when they collide with
one another, their potential energy is, for a short time, by no means
negligible. Now it is essential that there should be collisions between
gas molecules because this is the mechanism whereby a gas reaches
thermal equilibrium; if it were not for these collisions the gas could
never reach uniform temperature. "
Flowers & Mendoza, "Properties of Matter", Wiley, 1970, section 4.4
Hidden variables and physics don't blend.
All kinds of variables can vanish when you take a suitable limit. That
doesn't make them "hidden". The relevant one here is mean free path.
"mean free path" doesn't appear in PV = NRT for an ideal gas nor
can "mean free path" be obtained from it. [Old Man]
Of course not. What did you expect? Once you take something to its
limit, you can't undo the process.
Where do you think "P" and "T" came from? They are macroscopic
quantities, properties of a hypothetical continuous fluid. You want to
apply them to a gas of independent microscopic particles. That requires
approximations which can only be true for certain time and length
scales. The necessary approximations are "mean free path sufficiently
small" and "collision rate sufficiently large".
At the moment you write down P or T for a gas, you have _already_
implicitly applied those considerations and decided that the
approximation is valid.
--
Richard Herring
.
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| User: "Old Man" |
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| Title: Re: Transmitting Sound Through a Vacuum at c/3 |
15 Apr 2004 05:49:07 AM |
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"Richard Herring" <junk@[127.0.0.1]> wrote in message
news:EXCjufDc4kfAFwDw@baesystems.com...
In message <aYSdncuA0uop5eDdRVn-sw@prairiewave.com>, Old Man
<nomail@nomail.net> writes
"Richard Herring" <junk@[127.0.0.1]> wrote in message
news:WRhHffITaQfAFwAd@baesystems.com...
In message <hcadnR1FfsR_JeHdRVn-vw@prairiewave.com>, Old Man
<nomail@nomail.net> writes
"Richard Herring" <junk@[127.0.0.1]> wrote in message
news:w7Hu1mOPS7eAFw$n@baesystems.com...
In message <OPqdnYdrVfmnw-bdRVn-vg@prairiewave.com>, Old Man
<nomail@nomail.net> writes
[...]
The calculation for the speed of sound in a baryonic gas, is
according to the ideal gas law, wherein gas molecules do
not interact with each other. The gas interacts only with the
container walls.
That's a Knudsen gas, not an ideal gas.
Zero molecular volume, zero molecular interaction, except
with walls is precisely an ideal gas => PV = NRT. That's
way it's called IDEAL.
No. That's a typical dictionary definition, and it's wrong. A lot of
texts also seem to get this wrong. There's one more condition: *elastic
collisions*.
See for example
http://www.ucdsb.on.ca/tiss/stretton/chem1/gases9.html
"1. A gas consists of molecules in constant random motion.
2. Gas molecules influence each other only by collision; they exert
no
other forces on each other.
3. All collisions between gas molecules are perfectly elastic; all
kinetic energy is conserved.
4. The volume actually occupied by the molecules of a gas is
negligibly small; the vast majority of the volume of the gas is empty
space through which the gas molecules are moving."
This is inconsistent. It's metaphysics.
Take it up with the author of that web page. Or with Flowers and
Mendoza, the authors of the text I used as an undergraduate.
There can't be a finite
cross section for elastic scattering if the gas molecules occupy
zero volume. The volume derives from the scattering cross
section. If the gas molecules don't scatter, the volume is zero.
Just so. That's why point 4 above says "negligibly small", not "zero".
Richard would have use believe in hidden variables.
Don't put words into my mouth. What I said was that your definition of
"ideal gas" was incomplete.
A finite "mean-free-path" (as opposed to infinite) doesn't
lead to PV = NRT. "mean-free-path" is a metaphysical foot
note for an ideal gas. Real gasses do things that don't follow
from the ideal gas law, but sound propagation isn't amongst
them. [Old Man]
If sound
propagation depended on an elastic scattering cross section,
Then it's magnitude would appear in the result for the speed
of sound. It doesn't.
Bad syllogism. Did you miss the section on limits in Analysis 101?
It doesn't appear in the ideal gas law.
I'll tell you what *does* appear in the ideal gas law, and that's a term
usually denoted by T. For it to be meaningful, the gas has to be in
thermal equilibrium. How do you propose to achieve that if the particles
don't interact?
It doesn't appear in a rigorous derivation.
You haven't *shown* a rigorous derivation of the speed of sound. Hint:
at low enough pressures it's dispersive.
It's nothing but
words.
It's a necessity for thermal equilibrium.
It's a figment of Richard's metaphysical imagination.
"Most of the time, of course, the molecules are far apart and their
interatomic potential energy is negligible. But when they collide with
one another, their potential energy is, for a short time, by no means
negligible. Now it is essential that there should be collisions between
gas molecules because this is the mechanism whereby a gas reaches
thermal equilibrium; if it were not for these collisions the gas could
never reach uniform temperature. "
Flowers & Mendoza, "Properties of Matter", Wiley, 1970, section 4.4
Hidden variables and physics don't blend.
All kinds of variables can vanish when you take a suitable limit. That
doesn't make them "hidden". The relevant one here is mean free path.
The effects of a finite scattering cross section upon the
Ideal gas law and the propagation of sound can be and have
been measured. But in that case, the variables aren't hidden.
They appear explicitly in the theoretical predictions.
It supports sound propagation.
Only if the mean free path is much less than the container dimensions
is
pressure a meaningful quantity. You can't maintain pressure equilibrium
without a mechanism for exchanging momentum between particles.
If the mean free path is larger than the container you get a Knudsen
gas, which has many interesting properties, but is rather different
because pressure equilibrium is not one of them.
--
Richard Herring
.
|
|
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| User: "Richard Herring" |
|
| Title: Re: Transmitting Sound Through a Vacuum at c/3 |
15 Apr 2004 07:10:14 AM |
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|
In message <wNKdnYcFvJk79ePdRVn-sA@prairiewave.com>, Old Man
<nomail@nomail.net> writes
"Richard Herring" <junk@[127.0.0.1]> wrote in message
news:EXCjufDc4kfAFwDw@baesystems.com...
In message <aYSdncuA0uop5eDdRVn-sw@prairiewave.com>, Old Man
<nomail@nomail.net> writes
"Richard Herring" <junk@[127.0.0.1]> wrote in message
news:WRhHffITaQfAFwAd@baesystems.com...
In message <hcadnR1FfsR_JeHdRVn-vw@prairiewave.com>, Old Man
<nomail@nomail.net> writes
"Richard Herring" <junk@[127.0.0.1]> wrote in message
news:w7Hu1mOPS7eAFw$n@baesystems.com...
In message <OPqdnYdrVfmnw-bdRVn-vg@prairiewave.com>, Old Man
<nomail@nomail.net> writes
[...]
The calculation for the speed of sound in a baryonic gas, is
according to the ideal gas law, wherein gas molecules do
not interact with each other. The gas interacts only with the
container walls.
That's a Knudsen gas, not an ideal gas.
Zero molecular volume, zero molecular interaction, except
with walls is precisely an ideal gas => PV = NRT. That's
way it's called IDEAL.
No. That's a typical dictionary definition, and it's wrong. A lot of
texts also seem to get this wrong. There's one more condition: *elastic
collisions*.
See for example
http://www.ucdsb.on.ca/tiss/stretton/chem1/gases9.html
"1. A gas consists of molecules in constant random motion.
2. Gas molecules influence each other only by collision; they exert
no
other forces on each other.
3. All collisions between gas molecules are perfectly elastic; all
kinetic energy is conserved.
4. The volume actually occupied by the molecules of a gas is
negligibly small; the vast majority of the volume of the gas is empty
space through which the gas molecules are moving."
This is inconsistent. It's metaphysics.
Take it up with the author of that web page. Or with Flowers and
Mendoza, the authors of the text I used as an undergraduate.
There can't be a finite
cross section for elastic scattering if the gas molecules occupy
zero volume. The volume derives from the scattering cross
section. If the gas molecules don't scatter, the volume is zero.
Just so. That's why point 4 above says "negligibly small", not "zero".
Richard would have use believe in hidden variables.
Don't put words into my mouth. What I said was that your definition of
"ideal gas" was incomplete.
A finite "mean-free-path" (as opposed to infinite) doesn't
lead to PV = NRT.
PV=NRT requires the assumption (inter alia) of thermal equilibrium. That
in turn requires a mechanism.
Again:
"Most of the time, of course, the molecules are far apart and their
interatomic potential energy is negligible. But when they collide with
one another, their potential energy is, for a short time, by no means
negligible. Now it is essential that there should be collisions between
gas molecules because this is the mechanism whereby a gas reaches
thermal equilibrium; if it were not for these collisions the gas could
never reach uniform temperature."
Flowers & Mendoza, "Properties of Matter", Wiley, 1970, section 4.4
"mean-free-path" is a metaphysical foot
note for an ideal gas. Real gasses do things that don't follow
from the ideal gas law, but sound propagation isn't amongst
them.
I don't recall saying it was. But... do you think the interstellar
medium is anything like an ideal gas?
--
Richard Herring
.
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| User: "Sam Wormley" |
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| Title: Re: Transmitting Sound Through a Vacuum at c/3 | | | | | |
