The temperature of a photon ?
Peter Christensen
answer is no.)
It always applies that E = p*c (p is the momentum which depend only on the
wavelength) for a photon. But as the speed of light is always c, there can't
be a kinectic energy as for massive particles, where E_kin = 1/2 m*v^2 can
be expressed as a measure of temperature for a particle.
I know, that a radiation profile (photons) can express the temperature of
it's source, but I still don't think, that it will make sence to talk about
a temperature for single photons, as it would do when talking about massive
particles with E = E_kin = k_B*T.
Are there anybody who would like to try to define a temperature for a
photon?
PC
Sam Wormley
When a cosmologist talks about the 'temperature' of a photon they are
basically describing the equivalent energy of a photon.
http://www.physlink.com/education/askexperts/ae210.cfm
Planck's Law
http://scienceworld.wolfram.com/physics/PlanckLaw.html
Sue...
This is a classic calorimetry exercise. Simply measure the
temperature of the brain of a theorist, before and after he
muses about light particles, then apply the standard equations.
http://en.wikipedia.org/wiki/Calorimetry
http://nobelprize.org/physics/laureates/1965/feynman-lecture.html
http://nobelprize.org/physics/articles/ekspong/index.html
Sue...
Sue...
critten...@yahoo.com
And the state of quanta appears to be virtual photon inclusive.
Meaning the uncollapsed quanta truely have temperature. A collaspe
causes state.
A field appears temperature variable inclusive. A state of the field is
its age or temperature.
So quanta all have temperature making the individual collapse the
defect in example. A quantity of energy to appear to the next state
from field location. A location underfinable.
Making the temperature of vacumn equal to the photons. A space
temperature details the collapse potential. A space of virtual quanta,
uncollapsed.
So you equation number two is a correct statement. A source defines
virtual temperature for the location of origin is a known collapse,
i.e. particle state. And here particle state causes virtual
temperature. Invert temperature when defining photon source
quantification. A source difference in temperature appears the photon.
And state age then decays to a temperature quantification for any
temperature variable.
A unit temperature is five fold transform particle age. And the
distance before collapse appears the true age or temperature of all
true particles of this class. A physical necessity requires this
distinction of particle class temeprature measure.
Remember, the size of the universe defines its age. And temperature
cools always.
So the variable to test my exposition is temperature of the source
related to the energy change of the target. A gamma factor removed.
Peter Christensen
"Sam Wormley" <swor...@mchsi.com> skrev i en meddelelse
news:dAbLg.126441$FQ1.69668@attbi_s71...
From the link: "So when a cosmologist talks about the 'temperature' of a
photon they are basically describing the equivalent energy of a photon."
This means E = h*f = k_B*T. But as in this reference, I would also use the
word 'temperature' in ''. But it's a measure of the temperature of it's
source, no doubt about this.
> Planck's Law
> http://scienceworld.wolfram.com/physics/PlanckLaw.html
A good link...
PC
critten...@yahoo.com
This is the physical interpretation of gamma-5?
Peter Christensen
"Sue..." <suzyse...@yahoo.com.au> skrev i en meddelelse
news:1157449519.4...@h48g2000cwc.googlegroups.com...
As I understand it, those who are thinking should have a slightly higher
brain-temperature? -Maybe science can determine this someday out in the
future...
PC
Sue...
There are two reason people use photons.
1. They love statistics.
2. They hate Maxwells equations.
The latter are by far in the majority
.
Sue...
http://farside.ph.utexas.edu/teaching.html
http://web.mit.edu/8.02t/www/802TEAL3D/teal_tour.htm
>
> PC
Sue...
Why does that sound like movie stars that make movies about
move stars ? The 'collapse' is imaginary so further piling on
imaginaies and virtuals makes more heat than light. ;-)
Sue...
G=EMC^2 Glazier
as heat. Photon going through the vacuum of space are cold. When the
photons hit(absorbed by object they case its atoms to move around
faster,and this makes them warm up. A vacuum will not transfer heat. I
think I just invented the vacuum bottle. Go figure Bert
mme...@cars3.uchicago.edu
>Does it make sence to talk about the temperature of a photon? (-My own
>answer is no.)
>
>It always applies that E = p*c (p is the momentum which depend only on the
>wavelength) for a photon. But as the speed of light is always c, there can't
>be a kinectic energy as for massive particles,
Oh, of course there can be kinetic energy. All of photon's energy is
kinetic.
> where E_kin = 1/2 m*v^2 can
>be expressed as a measure of temperature for a particle.
>
Not quite.
>I know, that a radiation profile (photons) can express the temperature of
>it's source, but I still don't think, that it will make sence to talk about
>a temperature for single photons, as it would do when talking about massive
>particles with E = E_kin = k_B*T.
It doesn't make sense to talk about temperature for single massive
particels, either. Temperature is an ensamble property.
>
>Are there anybody who would like to try to define a temperature for a
>photon?
>
Check out cavity radiation.
Mati Meron | "When you argue with a fool,
me...@cars.uchicago.edu | chances are he is doing just the same"
Quantum_Ranger
Long wave Photons are/have finite temperatures, but as with the H.U.P
you cannot isolate it with sufficient accuracy to be specific?
I believe Photons have variable themonic quantities, at variable
locations along the EM spectrum?
Simplistically there are "COLD" photons and there are "HOT" photons.
Jeff…Relf
PhysLink.COM:
When a cosmologist talks about the 'temperature' of a photon they are
basically describing the equivalent energy of a photon.
PhysLink.COM/education/askexperts/ae210.cfm
Good links. There are lots of definitions of temperature.
Although ideal laser ( one that's perfectly coherent ) is at absolute zero,
strong ones can burn through steal ( thanks to their relativistic mass ).
How cool is that ?
Atom lasers employ Bose-Einstein condensates that are near absolute zero.
See:
WikiPedia.ORG/wiki/Bose-Einstein_condensate
WikiPedia.ORG/wiki/Atom_laser
A Bose-Einstein ( laser ) system's population inversion
has negative Kelvins. See:
Since we started with over half the atoms in the spin-down state,
initially this drives the system towards a 50/50 mixture,
so the entropy is _Increasing_ [ dissipated energy ],
corresponding to a positive temperature.
However, at some point more than half of the spins are in
the spin-up position. In this case, adding additional energy
_Reduces_ the entropy [ accumulated energy ]
since it moves the system further from a 50/50 mixture.
This reduction in entropy with the addition of energy
corresponds to a negative temperature.
This phenomenon can also be observed in many lasing systems,
wherein a large fraction of the system's atoms
( for chemical and gas lasers ) or electrons ( in semi-conductor lasers )
are in excited states. This is referred to as a population inversion.
__ WikiPedia.ORG/wiki/Negative_temperature
Eric Gisse
Jeff...Relf wrote:
[snip garbage]
Have you ever taken a course in physics...EVER?
Peter Christensen
I understand what you mean. Just some particle is just much easier than
some E-fields (electric) and B-fields (magnetic)...
But these fields are actually quite difficult to convert into quantums
of photons.
pet c
Peter Christensen
Jeff...Relf skrev:
> Hi Sam_Wormley, You wrote:
>
> PhysLink.COM:
>
> When a cosmologist talks about the 'temperature' of a photon they are
> basically describing the equivalent energy of a photon.
> PhysLink.COM/education/askexperts/ae210.cfm
>
> Planck's Law
> http://scienceworld.wolfram.com/physics/PlanckLaw.html
>
> Good links. There are lots of definitions of temperature.
>
> Although ideal laser ( one that's perfectly coherent ) is at absolute zero,
> strong ones can burn through steal ( thanks to their relativistic mass ).
> How cool is that ?
I was also thinking about lasers and also semiconductor devices like
LED's (ligth emitting diods). Here the wave-length has absolutely
nothing to do with the temperature of the emitting device. The energy
levels for electrons are defining the energies for the emitted photons,
and the electrons are in some non-equlibrium states, where the concept
'temperature' doesn't really make sence.
pet c
Peter Christensen
Eric Gisse skrev:
> Jeff...Relf wrote:
>
> [snip garbage]
>
> Have you ever taken a course in physics...EVER?
Personally, I don't really like the idea of negative temperatures. I've
been working with lasers and semiconductor physice for a few years,
without ever hearing about 'negative Kelvins'...
pet c
Peter Christensen
mme...@cars3.uchicago.edu skrev:
> In article <44fd41e7$0$75040$1472...@news.sunsite.dk>, "Peter Christensen" <Pe...@MailAPS.org> writes:
> >Does it make sence to talk about the temperature of a photon? (-My own
> >answer is no.)
> >
> A photon, no. Photons, yes.
Agree.
> >It always applies that E = p*c (p is the momentum which depend only on the
> >wavelength) for a photon. But as the speed of light is always c, there can't
> >be a kinectic energy as for massive particles,
>
> Oh, of course there can be kinetic energy. All of photon's energy is
> kinetic.
I don't think about it like that, but I understand what you mean. When
I think about photons, there aren't any 'velocity' for E = 1/2*m*v^2,
apart from c of course, to define a kinetic energy. -But if you insist
about 'kinetic energy', then that it's ok for me. I understand what you
mean. But I've learn't (recently), that one shold avoid both talking
about potential and kinetic energies when working with relativity or
'relastivic particles' like photons.
> > where E_kin = 1/2 m*v^2 can
> >be expressed as a measure of temperature for a particle.
> >
> Not quite.
Maybe from an old book in termodynamics. I assume, that you understand
what I'm thinking, but ok: 'Not quite'...
> >I know, that a radiation profile (photons) can express the temperature of
> >it's source, but I still don't think, that it will make sence to talk about
> >a temperature for single photons, as it would do when talking about massive
> >particles with E = E_kin = k_B*T.
>
> It doesn't make sense to talk about temperature for single massive
> particels, either. Temperature is an ensamble property.
I think, that you've right. -Once again I've been thinking about some
old book in termodynamics, where T was a measure for the speed, or
kinetic energy, of the individual particles...
> >
> >Are there anybody who would like to try to define a temperature for a
> >photon?
> >
> Check out cavity radiation.
A photon as a measure of a temperature, I understand it like that. But
I still wouldn't like to define the 'temperature of a photon'. Maybe
it's just a hypothetical question...
-I remember our discussion about the 'mass of the photon' (agree that
it's zero) some time ago. This time it isn't a posting for April 1st.
(Or do you think so? :-) ) -Thanks for the reply...
pet c
-I post this question, because I think that it could be a good thread.
(good subject or not?), not because I'm in doubt about it, but ok,
always nice to learn something new...
PD
Peter Christensen wrote:
> mme...@cars3.uchicago.edu skrev:
>
> > In article <44fd41e7$0$75040$1472...@news.sunsite.dk>, "Peter Christensen" <Pe...@MailAPS.org> writes:
> > >Does it make sence to talk about the temperature of a photon? (-My own
> > >answer is no.)
> > >
> > A photon, no. Photons, yes.
>
> Agree.
>
> > >It always applies that E = p*c (p is the momentum which depend only on the
> > >wavelength) for a photon. But as the speed of light is always c, there can't
> > >be a kinectic energy as for massive particles,
> >
> > Oh, of course there can be kinetic energy. All of photon's energy is
> > kinetic.
>
> I don't think about it like that, but I understand what you mean. When
> I think about photons, there aren't any 'velocity' for E = 1/2*m*v^2,
> apart from c of course, to define a kinetic energy. -But if you insist
> about 'kinetic energy', then that it's ok for me. I understand what you
> mean. But I've learn't (recently), that one shold avoid both talking
> about potential and kinetic energies when working with relativity or
> 'relastivic particles' like photons.
Terminology issue. You are taking KE to be *defined* as 1/2*m*v^2. This
is not a good definition. A better definition is the energy, minus the
rest energy, minus potential energy. In the case of the photon, the
energy it has is what's left.
>
> > > where E_kin = 1/2 m*v^2 can
> > >be expressed as a measure of temperature for a particle.
> > >
> > Not quite.
>
> Maybe from an old book in termodynamics. I assume, that you understand
> what I'm thinking, but ok: 'Not quite'...
>
> > >I know, that a radiation profile (photons) can express the temperature of
> > >it's source, but I still don't think, that it will make sence to talk about
> > >a temperature for single photons, as it would do when talking about massive
> > >particles with E = E_kin = k_B*T.
> >
> > It doesn't make sense to talk about temperature for single massive
> > particels, either. Temperature is an ensamble property.
>
> I think, that you've right. -Once again I've been thinking about some
> old book in termodynamics, where T was a measure for the speed, or
> kinetic energy, of the individual particles...
Again, a misuse of terminology. An individual atom with a speed does
not have a temperature. A *collection* of interacting particles has a
temperature, and T is a measure of the *mean* kinetic energy of that
collection.
Randy Poe
Peter Christensen wrote:
> mme...@cars3.uchicago.edu skrev:
>
> > In article <44fd41e7$0$75040$1472...@news.sunsite.dk>, "Peter Christensen" <Pe...@MailAPS.org> writes:
> > >Does it make sence to talk about the temperature of a photon? (-My own
> > >answer is no.)
> > >
> > A photon, no. Photons, yes.
>
> Agree.
>
> > >It always applies that E = p*c (p is the momentum which depend only on the
> > >wavelength) for a photon. But as the speed of light is always c, there can't
> > >be a kinectic energy as for massive particles,
> >
> > Oh, of course there can be kinetic energy. All of photon's energy is
> > kinetic.
>
> I don't think about it like that, but I understand what you mean. When
> I think about photons, there aren't any 'velocity' for E = 1/2*m*v^2,
> apart from c of course, to define a kinetic energy.
But E = 1/2*mv^2 is not the definition of kinetic energy.
Not even for trains and ping-pong balls..
> -But if you insist
> about 'kinetic energy', then that it's ok for me. I understand what you
> mean. But I've learn't (recently), that one shold avoid both talking
> about potential and kinetic energies when working with relativity or
> 'relastivic particles' like photons.
>
> > > where E_kin = 1/2 m*v^2 can
> > >be expressed as a measure of temperature for a particle.
> > >
> > Not quite.
>
> Maybe from an old book in termodynamics.
What you probably saw was that the AVERAGE kinetic
energy (per degree of freedom) over a collection
of particles is a measure of temperature.
Temperature by definition is an average over a population.
> > It doesn't make sense to talk about temperature for single massive
> > particels, either. Temperature is an ensamble property.
>
> I think, that you've right. -Once again I've been thinking about some
> old book in termodynamics, where T was a measure for the speed, or
> kinetic energy, of the individual particles...
Average speed of the particles, given a collection.
- Randy
mme...@cars3.uchicago.edu
>
>mme...@cars3.uchicago.edu skrev:
>
>> In article <44fd41e7$0$75040$1472...@news.sunsite.dk>, "Peter Christensen" <Pe...@MailAPS.org> writes:
>> >Does it make sence to talk about the temperature of a photon? (-My own
>> >answer is no.)
>> >
>> A photon, no. Photons, yes.
>
>Agree.
>
>> >It always applies that E = p*c (p is the momentum which depend only on the
>> >wavelength) for a photon. But as the speed of light is always c, there can't
>> >be a kinectic energy as for massive particles,
>>
>> Oh, of course there can be kinetic energy. All of photon's energy is
>> kinetic.
>
>I don't think about it like that, but I understand what you mean. When
>I think about photons, there aren't any 'velocity' for E = 1/2*m*v^2,
>apart from c of course, to define a kinetic energy. -But if you insist
>about 'kinetic energy', then that it's ok for me. I understand what you
>mean. But I've learn't (recently), that one shold avoid both talking
>about potential and kinetic energies when working with relativity or
>'relastivic particles' like photons.
>
With potential one should be careful.
>> > where E_kin = 1/2 m*v^2 can
>> >be expressed as a measure of temperature for a particle.
>> >
>> Not quite.
>
>Maybe from an old book in termodynamics. I assume, that you understand
>what I'm thinking, but ok: 'Not quite'...
>
>> >I know, that a radiation profile (photons) can express the temperature of
>> >it's source, but I still don't think, that it will make sence to talk about
>> >a temperature for single photons, as it would do when talking about massive
>> >particles with E = E_kin = k_B*T.
>>
>> It doesn't make sense to talk about temperature for single massive
>> particels, either. Temperature is an ensamble property.
>
>I think, that you've right. -Once again I've been thinking about some
>old book in termodynamics, where T was a measure for the speed, or
>kinetic energy, of the individual particles...
>
Context is important. In the framework of thermodynamics it is always
understood that:
1) We're talking about an ensamble (in fact, all thermodynamic
properties are in the limit of N-> infinity where N is the number of
particles).
2) The velocities are relative to the rest frame of the ensamble as a
whole.
Absent the second point you would get the "coonclusion" that the air
in a moving car is at room temperature relative to the car but at much
higher temperature relative to the ground:-)
So, the thing you quote, explicitly stated, would go like: "Given some
amount of gas at temperature T, the average kinetic energy of a
molecule, in the rest frame of the ensamble, is proportional to T".
That's all.
If you'll talk with accelerator people you'll find that they often use
the term "cooling the beam". And no, this doesn't mean lowering the
energy of the beam, just lowering the energy spread, i.e. the
dispersion of particle velocities relative to the mean velocity.
>> >
>> >Are there anybody who would like to try to define a temperature for a
>> >photon?
>> >
>> Check out cavity radiation.
>
>A photon as a measure of a temperature, I understand it like that. But
>I still wouldn't like to define the 'temperature of a photon'.
As I said, there ain't no such thing.
> Maybe it's just a hypothetical question...
>It is no more hypothetical than "temperature of gas molecule". There
>ain't no such thing either. There is "temperature of an ensamble of
gas molecules" and if you've such ensamble and it is at thermal
equilibrium, you can, from the temperature, to find the mean velocity,
the velocity dispersion etc. for an individual molecule. But this
doesn't mean that you can take an individual molecule and talk about
its temperature.
So, it is really no different with photons, except that the statistic
is different. Given an ensamble of photons in thermal equilibrium
with some body (the additional body is necessary in this case, do you
know why?), the ensamble has a temperature and from this temperature
you can find the average energy of a photon, the energy dispersion, in
fact the whole energy distribution. But there is no temperature for
an individual photon.
>
>-I remember our discussion about the 'mass of the photon' (agree that
>it's zero) some time ago. This time it isn't a posting for April 1st.
>(Or do you think so? :-) )
Oh no, not at all. First, it is not April and second, I remember the
previous exchange.
-Thanks for the reply...
>
Sure, you're welcome.
>pet c
>
>-I post this question, because I think that it could be a good thread.
>(good subject or not?), not because I'm in doubt about it, but ok,
>always nice to learn something new...
>
It is a good question, a very good one indeed. Provides an opening
for clarification of terms commonly used (and abused). The statement
"temperature is just the kinetic energy of particles" is way to often
used and abused. As it stands, without all the qualifiers mentioned
above, it is just plain false.
Jeff…Relf
to create the required Bose-Einstein condensate, so it does seem that
" temperature " has something to do with it.
Entropy is the Joules per Kelvin which can't be increased by an ignition.
During population inversion, Kelvins are negative because
adding Joules _Reduces_ entropy... puting quanta in spin-up ( excited ) states.
See: WikiPedia.ORG/wiki/Negative_temperature
Peter Christensen
> In article <1158142120....@h48g2000cwc.googlegroups.com>, "Peter Christensen" <Pe...@MailAPS.org> writes:
> >
> >mme...@cars3.uchicago.edu skrev:
> >
> >> In article <44fd41e7$0$75040$1472...@news.sunsite.dk>, "Peter Christensen" <Pe...@MailAPS.org> writes:
> >> >Does it make sence to talk about the temperature of a photon? (-My own
> >> >answer is no.)
> >> >
> >> A photon, no. Photons, yes.
> >
> >Agree.
> >
> >> >It always applies that E = p*c (p is the momentum which depend only on the
> >> >wavelength) for a photon. But as the speed of light is always c, there can't
> >> >be a kinectic energy as for massive particles,
> >>
> >> Oh, of course there can be kinetic energy. All of photon's energy is
> >> kinetic.
> >
> >I don't think about it like that, but I understand what you mean. When
> >I think about photons, there aren't any 'velocity' for E = 1/2*m*v^2,
> >apart from c of course, to define a kinetic energy. -But if you insist
> >about 'kinetic energy', then that it's ok for me. I understand what you
> >mean. But I've learn't (recently), that one shold avoid both talking
> >about potential and kinetic energies when working with relativity or
> >'relastivic particles' like photons.
> >
> Kinetic is OK, well defined. Just use E_k = E - sqrt(E^2 - p^2c^2).
> With potential one should be careful.
OK
> >> >I know, that a radiation profile (photons) can express the temperature of
> >> >it's source, but I still don't think, that it will make sence to talk about
> >> >a temperature for single photons, as it would do when talking about massive
> >> >particles with E = E_kin = k_B*T.
> >>
> >> It doesn't make sense to talk about temperature for single massive
> >> particels, either. Temperature is an ensamble property.
> >
> >I think, that you've right. -Once again I've been thinking about some
> >old book in termodynamics, where T was a measure for the speed, or
> >kinetic energy, of the individual particles...
> >
> Context is important. In the framework of thermodynamics it is always
> understood that:
>
> 1) We're talking about an ensamble (in fact, all thermodynamic
> properties are in the limit of N-> infinity where N is the number of
> particles).
> 2) The velocities are relative to the rest frame of the ensamble as a
> whole.
>
> Absent the second point you would get the "coonclusion" that the air
> in a moving car is at room temperature relative to the car but at much
> higher temperature relative to the ground:-)
>
> So, the thing you quote, explicitly stated, would go like: "Given some
> amount of gas at temperature T, the average kinetic energy of a
> molecule, in the rest frame of the ensamble, is proportional to T".
> That's all.
Using another frame, which is moving compared to the rest frame of the
ensamble, would definatively give some strange results. As this can't
really be handled, I'm sure, that we will never get a 4-vector for
temperature.
> If you'll talk with accelerator people you'll find that they often use
> the term "cooling the beam". And no, this doesn't mean lowering the
> energy of the beam, just lowering the energy spread, i.e. the
> dispersion of particle velocities relative to the mean velocity.
May I ask, if you have been working with accelerators?
> > Maybe it's just a hypothetical question...
>
> >It is no more hypothetical than "temperature of gas molecule". There
> >ain't no such thing either. There is "temperature of an ensamble of
> gas molecules" and if you've such ensamble and it is at thermal
> equilibrium, you can, from the temperature, to find the mean velocity,
> the velocity dispersion etc. for an individual molecule. But this
> doesn't mean that you can take an individual molecule and talk about
> its temperature.
No, that's right.
Another example: Sometimes in plasmaphysics for fusion (for example
www.ITER.org) people are talking about 'temperature' as a measure of
the kinetic energy of (individual) particles. But again, it must be
only on ensembles of particles, and not on individual particles. -I
think, that there is some confusion about 'con-fusion', as some people
are saying, even among the best scientists today... I'm interested in
plasma-physics for fusion myself. (Even though, that I'm not actually
working with the subject at all.)
> So, it is really no different with photons, except that the statistic
> is different. Given an ensamble of photons in thermal equilibrium
> with some body (the additional body is necessary in this case, do you
> know why?), the ensamble has a temperature and from this temperature
> you can find the average energy of a photon, the energy dispersion, in
> fact the whole energy distribution. But there is no temperature for
> an individual photon.
> >
> >-I remember our discussion about the 'mass of the photon' (agree that
> >it's zero) some time ago. This time it isn't a posting for April 1st.
> >(Or do you think so? :-) )
>
> Oh no, not at all. First, it is not April and second, I remember the
> previous exchange.
On Fools Day (April 1st): http://en.wikipedia.org/wiki/April_Fool's_Day
and second:
"It is widely celebrated on the Internet"
Just a stupid joke gives more attention than a serious posting, that's
worth to remember...
(But you've right, that I've actually got confused about something in
some of my replies.)
Here the joke on mass:
http://groups.google.com/group/sci.physics.relativity/browse_frm/thread/a588600b78c54da4/22535d3586b44eb0?lnk=st&q=pech%40mailaps.org&rnum=1&hl=en#22535d3586b44eb0
(417 replies)
Here a good question on temperature:
http://groups.google.com/group/sci.physics/browse_frm/thread/47626a9b8242be33/02cfe5a053cba16e?lnk=st&q=pech%40mailaps.org&rnum=5&hl=en#02cfe5a053cba16e
(so far 22 replies)
-So the stupid provocation actually got 20 times more replies.
That's worth to think about, I think... (:<)
> -Thanks for the reply...
> >
> Sure, you're welcome.
Thanks Mati.
> >pet c
> >
> >-I post this question, because I think that it could be a good thread.
> >(good subject or not?), not because I'm in doubt about it, but ok,
> >always nice to learn something new...
> >
> It is a good question, a very good one indeed. Provides an opening
> for clarification of terms commonly used (and abused). The statement
> "temperature is just the kinetic energy of particles" is way to often
> used and abused. As it stands, without all the qualifiers mentioned
> above, it is just plain false.
You know probably better than me here. But it's worth to think about,
that temperature is basically a measure of energy, when defined. I
agree.
> Mati Meron | "When you argue with a fool,
> me...@cars.uchicago.edu | chances are he is doing just the same"
Here a fool only on Fools Day :-)
pet c
mme...@cars3.uchicago.edu
Yes, indeed.
>
>> If you'll talk with accelerator people you'll find that they often use
>> the term "cooling the beam". And no, this doesn't mean lowering the
>> energy of the beam, just lowering the energy spread, i.e. the
>> dispersion of particle velocities relative to the mean velocity.
>
>May I ask, if you have been working with accelerators?
>
Oh, yeah. Van de Graafs in the remote past, synchrotrons over the
last two decades.
>> > Maybe it's just a hypothetical question...
>>
>> >It is no more hypothetical than "temperature of gas molecule". There
>> >ain't no such thing either. There is "temperature of an ensamble of
>> gas molecules" and if you've such ensamble and it is at thermal
>> equilibrium, you can, from the temperature, to find the mean velocity,
>> the velocity dispersion etc. for an individual molecule. But this
>> doesn't mean that you can take an individual molecule and talk about
>> its temperature.
>
>No, that's right.
>
>Another example: Sometimes in plasmaphysics for fusion (for example
>www.ITER.org) people are talking about 'temperature' as a measure of
>the kinetic energy of (individual) particles. But again, it must be
>only on ensembles of particles, and not on individual particles.
Exactly. Fully expanded, such statement means "in the rest frame of
this plasma, the mean kinetic energy of the particles is ...". Or,
inthe other words, the temperature is used as a measure of a mean
collision velocity. Not understanding this leads to such silliness as
describing the fusiun occuring when a beam of deutrons, from a desktop
accelator, is fired at tritium targets, as "room temeperature fusion".
> -I think, that there is some confusion about 'con-fusion', as some people
>are saying, even among the best scientists today... I'm interested in
>plasma-physics for fusion myself. (Even though, that I'm not actually
>working with the subject at all.)
I've been very interested in it, many years ago, but eventually I
learned that the field is long on promise, short on delivery. I'm
still hopeful, mind you.
Aha, yes:-)
>That's worth to think about, I think... (:<)
That's something advertising people are keenly aware of.
>
>> -Thanks for the reply...
>> >
>> Sure, you're welcome.
>
>Thanks Mati.
>
>> >pet c
>> >
>> >-I post this question, because I think that it could be a good thread.
>> >(good subject or not?), not because I'm in doubt about it, but ok,
>> >always nice to learn something new...
>> >
>> It is a good question, a very good one indeed. Provides an opening
>> for clarification of terms commonly used (and abused). The statement
>> "temperature is just the kinetic energy of particles" is way to often
>> used and abused. As it stands, without all the qualifiers mentioned
>> above, it is just plain false.
>
>You know probably better than me here. But it's worth to think about,
>that temperature is basically a measure of energy, when defined. I
>agree.
>
So here we end with full agreement again. You realize that goes
against the tradition of this ng, right:-)
>> Mati Meron | "When you argue with a fool,
>> me...@cars.uchicago.edu | chances are he is doing just the same"
>
>Here a fool only on Fools Day :-)
>
We need more (much more) of such "foolishness", to make sci.physics a
decent place.
G=EMC^2 Glazier
effects Bert