Re: The temperature of a photon ?

In article <1158743865.802669.63930@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>, "Peter Christensen" <PeCh@xxxxxxxxxxx> writes:
mmeron@xxxxxxxxxxxxxxxxxx wrote:
In article <1158142120.208363.61430@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>, "Peter Christensen" <PeCh@xxxxxxxxxxx> writes:

mmeron@xxxxxxxxxxxxxxxxxx skrev:

In article <44fd41e7$0$75040$14726298@xxxxxxxxxxxxxxx>, "Peter Christensen" <PeCh@xxxxxxxxxxx> 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.

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.

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.

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,
meron@xxxxxxxxxxxxxxxxx | 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.

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

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