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Thermal radiation question

Thermal radiation question - Physics Forum

Thermal radiation question - Physics Forum. Discuss and ask physics questions, kinematics and other physics problems.


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  #1  
Old 07-08-2005, 12:49 AM
Craig Franck
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Default Thermal radiation question



I've been reading that thermal radiation comes from two components:
state changes of electrons absorbing and emitting photons, and entire
atoms as moving-charges emitting radiation. Nothing I've read
addresses which component is more dominant at various temperatures,
say a human body or a white-hot black body. Also, when atoms
collide at room temperatures, does this cause them to emit photons
or does it just effect their motion?

It seems that if a radiating black body where such that no atoms
collided with each other or any photons, then the temperature should
continually increase with no heat added since all that would radiate
away would be what was given up by the moving atoms in the form
of ER. But if there were no particle collisions, how could the
temperature go up?

The reason I ask is it appears the density of a system of particles
somehow effects how well the 2nd law of thermodynamics functions.

--
Craig Franck
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Cortland, NY


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  #2  
Old 07-08-2005, 06:18 AM
tadchem
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Default Thermal radiation question


"Craig Franck" <[Only registered users see links. ]> wrote in message
news:74kze.9100$4U5.8630@trndny03...

Thermal radiation is always referenced to 'black-body radiation'
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Because the radiation is continually re-absorbed and re-emitted, the
assumption of thermal equilibrium is made - quite justifiably, as a
hypothetical 'non-equilibrium' photon moves at the speed of light and cannot
move further than the greatest linear dimension of the body before it is
reabsorbed.

All equilibrium have been re-emitted, so the identity of the original source
is lost. The only parameter that remains to describe the spectral
distribution (individual energies) of the photons is the temperature.


Colliding atoms absorb and emit photons. Photon energies associated with
molecular collisions are also black body radiation. Practically, the photon
energies associated with molecular collisions at room temperature are in the
far infrared part of the spectrum. You can 'see' them with a thermal
imaging system.


'No particle collisions' means a universe with only one particle. There
would be no motion and no 'temperature.' Such a body could not radiate.
Temperature is a statistical concept - it is a measure of the *average*
kinetic energy of the particles in a system.

The theory of black body radiation essentially says that photons are
particles with precisely determinable kinetic energy, and participate in the
averaging process use to calculate 'temperature.'

Your rationale for temperature increase with no added heat makes no sense.
Energy is energy. If temperature (average kinetic energy) is increasing,
the required energy must come from somewhere. If no energy is added from
the outside, then the only available energy source is the internal structure
of the particles themselves. If the particles are all in the appropriate
distribution of states for the temperature (thermal equilibrium), then there
can be no source of energy to increase the temperature.


The Second Law of Thermodynamics functions perfectly, regardless of the
density of the system. The exact values of the entropy and the Gibbs' Free
Energy of the system will always depend on the density, but the Law itself
works perfectly well regardless:
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Tom Davidson
Richmond, VA


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  #3  
Old 07-09-2005, 12:13 AM
Craig Franck
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Default Thermal radiation question

"tadchem" wrote



I'm thinking of a situation where all the particles are moving synchronously:
They all have exactly the same kinetic energy and never collide. (Perhaps
a perfectly-insulated gas that has existed for an infinite period of time might
get into this state.)


This touches on my point. It seems there are two kinds of heat energy in
the system. The kinetic energy of the atoms, and the photons. An atom
can collide with another atom, or a photon emitted by an atom can collide
with another atom.


Imagine you put an insulator around a glowing black body. Now no
energy escapes. It should get hotter. But -- suddenly -- all the atoms
start moving synchronously. And so do the photons. They never interact.

It seems like a static situation. So it appears the system needs chaotic
interaction between atoms and photons in order to operate according to
the Second Law of Thermodynamics. (It might be the work involved
in creating such an artificial situation makes the temperature
considerations a secondary issue.)

--
Craig Franck
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Cortland, NY


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  #4  
Old 07-09-2005, 10:35 PM
tadchem
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Default Thermal radiation question



Craig Franck wrote:


That state is the idealized Bose--Einstein condensate. It occurs at 0
Kelvins.


The photon energy *IS* kinetic energy. Each photon is moving at c (you
can't get any more kinetic than that). The energy is related to its
wavelength and is an inherent property of the photon. It cannot be
changed without absorbing the photon.

Photon kinetic energy is just as important to the temperature
calculation as particle kinetic energy.


No, it shouldn't. A blackbody (a theoretical construct to be sure) is
in thermal equilibrium with the photons that interact with it. In
fact, as a blackbody radiates its own energy, if it doesn't absorb any
new energy from the universe, it will get cooler.


How? Moving relative to what? Motion requires kinetic energy. Your
idealized situation has no kinetic energy, so for it to get some, it
would need an INPUT of kinetic energy.


Photons that never interact never lose or gain energy. We look at
phtons that are billions or years old and the energy they have lost has
resulted from interaction with gravitational fields or relativistic
doppler shift.

Tom Davidson
Richmond, VA

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  #5  
Old 07-10-2005, 12:45 AM
Craig Franck
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Default Thermal radiation question

"tadchem" wrote



I wasn't considering that. The more energetic the photon, the more kinetic
energy it has. So an X-ray is like a baseball hitting you as opposed to a
marble of visible light.

I thought when an atom absorbed a photon, all the energy from the packet
went into kicking an electron into a higher energy level, not affecting the
course of the atom itself. So the electrons acted like shock absorbers.


Okay, that solves one problem. If you were to insulate a blackbody,
it would maintain the same temperature since no energy was being put
into or allowed to escape the system. That is a static situation even if all
the particles are in a chaotic state.


My conception of this has totally changed, but now it might be a gas
at absolute zero somehow being put into orbit around planet. All the
particles would orbit together, but perhaps not move relative to one
another. So if an astronaut in the same orbit put his hand into the
gas, it would freeze regardless of how fast that orbit was.

--
Craig Franck
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Cortland, NY


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  #6  
Old 07-10-2005, 02:06 PM
tadchem
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Default Thermal radiation question


"Craig Franck" <[Only registered users see links. ]> wrote in message
news:Kc_ze.4903$rx4.3874@trndny05...

the

What an atom does with the energy it absorbs from a photon depends on how
much energy it absorbs. Generally (there *will* be exceptions) very low
energies are absorbed by molecules resulting in an increase in their angular
momentum and rotational energy. Higher energies (roughly infrared) will be
absorbed by molecules increasing the internal vibrations - an often the
rotation as well. Even higher energies (vis, UV) energize individual
electrons in either molecular orbitals or atomic orbitals. X-rays are
usually too energetic for molecules and are absorbed by individual electrons
in individual atoms - up to a point called the X-ray absorbtion edge
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At higher energies, the photon completely ejects electrons, and the excess
energy becomes kinetic energy of the ion-electron system.


Practically, when Bose-Einstein condensates are produced, it is in a massive
vacuum apparatus that compensates for the earth's gravity and rotation
through the fields used to confine the condensate. In the laboratory
carefully controlled beams of photons are used to 'steal' kinetic energy
from atoms once they enter the region of observation within the apparatus,
freezing them in place.


Tom Davidson
Richmond, VA
Tom Davidson
Richmond, VA


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  #7  
Old 07-10-2005, 02:51 PM
N:dlzc D:aol T:com \(dlzc\)
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Default Thermal radiation question

Dear tadchem:

"tadchem" <[Only registered users see links. ]> wrote in message
news:[Only registered users see links. ]...

Tom, I was not sure how linear momentum (the photon) can be
turned into angular momentum (the alteration of an electron's
orbital of a nucleus-electron system). Obviously there should be
some "mechanical kick", in both axes... linearly *and*
rotationally.

Do atoms actually recoil rotationally when a photon is
absorbed/emitted, and the electron doesn't
leave/arrive-from-outside the nucleus? (hopefully you can
decipher my question...)

David A. Smith


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  #8  
Old 07-10-2005, 04:47 PM
tadchem
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Default Thermal radiation question


"N:dlzc D:aol T:com (dlzc)" <N: dlzc1 D:cox T:[Only registered users see links. ]> wrote in
message news:yBaAe.23109$Qo.20148@fed1read01...


The photon has angular momentum of its own. The 'allowable' one-photon
transitions of electron states within an atom always involve a change in the
electron's angular momentum corresponding to the angular momentum of a
photon.

The linear momentum of a photon is pretty small: p = E/c

Linear momentum can become angular momentum in a two-body interaction when
the interaction is analyzed in a center-of-mass coordinate system. When one
body is a photon (rest mass 0), however, the linear momentum of the photon
makes no contribution to the angular momentum of the system.


When you are dealing with QM, anything that you may think is 'obvious' is
probably wrong, especially if it is based upon a mechanical analogy. In QM
you have to ignore the 'mechanics' of what is involved and look solely at
those quantities that are conserved in interactions: net charge, net linear
momentum, net angular momentum, total energy, etc.


Individual atoms (whether neutral or charged) are spherically symmetric.
There is no way to distinguish one spatial orientation from another. The
concept of 'rotation' is meaningless to a bare atom. For a given electron
transition *within* the atom that is a different story. One state *may* be
spherically symmetric (so-called S states, states with angular momentum J =
0) but the other state involved in a single photon interaction that changes
the electron's angular momentum will certainly have a non-zero J, and
'rotation' is a meaningful concept for the state of that electron.

Electron Spin Resonance is all about producing non-spherical states of
atoms/ions (by exciting their electrons with radio waves) and measuring
their interactions with magnetic fields - specifically the rate at which the
atoms/ions spin.


Tom Davidson
Richmond, VA



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