# Heat, Atoms and Photons

1. Jun 6, 2009

### treddie

What does it mean to heat an isolated atom with photons, from a hypothetical 0deg, to 100deg? What is happening in the atom's electron shells, and how long can the atom retain the heat?

2. Jun 6, 2009

### diazona

Well, pretty much the only way a single isolated atom can gain energy is to absorb a photon, which increases the principal quantum number of one of its electrons. But the electron would almost immediately drop back down to the ground state (that's a probabilistic process, but the lifetime is very short). And besides, that doesn't really count as "heat." If you're really going to heat an atom up, it needs to be in some sort of potential that allows it to vibrate, or at least move around without leaving a confined area.

3. Jun 6, 2009

### Bob S

At 0 and 100 deg C (273 and 373 kelvin), the only energy the atom (or molecule) is thermal (kinetic) energy. The average mean square velocity v2 is related to temperature by
(1/2)Mv2 = (3/2)RT
where M is mass per mole, R = 8317 meters2/sec2-kelvin and T in kelvin. So
v=sqrt(3RT/M)
Using nitrogen gram molecular mass M= 28
we get root mean square v = 493 meters per sec when it is in a box at 273 kelvin.
So velocity is proportional to square root of temperature.
[Added edit] Now if you put nitrogen molecules in an insulated perfectly reflecting box in thermal equilibrium with thermal photons at 273 kelvin, their rms velocity will not change. But if you put them in intergalactic space (where the temperature is several kelvin), they will maintain their velocities until they collide and exchange energy with other molecules or photons. Nitrogen can radiate microwave photons.

Last edited: Jun 7, 2009
4. Jun 7, 2009

### gonegahgah

If you have two isolated photons that are traveling separate paths which during their journey come close enough to "freeze" together will heat radiation be released?

5. Jun 7, 2009

### diazona

I'm not sure what you mean - photons can't "freeze" together.

It is possible for two photons to collide, and in doing so they can turn into any of several possible sets of particles (including just two photons).

6. Jun 7, 2009

### gonegahgah

Oops! I meant atoms. Sorry.
I should have written:
If you have two isolated atoms that are traveling separate paths which during their journey come close enough to "freeze" together will heat radiation be released?

Last edited: Jun 7, 2009
7. Jun 7, 2009

### treddie

Well, pretty much the only way a single isolated atom can gain energy is to absorb a photon, which increases the principal quantum number of one of its electrons. But the electron would almost immediately drop back down to the ground state (that's a probabilistic process, but the lifetime is very short). And besides, that doesn't really count as "heat." If you're really going to heat an atom up, it needs to be in some sort of potential that allows it to vibrate, or at least move around without leaving a confined area.

I pretty much came to the same conclusion but this leads to my confusion concerning what exactly is heat. For instance, heat is considered in one line of thought as vibration. But vibration of what? I can only surmise that it is really the "white noise" if you will, a slew of asynchronous events of electrons jumping out of and back into their ground states, as photons are absorbed and released.

At 0 and 100 deg C (273 and 373 kelvin), the only energy the atom (or molecule) is thermal (kinetic) energy. The average mean square velocity v2 is related to temperature by
(1/2)Mv2 = (3/2)RT
where M is mass per mole, R = 8317 meters2/sec2-kelvin and T in kelvin. So
v=sqrt(3RT/M)
Using nitrogen gram molecular mass M= 28
we get root mean square v = 493 meters per sec when it is in a box at 273 kelvin.
So velocity is proportional to square root of temperature.

Please correct me if I am wrong, but I can't see that velocity alone of an atom is really temperature since it has nothing to do with activity in the electron shells. The only time it seems to enter in is when that atom is NOT in isolation and collides with other atoms, transferring momentum that gets portioned out partly as linear motion, and partly as heat ("vibration" as asynchronous electron jumps).

The equation "(1/2)Mv2 = (3/2)RT" seems confusing to me since I can't imagine that the overall velocity of a heated atom would be anything other than mean zero. We know that photons transfer momentum to atoms, but unless all of the photons are coming from a given direction, their directions of collision with the atom are random and should sum to mean zero. And when the atom is NOT in isolation, velocity would at least in part, depend on pressure, which is absent from the equation.

Last edited: Jun 7, 2009
8. Jun 8, 2009

### Bob S

The total energy of a molecule is potential energy (excited state) plus kinetic energy E (heat) (mv2/2). Put your hand in a hot oven and feel convection of heat (kinetic energy) in air. The momentum of gas molecules (=sqrt(2mE)) hitting walls of enclosures is what creates gas pressure.
You are correct that the mean velocity is zero. If there are a billion atoms moving to the right, and a billion to the left, the average velocity is near zero. But the sum of the squares of the velocities of all molecules is a positive definite number, as is the average of the squares of all velocities. So the square root of the average is also positive definite. At low temperatures, nearly all the "heat" energy in gas molecules is kinetic energy. My equation above is just an extension of the ideal gas equation PV = nRT, covered in most high school and college freshmen physics books.

9. Jun 9, 2009

### treddie

Oh, OK...sum of squares makes complete sense...you want the ave SPEED, not the ave velocity. But I don't doubt your equation. And yes, it's basic physics. It just seems counter-intuitive. For instance, atoms moving together through space can be "hot" or "cold", regardless of how fast they are moving. If we say that a given value of heat is the process of colliding molecules at a given speed, then using that logic and applying it to an interstellar plasma at the incredibly low, low densities of those plasmas, means that the probability of collisions is enormously low compared to say, air at STP. Yet an interstellar plasma can have temperatures around 8000K. That's 8000K with relatively few collisions, AND plenty of time for individual atoms to return to their ground states. So where is the heat in an interstellar plasma? It should be near 0 deg K. Unless the intense heat has nothing to do with the speed of the atoms, but instead is due to constant bombardment by intense radiation from a suitable source.

10. Jun 9, 2009

### mikelepore

treddie, if I read you right, I think you said you expect a low temperature when the density is very low? That doesn't follow. The average kinetic energy of a group of particles doesn't become small just because there are few particles. Compare to other examples of averages -- the average speed of ten cars isn't necessarily smaller than the average speed of a thousand cars.

To refer to "the heat in" individual substances or locations is an incorrect use of the word. Things have internal energy. The word heat only refers to the transfer of energy from one body to another, for example, the conduction of heat from a higher temperature region to a lower temperature region.

11. Jun 9, 2009

### mikelepore

When you say atoms freeze together, do you mean nuclear fusion? A lot of heat is released. Fusion is what's happening in the sun. Both electromagnetic waves and high speed particles come out of it.

12. Jun 12, 2009

### treddie

Then here is my current reasoning:

An ice cube at rest in a vacuum with no heat source has a temperature of roughly 0deg K, and that same ice cube moving 500 miles per hour still has the same temperature (assuming that it magically got to 500 miles per hour without having ACCELERATED to do so). Now, I can see that there are in essence two different velocities here; the velocity of the mass as a whole, and the internal ave. velocities of all the molecules in the ice cube relative to one another. So it seems that the velocity being referred to is the internal ave. velocity in the equation, "(1/2)Mv2 = (3/2)RT", which directly relates v^2 to T. But here is where my confusion is. Instead of an ice cube in space, imagine a super hot plasma, typically found in nebulae being bombarded by intense proton radiation from a black hole let's say. The problem here, is that the plasma can have a density so incredibly low, that few of the atoms in that plasma actually physically interact, so the rate of collisions is very low. This being my assumption, even though when a "rare" collision occurs, momentum is transferred and photons released as a consequence (and a corresponding temperature produced), the energy density would be so low as to be practically undetectable. Yet we can read the temperatures of super-heated plasmas from thousands and millions of light years away. There is an 'invisible" heat in the system (the POTENTIAL collision energies in the system), but it will not register as temperature UNTIL collisions occur.

But maybe this is where my problem is...sure, the density of the plasma may be very low, and the corresponding intensity of its temperature very low, but the SIZE of the plasma cloud is astronomically huge and confined to viewing angles on the order of maybe a degree or less (due to the vast distances between us and the plasma). This means that those rare collision events are confined to a very small viewing area, which increases the APPARENT energy density relative to us over a given viewing area. In other words, distance compresses all of those widely spaced collisions into a small viewing area, thus compresses the amount of photons we see into that same area, so we see a much more easily detectable temperature. If we were very close to the cloud and took a temperature reading across the same angle of view, we would be sampling a much smaller area in the cloud and the energy density relative to us would be very low, perhaps undetectable.

So I can only surmise that there are two heats in the system:
1. The mutual velocities (invisible heat) of the particles in the system is potential energy that can not be read off as a temperature (visible heat), until,

2. A collision occurs causing the release of photons that CAN be read off as a detectable temperature (visible heat).

Temperature would, therefore, be the INDIRECT detection of the kinetic energy in the system ONLY after collisions occur. Therefore, since we are indirectly reading the energy in the system (the speed of the particles) only by the release of photons, we can INFER the kinetic energy in the system by the wavelength of the released photons.

Am I getting closer? Or farther away?