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George
2006-Aug-22, 02:46 AM
I didn't think electrons were capable of scattering visible light. However, based on a book I'm reading, light emission from the lower region of the solar corona is a result of scattering by electrons. Is it the extremely high temperatures which may be generating relativistic speeds and, thus, greater masses in electrons, thereby making them "large" enough to scatter light?

Jeff Root
2006-Aug-22, 03:50 AM
What else but electrons would scatter light?

That's not really a rhetorical question. Although I do expect that
protons and atomic nuclei can scatter light of the right frequency
as well as electrons do.

-- Jeff, in Minneapolis

George
2006-Aug-22, 03:52 AM
It was my understanding electrons, free elctrons, were too small to scatter light (at least visible light).

Ken G
2006-Aug-22, 04:41 AM
Free electrons have a constant cross section, independent of the light frequency, when treated nonrelativistically. This is called "Thompson scattering", and has a cross section roughly the square of the "classical radius of the electron" (the radius when e^2/r ~ mc^2). That radius is about the square of the fine structure constant smaller than a hydrogen atom, so about 10^(-4) angstroms. Much smaller than the wavelength of visible light, but still a significant size for scattering light.

George
2006-Aug-22, 12:58 PM
Much smaller than the wavelength of visible light, but still a significant size for scattering light.
Thanks, Ken. I can't recall where I read they were too small.

Jeff Root
2006-Aug-22, 08:06 PM
Maybe this be an appropriate thread in which to ask...

What techniques are available to determine the size of the
electron? What results do those techniques give?

-- Jeff, in Minneapolis

korjik
2006-Aug-22, 08:41 PM
dunno if they do it this way but:
1) get the radius of the neutron from a proton/neutron scattering or a neutron/neutron scattering experiment.

2) scatter electrons off the neutrons.

the cross sections involved should be dependent on the radius of the particles. I believe that current theory says the total cross section involved should equal the neutron cross section, leaving the electron as a point particle.

Anyone feel free to prove me wrong. This is just mostly a guess

grav
2006-Aug-22, 09:37 PM
I was going to ask about scattering of electrons and baryons, but let me make it simpler. Let's say a basketball lay on the ground and we threw a tennis ball at it, which has one fourth the radius. The effective perimeter in which acollision can take place would seem to be pi(Rb+2Rt)^2, whereas the radius to the outer perimeter would be that of the basketball plus the diameter of the tennis ball. On the other hand, if we threw the basketball at a stationary tennis ball, it would seem it should be pi(2*Rb+Rt)^2. Of course, the distance between their centers would be the same in each case, (Rb+Rt), but which effective perimeter for collision is correct? It would seem it should be the same in either case since in free space we could not tell whether the tennis ball was moving toward the basketball or vice versa.

George
2006-Aug-22, 11:40 PM
After reviewing a brief article, per Ken's answer, on Thompson scattering by electrons, I am still a bit puzzled. Do the electrons gain, or lose, momentum during the scattering event, thus disallowing elastic scattering (e.g. Rayleigh Scattering)?

[I think I was stuck on Rayleigh Scattering when I thought electrons could not scatter.]

Ken G
2006-Aug-23, 01:27 AM
Of course, the distance between their centers would be the same in each case, (Rb+Rt), but which effective perimeter for collision is correct?

Neither, the center of either ball must hit a circle of area pi*(Rb+Rt)^2. You're right it should be the same either way.

Ken G
2006-Aug-23, 01:51 AM
What techniques are available to determine the size of the
electron? What results do those techniques give?


The electron is considered a point particle because it has no internal energy modes that can vary, like internal excited states. But externally, it can manifest as though it has a size in a lot of ways. In an atom, it has an uncertainty in its location of order the Bohr radius (though that has more to do with its interaction with the proton). But you can also think of the Bohr radius as the scale on which the electron begins to exhibit obvious wavelike behavior (its deBroglie wavelength) for its interaction energy with the proton. That latter concept gives a size scale for the electron to be able to interfere with itself, that applies in any situation where you know its energy. Then on smaller scales, you have the Compton wavelength, which is kind of a minimum deBroglie wavelength (where the electron starts behaving relativistically). Finally, smaller still, you have the "classical radius" of the electron, which is the maximum scale it can interfere with itself constructively when it interacts with electromagnetic fields (so that controls its light scattering cross section when not relativistic). Those are all "sizes" for the electron's response, but they are not considered to be the actual size of the point electron.

Ken G
2006-Aug-23, 01:54 AM
Do the electrons gain, or lose, momentum during the scattering event, thus disallowing elastic scattering (e.g. Rayleigh Scattering)?


It depends on how much energy/momentum they started with. Ultimately, if exposed long enough to a thermal radiation field, they would equilibrate their temperature with the radiation, and vice versa (as per the CMB before it initially decouples.)

George
2006-Aug-23, 03:56 AM
Yes, like a isothermal gas, on average it's all the same temperature, but individually there is a little dancing going on. Light will encounter different electron momentums producing inelastic scattering, but the net result would be a statistical Planck curve given enough time and space. Is this pretty close?

I had forgotten that Thompson Scattering explains the opaqueness of light prior to recombination, as well as, the erratic random walk in the sun.

Ken G
2006-Aug-23, 04:21 AM
Right, on all counts.

George
2006-Aug-23, 04:28 AM
* oinks and shows off acorn * :)

grav
2006-Aug-23, 01:46 PM
Neither, the center of either ball must hit a circle of area pi*(Rb+Rt)^2. You're right it should be the same either way.
That would seem like it should be the case, for symmetry of either way we happen to look at it, but what if we placed each ball at the center of a well with a radius of 2Rb? If the tennis ball were dropped randomly with the basketball at the bottom, there would be plenty of places it could fall without colliding with the baketball, since the total collision area is much smaller than the area of the well in this case, pi(Rb+2Rt)^2<pi(2Rb)^2. But if the basketball were dropped in with the tennis ball at the bottom, the basketball would collide with it every time (if the bottom were water or soft mud so it could keep going), since the collision area is larger than the area of the well, where pi(2Rb+Rt)^2>pi(2Rb)^2. Now, it seems my placement of the ball in the center of the bottom of the well should be the key. But this would have nothing to do (as far as I can tell) with the sum of the radii of the two. It would have to do with the ratio of the effective target area to the total possible target area, and the consideration of all possible placements within that possible target area as well. But then, if we widened the total possible target area immensely, then the total effective collision area would once again depend on which ball is considered as the target. I know this can't be right, however. Please help me to visualize it.

Ken G
2006-Aug-23, 02:37 PM
But if the basketball were dropped in with the tennis ball at the bottom, the basketball would collide with it every time
Only if the tennis ball was placed in the center, as you surmised. That's not a fair comparison-- you have to place one ball at random and drop the other at random as well, since collisions are a random process. If you do it that way, you'll get the same collision probability either way you do it, as you realize it must.

Spaceman Spiff
2006-Aug-23, 04:31 PM
After reviewing a brief article, per Ken's answer, on Thompson scattering by electrons, I am still a bit puzzled. Do the electrons gain, or lose, momentum during the scattering event, thus disallowing elastic scattering (e.g. Rayleigh Scattering)?

[I think I was stuck on Rayleigh Scattering when I thought electrons could not scatter.]

Rayleigh scattering happens only to bound electrons. Thomson/Compton scattering is an effectively free electron scattering. I say "effectively" because if the photon energy is >> binding energy of a bound electron, then Compton scattering can indeed occur, the recoil of which then can unbind that electron.

George
2006-Aug-24, 12:09 AM
Thanks. Apparently, Compton scattering occurs much more often than Rayleigh scattering even when you have an equal number of protons and free electrons.

My interest was to understand the difference between them. Rayleigh scattering does not alter the light's wavelength, yet Compton does. My guess was a momentum exchange occurs between a photon and electron for inelastic scattering. Is this correct?

Ken G
2006-Aug-24, 01:00 AM
The difference is, in Rayleigh scattering, the electron is anchored to a proton (it happens to bound electrons, as the spaceman said), so it doesn't recoil. A free electron does recoil, especially if you hit it with relativistic energies.

George
2006-Aug-24, 03:48 AM
...and the change in motion of the electron makes the scattering inelastic. Is that a fair causal explanation?

Ken G
2006-Aug-24, 05:53 AM
Yes, if the electron picks up energy by recoiling, then the photon loses it.

George
2006-Aug-25, 04:18 AM
While we're here....

I don't see how scattering could explain the final photon distribution from the sun, when comparing them to the higher energy, original core photons. Do electrons recombine by loosing energy during scattering, or by some kind of quantum wiggle, or both? Thus, allowing the emission of weaker photons when they are ejected again?

Ken G
2006-Aug-25, 01:31 PM
There are many ways to downgrade photon energies. Free electrons can do it slowly with all their recoiling, or recombination can do it, as you say. The real point is, it doesn't matter the details of what does it, eventually the light will come to temperature equilibrium with the material through which it passes. In the case of the Sun, the temperature structure is dictated by the rules of radiation transport, and the matter comes to the temperature of the radiation field.

George
2006-Aug-25, 03:15 PM
What intrigues me is the difference in the number of photons that leave the sun vs. the number generated in the core. Although the total radiance is the same, there must be a mechanism which converts the greater energitic photons into multiple less energetic photons. There's a dance goin' on in there. :)

Ken G
2006-Aug-25, 04:59 PM
Yes, quite so. It is energy that is conserved, so high energy photons are downconverted into many low energy ones. The opposite also happens, but at any given temperature, there is simply many many more ways for the photons to have the distribution of energy we see than any other. The detailed processes that allow this are not important, just as the details of how you throw dice don't affect the result either (if that isn't true for you, take me to Vegas!). That's the beauty of thermodynamics, the "dance" is going on, but you don't need to know the steps.

George
2006-Aug-25, 05:56 PM
I suspect your dice analogy is a good one. A stastical approach is well respected, though I've forgotten most of this dance (it is kind fast paced). ;)

trinitree88
2006-Aug-26, 10:51 PM
George. When the photon loses energy to free electrons, it's Compton scattering. If the temperature of the electrons happens to be high, the reverse process may occur....the photon gains energy from the electrons. That's inverse Compton scattering. Statistically, it's less likely for a low temperature bunch of electrons, as their velocities are Maxwellian. Pete.

George
2006-Aug-27, 02:11 AM
George. When the photon loses energy to free electrons, it's Compton scattering. If the temperature of the electrons happens to be high, the reverse process may occur....the photon gains energy from the electrons. That's inverse Compton scattering. Statistically, it's less likely for a low temperature bunch of electrons, as their velocities are Maxwellian. Pete.
Thanks, Pete. Somehow I had in my mind that photons just weren't going to be bothered with electrons. This was when I was stuck on elastic scattering, however.

I had suspected photons could gain energy since it is easy to visualize an electron slowing down at times when encountering a photon. Would the statistical ups and downs match Planck's BB curve, ignoring absorption bands from atoms?

trinitree88
2006-Aug-27, 10:51 PM
Thanks, Pete. Somehow I had in my mind that photons just weren't going to be bothered with electrons. This was when I was stuck on elastic scattering, however.

I had suspected photons could gain energy since it is easy to visualize an electron slowing down at times when encountering a photon. Would the statistical ups and downs match Planck's BB curve, ignoring absorption bands from atoms?

George. You're welcome on the first part. Your question is a bit tricky for me. I don't know that directly. A black body curve is not quite the same shape as a maxwellian distribution of velocities, and the root equations are not identical, so they're only similar curves by inspection.
Since line spectra from atoms undergoing recombination can shift both up, and down, the stellar spectra exhibit broadening of these lines, then, when the photons pass through the outer regions of the "cooler" stellar atmospheres, some photons can be reabsorbed, causing dark line spectra.
It is interesting to note that Doppler broadening of hydrogen line spectra, due to Compton, and inverse Compton scattering, by electrons is how they infer the temperature of the plasma in Tokamak fusion reactors.:cool: Pete.

George
2006-Aug-28, 01:11 AM
Your question is a bit tricky for me. I don't know that directly. A black body curve is not quite the same shape as a maxwellian distribution of velocities, and the root equations are not identical, so they're only similar curves by inspection.
Hmmmm... I hadn't meant "up and down" to apply to electron velocities but it is an interesting thought. :)


Since line spectra from atoms undergoing recombination can shift both up, and down, the stellar spectra exhibit broadening of these lines, then, when the photons pass through the outer regions of the "cooler" stellar atmospheres, some photons can be reabsorbed, causing dark line spectra. Is the broadening greater with greater tempuratures, up to a point, I suppose?

It would be interesting to see someone do a portrayal of the random walk of a single progenitor gamma ray as it starts in the core of the sun and travels to the surface, passing through the various shells, resulting in its more than 5 thousand colorful offspring.

Ken G
2006-Aug-28, 01:15 AM
There are only three reasons that the Planck function and the Maxwellian distribution are different. One is in the way they fill in energy states, which is different partly because photons are relativistic and a Maxwellian is not. The second is in the fact that you generally have a fixed number of electrons that are sharing the energy, whereas photons appear and disappear permanently, in respose to the temperature. Finally, electrons are "fermions" and photons are "bosons", which means you can only put one electron in one state, but any number of photons. Put together, those three effects are the only differences, otherwise it's all just counting the states available at any total energy. Detailed mechanisms don't matter at all, given enough time-- that's the beauty of thermodynamics in a nutshell.

George
2006-Aug-28, 12:30 PM
It would just be a guess, but wouldn't the two distributrion curves (Planck and Maxwellian) be similar, nevertheless?

papageno
2006-Aug-28, 09:53 PM
It would just be a guess, but wouldn't the two distributrion curves (Planck and Maxwellian) be similar, nevertheless?
Only where e(h nu)/(kBT) >> 1.

trinitree88
2006-Aug-28, 10:28 PM
Hmmmm... I hadn't meant "up and down" to apply to electron velocities but it is an interesting thought. :)

Is the broadening greater with greater tempuratures, up to a point, I suppose?

It would be interesting to see someone do a portrayal of the random walk of a single progenitor gamma ray as it starts in the core of the sun and travels to the surface, passing through the various shells, resulting in its more than 5 thousand colorful offspring.



George. Yes, the Doppler broadening increases with temperature. I was taking a tour of MIT's plasma fusion center, and somebody asked the guide what kind of "thermometer" they used...as a joke...to measure the plasma temperature. A short discourse on lines becoming bands ensued...interesting. Pete.

George
2006-Aug-29, 02:57 AM
George. Yes, the Doppler broadening increases with temperature. I was taking a tour of MIT's plasma fusion center, and somebody asked the guide what kind of "thermometer" they used...as a joke...to measure the plasma temperature. A short discourse on lines becoming bands ensued...interesting. Pete.
That is pretty interesting and indicative of the complexity of those things. Too bad the surrounding air doesn't just glow blue as an indication of strength. :)