Erasmus said:
Which stat-mech book are you using,btw?
Fundamentals of Statistical and Thermal Physics
http://www.amazon.co...l/dp/0070518009 , by Frederick Reif.
Qfwfq said:
So, what about polarization dependence? I remember a basic exercise in my RQFT course, which explains it fine for the Compton effect.
Qfwfq,
I am not familiar with a QM polarization explanation in the Compton Effect. Do you have a reference or do you care to explain? How does the Compton effect vary with polarization? I have made some predictions about this and it would be very interesting to see if they are true.
Here is how this New Theory explains the Compton Effect for changed x-ray wavelengths at different angles:
This New Theory predicts that the most likely ejection angle for the Compton electrons would be at 90
o. However, since the Compton experiment uses a solid metal plate as the target, 90
o ejections would be impossible (a metal vapor would be better, allowing transverse ejections). The forward direction would be the most
unlikely for ejection, as the transverse
E field would again tend to eject the electrons sideways. Hence, the most likely ejection direction for a solid plate experiment would be somewhere in between 0
o and 90
o.
The maximum velocity would again be dependent on a
non-acceleration resonance, just like in the photoelectric effect. The maximum energy might not go as
hν, as higher harmonic resonances are probable, reducing the velocity. However, once this velocity is known, then the change in x-ray wavelength would simply be due to
a Doppler shift, as the receding electrons reflect the incident x-rays, shifting the frequency towards the red.
Let's try this out and see how this works. The max ejection velocities are known. The max velocity found for ejected electrons is approximately .07c for Compton's Experiment. (See
Recoil Electrons From Aluminum). We will take the average velocity to be approximately .05c, for a
rough estimate. Also as a
rough estimate to see if we are in the ballpark, we will try 45
o as the most likely ejection angle.
We wish to find the resulting retransmitted wavelength using the Doppler formula:
Now the electron would continuously Doppler-shift the incident wave, from no frequency shift at the beginning of its acceleration, to a maximum shift in frequency at its final velocity (at the end of its acceleration). Thus, the retransmitted wave’s frequency shift would be a broadened spike. The center of the spike would be associated with some electron velocity in between its initial and final velocity, and not the electron’s final velocity. We will make the reasonable assumption that the average retransmission velocity of the pulsating electron is ½ its average final velocity (again for a
rough estimate).
So finally, we have a rough estimate for the Doppler shifted wavelengths:
Not a bad rough estimate (obviously, we could change our very rough assumptions to match the data. We will avoid this, as to not be hypocritical, but some combinations give exact results.)
However, if one changes to a metal vapor target, instead of a metal plate target, things would really change. This would allow transverse ejections, changing the character of the Doppler Shift. It would be very interesting to redo this experiment with a vapor target to see if the Compton wavelengths change. The QM prediction would stay the same.
In addition, polarizations could be considered. For Compton's experiment, if the x-rays were vertically polarized, then the ejections would no longer be in the Compton plane. This would change things. It would be easy to make predictions on how this would change things with this New Theory.
For horizontal polarizations, the ejections would tend to stay in the Compton plane. It would be interesting to do an electron velocity distribution experiment for the Compton electrons to verify these polarization predictions. QM would not make such electron velocity distribution predictions, because QM treats this phenomena as a collision instead of a transverse force ejection. Here, Qfwfq, is where I would like to see your QM Compton polarization results.
Andrew A. Gray