pines-demon
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According to the paper (link above), they had a 48,8 times amplification. Which lead to a ##\omega\approx 6\times10^{-3}\,##rad/s (?) estimate. Maybe adjust for the parameters of the experiment, but you are in the ballpark.Charles Link said:The change in angular momentum I think is very small, even if you get every electron to go to the other spin state. Let's try to quantify that: ## mvr=\hbar \approx 1.0 E-34 ## joule-sec. Let's work with 22 lbs. of iron=10 kg=10^4 grams. atom weight=56, so we have about 200 moles ## \approx 1.2 E +26 ## atoms, (I'm going to assume one electron per atom, even though there may be more than one), so that ## mvr \approx 1.0 E-8 ## joule-sec. The ##10 ## kg of iron has volume ## \approx 1 E-3 ## m^3, so that ## r \approx .05 ##meters. Using ## I \approx Mr^2/2 \approx .01 ## gives ## \dot{\theta}=\omega=mvr/I=1E-6 ## radians/sec ##\approx 5E-5 ## degrees/sec. I didn't check the arithmetic carefully, but it appears (and I think this part is correct), any motion is extremely small and would be difficult to observe, even if you had one ## \hbar ## for each electron.