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Jan 18, 2016 at 21:35 comment added Charlie Ok, it works with $\Gamma=\frac{k_BT}{2\gamma}$, using the half-maximum convention for the $\Theta$ function.
Jan 18, 2016 at 14:32 comment added Charlie Got it. I implicitly used $\frac{1}{2}m\langle v^2\rangle=\frac{1}{2}k_B T$. But wouldn't $\langle \eta(t) \eta(s)\rangle=\gamma \delta(t-s)$ be wrong, dimensionally speaking?
Jan 18, 2016 at 14:20 history edited Tom-Tom CC BY-SA 3.0
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Jan 18, 2016 at 14:16 vote accept Charlie
Jan 17, 2016 at 21:41 history answered Tom-Tom CC BY-SA 3.0