- #1
Steve Zissou
- 49
- 0
Hello physicians.
Consider the following Brownian motion particle:
[tex] \dot{x}(t)=\alpha(t)+\beta(t)\eta(t) [/tex]
The kinetic energy of which would be
[tex] \frac{1}{2}v^2=\frac{1}{2}(\dot{x}(t))^2 [/tex]
(for some unit mass.)
The potential is...?
Consider the following Brownian motion particle:
[tex] \dot{x}(t)=\alpha(t)+\beta(t)\eta(t) [/tex]
The kinetic energy of which would be
[tex] \frac{1}{2}v^2=\frac{1}{2}(\dot{x}(t))^2 [/tex]
(for some unit mass.)
The potential is...?