From: \ (jonathan_at_ibt.unam.mx)
Date: Sun Jul 17 2005 - 12:09:41 CDT
I am not familiar with Langevin Dynamics, so I can only answer question two.
The only thing you missed here is that the Dirac delta does have dimensions, and
they are inverse to the ones of its variable.
This is easy to notice from the function's particular property that
Integral(DiracDelta(x-x0)*dx)=1 if x runs along an interval that contains x0.
Through dimensional analysis, we see that the right side is adimensional. On the
left side there is a "sum of products", so the dimensions are those of the
products. That is, each of the products DiracDelta(x-x0)*dx must be adimensional
(like 1 on the right hand). We know that dx has the dimensions of x, so the
dimensions of the delta must the inverse of these.
Thus, in this case, DiracDelta(t-t') has dimensions of (time)^(-1), so the
dimensions of 2*m*gamma*kB*T*delta(t-t') are
(mass)*(time)^(-1)*(force)*(distance)*(time)^(-1) or (force)^2.
Hope you get the rest of the answers.
Quoting Blake Charlebois <bdc_at_mie.utoronto.ca>:
> QUESTION TWO:
> It seems to me that the dimensions of the left- and right-hand sides
> <R(t)*R(t')> = 2*m*gamma*kB*T*delta(t-t') do not agree. What blunder am
> making here?
> The dimensions of gamma are (time)^(-1).
> The dimensions of 2*m*gamma*kB*T*delta(t-t') are
> (mass)*(time)^(-1)*(force)*(distance) or (force)^2*(time).
> The dimensions of <R(t)*R(t')> are (force)^2.
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