From: Sterling Paramore (paramore_at_hec.utah.edu)
Date: Fri Feb 09 2007 - 17:08:12 CST
In my notes, H refers to the energy of the system of interest (i.e., not
the cantilever). The work is then the work that the force does on the
system of interest. You could also calculate the work done on the total
system (system of interest + cantilever), by just adding in the
potential of the cantilever to the hamiltonian. In this case, I get
\dot{H} = F_SMD * \dot{R}, where R is the equilibrium position of the
SMD potential (if modelled harmonically). This is the work done on the
total system by moving the cantilever equilibrium position. It really
depends on what you want.
-Sterling
Marcos Sotomayor wrote:
>
> Are you sure you can describe the motion of the system when the SMD
> force is applied (\dot{H}) without taking into account the external
> force? I don't understand why you can claim that \dot{H} does not have
> the term dU_SMD(t)/dt?
>
> Marcos
>
> On Fri, 9 Feb 2007, Sterling Paramore wrote:
>
>> Changed my mind, the work should definitely be F*d. I wrote up a
>> short proof here:
>> http://www.cbms.utah.edu/~paramore/smd_work.jpg
>>
>> -Sterling
>>
>> Sterling Paramore wrote:
>>
>>
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