From: Sterling Paramore (paramore_at_hec.utah.edu)
Date: Fri Feb 09 2007 - 16:35:56 CST
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:
> Marcos,
>
> Well, I'd like to keep the First Law no matter what. So I'd say that
> for adiabatic extension, however the energy changed should be exactly
> equal to the work. I would think that it would end up being \int F*d,
> but I could be wrong.
>
> -Sterling
>
> Marcos Sotomayor wrote:
>
>>
>> Hi Sterling,
>>
>> Let me be more specific. First, when the external force applied to
>> your system is time-dependent, the Hamiltonian of your system is not
>> conserved anymore and does not represent the total energy. Moreover,
>> the expression F \times d, as far as I can tell, is no longer the
>> work done on the system (usually one assumes a time independent
>> hamiltonian/force to derive that expression). So my sentence should
>> have read like: " The quantity F \times d (previously work done)
>> should only be similar to the energy change". Please correct me if I
>> am wrong on that.
>>
>> I think you are right in that the total amount of work done by the
>> external force does match the energy change, but you cannot use Fd to
>> measure it. Therefore, the first law of thermodynamics is not
>> violated. There is a nice article on a similar topic that you may
>> want to check:
>>
>> "Invariants for time-dependent Hamiltonian systems"
>> Jurgen Struckmeier and Claus Riedel
>> Physical Review E 64 026503
>>
>> Regards,
>> Marcos
>>
>>
>>
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