Re: About the command "loadtotalforces"

From: Marc Q. Ma (qma_at_oak.njit.edu)
Date: Fri Aug 26 2005 - 09:25:58 CDT

Hi,

When objects are moving at constant velocity, their acceleration is
zero, which means the forces applied on them is zero. If the atom you
are applying steering force on also moves at constant velocity, the
forces due to other atoms (linear/angular springs, nonbonded forces
...) in the system would be equal to the steering force in the opposite
direction, giving total 0 force on that atom. The atoms would show
acceleration (positive or negative) from time to time, but on average
their velocities should approach a constant. Thus, as Jim pointed out,
you should expect average forces to be zero.

Marc
On Aug 25, 2005, at 7:14 PM, Jim Phillips wrote:

> Hi,
>
> You should consult your local physicist, who will point out that based
> on F=ma, since none of the atoms in your simulation are experiencing a
> consistent acceleration you should expect a net force of zero. If you
> did loadforces instead, then only the atoms you are directly applying
> forces to would have a nonzero average force, and this would be the
> negative of the steering forces you are applying.
>
> -Jim
>
>
> On Mon, 22 Aug 2005, LEWYN LI wrote:
>
>> Dear fellow NAMD users,
>>
>> I am using the command "loadtotalforces" in NAMD to print the
>> forces on selected atoms. I printed out the forces with
>> "loadtotalforces" during a simulation where I was stretching a
>> protein at a constant velocity. I fixed the N-terminal C-alpha atom
>> and moved the C-terminal C-alpha atom at 0.5 A/ps.
>>
>> I originally expected that the atoms near the moving C-terminus
>> would experience the applied force more, therefore the forces on them
>> should be higher than the atoms in the middle.
>>
>> Instead I found that most C-alpah atoms have an average force of
>> approximately zero, regardless of their position along the
>> polypeptide backbone!
>>
>> Has anyone seen something similar in their simulations? Any
>> insight or comment would be greatly appreciated. Thank you very
>> much.
>>
>>
>>
>> Yours faithfully,
>>
>>
>> LEWYN LI
>>
>

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