Re: Strange vdw_dU/dl values in thermodynamic integration

From: Floris Buelens (
Date: Tue Oct 26 2010 - 01:23:47 CDT

Hi Hugh,

What you say below sounds right, with alchVdwShiftCoeff=0 you have linear
scaling of your potential, which means for two potentuals U1 and U2, you have
something like
U = lambda * U1 + (1 - lambda) * U2
Differentiate with respect to lambda and you get dU/dlambda = U1 - U2 (like you
say, the difference between unscaled potentials)
Since there's no lambda dependence in U1 - U2, dU/dlambda is the same for any
given configuration of the system regardless of the lambda value. But at
different lambda values you're sampling a different energy landscape so the
ensemble average you collect (which is what you integrate) does depend on the
value of lambda.
With alchVdwShiftCoeff not equal to zero you lose the simple dependence on
lambda - dU/dl is no longer just the difference between unscaled potentials and
your instantaneous value of dU/dl does change when you change lambda.
Best wishes,


----- Original Message ----
From: Hugh Heldenbrand <>
To: Floris Buelens <>
Sent: Fri, 22 October, 2010 18:33:20
Subject: Re: namd-l: Strange vdw_dU/dl values in thermodynamic integration

  Hi Floris,

I think that I see the mistake that I am making. dU/dl works out to be
the difference in potentials (for linear coupling) of the two states
(and the second state is 0 for a simple decoupling). I have been
mistakenly thinking that it should be the difference in the potentials
after they have been scaled by lambda. But that is not correct, it is
the difference in potentials without any other factor at all. And for
the soft-core potential the derivative is something else entirely.

So dU/dl does not depend explicitly on lambda. But in a free energy
simulation the system in propagated on the potential energy surface of
the combined potential (scaled by lambda), so you will sample
configurations in a lambda-dependent way. So you will end up with dU/dl
averages that do change with lambda, even though the dependence is not

Is this correct? I'm sorry about all the confusion, and I don't mean for
you to have to teach me TI. I am just trying to understand the results
I am getting from NAMD.


On 10/22/2010 01:36 AM, Floris Buelens wrote:
> When applying a soft-core potential the derivative is no longer a linear
> function of lambda. The VDW potential should be equal to dU/dl if you set
> alchVdwShiftCoeff to 0, which reduces the soft core potential back to linear
> scaling of Lennard-Jones.
> ----- Original Message ----
> From: Hugh Heldenbrand<>
> To:
> Sent: Fri, 22 October, 2010 2:37:25
> Subject: namd-l: Strange vdw_dU/dl values in thermodynamic integration
> Hello-
> I am attempting to do thermodynamic integration calculations using parameter
> coordinate files in the AMBER format. Some of the vdw_dU/dl values that I was
> getting seemed strange, so I did a test on two isolated atoms with zero charge
> and the vdw parameters for the C atom type from the AMBER parm99 forcefield.
> my test the two atoms are separated by three angstroms and I decouple one of
> them. My understanding is that for a lambda value of 0 the two atoms should
> interact fully with a vdw_dU/dl value equal to their VDW potential energy.
> However, in my output file the VDW potential energy is 0.8144, and the
> is 4.7432 for the 0th timestep.
> Here are the settings that I use to make NAMD AMBER forcefield compatible:
> # AMBER settings
> rigidTolerance 0.0005
> cutoff 10
> switching off
> amber on
> exclude scaled1-4
> 1-4scaling 0.833333
> Although in this case there aren't any 1-4 terms or rigid bonds.
> Here are the TI settings that I used:
> alch on
> alchType ti
> alchLambda 0.00
> alchLambda2 0.00
> alchFile ./namd_data/cc.pdb
> alchCol B
> alchOutFreq 1
> alchOutFile ti_out
> alchVdwShiftCoeff 5
> alchVdwLambdaEnd 0.5
> alchElecLambdaStart 0.5
> Am I misunderstanding TI vdw interactions or is this a bug?
> -Hugh Heldenbrand
> University of Minnesota


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