I have a functional $\Omega[n,v] = \int \omega \mathrm{d}\boldsymbol{r}$ with $\omega\left(n(\boldsymbol{r},\boldsymbol{R}),v(\boldsymbol{r},\boldsymbol{R})\right)$ (where $n$ and $v$ are functions depending on position in space) and want to calculate the partial derivative with respect to some other vector $\boldsymbol{R}$. I want to know if it is ok to write:

$\frac{\mathrm{d} \Omega}{\mathrm{d}\boldsymbol{R}} = \int \frac{\delta \Omega}{\delta n}\nabla_\boldsymbol{R} n~\mathrm{d}\boldsymbol{r}+\int \frac{\delta \Omega}{\delta v}\nabla_\boldsymbol{R}v~\mathrm{d}\boldsymbol{r}$

(where $\frac{\delta \Omega}{\delta n}$ represent functional derivatives).

I think this should be correct, but I cannot find a proper derivation, can someone help me with that or present a derivation?

If one uses the intuitive chain rule one gets a different result. We can convert the above expression and then compare to the conventional chain rule:

$\frac{\mathrm{d} \Omega}{\mathrm{d}\boldsymbol{R}} = \int \frac{\delta}{\delta n}\left[\int \omega \mathrm{d}\boldsymbol{r}\right] \nabla_\boldsymbol{R}n ~\mathrm{d}\boldsymbol{r}+ \int \frac{\delta}{\delta v}\left[\int \omega \mathrm{d}\boldsymbol{r}\right]\nabla_\boldsymbol{R}v ~ \mathrm{d}\boldsymbol{r} = \int \left(\frac{\partial \omega}{\partial n}-\nabla\cdot \frac{\partial \omega}{\partial \nabla n}\right)\nabla_\boldsymbol{R}n ~ \mathrm{d}\boldsymbol{r}+\int \left(\frac{\partial \omega}{\partial v}-\nabla\cdot \frac{\partial \omega}{\partial \nabla v}\right)\nabla_\boldsymbol{R}v ~ \mathrm{d}\boldsymbol{r}\\=\int \frac{\partial \omega}{\partial n}\nabla_\boldsymbol{R}n ~ \mathrm{d}\boldsymbol{r}+\int \frac{\partial \omega}{\partial v}\nabla_\boldsymbol{R}v ~ \mathrm{d}\boldsymbol{r} \underbrace{- \int \left(\nabla\cdot \frac{\partial \omega}{\partial \nabla n}\right)\nabla_\boldsymbol{R}n ~ \mathrm{d}\boldsymbol{r}-\int \left(\nabla\cdot \frac{\partial \omega}{\partial \nabla v}\right)\nabla_\boldsymbol{R}v ~ \mathrm{d}\boldsymbol{r}}_{additional~terms}\\=\int \frac{\partial \omega}{\partial n}\nabla_\boldsymbol{R}n ~ \mathrm{d}\boldsymbol{r}+\int \frac{\partial \omega}{\partial v}\nabla_\boldsymbol{R}v ~ \mathrm{d}\boldsymbol{r} \underbrace{+ \int \frac{\partial \omega}{\partial \nabla n}\nabla_\boldsymbol{R} (\nabla n)~ \mathrm{d}\boldsymbol{r}+\int \frac{\partial \omega}{\partial \nabla v}\nabla_\boldsymbol{R} (\nabla v)~ \mathrm{d}\boldsymbol{r}}_{additional~terms}$

Where in the last step, I threw away the surface terms, which is valid in my case.

Do I understand it correctly, that if $\omega$ is also a function of $\nabla n$ and $\nabla v$, we necessarily have to include the $additional~terms$, so the conventional chain rule does not work, since the gradients are also considered as test functions?

Edit: we actual want the total derivative of $\Omega$ with respect to the nuclear positions not the partial derivative!


I just realized when looking at a simple example that of course the chain rule has to account for derivatives. Consider e.g. the function

$f(x) = \underbrace{x^2}_{g(x)}+\underbrace{2x}_{g'(x)} = g(x)+g'(x)$

To calculate the total differential, we need to calculate

$\frac{\mathrm{d} f}{\mathrm{d}x} = \frac{\partial f}{\partial g}\frac{\partial g}{\partial x}+\frac{\partial f}{\partial g'}\frac{\partial g'}{\partial x} = 2x+2$

This means for the case above that indeed we need the functional derivatives to calculate the forces.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.