3
$\begingroup$

Let $\mathbf{R} = [x,y,z]$ be a cartesian vector, $R_\alpha$ it's tensor representation with $\alpha = x,y,z$ and let $R=\sqrt{x^2 + y^2 + z^2}$ be its norm. I want to do tensor derivatives of the Coulomb potential $1/R$. The first derivative is $\frac{\partial}{\partial R_\alpha} \frac{1}{R} = -\frac{R_\alpha}{R^3}$ and the second derivative is $\frac{\partial}{\partial R_\beta} \frac{\partial}{\partial R_\alpha} \frac{1}{R}= \frac{\delta_{\alpha\beta}R^2 - R_\alpha R_\beta }{R^5}$. I want to make further derivatives in Mathematica.

I tried

R = Sqrt[x^2 + y^2 + z^2]

$\sqrt{x^2 + y^2 + z^2}$

rR = 1/R

$\frac{1}{\sqrt{x^2 + y^2 + z^2}}$

drR = Grad[rR, {x, y, z}, "Cartesian"]

$\{-\frac{x}{(x^2 + y^2 + z^2)^{3/2}}, -\frac{y}{(x^2 + y^2 + z^2)^{3/2}}, -\frac{z}{(x^2 + y^2 + z^2)^{3/2}} \}$

So can I make Mathematica identify the denominators as $R^3$ and the numerators as $R_\alpha$ and get it to the compact form $-\frac{R_\alpha}{R^3}$, or is there some other way to do tensor calculus/arithmetics compactly?

$\endgroup$
9
$\begingroup$

Perhaps something like the following will suffice?

R /: D[R, R[α_], NonConstants->{R}] := R[α]/R
R /: D[R[α_], R[β_], NonConstants->{R}] := KroneckerDelta[α, β]

R /: MakeBoxes[R[α_], fmt_] := MakeBoxes[Subscript[R,α], fmt]

Your examples:

D[1/R, R[α], NonConstants->{R}] //TeXForm

$-\frac{R_{\alpha }}{R^3}$

D[1/R, R[α], R[β], NonConstants->{R}] //TeXForm

$\frac{3 R_{\alpha } R_{\beta }}{R^5}-\frac{\delta _{\alpha ,\beta }}{R^3}$

$\endgroup$
  • $\begingroup$ This is beautiful, thanks a lot! Maybe you put in Clear[R] at the top, because I didn't get it to work first since $R$ was already defined in the notebook. $\endgroup$ – Jonatan Öström May 2 '17 at 10:06

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.