Consider a volume $V$ in $\mathbb{R}^3$ enclosed by a closed surface (orientable) $S$ with outgoing normal unit vector $\hat{n}$. Consider also a scalar field $f:\mathbb{R}^3 \to \mathbb{R}$ that depends only on the radial coordinate $r$, in spherical coordinates that is $f=f(r)$.

In a proof on textbook, under the previous definitions, in a passage it is written that


Which seems to me as a sort of "chain rule for directional derivatives". If these where total or partial derivatives (as the one with respect to $r$ is) I would not have problems, but here one derivative is directional in the direction fo $\hat{n}$ (and also the radial derivative can be seen as a directional dervative). I have two main questions about it:

  • What theorem / definition is equality $(A)$ based on? Is it an application of chain rule?
  • Since $r$ is the radial coordinate what is exactly $\frac{dr}{dn}$? Using a possible definition for the directional derivative $\frac{dr}{dn}=\nabla r \cdot \hat{n}$, but $\nabla r$ is $\hat{r}$, so is this term equal to the radial component of $\hat{n}$ (or the normal component of $\hat{r}$)?

My considerations:

The directional derivative of $f$ in the direction of $\hat{n}$ is denoted by $\frac{df}{dn}=\nabla f \cdot \hat{n}$, while, since $f=f(r)$ I can consider the derivative of $r$ with respect to $r$, which is a "total" derivative $\frac{df}{dr}$.

But also, since the radial component of $\nabla f$ in spherical coordinates is just $\frac{\partial f}{\partial r}$ I can consider the derivative of $f$ with respect to $r$ as a directional derivative and write $\frac{df}{dr}=\nabla f \cdot \hat{r}$.

So I have

$$\frac{df}{dn}=\nabla f \cdot \hat{n} \,\,\,\,\,\,\,,\,\,\,\,\,\,\frac{df}{dr}=\nabla f \cdot \hat{r} $$

Substituting in $(A)$ this gives

$$\nabla f \cdot \hat{n}=\nabla f \cdot \hat{r} \frac{dr}{dn}\implies \hat{n}=\hat{r}\frac{dr}{dn}\implies \hat{r} \,\,\, \mathrm{is \,\, always \,\, parallel \,\, to} \,\,\hat{n} ?!$$

Also, since they are unit vectors, from the previous follows that $ \frac{dr}{dn}=\pm 1$ always. So there are some consequence from $(A)$ that does no make sense to me.

  • $\begingroup$ @Chappers I apologize, if I'm not wrong again it should be $\nabla r=\hat{r}$ so I modified the question $\endgroup$
    – Gianolepo
    Aug 18, 2017 at 21:32

1 Answer 1


It's better to think of the directional derivative as $$ \frac{dF}{dv}(a) = \lim_{h \to 0} \frac{F(a+hv)-F(a)}{h} = \left. \frac{d}{dt}\right|_{t=0} F(a+tv) : $$ in other words, take the function given by $t \mapsto F(a+tv)$ (i.e. $F$ with argument along a straight line through $a$ with direction vector $v$), and then find the ordinary derivative of this function at $x=0$. This idea extends to more a more general context.

Since if $F$ is differentiable the directional derivative is linear in $v$, it then follows that it can be written as $\nabla F \cdot v$ for a particular vector $\nabla F$ called the gradient.

Now, $r$ is just another function $\mathbb{R}^3 \to \mathbb{R}$. Meanwhile, $f: \mathbb{R} \to \mathbb{R}$ is an ordinary function. So $df/dn$ has to mean something else: it should strictly speaking be $ d(f \circ r)/dn $, since $f \circ r$ is a function $\mathbb{R}^3 \to \mathbb{R}$. But by the definition of the directional derivative, $$ \frac{d(f \circ r)}{dn} = \left. \frac{d}{dt}\right|_{t=0} f(r(a+tv)) = \left. \frac{df}{dr} \right|_{r=r(a)} \left. \frac{d}{dt} \right|_{t=0} r(a+tv), $$ by the ordinary chain rule. The latter expression is simply $f'(r) dr/dn$.

Now, your other questions. Indeed, $dr/dn = \nabla r \cdot n = \hat{r} \cdot n$, so it is the $\hat{r}$-component of $n$, and by symmetry and since $n$ is normalised, the $n$-component of $\hat{r}$.

I think several of your other points are best answered by remembering the difference between $f$ and $f \circ r$. But lastly, indeed $df/dr = \nabla (f\circ r) \cdot \hat{r} $. So $$\nabla (f\circ r) \cdot n = \nabla (f\circ r) \cdot \hat{r} \frac{dr}{dn}.$$ But this does not mean that $n = \hat{r} (dr/dn)$: it only tells you about the component in the direction of $\nabla(f\circ r)$. But $\nabla (f \circ r) = f'(r) \nabla r = f'(r) \hat{r} $ by the chain rule, so all you really have is that $$ f'(r) \hat{r} \cdot n = f'(r) \hat{r} \cdot \hat{r} \frac{dr}{dn}, $$ or $ \hat{r} \cdot n = dr/dn $, which you knew anyway.


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