The following argument was found when I was reading a complex analysis book.

Let $F=\nabla u$ be a gradient field. $\gamma$ be a piecewise smooth curve.Let $\hat{n}=$unit normal vector to $\gamma$. Then $\int_{\gamma} \nabla u . \hat{n} ds$ represents path integral of normal component of $F$ along $\gamma$. Then $\displaystyle \int_{\gamma} \nabla u . \hat{n} ds=\int_{\gamma}\frac{\partial u}{\partial n}ds$. I am really confused on this equation. How can we derive it and what is the meaning of $\displaystyle \frac{\partial u}{\partial n}$ here?

  • $\begingroup$ By definition (see next comment!) $\nabla u \cdot \hat n = {\partial u \over \partial n}$. So it seems to me that you have an extra integral sign in $$ \int\int_{\gamma} \nabla u . \hat{n} ds$$ - i.e. you should have an equality of line integrals. $\endgroup$ – peter a g Sep 25 '16 at 14:48
  • $\begingroup$ Well, I take back the 'by definition'. By the chain rule, the directional derivative $\partial u \over \partial n$ is $\nabla u\cdot \hat n$. The quantity $\partial u \over \partial n$ is the derivative of $u$ along the normal to the curve $\gamma$. $\endgroup$ – peter a g Sep 25 '16 at 14:50
  • $\begingroup$ @peterag :Comments are to improve the quality of the OP's question, which your comment does not. The proper place for the answer you've written is the big, empty box below labelled "Your Answer". $\endgroup$ – Eric Towers Sep 25 '16 at 15:04
  • $\begingroup$ @EricTowers - agreed: I am in the process of writing one up. Thanks though $\endgroup$ – peter a g Sep 25 '16 at 15:04

Suppose that $P$ is a point on the (smooth) curve $\gamma$, and suppose that $\alpha$ is a (smooth) parametrized curve through $P$, with $P=\alpha(0)$, and is normal to $\gamma$ - i.e. tangent vectors (at $P$) to $\alpha$ are perpendicular to tangent vectors to $\gamma$.

Suppose that $\alpha$ is paramatrized by length (for the moment), i.e., $\alpha'(t)$ has norm one. In particular, the vector $\hat{n} = \alpha'(0)$ has norm $1$, and is perpendicular to $\gamma$ at $P$.

Then one can consider the function $f(t) = (u\circ \alpha) (t)$. By definition, $${\partial u \over \partial \hat n} = f'(0), $$ i.e., the LHS is (just) notation for the RHS.

On the other hand, by the chain rule, one has

$$ f'(0) = \nabla u (P) \cdot \hat n = {\partial u\over \partial x} \hat n_x + {\partial u\over \partial y} \hat n_y ,$$ where $\hat n_x$ and $ \hat n_y$ are the components of $\hat n$, and the partial derivatives (${\partial u/ \partial x} $ and ditto for $y$) of $u$ are evaluated at $P$.

Therefore, combining the two equations, $$ {\partial u \over \partial\hat n} =\nabla u (P) \cdot \hat n. $$


  • The choice of $\alpha$ depends only on $\hat n$ and $P$: one can use any curve $ t \mapsto \alpha (t) $ with $P =\alpha(0)$, and $\hat n =\alpha'(0)$: by the chain rule, say - once again: $$ f'(0)= \nabla u(P) \cdot \hat n.$$ So the notation $\partial u/\partial \hat n$ makes sense, as it does not - to repeat - depend on the choice of $\alpha$; for $\alpha$, you can even take a straight line through $P$: $$ \alpha (t) = P + t\, \hat n,$$ where $\hat n$ does not depend on $t$.

  • On the other hand, $\alpha$ depends on $P$ (of course!): $ \alpha= \alpha_P$. In any case, as $\gamma$ is smooth, $\hat n= \hat n_P$ varies with $P$ smoothly, so the quantity $\partial u\over \partial \hat n$ varies smoothly (assuming $u$ is smooth).

  • There are in fact two normals - of course... Implicit in the notation is a choice: e.g., an outward normal.

  • As you see, if $\hat n = (1,0)=\bf i$, $${\partial u\over \partial \hat n}= {\partial u\over \partial x}.$$

  • $\begingroup$ @EricTowers - see? My moral failings are not all encompassing - although a complete list of them probably could not fit into the big empty answer box - oh well. $\endgroup$ – peter a g Sep 25 '16 at 15:58
  • $\begingroup$ This suggests a scenario. I wonder on which Stack such a question would fit: "What are my moral failings? Be specific and complete." :-) (I suspect mine would challenge the ability of the backend database(s) to continue running...) $\endgroup$ – Eric Towers Sep 25 '16 at 16:10

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