0
$\begingroup$

I don't really get this identity : $$ \frac{\partial\Phi}{\partial n} = \vec n . \nabla \Phi $$ So gradient is: $$ \nabla \Phi = \frac{\partial\Phi}{\partial x} \vec i +\frac{\partial\Phi}{\partial y} \vec j +\frac{\partial\Phi}{\partial z} \vec k $$ Normal vector is:

$$ \vec n = n_x \vec i + n_y \vec j+ n_z \vec k $$ RHS of identity:

$$\vec n . \nabla \Phi = n_x \frac{\partial\Phi}{\partial x} + n_y \frac{\partial\Phi}{\partial y} + n_z \frac{\partial\Phi}{\partial z} $$

So must be:

$$ \frac{\partial\Phi}{\partial n} = n_x \frac{\partial\Phi}{\partial x} + n_y \frac{\partial\Phi}{\partial y} + n_z \frac{\partial\Phi}{\partial z} $$

Now is my question, what is $n_x $ , so the components of the normal vector

Can this be true: $$n_x = \frac{\partial x}{\partial n} $$

$\endgroup$
0
$\begingroup$

The inner product $\mathbf{A}\cdot \mathbf{\hat{B}}$ tells you how much of the vector $\mathbf{A}$ lies along the direction of the unitary vector $\mathbf{\hat{B}}$. If it is zero it means that they are perpendicular, if you get back $|\mathbf{A}|$ it means they were parallel.

Now, the gradient $\nabla \Phi$ is a vector and $\mathbf{\hat{n}}$ is usually an unitary vector. The product

$$ \mathbf{\hat{n}} \cdot \nabla \Phi $$

Is then how much of the vector $\nabla\Phi$ is projected in the direction of $\mathbf{\hat{n}}$, or say in another words, it tells you what is rate of change of $\Phi$ along $\mathbf{\hat{n}}$, hence the name: directional derivative, usually denoted as

$$ \frac{\partial \Phi}{\partial \mathbf{\hat{n}}} = \mathbf{\hat{n}} \cdot \nabla \Phi $$

or as $\nabla_{\mathbf{\hat{u}}}\Phi$. But just to emphasize, it does not mean you are taking the derivative with respect a vector, only means that you are calculating the projection of the derivative along said vector

$\endgroup$
  • $\begingroup$ Hi, thanks for feedback. But I still don't get how in the LHS the $\vec n $ pops in at $\frac {\partial \Phi} {\partial n} $ $\endgroup$ – dreckschweinhund Nov 23 '16 at 11:19
  • $\begingroup$ @dreckschweinhund Just a notation: the rate of change of $\Phi$ in the direction of $\mathbf{\hat{n}}$ $\endgroup$ – caverac Nov 27 '16 at 2:20

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.