What is the proof that the divergence is normal to the surface?

[swampwiz]

If I am given a function

f( x , y , z , ...) = C

then the normal direction to it is simply the (unit vector of the) divergence of the function. How has this been proven?

 

[BvU]

It can't be proven because it is not true. Divergence is a number.

Did you mean gradient ?

 

[stevendaryl]

Here's a handwavy proof that can be made rigorous:

Take two nearby points $A$ and $B$ on the surface $f(x,y,z,...) = C$, let $\Delta \vec{r}$ be the displacement from $A$ to $B$. Then the change in $f$ will be approximately given by $f(B) - f(A) = \nabla f \cdot \Delta \vec{r}$. Since $f$ is constant on the surface, $f(B) - f(A) = 0$. So we have:

$\nabla f \cdot \Delta \vec{r} = 0$

which implies that $\nabla f$ is orthogonal to $\Delta \vec{r}$.

 

[lavinia]

Here's a handwavy proof that can be made rigorous:

Take two nearby points $A$ and $B$ on the surface $f(x,y,z,...) = C$, let $\Delta \vec{r}$ be the displacement from $A$ to $B$. Then the change in $f$ will be approximately given by $f(B) - f(A) = \nabla f \cdot \Delta \vec{r}$. Since $f$ is constant on the surface, $f(B) - f(A) = 0$. So we have:

$\nabla f \cdot \Delta \vec{r} = 0$

which implies that $\nabla f$ is orthogonal to $\Delta \vec{r}$.

Equivalently, if c(t) is a curve on the surface $f(x,y,z,...)=C$ then $f(c(t) = C$ so $df/dt = 0$. By the Chain rule $0=df/dt = ∇f⋅dc/dt$. Since $dc/dt$ is tangent to the surface for all such curves $c(t)$, $∇f$ is normal to the surface.

 

[swampwiz]

It can't be proven because it is not true. Divergence is a number.

Did you mean gradient ?

Yes, I meant gradient.

 

[swampwiz]

Here's a handwavy proof that can be made rigorous:

Take two nearby points $A$ and $B$ on the surface $f(x,y,z,...) = C$, let $\Delta \vec{r}$ be the displacement from $A$ to $B$. Then the change in $f$ will be approximately given by $f(B) - f(A) = \nabla f \cdot \Delta \vec{r}$. Since $f$ is constant on the surface, $f(B) - f(A) = 0$. So we have:

$\nabla f \cdot \Delta \vec{r} = 0$

which implies that $\nabla f$ is orthogonal to $\Delta \vec{r}$.

This works for me. I presume that making this rigorous requires some tedious δ-ε analysis.

 

[Orodruin]

This works for me. I presume that making this rigorous requires some tedious δ-ε analysis.

Why? All you need is in #4. It is just the chain rule, which I assume you have already done your epsilon delta construction for.