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

• I
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?

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

Staff Emeritus
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}##.

• swampwiz, lavinia and fresh_42
Gold Member
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.

Last edited:
swampwiz
It can't be proven because it is not true. Divergence is a number.

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.

Staff Emeritus