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

• I

## Main Question or Discussion Point

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?

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BvU
Homework Helper
2019 Award
It can't be proven because it is not true. Divergence is a number.

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

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
Staff Emeritus