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

  • I
  • Thread starter swampwiz
  • Start date
  • #1
293
10

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?
 

Answers and Replies

  • #2
BvU
Science Advisor
Homework Helper
2019 Award
13,298
3,128
It can't be proven because it is not true. Divergence is a number.

Did you mean gradient ?
 
  • #3
stevendaryl
Staff Emeritus
Science Advisor
Insights Author
8,401
2,580
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}##.
 
  • Like
Likes swampwiz, lavinia and fresh_42
  • #4
lavinia
Science Advisor
Gold Member
3,204
604
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:
  • #5
293
10
It can't be proven because it is not true. Divergence is a number.

Did you mean gradient ?
Yes, I meant gradient.
 
  • #6
293
10
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.
 
  • #7
Orodruin
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
Gold Member
16,814
6,609
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.
 

Related Threads on What is the proof that the divergence is normal to the surface?

Replies
8
Views
3K
Replies
5
Views
3K
Replies
4
Views
1K
Replies
8
Views
3K
Replies
1
Views
3K
  • Last Post
Replies
9
Views
2K
Replies
5
Views
3K
Replies
9
Views
7K
  • Last Post
Replies
2
Views
3K
Replies
15
Views
3K
Top