Gradient of scalar field is zero everywhere given boundary conditions

  • B
  • Thread starter etotheipi
  • Start date
  • #1
etotheipi
Gold Member
2019 Award
2,703
1,623
I'm struggling with a few steps of this argument. It's given that we have a surface ##S## bounding a volume ##V##, and a scalar field ##\phi## such that ##\nabla^2 \phi = 0## everywhere inside ##S##, and that ##\nabla \phi## is orthogonal to ##S## at all points on the surface.

They say this is sufficient to deduce that ##\phi## equals a constant ##k## everywhere along the surface ##S## (I guess this is because it constrains the gradient vector to always be orthogonal to ##S##... is there a more mathematical way of putting that?) Then they perform a shift ##\phi' = \phi - k## so that ##\phi' = 0## everywhere on ##S##, and immediately write down the relation$$\int_{V} (\nabla \phi') \cdot (\nabla \phi') \propto \int_S \mathbf{n} \cdot (\phi' \nabla \phi')$$Since ##\phi' = 0## everywhere on ##S## the RHS is zero, and because the integrand on the LHS is non-negative it must be the case that ##\nabla \phi' = 0## everywhere inside ##V##, and consequently that ##\nabla \phi = 0## everywhere inside ##V##.

But I was wondering how they came up with that integral equation? Thanks
 

Answers and Replies

  • #2
PeroK
Science Advisor
Homework Helper
Insights Author
Gold Member
14,710
6,954
$$\int_{V} (\nabla \phi') \cdot (\nabla \phi') \propto \int_S \mathbf{n} \cdot (\phi' \nabla \phi')$$

But I was wondering how they came up with that integral equation? Thanks
This is the 3D equivalent of integration by parts, or Green's First Identity.
 
  • Informative
Likes etotheipi
  • #3
etotheipi
Gold Member
2019 Award
2,703
1,623
This is the 3D equivalent of integration by parts, or Green's First Identity.
Cool, I wasn't aware of this. Since ##\triangle \phi' = 0## the identity that you quoted does reduce to their statement (I suppose they were being a little sloppy by leaving out the ##dV## and ##dS##, and making up for it with a ##\propto## sign, but they are easy enough to re-insert). Thanks!
 
  • #4
PeroK
Science Advisor
Homework Helper
Insights Author
Gold Member
14,710
6,954
Cool, I wasn't aware of this. Since ##\triangle \phi' = 0## the identity that you quoted does reduce to their statement (I suppose they were being a little sloppy by leaving out the ##dV## and ##dS##, and making up for it with a ##\propto## sign, but they are easy enough to re-insert). Thanks!
Yes, especially when you are learning this stuff it pays to keep track of things like that.
 
  • Like
Likes etotheipi

Related Threads on Gradient of scalar field is zero everywhere given boundary conditions

Replies
14
Views
1K
  • Last Post
Replies
1
Views
621
Replies
3
Views
4K
Replies
2
Views
1K
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
7
Views
1K
  • Last Post
Replies
3
Views
3K
Replies
3
Views
788
Replies
1
Views
2K
Top