Gradient of scalar field is zero everywhere given boundary conditions

[etotheipi]

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

 

[PeroK]

$$\int_{V} (\nabla \phi') \cdot (\nabla \phi') \propto \int_S \mathbf{n} \cdot (\phi' \nabla \phi')$$

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.

 

[etotheipi]

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!

 

[PeroK]

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.