- #1

#### etotheipi

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