Gradient of scalar field is zero everywhere given boundary conditions

In summary, the conversation discusses a problem involving a surface and a scalar field, and explains how using certain conditions allows for the deduction that the field is constant along the surface. The conversation also introduces an integral equation and discusses its relation to Green's First Identity.
  • #1

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
 
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  • #2
etotheipi said:
$$\int_{V} (\nabla \phi') \cdot (\nabla \phi') \propto \int_S \mathbf{n} \cdot (\phi' \nabla \phi')$$

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.
 
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  • #3
PeroK said:
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
etotheipi said:
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.
 
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