0
$\begingroup$

Given a scalar function (field?) $f(\mathbf{x})$ (where $\mathbf{x}=[x_1,x_2,x_3]$) with gradient

$$\nabla f= [f_{x_1},f_{x_2},f_{x_3}]$$

Where

$$f_{x_i}=\frac{\partial f}{\partial x_i}$$

And given

$$\nabla f \cdot \mathbf{x}=c$$

where $c$ is some constant,

Is there a closed surface such that the outward pointing normal vector is everywhere equal to the argument vector $\mathbf{x}$, so that:

$$\int \int_S \nabla f \cdot \mathbf{x} \:dS=cA$$

where $A$ is the area of the closed surface. And can this closed surface, if it exists, be arbitrarily chosen such that $A=1$ or $A=4\pi r^2$ as convenient, or is it uniquely determined?

  • 1
    For what it's worth, the _only_ surfaces whose normal vector is everywhere proportional to the position vector $\mathbf{x}$ are (portions of) spheres centered at the origin. This has nothing to do with a choice of smooth function $f$. Given that, the condition $c = \nabla f \cdot \mathbf{x}$ on a sphere is quite restrictive. (Offhand, I'd guess it implies $f$ is constant on spheres centered at the origin, though I don't have a proof.) – 2017-01-09
  • 0
    @AndrewD.Hwang Worth a lot to this novice, thanks! And I agree that the condition implies constant $f$, but this is interesting (to me) since by the divergence theorem this will lead you to an inverse square law ($c=\frac{f}{4 \pi r^2}$). Correct me if I'm wrong. – 2017-01-09

0 Answers 0