Checking if a function is an equipotential surface

[patric44]

Homework Statement

checking if a function is could form an equipotential surface

Relevant Equations

laplacian(f)/ |grad(f)|^2 = constant

hi guys
I came across that theorem that could be used to check if a surface represented by the function f(x,y,z) = λ could represent an equipotential surface or not, and it states that if this condition holds:
$$\frac{\nabla^{2}\;f}{|\vec{\nabla\;f}|^{2}} = \phi(\lambda)$$
then f(x,y,z) could represent an equipotential surface, my question here is this theorem applicaple also in spherical cordinates as its or it should be modifed? also can some one explain the intuition behind it

 

[pasmith]

$f(x,y,z) = \lambda$ will be an equipotential surface if $\nabla f$ is everywhere parallel to $\nabla V$. Thus $$\nabla f = \alpha(\mathbf{x})\nabla V$$ where $\alpha \neq 0$. Then $$\begin{split} \nabla^2f &= \alpha\nabla^2 V + \nabla \alpha \cdot \nabla V &= \frac{\nabla \alpha}{\alpha} \cdot \nabla f \end{split}$$ assuming $\nabla^2 V = 0$. You can then get $$\nabla^2 f = \frac 1{\alpha(\lambda)} \frac{d\alpha}{d\lambda} \|\nabla f\|^2 = \phi(\lambda)\|\nabla f\|^2$$ if you assume that $\alpha$ depends on $\mathbf{x}$ only in the combination $f(\mathbf{x})$ but I don't see why that assumption is required.

As the result is obtained using vector notation, it is valid in any orthogonal coordinate system.

 

[patric44]

$f(x,y,z) = \lambda$ will be an equipotential surface if $\nabla f$ is everywhere parallel to $\nabla V$. Thus $$\nabla f = \alpha(\mathbf{x})\nabla V$$ where $\alpha \neq 0$. Then $$\begin{split} \nabla^2f &= \alpha\nabla^2 V + \nabla \alpha \cdot \nabla V &= \frac{\nabla \alpha}{\alpha} \cdot \nabla f \end{split}$$ assuming $\nabla^2 V = 0$. You can then get $$\nabla^2 f = \frac 1{\alpha(\lambda)} \frac{d\alpha}{d\lambda} \|\nabla f\|^2 = \phi(\lambda)\|\nabla f\|^2$$ if you assume that $\alpha$ depends on $\mathbf{x}$ only in the combination $f(\mathbf{x})$ but I don't see why that assumption is required.

As the result is obtained using vector notation, it is valid in any orthogonal coordinate system.

thanks its clear now, but can you explain the last step
$$\nabla^2 f = \frac 1{\alpha(\lambda)} \frac{d\alpha}{d\lambda} \|\nabla f\|^2 = \phi(\lambda)\|\nabla f\|^2$$

 

[pasmith]

$f(x,y,z) = \lambda$ will be an equipotential surface if $\nabla f$ is everywhere parallel to $\nabla V$. Thus $$\nabla f = \alpha(\mathbf{x})\nabla V$$ where $\alpha \neq 0$. Then $$\begin{split} \nabla^2f &= \alpha\nabla^2 V + \nabla \alpha \cdot \nabla V &= \frac{\nabla \alpha}{\alpha} \cdot \nabla f \end{split}$$ assuming $\nabla^2 V = 0$. You can then get $$\nabla^2 f = \frac 1{\alpha(\lambda)} \frac{d\alpha}{d\lambda} \|\nabla f\|^2 = \phi(\lambda)\|\nabla f\|^2$$ if you assume that $\alpha$ depends on $\mathbf{x}$ only in the combination $f(\mathbf{x})$ but I don't see why that assumption is required.

Taking the curl of the first equation gives $$0 = \nabla \alpha \times \nabla V = \alpha(\nabla \alpha \times \nabla f)$$ so $\nabla \alpha$ and $\nabla f$ are parallel. Thus $\alpha$ is constant on surfaces of constant $f$, so indeed $\alpha = \alpha(\lambda)$.