Checking if a function is an equipotential surface

AI Thread Summary
The discussion centers on determining if a function f(x,y,z) = λ represents an equipotential surface using the condition involving the Laplacian and gradient. It is established that f is an equipotential surface if the gradient of f is parallel to the gradient of the potential V, leading to the relationship between their gradients. The theorem's applicability in spherical coordinates is confirmed, as the results derived using vector notation hold in any orthogonal coordinate system. Clarification is sought on the last step of the derivation, which shows that the gradients of α and f are parallel, indicating that α is constant on surfaces of constant f. This understanding reinforces the relationship between the gradients and the nature of equipotential surfaces.
patric44
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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
 
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f(x,y,z) = \lambda will be an equipotential surface if \nabla f is everywhere parallel to \nabla V. Thus <br /> \nabla f = \alpha(\mathbf{x})\nabla V where \alpha \neq 0. Then <br /> \begin{split}<br /> \nabla^2f &amp;= \alpha\nabla^2 V + \nabla \alpha \cdot \nabla V<br /> &amp;= \frac{\nabla \alpha}{\alpha} \cdot \nabla f<br /> \end{split} assuming \nabla^2 V = 0. You can then get <br /> \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.
 
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pasmith said:
f(x,y,z) = \lambda will be an equipotential surface if \nabla f is everywhere parallel to \nabla V. Thus <br /> \nabla f = \alpha(\mathbf{x})\nabla V where \alpha \neq 0. Then <br /> \begin{split}<br /> \nabla^2f &amp;= \alpha\nabla^2 V + \nabla \alpha \cdot \nabla V<br /> &amp;= \frac{\nabla \alpha}{\alpha} \cdot \nabla f<br /> \end{split} assuming \nabla^2 V = 0. You can then get <br /> \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
<br /> \nabla^2 f = \frac 1{\alpha(\lambda)} \frac{d\alpha}{d\lambda} \|\nabla f\|^2 = \phi(\lambda)\|\nabla f\|^2
 
pasmith said:
f(x,y,z) = \lambda will be an equipotential surface if \nabla f is everywhere parallel to \nabla V. Thus <br /> \nabla f = \alpha(\mathbf{x})\nabla V where \alpha \neq 0. Then <br /> \begin{split}<br /> \nabla^2f &amp;= \alpha\nabla^2 V + \nabla \alpha \cdot \nabla V<br /> &amp;= \frac{\nabla \alpha}{\alpha} \cdot \nabla f<br /> \end{split} assuming \nabla^2 V = 0. You can then get <br /> \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 <br /> 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).
 
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