# Growth of Radially Symmetric Potential Fields

Suppose I have a function $F: R^3 \to R^3$ which satisfies:

1) There exists $\Psi: R^3 \to R$ such that $F = \nabla \Psi$ and

2) $F(x)$ depends only on $\|x\|$

Can I conclude that $\|F(x)\| = C\frac{1}{\|x\|^2}$ for some constant $C$? Or maybe at least $\|F(x)\| \le C\frac{1}{\|x\|^2}$? Maybe there is another condition to add that would let me make the conclusion?

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How about $F=Const$? (any constant vector field is the gradient of the corresponding linear function) – fedja Jul 22 '11 at 2:19
I think @fedja's comment can be made stronger: Only a constant vector field can satisfy those conditions. Consider what $\nabla\times F = 0$ implies for $F_y$ and $F_z$ at $x = (1,0,0)$. (Unless you mean it's $\lVert F(x) \rVert$ and not $F(x)$ that depends only on $\lVert x \rVert$.) – Rahul Jul 22 '11 at 2:47

The question is, as others pointed out with their good natured jokes, indeed, a bit ill-formulated. I guess that you meant: 2) $\Psi(x)$ only depends on $||x||$. But my somewhat educated guess is that you wonder, whether a spherically symmetric potential leads to a force field that decays according to the familiar quadratic reciprocal law.

The counterexample that you may be most familiar with would be a 3-dimensional harmonic oscillator. Its potential is $\Psi(x,y,z)=k(x^2+y^2+z^2)$, which surely has spherical symmetry, and $\nabla \Psi=2k(x,y,z)=2k\vec{r}$ does not decay but increases in strength as the distance from the origina grows (the force really is the negative gradient of the potential, but let's ignore that).

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