# Showing a vector field is not conservative

Let $f: R^n \to R^n$. $||x||$ is Euclidean norm. Define $f(x) = g(||x||)$. where $g: [0, \infty) \to R^n$ is differentiable on $(0, \infty)$. I want to show that $f$ is not conservative on $R^n$. ($g$ is not constant)

### MY Effort:

Write $f(x) = (f_1(x),...,f_n(x))$ and $g(||x||) = (g_1(||x||),...,g_n(||x||))$ Hence,

$$f_i = g_i(||x||)$$. In particular

$$\frac{ \partial f_i}{\partial x_j} = \frac{x_j}{\sqrt{x_1^2+...+x_n^2}} \cdot \frac{d g_i }{d x_j}(||x||)$$

and

$$\frac{ \partial f_j}{\partial x_i} = \frac{x_i}{\sqrt{x_1^2+...+x_n^2}} \cdot \frac{d g_j }{d x_i}(||x||)$$

Is this enough to conclude that mixed partials are not equal, therefore $f$ is not conservative?

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How do you know that they are not equal? $g$ might be constant. – John Ma May 4 '14 at 9:43
Also, is $F=f$? – John Ma May 4 '14 at 9:45
I have edited. yes, $f =F$. Also, I assume $g$ is not constant. – Math can be Fun May 4 '14 at 9:46
Also, it should be $\frac{dg_i}{dt}$ instead. – John Ma May 4 '14 at 9:51
@GiuseppeNegro: It seems that we cannot write the field as something like $f(x) = g(||x||)$. The magnitude are the same but the direction are different – John Ma May 4 '14 at 9:54

We are given a vector-valued function $${\bf g}:\qquad t\mapsto{\bf g}(t)=\bigl(g_1(t),\ldots,g_n(t)\bigr)\qquad(t>0)$$ and define a vector field ${\bf F}=(F_1,\ldots,F_n)$ on $\dot{\mathbb R}^n$ by $${\bf F}({\bf x}):={\bf g}\bigl(|{\bf x}|\bigr)\ .$$ The integrability condition for this field reads $$F_{i.j}({\bf x})-F_{j.i}({\bf x})\equiv 0\qquad \forall i\ne j\ .\tag{1}$$ Now according to the chain rule we have $$F_{i.j}({\bf x})=g_i'\bigl(|{\bf x}|\bigr)\>{\partial|{\bf x}|\over\partial x_j}=g_i'\bigl(|{\bf x}|\bigr){x_j\over{|\bf x}|}\ .$$ The condition $(1)$ can therefore be interpreted as follows: The matrix $$\left[\matrix{g_1'\bigl(|{\bf x}|\bigr)&g_2'\bigl(|{\bf x}|\bigr)&\ldots&g_n'\bigl(|{\bf x}|\bigr)\cr x_1&x_2&\ldots&x_n\cr}\right]\tag{2}$$ must have all subdeterminants of order $2$ equal zero, in other words: have rank $\leq1$, for all ${\bf x}\in\dot{\mathbb R}^n$.

Assume now that ${\bf g}'(r)\ne{\bf 0}$ for some $r>0$. Then the rank condition on the matrix $(2)$ would imply that all vectors ${\bf x}$ on the sphere $S^{n-1}_r$ of radius $r$ are multiples of ${\bf g}'(r)$, which is absurd.

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This is a comment with unwieldy formulas.

The correct formulas: $$\frac{ \partial f_i}{\partial x_j} = \frac{x_j}{\sqrt{x_1^2+...+x_n^2}} \cdot g_i'(||x||)$$ ($g$ is a vector-valued 1-var function)

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the derivative of $g_i$ with respect to what variable ? – Math can be Fun May 4 '14 at 11:03
Respect to the only variable. Isn't a partial derivative. – Martín-Blas Pérez Pinilla May 4 '14 at 13:17