What are the unknown steps to show $\mathbf{F}=-\nabla\psi$?

Question

Given: $\mathbf{F}=F(r)\ \mathbf{\hat{r}}$ is a spherically symmetric radial vector field which is continuous everywhere except at origin.

To prove: $\mathbf{F}=-\nabla \psi$

Proof:

\begin{align} \mathbf{F} \cdot \mathbf{\hat{r}}&=F(r)\ \mathbf{\hat{r}} \cdot \mathbf{\hat{r}}=F(r)=\dfrac{d(\int F\ dr)}{dr}\\ &=\dfrac{d\psi}{dr}=\nabla \psi \cdot \mathbf{\hat{r}} \end{align}

From here we cannot straight away deduce that: $\mathbf{F}=\nabla\psi$

UNKNOWN STEPS ???

$$\mathbf{F}=-\nabla\psi$$

$\mathbf{F}$ is conservative.

Question: From what steps (unknown to me) can we get $\mathbf{F}=-\nabla\psi\ $?

Answer 1

Let's write $$ \newcommand{\rr}{{\mathbf{\hat{r}}}} \newcommand{\th}{{\mathbf{\hat{\theta}}}} \newcommand{\FF}{{\mathbf{F}}} \FF(r, \theta) = F(r) \rr + 0 \th $$ And define $$ \psi(r, \theta) = -\int_0^r F(t) ~ dt $$ Then we have \begin{align} \frac{d\psi}{dr}(r, \theta) &= -F(r)\\ \frac{d \psi}{d\theta}(r, \theta) &= 0\\ \nabla \psi (r, \theta) &= \frac{1}{r} \frac{d \psi}{d\theta} \th + \frac{d \psi}{dr}\rr\\ &= \frac{1}{r} 0 ~ \th + -F(r)\rr\\ &= -F(r)\rr \tag{*}\\ \end{align} so that $\nabla \psi$ is again a radial vector field. Two radial vector fields are equal if the coefficient of $\rr$ is the same in each.

As you observe, \begin{align} \FF(r, \theta) \cdot \rr &= F(r) \rr \cdot \rr \\ &= F(r) \\ &= -\frac{d\psi}{dr} (r, \theta) \\ &= -\nabla \psi (r, \theta) \cdot \rr \end{align}

Alternatively, you could just look at equation (*) and compare it to the definition of $\FF$ and see that $\FF = -\nabla \psi$.

Summary:

1. You needed to include a minus-sign in the definition of $\psi$ to get the answer you were asked for.

2. You needed to make the observation that the gradient of $\psi$ was a radial field, and then the computation you did would suffice.

Answer 2

You have a differential equation here in $-\operatorname{grad}\psi=\mathbf F=f(\mathbf r) \frac{\bf r}{\|\mathbf r\|}$, where $f$ is a scalar function and $\mathbf F$ a vector function.

One way to solve it for $\psi$ is by guessing a solution and then checking if its right.

By definition :

$$\mathbf r\cdot\operatorname{grad}\psi=\lim_{t\to0}\frac{\psi(\mathbf x+t\mathbf r)-\psi(\mathbf x)}t$$

where $t$ is a scalar and $\mathbf x, \mathbf r$ vectors.

From this follows :

$$\operatorname{grad}(\|\mathbf r\|)=\frac{\mathbf r}{\|\mathbf r\|}$$

And chain rule :

$$\operatorname{grad}(f(g(\mathbf r)))=\frac{\mathrm df}{\mathrm dg}\operatorname{grad}(g)$$

Now guessing $\psi(\mathbf r)=-\left(\int f(\mathbf r)\,\mathrm d\|\mathbf r\|\right)\cdot\|\mathbf r\|$

we get $-\operatorname{grad}\psi=f(\mathbf r)\frac{\mathbf r}{\|\mathbf r\|}$ so this is the right solution to the differential equation.