Scalar potential vector

Question

I am trying to find the scalar potential, $\phi(\vec r)$, of a conservative vector field $\vec a(\vec r)$. I am integrating along a straight line from $\vec r_0$ to $\vec r$ which is parametrised by $\vec r^\prime = \lambda\vec r$ with $0\le\lambda\le1$. Thus $d\vec r^\prime=d\lambda\vec r$ and hence:$$\phi(\vec r)=\int_0^\vec r \vec a(\vec r^\prime)\cdot d\vec r^\prime=\int_{\lambda=0}^{\lambda=1} \vec a(\lambda\vec r)\cdot \vec r d\lambda$$

The conservative vector field I am using is $\vec a=\vec rf(r)$ where $f(r)$ is an arbitrary function of $r=|\vec r|$.

Therefore, $$\phi(\vec r)=\int_{\lambda=0}^{\lambda=1} \vec a(\lambda\vec r)\cdot \vec r d\lambda=\int_{0}^{1} \lambda\vec rf(\lambda r)\cdot \vec r d\lambda\: [\star]$$ $$=\int_{0}^{1} \lambda^2(\vec r\cdot \vec r)f(r) d\lambda\:[\star\star]=(\vec r\cdot \vec r)f(r)\int_{0}^{1} \lambda^2 d\lambda $$ $$=\frac{1}{3}(\vec r\cdot \vec r)f(r)=\frac{1}{3}r^2f(r)$$

Next, in order to verify that I have obtained the correct scalar potential I use the fact that: $$\vec a=\nabla \phi (\vec r) $$ $$\vec a=\nabla(\frac{1}{3}r^2f(r))=\frac{1}{3}((\nabla r^2)f(r)+(\nabla f(r))r^2)$$ $$=\frac{1}{3}(\vec r f(r) + f^\prime (r)(\nabla r)r^2)=\frac{1}{3}(\vec r f(r) + f^\prime (r)(r^{-1}\vec r)r^2)$$ $$=\frac{1}{3}(\vec r f(r) + f^\prime (r)r^{-1}\vec r)$$

I am supposed to get back my vector $\vec a=\vec rf(r)$ but this is not the case. I believe that there is potentially something wrong at $[\star]$ or $[\star\star]$.

I would like to know where I have gone wrong in order to get back my original vector.

Thanks

Answer 1

I would like to know where I have gone wrong in order to get back my original vector.

You went wrong at the step $[\star]$, when you substituted $f(\lambda r)\rightarrow \lambda f(r)$, because in principle we don't know if $f$ is linear or not. Therefore the only thing you can do is leave the potential as $$\phi(\vec{r})=r^2\int_0^1\lambda f(\lambda r)d\lambda\ .$$

From here you can recover the field taking the gradient. Don't let the fact that the expression depends on $\lambda$ intimidate you.

Since the potential actually depends only on $r=|\vec{r}|$, the gradient is $\nabla\phi=\hat{r}\displaystyle\frac{d\phi}{dr}$. Hence remembering the product rule for derivatives and the chain rule

\begin{align} \vec{a}&=\hat{r}\left[2r\int_0^1\lambda f(\lambda r)d\lambda+r^2\frac{d}{dr}\int_0^1\lambda f(\lambda r)d\lambda\right]\\ &=\hat{r}\left[r\int_0^12\lambda f(\lambda r)d\lambda+r^2\int_0^1\lambda^2 f'(\lambda r)d\lambda\right] \end{align}

If we now notice that $$\frac{d}{d\lambda}\left[\lambda^2 f(\lambda r)\right]=2\lambda f(\lambda r)+\lambda^2rf'(\lambda r)$$

it follows immediately that $$\vec{a}=\vec{r}\int_{\lambda=0}^{\lambda=1}d[\lambda^2f(\lambda r)]=\vec{r}f(r)\ .$$

Answer 2

Since $\vec a$ a conservative vector field, $\nabla \times \vec a=0 \iff$ $\vec a=\nabla \phi(\vec r)$ - and the intergal of $\nabla \phi(\vec r)$ is path independent: $\int_\gamma\nabla\phi(\vec r)\cdot d\vec r =\phi(\vec r_b)-\phi(\vec r_{a})$. The integral only depends upon the beginning and ending points.

Hence, $$\int^{r_b}_{r_a}\vec r f(r)\cdot d\vec r=\int^{r_b}_{r_a}\nabla\phi(\vec r)\cdot d\vec r =\phi(r_b)-\phi(r_{a})$$ or $$\int^{r_b}_{r_a} r f(r) dr=\int^{r_b}_{r_a}{\partial}_r\phi(r)dr=\phi(r_b)-\phi(r_{a})$$

where $f(r)$ is unknown.