# Scalar potential vector

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

• It seems that you lost a factor $\lambda$: $f(\lambda r) \rightarrow f(r)$ Apr 6, 2020 at 2:26
• @Vadim I took the $\lambda$ out and put it into the $\lambda^2$ factor to be integrated. Have I treated the $\lambda$ wrong? How should I treat it? Apr 6, 2020 at 2:33
• @Vadim Can you give me further guidance? Apr 6, 2020 at 3:44
• Generally speaking this is incorrect, since $f(r)$ is an arbitrary function, $f(\lambda r)\neq \lambda f(r)$ Apr 6, 2020 at 5:32
• @Vadim So do you know what I should do instead? I do not know how to correctly perform $$\int f(\lambda r)\:d\lambda$$. Apr 6, 2020 at 5:33

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

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.

• When I perform the check on $rf(r)$ via $\vec a=\nabla \phi(\vec r)$ I get $f(r)\vec r r^{-1} + f^\prime (r)\vec r$, which is not the same as the original $\vec a$. Hence $\phi(\vec r) \ne rf(r)$. Can you explain how you got that potential? Apr 6, 2020 at 3:09
• My main problem is knowing how to correctly treat the $f(\lambda r)$ when integrating Apr 6, 2020 at 3:33
• @Σbaryon: I showed you how I got the potential. It's a piece of cake - look up the curl of conservative vector field. It gives you the potential directly. Apr 6, 2020 at 4:10
• @CinaedSimoson I don't think your answer is correct - taking the gradient shows it: $\nabla (rf(r))= \mathbf{r}/r\partial_r(rf(r))= \mathbf{r}/r[f(r)+f'(r)]$ Apr 6, 2020 at 5:39
• @CinaedSimson My understanding is that given is vector field: $\vec{a}=\vec{r}f(r)$. It is certainly a gradient, but not the gradient of $rf(r)$. Apr 6, 2020 at 8:49