# Line integral of a point charge

I am trying to teach myself Electrodynamics through self-study of Griffiths' Introduction to Electrodynamics, and I am having difficulty with a calculation that involves a line integral of a point charge. I posted this question to Math Stack Exchange yesterday, but I was unable to get an answer. I'm not really sure that my question was understood. I have attempted to clarify it below.

I have been stuck on this for quite some time, and would appreciate any help that could be offered!

Cheers!

This question concerns a calculation in Section 2.2.4 of Griffiths' book on an Introduction to Electrodynamics (4th Ed.), in which he shows that the field of a point charge is curl-free.

The field of a point charge at the origin is given in spherical coordinates by $$\mathbf{E}=\frac{1}{4\pi\epsilon_0}\frac{q}{r^2}\ \hat{\mathbf{r}}.$$

Here $$\epsilon_0$$ is the permittivity of free space, $$q$$ is the magnitude of the charge, $$r$$ is the radius from the origin, and $$\hat{\mathbf{r}}$$ is radial spherical basis vector. He shows that this field is curl-free by demonstrating that its line integral around any closed loop is zero. So he starts by considering the integral $$\int_{\mathbf{a}}^{\mathbf{b}}\mathbf{E}\cdot d\mathbf{l}$$ over an arbitrary path in $$\mathbb{R}^3$$ (that presumably doesn't include the origin). This integral is calculated in spherical coordinates, so observing that the infinitesimal displacement vector is given by $$d\mathbf{l}=dr\ \hat{\mathbf{r}}+r\ d\theta\ \hat{\boldsymbol\theta}+r\ \sin\theta\ d\phi\ \hat{\boldsymbol\phi}$$ leads to the conclusion $$\mathbf{E}\cdot d\mathbf{l}=\frac{1}{4\pi\epsilon_0}\frac{q}{r^2}\ dr$$
(in Griffiths' notation $$\phi$$ is the azimuthal angle, and $$\theta$$ is the polar angle). Griffiths proceeds to evaluate the integral $$\int_\mathbf{a}^\mathbf{b}\mathbf{E}\cdot d\mathbf{l}=\frac{1}{4\pi\epsilon_0}\int_\mathbf{a}^\mathbf{b}\frac{q}{r^2}\ dr=\frac{-1}{4\pi\epsilon_0}\frac{q}{r}\bigg|^{r_b}_{r_a}=\frac{1}{4\pi\epsilon_0}\bigg(\frac{q}{r_a}-\frac{q}{r_b}\bigg).$$ Here $$r_a$$ and $$r_b$$ are the radii associated with $$\mathbf{a}$$ and $$\mathbf{b}$$. It is then argued that the integral around any closed path is zero, so that $$\mathbf{E}$$ is curl-free by Stokes theorem. I cannot understand how he got from $$\frac{1}{4\pi\epsilon_0}\int_\mathbf{a}^\mathbf{b}\frac{q}{r^2}\ dr$$ to $$\frac{-1}{4\pi\epsilon_0}\frac{q}{r}\bigg|^{r_b}_{r_a}.$$ This step seems to assume path independence, which is equivalent to what he is trying to prove. Thus this logic seems a bit circular. For suppose we do not use the fact that the integral is path-independent. Then it must be the case that we need to parameterize $$r$$ as a function of $$\phi$$ and $$\theta$$, so that $$r=r(\phi,\theta)$$. This parameterization would in turn change the infinitesimal displacement. This would make the integral considerably more difficult to evaluate.

How does this step not assume path-independence?

And also, this seems to assume that the path does not cross the origin. Does mean that our field $$\mathbf{E}$$ is curl-free everywhere but the origin, at which it is undefined?

I would be very grateful to anyone who could help clear up my confusion!

• The electric field is a gradient of a potential $\phi(r)\propto1/r$, so the integral is indeed path-independent and $curl(\vec{E})$ is zero. May 31 '20 at 12:35
• Do you understand why $$\mathbf{E}\cdot d\mathbf{l}=\frac{1}{4\pi\epsilon_0}\frac{q}{r^2}\ dr$$ ? May 31 '20 at 12:36
• @BioPhysicist Yes, I do. May 31 '20 at 12:37

Maybe it would have helped you if Griffiths had left in the other two terms in the path integral? $$\int {\bf E} \cdot d{\bf l} = \frac{1}{4\pi \epsilon_0} \int \frac{q}{r^2}\ dr + \int E_{\theta} r\ d\theta + \int E_{\phi} r \sin \theta\ d\phi,$$ but since $$E_{\theta} = E_\phi=0$$ in this case, then the last two terms disappear. The remaining integral is indeed path independent in the sense that it does not matter what excursions you take in $$\theta$$ and $$\phi$$ because the integrand does not depend on $$\theta$$ or $$\phi$$, all that matters is the starting and ending values of $$r$$.

That the line integral of a closed path is zero does indicate that the field is curl free (unless you hve chosen a pathological path where the E-field is always perpendicular to the path.

• Why does it not matter what excursions you take in $\theta$ and $\phi$? I think this is where my misunderstanding is. Don't the excursions one takes in $\theta$ and $\phi$ make $r$ a function of $\phi$ and $\theta$, so that $r=r(\phi,\theta)$? May 31 '20 at 12:43
• @jackrodgers1554 That would matter in the other two (disappearing) integrals where you would have to rewrite $r$ in terms of $\theta$ or $\phi$ respectively. May 31 '20 at 12:46
• Why doesn't it matter for the integral that does not disappear? I apologize for the obtuse question. May 31 '20 at 12:50
• $\int {\bf E}\cdot d{\bf l} = \int E_r + E_\theta r (d\theta/dr) + E_{\phi} r\sin\theta (d\phi/dr)\ dr$ May 31 '20 at 12:51
• Is it that you are confused between the scalar line integral and a vector line integral which involves the scalar product? @jackrodgers1554 May 31 '20 at 13:00

This part $$\int_\mathbf{a}^\mathbf{b}\mathbf{E}\cdot d\mathbf{l}=\frac{1}{4\pi\epsilon_0}\int_\mathbf{a}^\mathbf{b}\frac{q}{r^2}\ dr=\frac{-1}{4\pi\epsilon_0}\frac{q}{r}\bigg|^{r_b}_{r_a}$$ doesn't assume path independence. It follows from $$\mathbf{E}\cdot d\mathbf{l}=\frac{1}{4\pi\epsilon_0}\frac{q}{r^2}\ dr$$

The integrand only depends on the $$r$$ coordinate, so the integral just becomes an integral over $$r$$. This doesn't assume path independence, it actually shows that the integral only depends on the starting and ending $$r$$ coordinates.

Therefore, if you understand why $$\mathbf{E}\cdot d\mathbf{l}=\frac{1}{4\pi\epsilon_0}\frac{q}{r^2}\ dr$$ then you're good to go.

• But if we are integrating over a path, doesn't the $r$ coordinate depend upon $\theta$ and $\phi$? Isn't it the case that, over the path, $r=r(\theta,\phi)$? If not, can you please explain why this is not the case? May 31 '20 at 12:46

Let $$r$$ be a function of $$\theta$$ and $$\phi$$ (I'll write $$r$$ as $$r$$ itself and not $$r(\theta,\phi)$$ just to make things less cluttered). Now, we know that

$$\mathrm d r=\frac{\partial r}{\partial \theta}\mathrm d \theta+\frac{\partial r}{\partial \phi}\mathrm d\phi$$

Let's represent $$\displaystyle \frac{\partial r}{\partial \theta}=\dot r_{\theta}$$ and $$\displaystyle \frac{\partial r}{\partial \phi}=\dot r_{\phi}$$. Then, your integral becomes (I am ignoring the constants for now):

$$\int_\mathrm{a}^\mathrm{b}\frac{1}{r^2}\ dr= \int_\mathrm{a}^\mathrm{b} \left(\frac{\dot r_{\theta}\mathrm d \theta}{r^2}+ \frac{\dot r_{\phi}\mathrm d \phi}{r^2} \right)\tag{1}$$

However, we know that

$$\mathrm d\left(-\frac{1}{r}\right)=\frac{\dot r_{\theta}\mathrm d \theta}{r^2}+ \frac{\dot r_{\phi}\mathrm d \phi}{r^2}$$

Therefore, the integral in equation $$(1)$$ is converted to

$$\int_\mathrm{a}^\mathrm{b} \mathrm d\left(-\frac{1}{r}\right)=\left|-\frac{1}{r}\right|_{\mathrm a}^{\mathrm b}=\frac 1 a -\frac 1 b$$

Thus even if you explicitly involve path dependence, you will get the same result.

• Thank you very much, I think this makes sense. How are you getting the formula for $\mathrm{d}(-1/r)$? Is this just the chain rule? May 31 '20 at 12:56
• @jackrodgers1554 Yes. If you want more details, have a look at this Wikipedia section.
– user258881
May 31 '20 at 12:57
• Not sure if it's worth it, but not all paths can be written as $r(\theta,\phi)$ May 31 '20 at 13:44
• @BioPhysicist Yes, but I suppose that we are only considering paths which can be represented as a function of $\theta$ and $\phi$ (we are assuming that $r$ can be uniquely determined by any given value of $\theta$ and $\phi$ (as long as that value lies in the domain of the respective variables)).
– user258881
May 31 '20 at 13:49