# Griffiths page 80

1. Feb 11, 2008

### ehrenfest

1. The problem statement, all variables and given/known data
Please stop reading unless you have Griffith's E and M book (3rd Edition).

Maybe I am nitpicking, but I think the sign is wrong on the last equation on page 80. The formula for the potential was

$$V(\vec{r}) = -\int_{O}^{\vec{r}}\vec{E}\cdot d\vec{l}$$

Thus when you come in from z = infinity along the z axis $\vec{E}\cdot d\vec{l}$ becomes E(-dz) right ?

2. Relevant equations

3. The attempt at a solution

2. Feb 11, 2008

### Ben Niehoff

Suppose we have a line integral of some vector field along the curve $\vec \gamma (t)$:

$$\Phi = \int_{\vec a}^{\vec b} \vec F \cdot d\vec \ell.$$

Formally, this is given by

$$\Phi = \int_{t_a}^{t_b} \vec F \cdot \frac{d\vec \gamma}{dt} \; dt$$

where $d\vec\ell = (d\vec \gamma / dt) \; dt$.

However, suppose we were to integrate $\vec F$ along the same path, but in the reverse direction, from $\vec b$ to $\vec a$? We would expect to get the negative of our original result. This is identical to taking the path $\vec \gamma (t)$ in the opposite direction,

$$\vec \gamma (t) \rightarrow \vec \gamma (-t).$$

But in that case, we also have

$$dt \rightarrow -dt$$

and

$$\frac{d\vec \gamma}{dt} \rightarrow -\frac{d\vec \gamma}{dt}.$$

Therefore,

$$\Phi = -\int_{\vec b}^{\vec a} \vec F \cdot d\vec \ell = -\int_{t_b}^{t_a} \vec F \cdot \left(-\frac{d\vec \gamma}{dt}\right) \; (-dt) = -\int_{t_b}^{t_a} \vec F \cdot \frac{d\vec \gamma}{dt} \; dt.$$

Or in other words, apparently a reversal of path direction,

$$t \rightarrow -t$$

results in

$$d\vec\ell \rightarrow d\vec\ell.$$

That is, $d\vec\ell$ doesn't transform like an ordinary vector.

So, maybe as an exercise, try to resolve the apparent inconsistency in Griffiths by defining a formal path $\vec\gamma(t)$ which happens to lie on the z-axis. Do the formal substitution to reduce the path integral to an ordinary integral, and you should find that you pick up two sign changes; one from $d\vec\gamma / dt$ and one from $dt$.

3. Feb 12, 2008

### ehrenfest

Let $\gamma(t) = (0,0,-t)$.

Then $$V(\vec{r}) = -\int_{O}^{\vec{r}}\vec{E(\gamma)}\cdot d\vec{l} = -\int_{t=-\infty}^{-z}\vec{E(\gamma)}\cdot \frac{d\gamma}{dt} dt = -\int_{t=-\infty}^{-z}\vec{E(\gamma)}\cdot \frac{d\gamma}{dt} dt = \int_{t=-\infty}^{-z}\vec{E_z((0,0,-t))} dt = -\int_{t=\infty}^{z}\vec{E_z((0,0,t))} dt$$

$$= -(V((0,0,z))-V((0,0,\infty))) = V((0,0,\infty))-V((0,0,z))$$

So, Griffiths was right. That is so unintuitive! But if you go from z to infinity, then you can just use -dz and integrate from z to infinity !?!? Why does that make any sense when you are going "outward"!?!?

Last edited: Feb 12, 2008