Griffiths 3rd Ed. Page 80 Homework: V(\vec{r}) Wrong?

[ehrenfest]

Homework Statement

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 ?

Homework Equations

The Attempt at a Solution

 

[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.

 

[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"??