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

• ehrenfest
In summary, the conversation discusses an apparent inconsistency in Griffiths' E and M book regarding the transformation of a line integral along a path in the reverse direction. The conversation presents a formal path on the z-axis and concludes that there are two sign changes that result in the path integral being equivalent to an ordinary integral. This resolves the inconsistency and confirms that Griffiths was correct. However, the logic behind the transformation may seem unintuitive.
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 ?

## The Attempt at a Solution

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 $dt$.

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:

## 1. What is the reason for the mistake in the homework?

The mistake in the homework is most likely due to a miscalculation or misunderstanding of the concept being studied. It is important to carefully review the steps and equations used in the problem to identify where the mistake was made.

## 2. How can I correct the mistake in my homework?

To correct the mistake, carefully review the problem and the steps used to solve it. Check for any errors in calculations or misunderstandings of the concept. If necessary, seek clarification from the textbook or a teacher.

## 3. Will making this mistake affect my overall understanding of the material?

Making a mistake in one homework problem is not likely to significantly impact your overall understanding of the material. However, it is important to identify and correct the mistake to prevent any further misunderstandings.

## 4. Can I still get a good grade if I make a mistake in this homework problem?

One mistake in a homework problem is not likely to significantly affect your grade. However, it is important to identify and correct the mistake to ensure a solid understanding of the material and to prevent any further mistakes in future assignments.

## 5. Is it common for students to make mistakes in this particular homework problem?

It is possible for students to make mistakes in any homework problem, as it is part of the learning process. However, some problems may be more challenging or prone to mistakes, so it is important to carefully review and double-check your work to avoid errors.

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