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Griffiths page 80

  1. Feb 11, 2008 #1
    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

    [tex] V(\vec{r}) = -\int_{O}^{\vec{r}}\vec{E}\cdot d\vec{l} [/tex]

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

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Feb 11, 2008 #2

    Ben Niehoff

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

    [tex]\Phi = \int_{\vec a}^{\vec b} \vec F \cdot d\vec \ell.[/tex]

    Formally, this is given by

    [tex]\Phi = \int_{t_a}^{t_b} \vec F \cdot \frac{d\vec \gamma}{dt} \; dt[/tex]

    where [itex]d\vec\ell = (d\vec \gamma / dt) \; dt[/itex].

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

    [tex]\vec \gamma (t) \rightarrow \vec \gamma (-t).[/tex]

    But in that case, we also have

    [tex]dt \rightarrow -dt[/tex]


    [tex]\frac{d\vec \gamma}{dt} \rightarrow -\frac{d\vec \gamma}{dt}.[/tex]


    [tex]\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.[/tex]

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

    [tex]t \rightarrow -t[/tex]

    results in

    [tex]d\vec\ell \rightarrow d\vec\ell.[/tex]

    That is, [itex]d\vec\ell[/itex] 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 [itex]\vec\gamma(t)[/itex] 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 [itex]d\vec\gamma / dt[/itex] and one from [itex]dt[/itex].
  4. Feb 12, 2008 #3
    Let [itex]\gamma(t) = (0,0,-t)[/itex].

    Then [tex] 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 [/tex]

    [tex]= -(V((0,0,z))-V((0,0,\infty))) = V((0,0,\infty))-V((0,0,z)) [/tex]

    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
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