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Vector calculus

  1. May 6, 2006 #1
    Hi, can someone provide some suggestions? I'm stuck on the following questions.

    Q. In this question we will consider the consequences if photons had mass. For massive photons the Laplace equation for the electric potential is replaced by [itex]\nabla ^2 \Phi = m^2 \Phi [/itex]. (*)

    a) Use spherical coordinates to show that the Yukawa potential given by

    [tex]\Phi = \frac{{qe^{ - mr} }}{r}[/tex]

    is a solution of (*). Here q and m are constants. (Note: The Yukawa potential thus replaces the Coulomb potential for a point charge at the origin.)

    b) Find the corresponding Yukawa electric field

    [tex]\mathop E\limits^ \to = E\left( r \right)\mathop r\limits^ \to = - \nabla \Phi [/tex]. (The r is supposed to be a unit vector)

    c) Is div(E) = 0 in regions excluding the origin, i.e. in regions without charge? Is curl(E) = 0?

    d) Calculate:

    [tex]\int\limits_C^{} {\mathop E\limits^ \to \bullet d\mathop s\limits^ \to } [/tex] where C is the curve paramterised by

    [tex]\left( {r,\theta ,\phi } \right) = \left( {1 + 2\cos t,3t,2t} \right),t \in \left[ {0,2\pi } \right][/tex].

    Here is what I did.

    In part a, I worked out the Laplacian in spherical coordinates, said something about the independence of [itex]\Phi [/itex] on theta and phi, calculated the derivative of [itex]\Phi [/itex] with respect to r and showed that it equaled the RHS. Is that correct?

    Part b: In part 'a' I calculated the derivative of [itex]\Phi [/itex] with respect to r (the derivatives with respect to theta and phi are zero).

    [tex]
    \frac{{\partial \Phi }}{{\partial r}} = - q\left( {\frac{{mr + 1}}{{r^2 }}} \right)e^{ - mr}
    [/tex]

    So

    [tex]
    \mathop E\limits^ \to = - \nabla \Phi = \left( {q\left( {\frac{{mr + 1}}{{r^2 }}} \right)e^{ - mr} } \right)\mathop r\limits^ \to
    [/tex] ?

    c) I did this by actually calculating the curl and divergence. Is there an easier way? (I found that the divergence is non-zero and the curl is zero)

    d) Ok at first glance, two problems.

    (1) I don't get what's with the [itex]d\mathop s\limits^ \to [/itex]. Isn't ds normally reserved for arc length? If its a line integral then shouldn't the ds be a dr or a dx?

    (2) Ignoring (1) and taking the ds to be dx = c'(t)dt then we have a line integral. Evaluating E(c(t)) and calculating the line integral looks impossible. But since E looks like it's conservative (it's the gradient of a scalar function), I was thinking maybe make use of the function [itex]d\mathop s\limits^ \to [/itex].

    To this end, I would calculate c(0) = (3, 0, 0) and c(pi/2) = (2, ..., ...). Can I now plug these values the following?

    [itex]\Phi = \frac{{qe^{ - mr} }}{r}[/itex]

    It's just that this question involves spherical coordinates so I'm not sure if the [itex]\Phi \left( {final} \right) - \Phi \left( {initial} \right)[/itex] approach is valid.

    Any help would be good thanks.

    Note: This is not from a physics subject so no knowledge of physics should be required.
     
    Last edited: May 6, 2006
  2. jcsd
  3. May 7, 2006 #2
    It is a line integral, but along the arclength. So, technically speaking [itex]d\mathop s\limits^ \to [/itex] is a vector. In rectangular coordinates it would simply be <dx,dy,dz>. In spherical coordinates, it would be a little bit more complex in the theta and phi directions. But since in the integral you are projecting it along E, and E is entirely in the r direction, you don't have to worry about the theta and phi directions. It simply becomes the integral of E times dr. So you would integrate Edr from the starting to ending values of r.
     
  4. May 7, 2006 #3
    Ok thanks.
     
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