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Integration over a disc (SR)

  1. May 30, 2009 #1
    1. The problem statement, all variables and given/known data

    The planes x= ±a are charged to uniform surface density ±σ respectively.
    Find the charge and current densities in a frame moving with velocity (0,v,0) - [done]
    Find also the electromagnetic field in the moving frame by solving the problem in the moving frame


    2. Relevant equations
    Note that my lecturer reconfigures the problem (wlog) to a more familiar scenario from lectures/other problems, i.e. (with y-axis pointing up) the planes are at y=±a, and there is a frame [tex]\,\Sigma\,'[/tex] moving with velocity [tex]\,v\hat{x}[/tex].
    We obtain

    [tex]\rho = \sigma \delta(y-a) - \sigma \delta(y+a) [/tex]

    where [tex]\rho[/tex] is charge density.

    [tex] \vec{j} = \rho \vec{v} = \sigma v \left(\delta(y-a) - \delta(y+a)\right) \hat{x} [/tex]

    where [tex]\vec{j}[/tex] is current density.

    Also required is [tex]\nabla.\vec{D} = \rho \quad \mbox{where} \;\; \vec{D} = \epsilon_0 \vec{E}[/tex]


    3. The attempt at a solution
    Since [tex](c \rho , \vec{j})[/tex] is a 4-vector, using the Lorentz transformation matrix, we can derive

    [tex]\rho\,' = \frac{\rho}{\gamma}[/tex]
    (usual defn of gamma)

    [tex]j_1\,' = 0[/tex]

    Hence
    [tex]\rho\,' = \sigma\,'\left(\delta(y' - a) - \delta(y' + a)\right)[/tex]

    [tex]\sigma\,' = \frac{\sigma}{\gamma}[/tex]

    ---
    For the next bit, we start with one of Maxwell's equations in the [tex]\,\Sigma\,'[/tex] frame: [tex]\nabla'.\vec{E\,'} = \frac{\rho\,'}{\epsilon_0}[/tex]

    Now the solutions say we should integrate over two discs containing the planes y = ±a , and then use the divergence theorem to find

    [tex]\vec{E\,'} = -\frac{\sigma\,'}{\epsilon_0} \hat{y} \quad \mbox{where} \;\; -a < y < a \; \mbox{, else 0}[/tex]


    I can handle integration over a sphere/cylinder, but am not sure what to do with a disc, and also what happens on the RHS (which is a sum of delta fns, from derived equation for [tex]\rho\,'[/tex]).

    Thanks.
     
  2. jcsd
  3. May 30, 2009 #2

    Hao

    User Avatar

    I believe that by 'disc', they refer to a Gaussian pillbox.

    Since this is an infinite plane, you could replace 'disc' with square, rectangle, or any arbitrary shape that is parallel to the planes since all electric field lines point in the same direction (translational symmetry).

    The integral of a delta function is:
    [tex]f(c) = \int^{+\infty}_{-\infty}f(x) \delta(x - c) dx[/tex]
    The delta function 'picks out' the value of the function f(x).

    Minor point: Suppose [tex]\rho = Charge / Volume[/tex] is the charge density in the rest frame. In a frame moving relative to this rest frame, all distances in the rest frame are lorentz contracted: [tex]Volume' = Volume / \gamma[/tex]. Hence, the charge density in the moving frame would be [tex]\rho' = Charge / Volume' = \rho \gamma[/tex]. One example of this is that the electric field of a moving charge is strongest perdendicular to its direction of travel.
     
    Last edited: May 30, 2009
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