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Covariant LW fields derivation help

  1. Apr 23, 2015 #1
    Hi I am doing some exam revision for an EM class and I'm trying to understand a few things about this derivation. Specifically in equation 18.23 why do we not consider the derivative of the four velocity i.e ##\partial^{\alpha}U^\beta##

    Then going from 18.23 to 18.25 why is ##\frac{\partial(x-r(\tau))}{\partial x_\alpha}=2(x-r(\tau)^\alpha)##, where explicitly does the upper alpha come from?

    Also in computing these fields he goes back and uses the integral representation of the potential why is this done as opposed to calculating ##\partial^\alpha \frac{e\mu_0c}{4\pi}\frac{U^\beta(\tau)}{U\cdot [x-r(\tau)]}|_{\tau_0}##
     
  2. jcsd
  3. Apr 23, 2015 #2

    stevendaryl

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    In the expression

    [itex]A^\alpha = C \int d\tau U^\alpha \theta(x_0 - r_0(\tau)) \delta([x - r(\tau)]^2)[/itex]

    there are two positions involved. You are considering [itex]A^\alpha[/itex] at some position [itex]x^\mu[/itex], and you are considering the contribution due to a point-mass at position [itex]r^\mu[/itex]. [itex]U^\alpha[/itex] is equal to [itex]\frac{d}{d\tau} r^\alpha[/itex]. It doesn't depend on [itex]x^\mu[/itex].

    Because of the weird metric used in SR, if [itex]Q[/itex] is a 4-vector, then

    [itex]Q^2 = (Q^0)^2 - (Q^1)^2 - (Q^2)^2 - (Q^3)^2[/itex]

    So [itex](x - r(\tau))^2 = (x^0 - r^0(\tau))^2 - (x^1 - r^1(\tau))^2 -(x^2 - r^2(\tau))^2 -(x^3 - r^3(\tau))^2[/itex]

    So we have 4 equations:
    1. [itex]\frac{\partial}{\partial x^0} (x - r(\tau))^2 = 2 (x^0 - r^0(\tau))[/itex]
    2. [itex]\frac{\partial}{\partial x^1} (x - r(\tau))^2 = -2 (x^1 - r^1(\tau))[/itex]
    3. [itex]\frac{\partial}{\partial x^2} (x - r(\tau))^2 = -2 (x^2 - r^2(\tau))[/itex]
    4. [itex]\frac{\partial}{\partial x^3} (x - r(\tau))^2 = -2 (x^3 - r^3(\tau))[/itex]

    Those can be summarized by:

    [itex]\frac{\partial}{\partial x^\alpha} (x - r(\tau))^2 = 2 (x_\alpha - r_\alpha(\tau))[/itex]

    where [itex]x_\alpha = \pm x^\alpha[/itex], with the plus sign only in the case [itex]\alpha = 0[/itex]

    Now, if we instead do [itex]\frac{\partial}{\partial x_\alpha}[/itex], then we get, instead
    1. [itex]\frac{\partial}{\partial x_0} (x - r(\tau))^2 = 2 (x^0 - r^0(\tau))[/itex]
    2. [itex]\frac{\partial}{\partial x_1} (x - r(\tau))^2 = +2 (x^1 - r^1(\tau))[/itex]
    3. [itex]\frac{\partial}{\partial x_2} (x - r(\tau))^2 = +2 (x^2 - r^2(\tau))[/itex]
    4. [itex]\frac{\partial}{\partial x_3} (x - r(\tau))^2 = +2 (x^3 - r^3(\tau))[/itex]
    That's because again [itex]x_\alpha = \pm x^\alpha[/itex], so the signs change for all cases except for the [itex]\alpha = 0[/itex] case. That can be summarized by:

    [itex]\frac{\partial}{\partial x_\alpha} (x - r(\tau))^2 = 2 (x^\alpha - r^\alpha(\tau))[/itex]

    I don't know, he just thought it was easier that way.
     
    Last edited: Apr 23, 2015
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