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Differentiating the relativistic charge wrt velocity

  1. Jun 22, 2015 #1
    1. The problem statement, all variables and given/known data

    Given that ##Q' = \int d^{3}x \bigg[ \rho \big( \frac{vx}{c^{2}},x,y,z \big) - \frac{v}{c^{2}} j^{x} \big( \frac{vx}{c^{2}},x,y,z \big) \bigg]##, show the following:

    ##\frac{dQ'}{dv} = \frac{1}{c^{2}} \int d^{3}x \bigg[ x \big( \frac{\partial \rho}{\partial t} - \frac{v}{c^{2}} \frac{\partial j^{x}}{\partial t} \big) - j^{x} \bigg] \bigg|_{t = \frac{vx}{c^{2}}, x, y, z}##

    2. Relevant equations

    3. The attempt at a solution


    I understand that the second term can be easily obtained by differentiating ##- \frac{v}{c^{2}} j^{x}## with respect to ##v##. However, I am having trouble differentiating the first term with respect to ##v##.

    In essence, I need to show that ##\frac{\partial \rho}{\partial v} = \frac{1}{c^{2}} x \bigg( \frac{\partial \rho}{\partial t} - \frac{v}{c^{2}} \frac{\partial j^{x}}{\partial t} \bigg)##.

    Any hints?
     
  2. jcsd
  3. Jun 22, 2015 #2

    RUber

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    Homework Helper

    You have ##\frac{\partial }{\partial v}\left( p(t(v),x,y,z) - \frac {v}{c^2}j^x(t(v),x,y,z)\right) ##
    Using the chain rule and product rule, you should be able to work this out.
    ##\frac{\partial }{\partial v} p(t(v),x,y,z) = \frac{\partial p }{\partial t}\frac{\partial t }{\partial v} ##
    Then start with the product rule for the second part,
    ##\frac{\partial }{\partial v}\frac {v}{c^2} j^x(t(v),x,y,z)=\left( \frac{\partial }{\partial v}\frac {v}{c^2}\right)j^x+\frac {v}{c^2}\frac{\partial j^x}{\partial v} ##
    Notice that you will need the chain rule one more time for that last derivative.
     
  4. Jun 22, 2015 #3
    Thanks!! Got it!
     
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