Differentiating the relativistic charge wrt velocity

[spaghetti3451]

Homework Statement

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:

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

Homework Equations

3. The Attempt at a Solution [/B]

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?

 

[RUber]

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.

 

[spaghetti3451]

Thanks! Got it!