Work done by electromagnetic forces when considering continous charge distributions

Question

Here’s something that has always bothered me in my physics lectures. Suppose I want to calculate the work done by electromagnetic forces onto one charge

$$ \begin{equation} dW = \vec{F} \cdot d\vec{s} = Q \vec{E}\cdot\vec{v} dt \end{equation} $$

Now if I want to expand this idea to a continuous charge distributions $\rho$ my textbook teels me to do the following using $Q =\rho dV$ and $\vec{J}=\rho \vec{v}$.

$$ \begin{equation} dW = \rho dV \vec{E}\cdot \frac{1}{\rho} \vec{J} dt = \vec{E}\cdot\vec{J} dV dt \end{equation} $$

Now up until here I am thinking about the values $dW, dV, dt$ as finite differences. Now making the switch to Infinitesimal values is what causes problems for me. In my textbook the next line simply states $$ \begin{equation} \frac{dW}{dt} = \int_{\Omega}^{} \vec{E}\cdot\vec{J} \,dV ' \end{equation} $$

Now I have a couple of ussues here: Where do I suddenly pull out the integral? Why can i just divide by $dt$ as if it's a regular variable?

Answer 1

It's unclear why your integral at the end is not integrating over $dV$ (the name for the volume element $dV$ seems to be suddenly changed to $d\Omega'$, but I assume that is not the real question here).

Also, working with these separate "$d$" infinitesimals is a bit sloppy, mathematicians will probably tell you that you always need to combine them in some derivative of something with respect to something else, in a fraction like $dA/dB$.

But apart from that, you actually can can divide by $dt$ here, but that would not finish the job, because the "$dW$" in fact should have had two $d$'s, since you want it to be a finite difference in time and also in volume (there you see how things easily get messed up with this notation!)

So if you start with $$ \begin{equation} d(dW) = \vec{E}\cdot\vec{J} dV dt \end{equation} $$ and divide by $dt$ that would leave you with $$ \begin{equation} d\left(\frac{dW}{dt}\right) = \vec{E}\cdot\vec{J} dV \end{equation} $$ and applying an integral over space would remove one $d$ from the left-hand side, since the right-hand side is combining an integral with a $d$ which means summing over finite differences so it gives a total result: $$ \begin{equation} \frac{dW}{dt} = \int_{\Omega}^{} \vec{E}\cdot\vec{J} \,dV \end{equation} $$ At least that is the total result for $dW/dt$, not yet for $W$ because that would require another integral, over time (but to do that the division by $dt$ should not have been done, instead $dt$ should have remained on the right, and combined with a second integral: $$ \begin{equation} W = \int dt\ \int_{\Omega}^{} \vec{E}\cdot\vec{J} \,dV \end{equation} $$

So there are two games you can play with the $d$'s. Not only moving them around (e.g. by dividing) but also contracting them with $\int$ to create total values.