Problem - Demonstration of electromagnetic energy conservation

I want to solve the following problem from my Classical Electrodynamics' book:

Consider a pontual mass body, $m$, with charge $q$, which moves with an uniform velocity $v$ in a region where an electromagnetic field exists. Demonstrate that the total energy of the particle and the electromagnetic field:

$$W=\frac{1}{2}mv^2+\frac{1}{2}\int_{\infty}\left(\varepsilon E^2+\frac{B^2}{\mu}\right)dV$$

is a constant of movement.

And now, I will show you some sections of the resolution from the book that I didn't understand:

We start derivating the energy expression in order to the time, assuming that $V$ is fixed and the mean's caracteristics are independent of the time:

$$\frac{dW}{dt}=mv\frac{dv}{dt}+\int_{V} \left(\varepsilon E\frac{\partial E}{\partial t}+\frac{B}{\mu}\frac{\partial B}{\partial t}\right) dV=$$ $$m\left(\vec{v}.\frac{d\vec{v}}{dt}\right)+\int_{V}\left [ \left(\vec{E}.\frac{\partial \vec{D}}{dt}\right)+\left(\vec{H}.\frac{\partial \vec{B}}{dt}\right)\right]dV$$

$$...$$

$\vec{J}=\rho \vec{v}$ and $q=\int_{V} \rho$ $dV$ so:

$$\int_V \left(\vec{E}.\vec{J}\right)dV=(\vec{E}.\vec{v})q$$

$$...$$

I have the following doubts:

1. Can we write for example $\frac{dE}{dt}=\frac{\partial E}{ \partial t}$? The electromagnetic field also depends on the position vector $\vec{r}$. So from chain rule we have: $\frac{dE}{dt}=\frac{\partial E}{ \partial r}v+\frac{\partial E}{ \partial t}$
2. I know that the enunciation says that $v$ is uniform. But if they did all the work that I showed then I think that they assumed that $\frac{dv}{dt}$ isn't necessarily $0$. Is it right to write $v\frac{dv}{dt}=\left(\vec{v}.{\frac{d\vec{v}}{dt}}\right)$? For example if $v=(\sin(t),\cos(t),0)$ then $\frac{dv}{dt}=(\cos(t),-\sin(t),0)$. So, $v\frac{dv}{dt}=1$ and in other hand $\left(\vec{v}.{\frac{d\vec{v}}{dt}}\right)=0$ which is different.
3. Is it right to write $\int_V \left(\vec{E}.\vec{J}\right)dV=(\vec{E}.\vec{v})q$? It means that $\vec{E}.\vec{v}$ is a constant which could not be entirely true.

[PS: I'm just posting this question with all those details, because on the exam if I saw this question I would start to solve it by the most general way, and not assuming all thoses conditions that the resolution has. But the truth it's that if I didn't assumed those conditions I would fail. So, I just need to understand those details.]

1. Can we write for example $\frac{dE}{dt}=\frac{\partial E}{ \partial t}$?

Try writing an expression for $W(t_2)$ and $W(t_1)$ and subtracting them and then dividing the result by $t_2-t_1.$ If you then push the parts with the fields inside a common integral then will you see something that looks like $\partial/\partial t$?

Don't ever let symbols confuse you, they notation to express ideas. The result: $\frac{d\vec E}{dt}=\frac{\partial \vec E}{ \partial x}\hat x\cdot \vec v+\frac{\partial \vec E}{ \partial y}\hat y\cdot \vec v+\frac{\partial \vec E}{ \partial z}\hat z\cdot \vec v+\frac{\partial \vec E}{ \partial t}$ is computing the time rate of change of $\vec G(t)=\vec E (\vec r(t), t).$ But you are evaluating the field at every single point, everywhere in space, to do an integral over all space. So just compute that integral for two different times, subtract and scale, then take the limit.

Is it right to write $v\frac{dv}{dt}=\left(\vec{v}.{\frac{d\vec{v}}{dt}}\right)$?

The first one is just a mistake. To compute $\frac{d }{dt}\left(\vec v \cdot \vec v\right)$ you get $\left(\vec v \cdot \frac{d }{dt}\vec v\right)+\left(\frac{d }{dt} \vec v\right) \cdot \vec v$ by the product rule which equals $2\left(\vec v \cdot \frac{d }{dt}\vec v\right)$ from the symmetry of the dot product. So the dot should have always been there. If you don't know this rule just let $a=\hat x \cdot \vec v,$ $b=\hat y \cdot \vec v,$ and $c=\hat z \cdot \vec v$ then $$\frac{d }{dt}\left(\vec v \cdot \vec v\right)=\frac{d }{dt}\left(a^2+b^2+c^2\right)=2\left(a \frac{da}{dt} +b\frac{db}{dt}+c\frac{dc}{dt}\right).$$

If your reference is going to omit dots and then add them back as if that is a step, then make sure you know what you'd do, and learn to read the text as some kind of bad copy of the correct work, which you should be able to do yourself.

1. Is it right to write $\int_V \left(\vec{E}.\vec{J}\right)dV=(\vec{E}.\vec{v})q$? It means that $\vec{E}.\vec{v}$ is a constant which could not be entirely true.

Please make sure you can articulate the difference between constant (in time) and uniform (in space) and the velocity needs to be neither. If $\vec J=q\vec v \delta(\vec r(t))$ then $\int_V \left(\vec{E}\cdot\vec{J}\right)dV=\vec{E}(\vec r (t),t)\cdot(q\vec{v}(t))$ by evaluating the Dirac delta.

• Could you explain better the 3rd point? – Élio Pereira Sep 22 '15 at 17:27
• @ÉlioPereira Have you studied Dirac delta distributions yet? – Timaeus Sep 22 '15 at 18:13
• No, I didn't. I'm just beginning Mathematical Physics this semester – Élio Pereira Sep 22 '15 at 19:35
• @ÉlioPereira Then I don't know how you were reading $\vec J(\vec x,t)=\rho(\vec x, t) \vec v(t)$ since the former is a field and the latter is a single vector, you need a current that is zero everywhere except where the charge is and gives $\int\rho(\vec x,t) dv=q,$ if you imagine a limit of Gaussians with smaller and smaller standard deviations you are mostly there. If you are going deep into mathematical physics you might want to learn about weak derivatives and then distributions as bounded functionals on test functions. Then the Dirac distribution appears there as well. – Timaeus Sep 22 '15 at 23:51