Conservation of momentum in electromagnetics I'm trying to prove the conservation of momentum of a set of charged particles, but I'm trying to do it without using the momentum tensor, as it does the Jackson electrodynamics book.
If we start with the momentum and its derivative,
$$P=\sum m_j v_j+\epsilon_0\int E\times Bd^3r$$
$$\frac{dP}{dt}=\sum m_j \dot{v}_j+\epsilon_0\int \left(\dot{E}\times B+E\times \dot{B}\right)d^3r$$
Substituting the right part derivatives from Maxwell equations, and the left part from Lorentz equation,
\begin{align}\frac{dP}{dt}&=\sum q_j(E+v_j\times B)+\epsilon_0\int \left[\left(c^2 \nabla \times B-\frac{1}{\epsilon_0}J \right)\times B+E\times \left( -\nabla \times E \right)\right]d^3r \\
&=\sum q_j(E+v_j\times B)+\epsilon_0 c^2\int (\nabla \times B)\times Bd^3r-\int J\times Bd^3r-\epsilon_0 \int E\times (\nabla \times E) d^3r
\end{align}
Now substituting $J$ for a set of particles $J=\sum q_j v_j \delta(r-r_j)$, the integral vanishes with the Dirac delta,
\begin{align}\frac{dP}{dt}&=\sum q_j(E+v_j\times B)+\epsilon_0 c^2\int (\nabla \times B)\times Bd^3r-\sum q_j v_j\times B-\epsilon_0 \int E\times (\nabla \times E) d^3r \\
&=\sum q_jE+\epsilon_0 c^2\int (\nabla \times B)\times Bd^3r-\epsilon_0 \int E\times (\nabla \times E) d^3r
\end{align}
Now I use the property, $$A \times (\nabla \times A)=\frac{1}{2}\nabla (A \cdot A)-(A \cdot\nabla)A$$
So (using $\nabla(A \cdot A)=(\nabla A) A + A (\nabla A)$) the expression ends up being,
\begin{align}\frac{dP}{dt}&=\sum q_jE\\
 &-\epsilon_0 c^2\int \left(\frac{(\nabla B) B + B (\nabla B)}{2}-(B\cdot \nabla)B\right)d^3r\\ 
&-\epsilon_0 \int \left(\frac{(\nabla E) E + E (\nabla E)}{2}-(E\cdot \nabla)E\right)d^3r 
\end{align}
And from Maxwell equations, $\nabla \cdot B=0$ and $\nabla \cdot E= \rho/\epsilon_0$, and also for charged particles $\rho=\sum q_j \delta(r-r_j)$
\begin{align}\frac{dP}{dt}&=\sum q_jE
+\epsilon_0 c^2\int (B\cdot \nabla)Bd^3r- \int \rho Ed^3r+\epsilon_0 \int (E\cdot \nabla)Ed^3r \\
&=\sum q_jE
+\epsilon_0 c^2\int (B\cdot \nabla)Bd^3r- \int \sum q_j \delta(r-r_j)Ed^3r+\epsilon_0 \int (E\cdot \nabla)Ed^3r \\
&=\sum q_jE
+\epsilon_0 c^2\int (B\cdot \nabla)Bd^3r-\sum q_jE+\epsilon_0 \int (E\cdot \nabla)Ed^3r \\
&=\epsilon_0 c^2\int (B\cdot \nabla)Bd^3r+\epsilon_0 \int (E\cdot \nabla)Ed^3r
\end{align}
Arriving to \begin{equation}\boxed{\frac{dP}{dt}=\epsilon_0 \int\left(c^2 (B\cdot \nabla)B+ (E\cdot \nabla)E\right)d^3r}\end{equation}
And now I don't know at all how to follow, because if I apply the definition of the Nabla operators, we arrive to
\begin{align}\frac{dP}{dt}&=\epsilon_0 \sum_i \int\left(c^2B_i \frac{\partial B_i}{dx_i}+E_i \frac{\partial E_i}{dx_i} \right)d^3r \\
&=\epsilon_0 \sum_i \frac{1}{2}\frac{\partial}{\partial x_i}\int\left(c^2B_i^2 +E_i^2 \right)d^3r
\end{align}
But at this point I don't have idea of what I need to do next, I only know that we can change to the reciprocal space, where we have the variables discretized,
\begin{align}\mathscr{E}&=iN(k)\left(\alpha(k)-\alpha(-k)^*\right)\\
\mathscr{B}&=\frac{iN(k)}{c}\left(\hat{n}\times\alpha(k)+\hat{n}\times\alpha(-k)^*\right)
\end{align}
But I can't apply this to the components $E_i$ of the fields... so if someone could give me some help or some hint I will be very grateful
 A: Starting from this equation for the components:
$$
\frac{dP_i}{dt}=\sum q_jE_i+\epsilon_0 c^2\int \left((\nabla \times B)\times B\right)_i d^3r-\epsilon_0 \int \left(E\times (\nabla \times E)\right)_i d^3r
$$
Use the identity:
$$
\epsilon_{ijk}\epsilon_{ilm} = \delta_{jl}\delta_{km} - \delta_{jm}\delta_{kl}
$$
to rewrite the starting equation
$$
\frac{dP_i}{dt}=\sum q_jE_i+\epsilon_0 c^2\int \left(\epsilon_{ijk}(\nabla_l B_m)\epsilon_{jlm}B_k\right) d^3r - \epsilon_0\int \left(\epsilon_{ijk}E_j(\nabla_l E_m)\epsilon_{klm}\right) d^3r
$$
as:
$$
\frac{dP_i}{dt}=\sum q_jE_i+\epsilon_0 c^2\int \left((\delta_{kl}\delta_{im} - \delta_{km}\delta_{il})(\nabla_l B_m)B_k\right) d^3r-\epsilon_0 \int \left((\delta_{il}\delta_{jm}-\delta_{im}\delta_{jl})E_j(\nabla_l E_m)\right) d^3r
$$
Now, we can use the fact the the fields vanish at infinity to integrate by parts and we have:
$$
\frac{dP_i}{dt}=\sum q_jE_i-\epsilon_0 c^2\int \left((\delta_{kl}\delta_{im} - \delta_{km}\delta_{il})B_m(\nabla_l B_k)\right) d^3r+\epsilon_0 \int \left((\delta_{il}\delta_{jm}-\delta_{im}\delta_{jl})E_m(\nabla_l E_j)\right) d^3r
$$
Then, Maxwell's equations give us:
$$
\frac{dP_i}{dt}=\epsilon_0 c^2\int \delta_{km}\delta_{il}B_m\nabla_l B_k d^3r+\epsilon_0 \int \delta_{il}\delta_{jm}E_m\nabla_l E_j d^3r
$$
Or:
$$
\frac{dP_i}{dt}=\epsilon_0 c^2\int B_m\nabla_i B_m d^3r+\epsilon_0 \int E_m\nabla_i E_m d^3r\;,
$$
where the repeated indices (m) are summed over.
But it turns out that each of the two terms on the RHS are individually equal to zero. This is true because, as we have already assumed, the fields vanish at infinity. So we can integrate by parts to our heart's content.
For example:
$$
\int B_m(\nabla_i B_m) d^3r = -\int (\nabla_iB_m)(B_m) d^3r\;,
$$
which says that this integral is equal to the negative of itself--that is, zero! (You can also see that it is zero by the reasoning of the other answer, namely, that it is half the total derivative of the square of the field.)
And similarly for the electric field integral.
Therefore the time derivative of the total momentum is zero (total momentum is conserved).
A: You are almost there.
Let's continue with the boxed volume integral near the end of your question.
$$\begin{align}
\frac{d\mathbf{P}}{dt}
&=\epsilon_0 \int_\text{whole space}\left(c^2(\mathbf{B}\cdot\nabla)\mathbf{B}+(\mathbf{E}\cdot\nabla)\mathbf{E}\right)dV \\
&=\epsilon_0 \int_\text{whole space}\frac{1}{2}\nabla\left(c^2\mathbf{B}^2+\mathbf{E}^2\right)dV \\
&\quad\text{use Gauss's gradient theorem to convert the volume integral}\\ &\quad\text{ to a surface integral over a scalar integrand} \\
&=\epsilon_0\int_\text{infinitely big surface}\frac{1}{2}\left(c^2\mathbf{B}^2+\mathbf{E}^2\right)d\mathbf{S} \\
&\quad\text{on the infinitely far away surface the integrand vanishes (because $\mathbf{B}=\mathbf{0}$ and $\mathbf{E}=\mathbf{0}$}) \\
&=\mathbf{0}
\end{align}$$
