# What happens with the divergence of maxwell stress tensor when EM fields in vacuum?

The conservation momentum can be derived from the total electromagnetic force on the charges in volume $$V$$ and is written in this way:

$$\frac{d\vec{p}_{mech}}{dt} + \frac{d}{dt} \int_V (\vec{E}\times\vec{B}) = \int_V \nabla \cdot \hat{T}$$

In many textbooks argument that the second term is the total momentum stored in the fields, but I don't understand why. This term is proportional to the Poynting vector(energy/area) but how is related to the momentum of fields?

On the other hand, Can the density of momentum carried by the electromagnetic field in vacuum be derivated by Lorentz force? I was trying the follows

$$\vec{F}= \frac{d\vec{p}_{mech}}{dt} = \frac{d}{dt} (q\vec{E}- q\vec{v} \times \vec{B})$$

$$\vec{p}_{mech} = \int (q\vec{E}- q\vec{v} \times \vec{B}) dt$$

Now, I changed $$\vec{p}$$ to density momentum.

$$\vec{P}_{mech} = \int (\rho \vec{E}- \rho\vec{v} \times \vec{B}) dt$$

$$\vec{P}_{mech} = \int (\rho \vec{E}- \vec{J} \times \vec{B}) dt$$

$$\vec{P}_{mech} = \int (\rho \vec{E} - \epsilon_o \frac{\partial{E}}{{\partial t}} \times \vec{B}) dt$$

Then, in the vacuum $$\rho = 0$$

$$\vec{P}_{mech} = - \epsilon_o \int (\frac{\partial{(E \times B)}}{{\partial t}} ) dt$$

$$\vec{P}_{mech} = - \frac{ (\vec{E} \times \vec{B})}{\mu_o c^2 }$$

but why the density momentum carried by electromagnetic fields in vacuum is given by:

$$\vec{P}_{fields} = - \vec{P} _{mech} = \frac{ (\vec{E} \times \vec{B})}{\mu_o c^2 }$$

So, the previous equation means that the divergence of Maxwell stress tensor is zero, what means it?

• @mycts so, why the Poynting vector is associated with the momentum field ? $\frac{d( \vec{p}_{mech}+ \vec{p} _{field} )}{dt} = \nabla \cdot \hat{T}$, so in the vacuum $\nabla \cdot \hat{T} =0$ ? – PCat27 Jun 2 '19 at 13:58