# 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?

## 2 Answers

The Poynting vector gives the momentum of the field. What you have calculated is the momentum of the matter. The sum of matter and field momentum is conserved and in your example is zero, so the two quantities have opposite sign.

• @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

You need to read some graduate course on EM theory, Griffiths, Feynman, or Lifshitz & Landau, or Frenkel (in German, but the best). EM energy and EM momentum are concepts that were introduced and defined based on particular interpretation of the Poynting theorem and Maxwell's stress tensor theorem.

You can combine Maxwell's equations in such a way that the result is very similar to a work-energy theorem from pure mechanics (Poynting's theorem). You can combine them in a different way, so that the result is very similar to an impulse-momentum theorem from pure mechanics (Maxwell's stress tensor theorem).

In both cases, there are terms (functions of fields and sources) that make those EM equations differ from those in mechanics. These terms are grouped and the equations written in such a way that the equations can be interpreted as conservation laws. This (the interpretation) is not always possible, but for continuous charge and current density in vacuum it is possible. Then energy of EM field and momentum of EM field can be defined, based on those equations in vacuum.

Conservation of energy or momentum or how the expression for EM energy or EM momentum looks like cannot be derived from anything simpler. The Poynting/Maxwell stress equations can be derived from Maxwell's equations, but anything more is an interpretation. Those derived equations are interpreted as local conservation laws because that is a desirable end result, a convenient picture. We do that and the end by-product is a definition of energy and momentum of macroscopic EM field in vacuum.