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?


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.

  • $\begingroup$ @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 $ ? $\endgroup$ – PCat27 Jun 2 at 13:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.