Impulse of electromagnetic field

[Petar Mali]

We have

$$\vec{F}=\int_V\vec{f}dV=-\frac{d}{dt}\int_V(\vec{D}\times \vec{B})dV$$

$$\vec{g}=\vec{D}\times \vec{B}$$

$$\vec{F}=-\frac{d}{dt}\int_V\vec{g}dV$$

$$\vec{F}=\frac{d\vec{p}_{mech}}{dt}$$

$$\frac{d}{dt}(\vec{p}_{mech}+\int_V\vec{g}dV)=0$$

$$\vec{p}_{mech}+\int_V\vec{g}dV=\vec{const}$$

In total field law of conservation of impulse

$$\vec{p}_{mech}$$ - mechanical impulse of particles in field

In one book I found that $$\int_V\vec{g}dV=\vec{const}$$ is necessary but not always sufficient condition for law of action and reaction in electrodynamics.

My question is when is $$\int_V\vec{g}dV=\vec{const}$$ necessary and sufficient condition for this law? Thanks for your answer!

 

[Petar Mali]

Does anyone know this?

 

[Meir Achuz]

$$\vec{p}_{mech}+\int_V\vec{g}dV=\vec{const}$$

In total field law of conservation of impulse

$$\vec{p}_{mech}$$ - mechanical impulse of particles in field

In one book I found that $$\int_V\vec{g}dV=\vec{const}$$ is necessary but not always sufficient condition for law of action and reaction in electrodynamics.

My question is when is $$\int_V\vec{g}dV=\vec{const}$$ necessary and sufficient condition for this law? Thanks for your answer!

If you believe
$$\vec{p}_{mech}+\int_V\vec{g}dV=\vec{const}$$,
then isn't the answer obviously, yes?
I assume you mean NIII for the forces on objects.

 

[Petar Mali]

NIII?

No! It's not obviously, I think. What is your idea?