- #1
Petar Mali
- 283
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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!
\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!
