- #1
Petar Mali
- 290
- 0
We have
[tex]\vec{F}=\int_V\vec{f}dV=-\frac{d}{dt}\int_V(\vec{D}\times \vec{B})dV[/tex]
[tex]\vec{g}=\vec{D}\times \vec{B}[/tex]
[tex]\vec{F}=-\frac{d}{dt}\int_V\vec{g}dV[/tex]
[tex]\vec{F}=\frac{d\vec{p}_{mech}}{dt}[/tex]
[tex]\frac{d}{dt}(\vec{p}_{mech}+\int_V\vec{g}dV)=0[/tex]
[tex]\vec{p}_{mech}+\int_V\vec{g}dV=\vec{const}[/tex]
In total field law of conservation of impulse
[tex]\vec{p}_{mech}[/tex] - mechanical impulse of particles in field
In one book I found that [tex]\int_V\vec{g}dV=\vec{const}[/tex] is necessary but not always sufficient condition for law of action and reaction in electrodynamics.
My question is when is [tex]\int_V\vec{g}dV=\vec{const}[/tex] necessary and sufficient condition for this law? Thanks for your answer!
[tex]\vec{F}=\int_V\vec{f}dV=-\frac{d}{dt}\int_V(\vec{D}\times \vec{B})dV[/tex]
[tex]\vec{g}=\vec{D}\times \vec{B}[/tex]
[tex]\vec{F}=-\frac{d}{dt}\int_V\vec{g}dV[/tex]
[tex]\vec{F}=\frac{d\vec{p}_{mech}}{dt}[/tex]
[tex]\frac{d}{dt}(\vec{p}_{mech}+\int_V\vec{g}dV)=0[/tex]
[tex]\vec{p}_{mech}+\int_V\vec{g}dV=\vec{const}[/tex]
In total field law of conservation of impulse
[tex]\vec{p}_{mech}[/tex] - mechanical impulse of particles in field
In one book I found that [tex]\int_V\vec{g}dV=\vec{const}[/tex] is necessary but not always sufficient condition for law of action and reaction in electrodynamics.
My question is when is [tex]\int_V\vec{g}dV=\vec{const}[/tex] necessary and sufficient condition for this law? Thanks for your answer!