pellman
- 683
- 6
The continuous version of the Lorentz force is
[tex]\mathbf{f}=\rho\mathbf{E}+\mathbf{j}\times\mathbf{B}[/tex]
but what does f mean?
In the discrete version F is the force on the charged particle appearing on the RHS. So if x is the position of the charged particle we have that its equation of motion is
[tex]m\frac{d^2\mathbf{x}}{dt^2}=\mathbf{F}=q\mathbf{E}+q\frac{d \mathbf{x}}{dt}\times\mathbf{B}[/tex]
So I guess what I am asking is how does f relate to the equation of motion for the charge density? (for that matter, what is the equation of motion of a continuous charge density?)
[tex]\mathbf{f}=\rho\mathbf{E}+\mathbf{j}\times\mathbf{B}[/tex]
but what does f mean?
In the discrete version F is the force on the charged particle appearing on the RHS. So if x is the position of the charged particle we have that its equation of motion is
[tex]m\frac{d^2\mathbf{x}}{dt^2}=\mathbf{F}=q\mathbf{E}+q\frac{d \mathbf{x}}{dt}\times\mathbf{B}[/tex]
So I guess what I am asking is how does f relate to the equation of motion for the charge density? (for that matter, what is the equation of motion of a continuous charge density?)