Making Lorentz equation dimensionless for simulations \begin{equation}
\frac{dv}{dt}=\frac{e}{m}(E+\frac{v}{c}\times B)
\end{equation}
I'm making a numerical simulation and I need to make the equation dimensionless. I'm having trouble in doing so.
I was thinking to use the fundamental units:

*

*setting the time as inverse of gyrofrequency

*the mass equal to $m_{e}$

*using the Debye length

But I can't proceed further. I also tried using $c=1$ (what I need to scale up my relativistic simulation) without success.
 A: Wikipedia provides a good process of nondimensionalization:

*

*Identify all the independent and dependent variables;

*Replace each of them with a quantity scaled relative to a characteristic unit of measure to be determined;

*Divide through by the coefficient of the highest order polynomial or derivative term;

*Choose judiciously the definition of the characteristic unit for each variable so that the coefficients of as many terms as possible become 1;

*Rewrite the system of equations in terms of their new dimensionless quantities.

In your case, the independent variable is $t$ (since it is in the differential) and the dependent is $v$. Hence, we let $v=\eta v_c$ and $t=\tau t_c$ where $v_c$ & $t_c$ are the characteristic units. Replacing these into your equation,
$$
\frac{d(\eta v_c)}{d(\tau t_c)}=\frac{v_c}{t_c}\frac{d\eta}{d\tau}=\frac{e}{m}\left(E+\frac{\eta v_c}{c}\times B\right)
$$
We then divide through by $v_c/t_c$,
$$
\frac{d\eta}{d\tau}=\frac{et_c}{mv_c}E+\frac{et_c}{mc}\eta\times B
$$
Since the second term on the right has only $t_c$ there, we use that to find $t_c=mc/e$ and then,
$$\frac{et_c}{m}=v_c\implies c=v_c,$$
which should be sensible. We then have a dimensionless Lorentz equation,
$$
\frac{d\eta}{d\tau}=E+\eta\times B
$$
from your two scales, $t_c=m_ec/e$ (I am assuming you are modeling an electron here) and $v_c=c$.
