0
$\begingroup$

I am referring to this paper, on page 3 it is given that:

The telegrapher's equation is given as $$\partial_t P_+=D\partial_x^2P_+-v\partial_xP_+-\gamma P_++\gamma P_-,$$ $$\partial_t P_-=D\partial_x^2P_-+v\partial_xP_-+\gamma P_+-\gamma P_-.$$ where $P_+(x; t)$ and $P_-(x; t)$ are the probability density for the particle to be at position x with velocities $+v$ and $-v$, respectively. Furthermore, choosing $\gamma^-1$ as the unit of time and $v\gamma^-1$ as the unit of length to recast the above given equations in the dimensionless form, we get $$\partial_t P_+=\mathcal{D}\partial_x^2P_+-\partial_xP_+- P_++P_-,$$ $$\partial_t P_-=\mathcal{D}\partial_x^2P_-+\partial_xP_-+ P_+- P_-$$ where $\mathcal{D}=D\gamma/v^2$.

This transition is not a simple subsitution, I want to know how they they have proceeded with recasting equations to dimensionless form. I tried to check in wikipedia, but couldn't find it.

$\endgroup$
0
$\begingroup$

I am able to answer the question using help from wikipedia link: taking $P_+=aP_+$,$P_-=bP_-$, $t=t_c\tau$, $x=x_c\zeta$, where $\tau , \zeta$ are the new independent variables. We can solve the given equation, for dimensionality take $\gamma/v=1/x_c$, $b/a=1$ and $t_c\gamma=1$ all these are dimensionally compatible leading to the given normalization. :)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.