Scaling in heat transfer PDE and dimensionless group I found the following in my Scaling in Heat Transfer notes:

Rod Conduction
A rod of length $L$ initially at $T_0$ then (at $t = 0$) one end is raised to $T_1$. Find $T(x, t)$.
$$T_t = \kappa T_{x x}$$
Scaling
We write
$$x = Lx', \ t = \tau t', \ T = T_0 + (T_1 - T_0) T'(x', t')$$
with $\tau$ as yet unknown. This gives
$$T'_{t'} = \left[ \dfrac{\kappa \tau}{L^2} \right] T'_{x' x'}$$
There is one dimensionless group $[\cdot]$ and if we choose $\tau = \dfrac{L^2}{\kappa}$ we get the dimensionless problem.

What I'm confused about is how they got $T'_{t'} = \left[ \dfrac{\kappa \tau}{L^2} \right] T'_{x' x'}$? We do have $T_t = \kappa T_{x x}$, but how did $\kappa$ get replace with the dimensionless group $\left[ \dfrac{\kappa \tau}{L^2} \right]$? How is this justified? What is the reasoning behind this change?
I would greatly appreciate it if people could please take the time to clarify this.
 A: The original PDE is
$$\frac{\partial T}{\partial t} = [\kappa] \frac{\partial^2 T}{\partial x^2}$$
By the chain rule, the left hand side is:
$$\frac{\partial T}{\partial t} = \frac{\partial T}{\partial t'} \frac{\partial t'}{\partial t} = \frac{\partial T}{\partial t'} \frac{1}{\tau}$$
Applying the chain rule twice to the right hand side:
$$\frac{\partial T}{\partial x} = \frac{\partial T}{\partial x'} \frac{\partial x'}{\partial x} = \frac{\partial T}{\partial x'} \frac{1}{L}$$
$$\frac{\partial^2 T}{\partial x^2} = \frac{\partial}{\partial x}(\frac{\partial T}{\partial x}) = \frac{1}{L} \frac{\partial}{\partial x'}(\frac{\partial T}{\partial x}) = \frac{1}{L^2} \frac{\partial}{\partial x'}(\frac{\partial T}{\partial x'})  = \frac{1}{L^2} (\frac{\partial^2 T}{\partial x'^2})$$
Subtituting into the PDE
$$\frac{1}{\tau} \frac{\partial T}{\partial t'} = \left[\frac{\kappa}{L^2}\right] \frac{\partial^2 T}{\partial x'^2}$$
Or
$$\frac{\partial T}{\partial t'} = \left[\frac{\kappa \tau}{L^2}\right] \frac{\partial^2 T}{\partial x'^2}$$
Hope that helps
