Why are these equations equal to a constant? I am reading this part of a research paper

where the author states that the left hand side of equations (12) and (13) must be equal to a constant. However I could not understand the explanation he gives.
Does anyone know what the author means for why the equations are equal to a constant?
Thanks.
 A: Note that transitivity of equality implies the LHS of both equations are equal.
Now imagine you could vary $r$ and $t$ independently to any values you would please. The two LHS's should change (likely in a continuous manner), but must equal each other no matter what values you pick.
The only way this would be possible is if both LHS's equalled the same constant.
A: If you have the identity $f(r)=g(t)$ for all $r$ and $t$, then the functions must be constant. If you want, simply take derivatives on both sides with respect say to $r$. Then $f'(r)=0$ because $g$ does not depend on $t$. Hence, $f$ is constant and the same happens to $g$.
A: The equation (13) can be written on a simpler form :
$$f(r)=g(t)$$
which is true any $r$ and any $t$.
Given a value $t=t_1$ then $f(r)=g(t_1)$ any $r$. Hense $f(r)$ doesn't change while $r$ changes. Thus $f(r)=\text{constant}$.
Similary, $g(t)=\text{constant}$. Both constant are equal since $f(r)=g(t)$
$$f(r)=g(t)=C$$
Of course $C$ can be any constant.
So you get a system of two equations  : $\begin{cases} f(r)=C \\ g(t)=C \end{cases}$
in which $C$ is any constant. This means that $C$ is a parameter. 
The wording of the question lacks definition of functions and respective variables. Nevertheless, if my interpretation is correct, the argument is summarized below :

