Consider an infinite horizon cake eating differential game described by \begin{align} &\max_{u_1(t)} \int_0^\infty{e^{-r_1 t}\ln(u_1(t))dt}\\ &\max_{u_2(t)} \int_0^\infty{e^{-r_2 t}\ln(u_2(t))dt}\\ \text{s.t.} \quad &\dot x(t) = -u_1(t)-u_2(t) \end{align} where $t \in [0,\infty)$ refers to the current time, $u_i(t)$ to the consumption rate of player $i \in \{1,2\}$ and $x(t)$ to the size of the cake.
I'm interested in the time preference rates $r_i > 0$. Without further explanations I found in Clemhout and Wan, 1989, On Games of cake-eating, p. 131
Since one can always choose a unit for time without losing generality, we suppose that the length of a unit of time is such that the preference rates of the two players sum up to unity.
That is, we can normalize $r_1 + r_2 = 1$. But what does this mean actually? I'm particular puzzled about the without loss of generality part.
First note that $r_1>0$ and $r_2>0$. Let $r_i$ be such that $r_1+r_2=\rho$. By introducing a new independent variable $\tau=\rho t$ we get $$\max_{u_1(\tau)} \frac{1}{\rho}\int_0^\infty{e^{-\frac{r_1}{\rho}\tau}\ln(u_1(\tau))d\tau}.$$ The factor $\frac{1}{\rho}$ in front of the integral can be dropped as it doesn't influence the result. And the new discount rates $\frac{r_i}{\rho}$ now sum to 1.