# How did the author get these three expressions?

I am reading Nonplanar Relativistic Flow by Peter G. Eltgroth, Phys. Fluids 15, 2140–2144 (1972) doi.

I do not understand how the author arrives at expressions (12), (13) and (14) at page 2. Equation (12) in particular baffles me, since it says:

$$F(\beta) = -\dfrac{r}{ct}$$

As you can see the LHS is a function of $$\beta$$, the relativistic speed, while the RHS is independent of it.

EDIT Here are the equations:

We start from the special relativistic conservation equation:

$$\partial_\mu T^{\mu\nu}=0$$

and get (for an ultra-relativistic fluid):

$$\dfrac{\partial}{\partial ct}\left( \dfrac{E_1(1+\beta^2/3)}{1-\beta^2} \right) + \dfrac{\partial}{\partial r} \left( \dfrac{4}{3} \, \dfrac{E_1\beta}{(1-\beta^2)} \right) = 0$$

$$\dfrac{\partial}{\partial ct} \left( \dfrac{4}{3} \, \dfrac{E_1\beta}{(1-\beta^2)} \right) + \dfrac{\partial}{\partial r} \left( \dfrac{E_1(1+\beta^2/3)}{1-\beta^2} \right) = \dfrac{\nu}{3}\,\dfrac{E_1}{r}$$

where $$E_1$$ is the energy density. We assume that $$E_1=E_2(\beta(t,r))\,E_3(r)$$. We get (eqn (10) in the paper):

$$\dfrac{\partial}{\partial ct}\left( \dfrac{E_2(1+\beta^2/3)}{1-\beta^2} \right) + \dfrac{\partial}{\partial r} \left( \dfrac{4}{3} \, \dfrac{E_2\beta}{(1-\beta^2)} \right) + \dfrac{4}{3} \, \dfrac{\beta\,E_2}{(1-\beta^2)}\, \dfrac{E^\prime_3}{E_3}= 0 \tag{10}$$

and (eqn (11) in the paper)

$$\dfrac{\partial}{\partial ct} \left( \dfrac{4}{3} \, \dfrac{E_2\beta}{(1-\beta^2)} \right) + \dfrac{\partial}{\partial r} \left( \dfrac{E_2(1+\beta^2/3)}{1-\beta^2} \right) + E_2 \, \dfrac{1/3+\beta^2}{1-\beta^2}\, \dfrac{E^\prime_3}{E_3} = \dfrac{\nu}{3}\,\dfrac{E_2}{r} \tag{11}$$

We find from (10) that in order to have a consistent expression, we need:

$$\dfrac{\partial \beta/\partial t}{\partial \beta \partial r} = F(\beta)$$

$$\dfrac{E^\prime_3}{E_3}\, \left( \dfrac{\partial\beta}{\partial r} \right)^{-1} = G(\beta)$$

Now the part I don't get: the paper says that from the above expressions and eqn (11) it is possible to derive:

$$F(\beta) = -\dfrac{r}{ct} \tag{12}$$

$$G(\beta) = a\, \dfrac{F^\prime(\beta)}{\beta} \tag{13}$$

$$E_3 = E_0 \, r^a \tag{14}$$

I do not see how these last three expressions come from equation (11).