I have some questions regarding the author's solution to parts $(\mathrm{ii})$ and $(\mathrm{iii})$ to the following question on the Rindler metric:
The Rindler metric is defined by $$ds^2=\alpha^2x^2dt^2-dx^2\tag{1}$$ where $\alpha$ is a constant. $(\mathrm{i})$ Show that the non-zero Christoffel symbols are $$\Gamma_{xt}^t=\Gamma_{tx}^t=\frac1x,\quad\Gamma_{tt}^x=\alpha^2x\tag{2}$$ $(\mathrm{ii})$ Write down the geodesic equations for $t$ and $x$ and show that $x^2dt/ds$ is constant along geodesics.
$(\mathrm{iii})$ Solve the geodesic equations. Hint: do not attempt to solve the equations directly. Use $g_{ab}u^au^b=1$
For part $(\mathrm{i})$, since I'm not that interested in this part of the question I will just show the technique I used to derive the second Christoffel symbol in $(2)$ - Using the Levi-Civita connection: $$\Gamma_{bc}^a=\frac{1}{2}g^{ad}\left(\partial_bg_{dc} + \partial_cg_{bd}-\partial_dg_{bc}\right)\tag{3}$$ Reading the components of the metric off from $(1)$, I indentify $g_{tt}=\alpha^2x^2,\,g_{xx}=-1$, since off-diagonal components of the matrix representation of the metric are zero, $g(x)=\begin{pmatrix}\alpha^2x^2 & 0 \\0 & -1 \\\end{pmatrix}.$ Therefore, components of the inverse metric are easliy found and are $g^{tt}=\frac{1}{\alpha^2x^2},\,g^{xx}=-1$ such that $g^{-1}(x)=\begin{pmatrix} \frac{1}{\alpha^2x^2} & 0 \\ 0 & -1 \\ \end{pmatrix}$. Using $(3)$, there is only a non-zero contribution when $a=d=x$ and $b=c=t$: $$\begin{align}\Gamma_{tt}^x &=\frac{1}{2}g^{xx}\left(\partial_bg_{xt} + \partial_tg_{tx}-\partial_xg_{tt}\right)\\&=-\frac12\left(0+0-\partial_x\left[\alpha^2x^2\right]\right)\\&=\alpha^2 x\end{align}$$ By a similar technique using the Levi-Civita connection, I can also show that $$\Gamma_{xt}^t=\Gamma_{tx}^t=\frac1x$$
For part $(\mathrm{ii})$, using the geodesic equation, $$\frac{d^2x^a}{ds^2}+\Gamma_{bc}^a\frac{dx^b}{ds}\frac{dx^c}{ds}=0\tag{4}$$ and the Christoffel symbols in part $(\mathrm{i})$, and substituting the coordinates $\left(x^a=x^c,x^b\right)=(t, x)$ into $(4)$ yields:
$$\begin{align}\frac{d^2t}{ds^2}+\Gamma_{xt}^t\frac{dx}{ds}\frac{dt}{ds}+\Gamma_{tx}^t\frac{dt}{ds}\frac{dx}{ds}&=\frac{d^2t}{ds^2}+\frac1x\frac{dx}{ds}\frac{dt}{ds}+\frac1x\frac{dt}{ds}\frac{dx}{ds}\\&=\frac{d^2t}{ds^2}+\frac2x\frac{dx}{ds}\frac{dt}{ds}=0\end{align}$$ This is the first of the two geodesic equations, the other geodesic equation can be found by setting the coordinates of $(4)$ to $\left(x^a=x^b, x^c\right)=(x, t)$, which gives $$\frac{d^2x}{ds^2}+\Gamma_{tt}^x\frac{dt}{ds}\frac{dt}{ds}=\frac{d^2x}{ds^2}+\alpha^2x\left(\frac{dt}{ds}\right)^2=0$$ So in summary, the two geodesic equations are: $$\frac{d^2t}{ds^2}+\frac2x\frac{dx}{ds}\frac{dt}{ds}=0\tag{5}$$ $$\frac{d^2x}{ds^2}+\alpha^2x\left(\frac{dt}{ds}\right)^2=0\tag{6}$$ Now I need to show that $x^2\frac{dt}{ds}$ is constant for both $(5)$ and $(6)$, since both $(5)$ and $(6)$ have no explicit $t$ dependence then the following result may be used $$\frac{d}{ds}\left(x^2\ \frac{dt}{ds}\right)=0\tag{7}$$ Using the product rule to differentiate $(7)$ and the chain rule, $\frac{d}{ds}=\frac{dx}{ds} \frac{d}{dx}$, $$\begin{align} \frac{d}{ds}\left(x^2\ \frac{dt}{ds}\right)&=\frac{dx}{ds}\left(\frac{d}{dx}x^2\right)\frac{dt}{ds}+x^2\frac{d^2t}{ds^2}\\&=x^2\frac{d^2t}{ds^2}+2x\frac{dx}{ds}\frac{dt}{ds}=0\tag{8}\end{align}$$ Then dividing eqn $(8)$ by $x^2$ leads directly to equation $(5)$ as expected.
Now here is the problem, the question asks to "show that $\frac{d}{ds}\left(x^2\ \frac{dt}{ds}\right)$ is constant along geodesics". I have shown that for eqn $(5)$, $x^2\frac{dt}{ds}$ is constant, but how do I show that $x^2\frac{dt}{ds}$ is also constant for eqn $(6)$? The author's solution simply states that:
As the components of the metric do not depend on $t$ $$\alpha^2 k g_{ta}\frac{dx^a}{ds}= \alpha^2x^2\frac{dt}{ds}\tag{9}$$ is constant along geodesics.
I don't understand the logic behind the author's 'solution', the quote above doesn't even address the question; "show that $x^2\frac{dt}{ds}$ is constant along geodesics".
When I see the word "geodesics" and not "geodesic" I am making the assumption that the author is referring to eqn $(5)$ and $(6)$. I hope this assumption is correct.
To summarize, does anyone know what the author is trying to say with that eqn $(9)$ and then claiming that it is constant along eqns $(5)$ and $(6)$? Perhaps it's just me not understanding, but there doesn't seem to be any way of obtaining equations $(5)$ and $(6)$ with this eqn $(9)$.
Remark
I will ask about the solution to part $(\mathrm{iii})$ in a separate question as this post is becoming too long.