Another question about the Schwarzschild solution of General Relativity:
In the derivation (shown below) of the Schwarzschild metric from the vacuum Einstein Equation, at the step marked "HERE," we have the freedom to rescale the time coordinate from $dt \rightarrow e^{-g(t)}dt$. However, later in the derivation, when we pick up an integration constant at the step marked "HERE2," we are unable to absorb that constant into the metric. Actually this constant becomes a fundamental part of the metric itself. Why can't we incorporate this constant into the differentials of the metric as well?
Here's the derivation:
$$R_{\mu\nu}-\frac{1}{2}Rg_{\mu\nu} = 0$$
Assume the most general spherically symmetric metric solution to the above:
$$g_{\mu\nu} = \begin{pmatrix} -e^{2\alpha(r,t)} & 0 & 0 & 0\\ 0 & e^{2\beta(r,t)}&0&0\\0&0&r^2 & 0\\0 & 0 &0 & r^2\sin^2(\theta) \end{pmatrix}$$
Take the trace of both sides:
$$R + 2R = 0\\ \Rightarrow R = 0\\ R_{\mu\nu} - \frac{1}{2}(0)= 0\\ R_{\mu\nu}=0$$
We are free to say:
$$R_{01} = \frac{2}{r}\dot{\beta}=0\\ \Rightarrow \dot{\beta} = 0\\ \therefore \beta = \beta(r) $$
Now, we can take a time derivative of $R_{22}$,
$$\dot{R_{22}} = \frac{d\left(e^{-2\beta}[r(\beta'-\alpha') - 1] + 1\right)}{dt} = 0 = e^{2\beta}r\dot{\alpha}'$$ (Because $\dot{\beta} = 0$, $r$ is just a spacetime coordinate (and is therefore independent of other spacetime coordinates), and $\dot{\beta}' = \partial_r\partial_t\beta = \partial_r(0)$.) $$\dot{\alpha}' = 0\\ \therefore \alpha = f(r) + g(t)$$ because this is the only functional form for which $\partial_r\partial_t\alpha = \partial_t\partial_r\alpha = 0$. Now, the metric becomes:
$$g_{\mu\nu} = \begin{pmatrix} -e^{2f(r)}e^{2g(t)} & 0 & 0 & 0\\ 0 & e^{2\beta(r)}&0&0\\0&0&r^2 & 0\\0 & 0 &0 & r^2\sin^2(\theta) \end{pmatrix}$$
HERE, at this point, we can rescale time: $dt \rightarrow e^{-g(t)}dt$, which makes the metric:
$$g_{\mu\nu} = \begin{pmatrix} -e^{2f(r)} & 0 & 0 & 0\\ 0 & e^{2\beta(r)}&0&0\\0&0&r^2 & 0\\0 & 0 &0 & r^2\sin^2(\theta) \end{pmatrix}$$
Setting $R_{11} = R_{00}$, we find:
$$e^{2(\beta-\alpha)}R_{00} + R_{11} = 0\\ \frac{2}{r}\alpha' + \frac{2}{r}\beta' = 0\\ \Rightarrow \alpha = -\beta + const.$$
This constant is absorbed into $dr^2$.
HERE2, Now comes the crucial part, which I don't understand: looking back at $R_{22} = 0$, we have
$$(re^{2\alpha})' = 1\\ \int (re^{2\alpha})'= \int 1\\ re^{2\alpha} = r + \underbrace{\mu}_{\text{Constant of integration}}\\ e^{2\alpha} = 1 + \frac{\mu}{r}$$
So the metric finally ends up being:
$$ds^2 = -(1 + \frac{\mu}{r})dt^2 + (1 + \frac{\mu}{r})^{-1}dr^2 + r^2 d\Omega^2.$$
Why can't $\frac{\mu}{r}$ be absorbed into $dt^2$ and $dr^2$ like $g(t)$ and the other integration constant in this derivation were?? I don't understand the distinction at all, any help would be greatly appreciated!!