A Solving Schwarzschild Field Equations in this Form

[bolbteppa]

Applying Cartan's first and second structural equations to the vielbein forms
\begin{align}
e^t = A(r) dt , \ \ \ \ \ e^r = B(r) dr , \ \ \ \ \ e^{\theta} = C(r) d \theta , \ \ \ \ \ e^{\phi} = C(r) \sin \theta d \phi ,
\end{align}
taken from the metric
\begin{align}
ds^2 = A^2(r) dt^2 - B^2(r) dr^2 - C^2(r) d \theta^2 - C^2(r) \sin^2 (\theta) d \phi^2 , \ \ \ \ C(r) = r,
\end{align}
we end up with the Ricci curvature in the form
\begin{align}
R^t \, _t &= - \frac{1}{B^2}[\frac{A''}{A} - \frac{A'B'}{AB} + 2 \frac{A'}{Ar}] \\
R^r \, _r &= - \frac{1}{B^2}[\frac{A''}{A} - \frac{A'B'}{AB} - 2\frac{B'}{Br}] \\
R^{\theta} \, _{\theta} &= - \frac{1}{B^2}[- \frac{B'}{rB} + \frac{A'}{Ar} + \frac{1}{r^2}] + \frac{1}{r^2}= R^{\phi} \, _{\phi}.
\end{align}
This is the result in Zee's gravity, page 611, and at around 46 mins in this video (on setting $C = r$).

My question is solving the equation
\begin{align}
R_{\mu \nu} = 0
\end{align}
with the Ricci curvature in this form to find the Schwarzschild line element, noting the curvature is a bit different from the usual form. By subtracting $R^r \, _r$ from $R^t \, _t$ we find
\begin{align}
\frac{A'}{A} + \frac{B'}{B} = 0,
\end{align}
but now using this in the $R^{\theta} \, _{\theta}$ curvature term, we get
\begin{align}
0 &= - \frac{1}{B^2}[- \frac{B'}{rB} + \frac{A'}{Ar} + \frac{1}{r^2}] + \frac{1}{r^2} \\
&= - \frac{1}{B^2}[- \frac{B'}{rB} - \frac{B'}{Br} + \frac{1}{r^2}] + \frac{1}{r^2} \\
&= - \frac{1}{B^2}[- 2 \frac{B'}{rB} + \frac{1}{r^2}] + \frac{1}{r^2} \\
&= \frac{1}{B^2}[2r \frac{B'}{B} - 1] + 1 \\
&= 2r \frac{B'}{B} - 1 + B^2 \\
&= 2rB' - B + B^3
\end{align}
which does not reduce to a total derivative (as it usually should) allowing us to solve the equations.

What has gone wrong and how does one solve the equations in this form?

 

[martinbn]

If all the calculations a correct, the last equation can be solved. You can separate the variables.

 

[bolbteppa]

Thank you, yes this does seem to give the right solution

https://www.symbolab.com/solver/ordinary-differential-equation-calculator/2xy'-y+y^{3}=0

I was more expecting it to be literally the exact same as the method in the link (ending up with a total derivative) and didn't even think to solve it directly yet, wonder why it doesn't (perhaps need to make it an exact ode using integrating factors or something).

 

[martinbn]

In the link that you call the usual form the equations is $R_{\theta\theta}=0$, here you are looking at $R^\theta_{\;\;\theta}=0$.