Editing Weyl transformation (section)

==Formulas==
For the transformation
:<math> 
    g_{ab} = f(\phi(x)) \bar{g}_{ab} 
</math>    
We can derive the following formulas
:<math> 
\begin{align}
    g^{ab} &= \frac{1}{f(\phi(x))} \bar{g}^{ab}\\
    \sqrt{-g} &= \sqrt{-\bar{g}} f^{D/2} \\
    \Gamma^c_{ab} &= \bar{\Gamma}^c_{ab} + \frac{f'}{2f} \left(\delta^c_b \partial_a \phi + \delta^c_a \partial_b \phi - \bar{g}_{ab} \partial^c \phi \right) \equiv \bar{\Gamma}^c_{ab} + \gamma^c_{ab} \\
     R_{ab} &= \bar{R}_{ab} + \frac{f'' f- f^{\prime 2}}{2f^2} \left((2-D) \partial_a \phi \partial_b \phi - \bar{g}_{ab} \partial^c \phi \partial_c \phi \right) + \frac{f'}{2f} \left((2-D) \bar{\nabla}_a \partial_b \phi - \bar{g}_{ab} \bar{\Box} \phi\right) + \frac{1}{4} \frac{f^{\prime 2}}{f^2} (D-2) \left(\partial_a \phi \partial_b \phi - \bar{g}_{ab} \partial_c \phi \partial^c \phi \right) \\
     R &= \frac{1}{f} \bar{R} + \frac{1-D}{f} \left( \frac{f''f - f^{\prime 2}}{f^2} \partial^c \phi \partial_c \phi + \frac{f'}{f} \bar{\Box} \phi \right) + \frac{1}{4f} \frac{f^{\prime 2}}{f^2} (D-2) (1-D) \partial_c \phi \partial^c \phi 
\end{align}
</math>
Note that the Weyl tensor is invariant under a Weyl rescaling.