Graduate Geodesic Expansion: Finding the $\theta_{\pm}$ Factor

[ergospherical]

\begin{align*}
\mathrm{\mathbf{(a)}} \quad U_{\pm} \cdot U_{\pm} &= \dfrac{1}{2} (n_a n^a \pm 2 n_a m^a + m_a m^a) = \pm n_a m^a = 0 \\
U_+ \cdot U_- &= \dfrac{1}{2} (n_a n^a - m_a m^a) = \dfrac{1}{2} (-1-1) = -1 \\ \\

\mathrm{\mathbf{(b)}} \quad P^a_b &= \delta^a_b + U_{\mp}^a (U_{\pm})_b + U_{\pm}^a (U_{\mp})_b \\

&= \delta^a_b + \dfrac{1}{2} (n^a \mp m^a)(n_b \pm m_b) + \dfrac{1}{2} (n^a \pm m^a)(n_b \mp m_b) \\

&= \delta^a_b + n^a n_b - m^a m_b \\

&= h^a_b - m^a m_b \\ \\

\mathrm{\mathbf{(c)}} \quad \theta_{\pm} &= P^{ab} \nabla_a (U_{\pm})_b \\

&= \dfrac{1}{\sqrt{2}} (h^{ab} - m^a m^b) \nabla_a (n_b \pm m_b) \\

&= \dfrac{1}{\sqrt{2}} (h^{ab} - m^a m^b) (K_{ab} - n_a n^c \nabla_c n _b) \pm \dfrac{1}{\sqrt{2}} (h^{ab} - m^a m^b) (k_{ab} - m_a m^c \nabla_c m _b) \\

&= \dfrac{1}{\sqrt{2}} (h^{ab} - m^a m^b) K_{ab} - \dfrac{1}{\sqrt{2}} \underbrace{h^{ab} n_a}_{= \, 0} n^c \nabla_c n_b + \dfrac{1}{\sqrt{2}} m^b \underbrace{m^a n_a}_{= \, 0} n^c \nabla_c n_b \\

&\hspace{45pt} \pm \dfrac{1}{\sqrt{2}} (h^{ab} - m^a m^b) (k_{ab} - m_a m^c \nabla_c m _b) \\ \\

&= \dfrac{1}{\sqrt{2}} (h^{ab} - m^a m^b) K_{ab} \pm \dfrac{1}{\sqrt{2}} (h^{ab} - m^a m^b) (k_{ab} - m_a m^c \nabla_c m _b)
\end{align*}Why is there a factor of $\dfrac{1}{\sqrt{2}}$, and how do you re-write the second term?

 

[robphy]

Is the \frac{1}{\sqrt{2}} from the definition of U^a_{\pm}?

For background, it seems this question
is related to http://www.damtp.cam.ac.uk/user/hsr1000/black_holes_lectures_2020.pdf
and http://www.damtp.cam.ac.uk/user/hsr1000/blackholes2020_examples2.pdf

 

[ergospherical]

Is the \frac{1}{\sqrt{2}} from the definition of U^a_{\pm}?

Yeah, one thing I don't understand is why the $\dfrac{1}{\sqrt{2}}$ doesn't appear in the equation $\theta_{\pm} = (h^{ab} - m^a m^b)K_{ab} \pm k$ in the question.