# Proof of a solution of Klein-Gordon equation with Killing vector

In this (very good) notes: http://people.physics.tamu.edu/pope/geomlec.pdf it is set as an exercise to proof that if $u_i$ solves the Klein-Gordon equation:

$$(\Box -m^2 )u_i = 0$$

then you can proof, appealing to the properties of the Killing vectors, that the Killing vector $K^\mu \partial_\mu$ also solves

$$(\Box -m^2 )K^\mu \partial_\mu u_i = 0$$

and that the key point is to prove that $\Box(K^\mu \partial_\mu u_i ) = K^\mu \partial_\mu\Box u_i$

Honestly, I don't get with the key. Any help will be appreciated.

• Can't you just use Eq (5.52) to write $\square K_{\mu} = -R_{\mu\nu}K^{\nu}$, and then manipulate everything from there? – Alex Nelson Jul 26 '17 at 17:10
• Thanks, Alex. Yes, I thought in using that equality, but I can't get anything from it. – David Jul 26 '17 at 18:28

If $K$ is a Killing vector, it satisfies $$\nabla_\mu K_\nu + \nabla_\nu K_\mu = 0 \quad \implies \quad \nabla^\mu K_\mu = 0 ~.$$ Another property we will need can be proved by acting on the above euqation with $\nabla^\nu$. We find \begin{align} \Box K_\mu &= - \nabla^\nu \nabla_\mu K_\nu \\ &= [ \nabla_\mu , \nabla_\nu ]K^\nu \\ &= R^\nu{}_{\lambda\mu\nu} K^\lambda \\ &= - R_{\mu\nu} K^\nu~. \end{align} Now, we have $$( \nabla^\mu \nabla_\mu - m^2 ) K^\nu \nabla_\nu \phi = \Box K^\nu \nabla_\nu \phi + 2 \nabla^\mu K^\nu \nabla_\mu \nabla_\nu \phi + K^\nu \nabla^\mu \nabla_\mu \nabla_\nu \phi - m^2 K^\nu \nabla_\nu \phi ~.$$ Note $$\nabla^\mu K^\nu \nabla_\mu \nabla_\nu \phi =\frac{1}{2} ( \nabla^\mu K^\nu + \nabla^\nu K^\mu ) \nabla_\mu \nabla_\nu \phi = 0 ~,$$ and \begin{align} K^\nu \nabla^\mu \nabla_\mu \nabla_\nu \phi &= K^\nu \nabla^\mu \nabla_\nu \nabla_\mu \phi \\ &= K^\nu \nabla_\nu \Box \phi + K^\nu [ \nabla^\mu , \nabla_\nu ] \nabla_\mu \phi \\ &= m^2 K^\nu \nabla_\nu \phi + K^\mu R_{\mu\nu} \nabla^\mu \phi ~. \end{align} Putting all this together, we find \begin{align} ( \nabla^\mu \nabla_\mu - m^2 ) K^\nu \nabla_\nu \phi &= \Box K^\nu \nabla_\nu \phi + m^2 K^\nu \nabla_\nu \phi + K^\mu R_{\mu\nu} \nabla^\mu \phi - m^2 K^\nu \nabla_\nu \phi \\ &= 0 ~. \end{align} QED.
• I have a couple of questions if you don't mind: why are you writing the Killing vector as $K^\mu \nabla_\mu$ instead of $K^\mu \partial_\mu$? And the $\Box$ operator over the Killing vector shouldn't be only $(\Box K^\mu )\partial_\mu \phi + K^\mu \Box(\partial_\mu \phi)$? – David Jul 27 '17 at 22:17
• When acting on a scalar field, $\nabla_\mu \phi = \partial_\mu \phi$. $\Box = \nabla^\mu \nabla_\mu$ so it is a square of a derivative operator and it does not distribute in the way that you write. Is $\frac{d^2}{dx^2} (fg) = \frac{d^2f}{dx^2} g + f \frac{d^2g}{dx^2}$????? – Prahar Jul 27 '17 at 22:19
• Thanks. I didn't notice the square nature of the $\Box$ operator. – David Jul 28 '17 at 0:03