In a stationary and axisymmetric spacetime, there are two Killing vectors, say $\zeta^\mu$ and $\xi^\mu$, one timelike and one space like.

I understand that for a real scalar field, $\phi$, that obeys the symmetry of this spacetime one can say

$$\zeta^\mu \nabla_\mu \phi = \xi^\mu \nabla_\mu \phi = 0 $$

and therefore since the gradient of $\phi$ is orthogonal to both Killing vectors, $\nabla_\mu \phi \nabla^\mu \phi \geq 0$.

Now, my question is in regards to how this logic translates across if $\phi$ becomes a complex scalar field, where $\Phi = \phi_1 + i\phi_2$.

$\nabla_\mu \Phi \nabla^\mu \Phi \geq 0$ no longer makes sense to me, since the gradient is complex, however is it a valid comment to say that the real components of the complex field must be spacelike or null, if the complex field is to still obey the spacetime symmetries, i.e. $\nabla_\mu \phi_i \nabla^\mu \phi_i \geq 0$


You're right to conclude $\nabla_\mu\phi_j\nabla^\mu\phi_j\ge0$ for $j\in\{1,\,2\}$, so $\nabla_\mu\Phi^\ast\nabla^\mu\Phi=\sum_{j=1}^2\nabla_\mu\phi_j\nabla^\mu\phi_j\ge0$.

  • $\begingroup$ Okay thanks, just as a (potentially trivial) follow up, can I check that can that be generalised to $$\sum^2_{i,j=1} \nabla_{\mu} \phi_i \nabla^{\mu} \phi_{j} \geq 0$$. $\endgroup$ – Smp03 Aug 15 at 22:37
  • $\begingroup$ @Smp03 You can rewrite that as $\nabla_\mu\chi\nabla^\mu\chi\ge0$ with $\chi:=\sum_j\phi_j$. $\endgroup$ – J.G. Aug 16 at 7:05

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