Preserving Covariant Derivatives of Null Vectors Under Variation

[sarriiss]

Having two null vectors with $$n^{a} l_{a}=-1, \\ g_{ab}=-(l_{a}n_{b}+n_{a}l_{b}),\\ n^{a}\nabla_{a}n^{b}=0$$ gives $$\nabla_{a}n_{b}=\kappa n_{a}n_{b},\\ \nabla_{a}n^{a}=0,\\ \nabla_{a}l_{b}=-\kappa n_{a}l_{b},\\ \nabla_{a}l^{a}=\kappa$$.
How to show that under the variation of the null vectors, above covariant derivatives are preserved? In other words how to get the conditions on variation of null vectors which preserve the above covariant derivatives. To be specific, I need some hints to get equation (2.5) in Carlip's paper https://arxiv.org/abs/1702.04439. I started with $$l^{a}\rightarrow l^{a}+\delta l^{a},\\ n^{a}\rightarrow n^{a}+\delta n^{a}$$ but couldn't get equation (2.5) in Carlip's paper.

 

[anuttarasammyak]

Light geodesic is null, ds=0. Sometimes coordinate time t is used not s as parameter of geodesic equation. Is this prescription helpful in you case?

 

[Antarres]

Not sure if you solved your problem by now, but it's just a matter of combining terms in a Leibniz rule. Let's vary both the equations for covariant derivatives:
$$\nabla_a(\delta l_b) = -\delta \kappa n_a l_b - \kappa \delta n_a l_b - \kappa n_a\delta l_b$$
$$\nabla_a(\delta n_b) = \delta \kappa n_a n_b + \kappa(\delta n_a n_b + n_a \delta n_b)$$

Contract the second equation by ##l^b##:
$$\nabla_a(\delta n_b)l^b = -\delta\kappa n_a + \kappa(-\delta n_a + n_a l^b\delta n_b)$$
Apply Leibniz rule for covariant derivative:
$$\nabla_a(l^b\delta n_b) = \nabla_a(\delta n_b)l^b + \nabla_a(l^b)\delta n_b = -\delta \kappa n_a - \kappa\delta n_a$$
Now contract with ##n^a##, you find:
$$\bar{D}(l^b\delta n_b) = -\kappa n^a\delta n_a$$

Secondly, you contract the 2nd variation from the beggining, this time with ##n^b##:
$$\nabla_a(\delta n_b)n^b = \kappa n_a n^b\delta n_b$$
Apply the Leibniz rule as before:
$$\nabla_a(n^b \delta n_b) = 2\kappa n_a n^b \delta n_b$$
And finally contract by ##l^a##, you find:
$$D(n^b \delta n_b) = -2\kappa n^b\delta n_b$$

Finally you combine these two results to see that:
$$\bar{D}(l^b\delta n_b) = (D+\kappa)(n^b \delta n_b)$$

This is the first formula you find below. Similarly you will contract the first variation equation, and I'd assume you will find the second relation in (2.5) of that paper. Hope that helps.

 

[sarriiss]

Thank you so much! I derived the second relation.