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In terms of spin coefficients, shear is defined as

$$\sigma=(m_a)(m_b)(\nabla_al^b)$$

where $m,l$ denotes null tetrad vectors. I came across the following paragraph from one of Stephen Hawking's papers:

What happens is that the perturbation produces tidal forces which distort the horizon. If the black hole is not rotating, the distortion of an element of the horizon will be constant in time and the shear will be zero. If the black hole is rotating, however, distortion will be periodic in time and so there will be shear.

So distortion directly does not mean shear? In a nonrotating black hole, though distortion is constant in time, still, distortion is there, then why shear is zero?

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  • $\begingroup$ Hello! I have edited your question using MathJax (LaTeX) math typesetting. For future questions, you can refer to MathJax basic tutorial and quick reference. Thanks! $\endgroup$
    – jng224
    Commented Aug 24, 2021 at 11:04
  • $\begingroup$ Shear is not the only type of distortion, there are vorticity and divergence associated to null rays. Even other spin connections also have some interpretations in terms of curvature. $\endgroup$
    – KP99
    Commented Aug 25, 2021 at 14:54

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