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I have 3 questions:

  1. Is a vector perpendicular to any tangent vector of any null geodesic also a null vector?

  2. How can we find hypersurface to a null geodesic?

  3. Suppose we have a null geodesic in 4d, the transverse metric to it is 2d (You can look it up online. It is 2 dimensional) Intuitively, I don't understand how is this possible.

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  • $\begingroup$ I think that the answer to 3 should be that the hypersurface is indeed 3 dimensional but it has a null eigen direction and hence, it's contribution to the metric on the hypersurface is zero. Hence, the transverse metric is 2 dimensional but the transverse space is 3 dimensional $\endgroup$ – Tushar Gopalka May 7 '18 at 6:15
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1.

Nope. For example, in Minkowski spacetime, consider the vector $(v^\mu)=(1,0,0,1)$, which is a null vector. The vector $ (u^\mu)=(0,1,0,0) $ is orthogonal to it, yet, it is not a null vector.

2.

I don't understand this question.

3.

Consider a null geodesic with tangent vector $u^\mu$ ($u^\mu u_\mu$=0). Let $\lambda$ be the parameter along the null geodesic. Let $\Sigma_p<T_pM$ be the orthogonal complement to $u^\mu$ at $p\in M$. Note that because $u^\mu$ is a null vector, it is orthogonal to itself, hence $u_p\in\Sigma_p$. Let us choose two additional vectors in $\Sigma_p$, $e^\mu_1$ and $e^\mu_2$. We can choose these vectors such that $e^\mu_Ae^\nu_Bg_{\mu\nu}=\delta_{AB}$ ($A,B=1,2$), and because $u^\mu$ is a normal vector, we have $u^\mu e^\nu_A g_{\mu\nu}=0$.

The line element at $p$ can be expressed as $$ ds^2(p)=g_{\mu\nu}(p)u^\mu u^\nu d\lambda^2+2g_{\mu\nu}(p)e^\mu_A u^\nu d\sigma^Ad\lambda+g_{\mu\nu}(p)e^\mu_A e^\nu_B d\sigma^A d\sigma^B, $$ where $\sigma^A$ are some coordinates for which at $p$, $e^\mu_A$ are the coordinate basis vectors.

Writing in the relations between the basis vectors gives $$ ds^2(p)=g_{\mu\nu}(p)e^\mu_A e^\nu_B d\sigma^A d\sigma^B\equiv \delta_{AB}d\sigma^A d\sigma^B, $$ since the other contractions are all vanishing.

We can then see that if you make an infinitesimal displacement $d\xi=(d\lambda,d\sigma^1,d\sigma^2)$ that is orthogonal to the null curve, the contribution from $d\lambda$ is zero, hence it doesn't matter. After all, $\lambda$ is a null parameter, along which there is vanishing arc length, and so the physical/geometric displacement corresponding to $d\xi$ depends only on $d\sigma^1$ and $d\sigma^2$, which is why the metric is effectively two dimensional.

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  • $\begingroup$ In the 2nd question, I am asking how can we find an orthogonal hypersurface to a congruence of null geodesics? $\endgroup$ – Tushar Gopalka May 7 '18 at 15:21

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