Riemannian geometry - worldline meets nullcone.

Question

I've been studying the book "Semi-Riemannian Geometry" by B. O'Neill and doing some of the excersises. Chapter 6 (special relativity) includes the following one:

If $p$ is an event not on the world line of a freely falling observer $\omega$, then $\omega$ meets the nullcone $\Lambda(p)$ of $p$ in exactly two events: $\omega(\tau^{-})$ and $\omega(\tau^{+})$. Furthermore, if $q$ is an event $q=\omega((\tau^{+}+\tau^{-})/2)$, then $\vec{pq} \bot \omega$ and $\vec{pq}=q\omega(\tau^{-})=q\omega(\tau^{+})$

The first part I can see easily: Since the nullcone $\Lambda(p)$ of $p$ stretches to infinity, it must meet $\omega$ in at least two places. It cannot intersect in more than two places, since $\omega$ would then have to move faster than the speed of light. It is the last part that is troubling me.

I've tried drawing the problem as I see it. $q$ must be between $\omega(\tau^{+})$ and $\omega(\tau^{-})$, but showing that the vector $\vec{pq}$ spanning from $p$ to $q$ is perpendicular to the worldline $\omega$ (or lies in the restspace of the worldline, if you will) is giving me some trouble. My guess is that there is some trig identity I don't know. If anyone has a hint, I'd be happy :)

Answer 1

One thing that immediately stands out from your picture is that you seemed to have ignored the "freely falling" condition.

The world line of a freely falling observer is always a geodesic, and in the context of special relativity, is a straight line.

Also pay attention to the fact that "perpendicular" here is used in the Lorentzian sense, so in your picture it will not look "perpendicular".

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Finally, the hint. If $\Lambda(p)$ intersects the worldline $\omega$ at the proper times $\tau_\pm$, the position vector relative to an arbitrary origin of $\omega$ can be written as (since it is a straight line)

$$ \omega(\tau) = \omega(\tau_-) + \frac{\tau - \tau_-}{\tau_+-\tau_-} \left[ \omega(\tau_+) - \omega(\tau_-)\right] $$

in slope-intercept form. The term in brackets is a tangent vector of the world line (the slope). You know that $\|\omega(\tau_\pm) - p\| = 0$ by assumption. So next you just need to plug it in and compute

$$ \langle \omega(\frac12(\tau_+ + \tau_-)) - p, \omega(\tau_+)- \omega(\tau_-) \rangle $$

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Addendum: let me include the computation here.

$$\begin{align} \langle \omega(\frac12(\tau_+ + \tau_-)) - p, \omega(\tau_+) - \omega(\tau_-) \rangle & = \langle \frac12 (\omega(\tau_+) + \omega(\tau_-)) - p , \omega(\tau_+) - \omega(\tau_-) \rangle \\ & = \frac12 \langle (\omega(\tau_+) - p) + (\omega(\tau_-) - p) , (\omega(\tau_+) - p) - (\omega(\tau_-) - p) \rangle \\ &= \frac12 \left[ \langle \omega(\tau_+) - p, \omega(\tau_+) - p\rangle - \langle \omega(\tau_-) - p, \omega(\tau_-) - p\rangle \right] \end{align} $$ noting that the cross terms cancel each other. The term on the final line is equal to $\frac12 ( \|\omega(\tau_+) - p\|^2 - \|\omega(\tau_-)-p\|^2) = 0$.

Answer 2

Playing really fast and loose with some notation, in units where $c=1$, the triangle $(\tau^{-},p,\tau^{+})$ is a right triangle, with the right angle at $p$. $pq$ bisects $\tau^{-} \tau^{+}$ and produces two new triangles. Geometry will show that these two triangles are symmetric. Geometry will also show that if you bisect the angle $p q \tau^{+}$, the bisector will be parallel to $\tau^{-} p$. That is, it will be parallel to a light ray. Geometry will also show that the angle between $pq$ and this light ray is equal to the angle between the $\omega$ and this same light ray. This means that $pq$ and $\omega$ are normal in the relativistic sense (not "perpendicular" in the Euclidean sense).