Is there a notion of spatial angles in general relativity?

Example: The world line of a photon is given by $x^{\mu}(\lambda)$. Suppose it flies into my lab where I have a mirror. I align the mirror in such a way, that I measure a right angle between the incoming and outgoing photon. How can I now calculate the wordlline of the outgoing photon?

Of course I cheated a bit since I just used the old euclidean notion of an angle when I say "I measure a right angle between the incoming and outgoing photon". But I hope that this is allowed locally.

  • $\begingroup$ Locally spacetime is flat and you can measure angles just as Euclid did. I don't see the problem with calculating the worldline of the outgoing photon. Why is this hard? $\endgroup$ – John Rennie Apr 14 '14 at 15:03
  • $\begingroup$ Say the incoming photon comes from above on a radial geodesic in Schwarzschild: $x_{\text{in}}^{\mu}(t)=(t,r_{\text{in}}(t),\frac \pi 2, 0)$ where $\frac{dr_{\text{in}}}{dt} = 1-\frac{r_s}{r_{\text{in}}}$. It hits the mirror at $t=T$,$r=R$, $\theta=\frac \pi 2$, $\varphi = 0$ so that the light beam makes a right angle. Assume euclidean notions holds locally I write $x_{\text{out}}^{\mu}(t)=(T+t,r_{\text{out}}(t),\frac \pi 2, \varphi_{\text{out}}(t))$. Knowing $r_{\text{out}}(0)=R$, $\varphi_{\text{out}}(0)=0$, and $\frac{d\varphi_{\text{out}}}{dt}(0)>0$ is now enough to calculate the geodesic? $\endgroup$ – thyme Apr 14 '14 at 15:46

If it is a photon, then you know that $x^{a}x_{a}$ is always zero. You have measured the spatial direction of the photon with your mirror apparatus, so you know the values of the $x^{i}(0)$ at some time${}^{1}$, which we will call zero. Furthermore, we can infer the value of $x^{0}$ at this time from the fact that $x^{a}x_{a} = 0$.

Then, all we need is to know the metric tensor, and we can simply solve the set of ODEs given by,

$$\frac{\mathrm{d}^2 x^a (\lambda)}{\mathrm{d}\lambda^2} + \Gamma^a_{bc} \, \frac{\mathrm{d} x^b(\lambda)}{\mathrm{d}\lambda} \frac{\mathrm{d}x^c (\lambda)}{\mathrm{d}\lambda} = 0$$

for all subsequent times${}^{2}$. Typically, this will have to be done numerically, but there are some known exact solutions for simplified cases. $\lambda$ is the null affine parameter, and the $\Gamma$'s are the Christoffel symbols, which are defined from the metric tensor. For the familiar example of a spherically symmeteric gravitating body AND for weak gravity in nonrelativistic motion, the only relevant component of the metric tensor is $g_{tt} = -\left(1-\frac{2M}{r}\right)$, and the only relevant Christoffel symbol is $\Gamma_{tt}{}^{r} = \frac{M}{r^{2}}$, and our theory becomes Newtonian gravity.




${}^{1}$Here, I take letters from the beginning of the alphabet to be spacetime indices, while letters from the middle of the alphabet are sums over only the spatial components.

${}^{2}$ note that "time" (given by $\lambda$) here doesn't really mean time in the proper sense. Null geodesics don't experience any clock motion. Null geodesics, however, DO trace out "null affine parameter", which you can interpret as how many perpendicular null geodesics they cross as the photon moves.

  • $\begingroup$ And, I should add that the notion of a spatial angle is always definable so long as you have a local notion of time at a point. You just have to take your two vectors, remove the time component from both, take the inner product of the new vectors, and then you have $\cos\theta = \frac{a^{i}b_{i}}{\sqrt{a^{i}a_{i}b^{i}b_{i}}}$. Note that this will depend on the choice of time, but it will have to, because relative spatial angles of vectors are not preserved by Lorentz boosts, much less general coordinate transformations. $\endgroup$ – Jerry Schirmer Apr 14 '14 at 15:14

As a geometric quantity, the value of an "angle" can be determined and expressed in a coordinate-free way:

Given three pairwise space-like events, "$A$", "$B$" and "$C$", and given the positive real numbers
$\frac{s^2[ A C ]}{s^2[ A B ]}$, $\frac{s^2[ A C ]}{s^2[ B C ]}$ and $\frac{s^2[ B C ]}{s^2[ A B ]} = \frac{s^2[ A C ]}{s^2[ A B ]} / \frac{s^2[ A C ]}{s^2[ B C ]}$
as ratios between squares of interval magnitudes (or minimum arc lengths) between event pairs,
then the value of "angle at $B$, between $A$ and $C$" may either be expressed (directly) as

$\angle [ A B C ] := $ $\text{ArcSin} \left[ \frac{1}{2} \sqrt{ 2 + 2 \frac{s^2[ A C ]}{s^2[ A B ]} + 2 \frac{s^2[ A C ]}{s^2[ B C ]} - \frac{s^2[ B C ]}{s^2[ A B ]} - \frac{s^2[ A B ]}{s^2[ B C ]} - \frac{s^2[ A C ]}{s^2[ A B ]} \frac{s^2[ A C ]}{s^2[ B C ]} } \right];$

or, more generally, in terms of squares of interval magnitudes (or minimum arc lengths) involving additional events "$F$", "$G$" (named in the sense of variables) which are space-like to each other as well as to "$A$", "$B$" and "$C$", as

$\angle [ A B C ] := \text{Limit}_{\{ F, G \}} {\huge[} $
${\Large \{ } \frac{s^2[ B F ]}{s^2[ A B ]} \rightarrow 0, \frac{s^2[ B G ]}{s^2[ B C ]} \rightarrow 0, $
$2 + 2 \frac{s^2[ B F ]}{s^2[ A B ]} + 2 \frac{s^2[ B F ]}{s^2[ A F ]} - \frac{s^2[ A F ]}{s^2[ A B ]} - \frac{s^2[ A B ]}{s^2[ A F ]} - \frac{s^2[ B F ]}{s^2[ A B ]} \frac{s^2[ B F ]}{s^2[ A F ]} \rightarrow 0, $
$2 + 2 \frac{s^2[ B G ]}{s^2[ B C ]} + 2 \frac{s^2[ B G ]}{s^2[ C G ]} - \frac{s^2[ C G ]}{s^2[ B C ]} - \frac{s^2[ B C ]}{s^2[ C G ]} - \frac{s^2[ B G ]}{s^2[ B C ]} \frac{s^2[ B G ]}{s^2[ C G ]} \rightarrow 0 {\Large \} }; $
$\text{ArcSin} \left[ \frac{1}{2} \sqrt{ 2 + 2 \frac{s^2[ F G ]}{s^2[ B F ]} + 2 \frac{s^2[ F G ]}{s^2[ B G ]} - \frac{s^2[ B G ]}{s^2[ B F ]} - \frac{s^2[ B F ]}{s^2[ B G ]} - \frac{s^2[ F G ]}{s^2[ B F ]} \frac{s^2[ F G ]}{s^2[ B G ]} } \right] {\huge]} . $

If the region containing the events $A$, $B$ and $C$ is flat then these two "angle" values are equal.


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