# Null geodesics and rotation

1. Jul 8, 2014

### WannabeNewton

Hi all. It is well known that in Schwarzschild space-time, a torque-free gyroscope in circular orbit at any permissible angular velocity at the photon radius (also known as the photon sphere i.e. $r = 3M$) will, if initially tangent to the circle, remain tangent to the circle everywhere along its world-line; see e.g. http://arxiv.org/pdf/0708.2490.pdf. This is usually explained as follows. Say I'm described by an arbitrary worldline in an arbitrary space-time and I orient a gyroscope and a photon gun along some direction at an initial event; there is a swarm of mirrors surrounding me in my infinitesimal neighborhood and at each instant of my proper time, I shoot out a photon in the direction of the gyroscope axis at that instant and reorient the gyroscope axis in the direction of the reflected photon when it arrives back to me, with the direction being relative to my worldline of course. In doing this, the gyroscope axis gets Fermi-Walker transported along my worldline; see section 2 of https://www.zarm.uni-bremen.de/uploads/tx_sibibtex/2001LaemmerzahlNeugebauer.pdf and problem 5 on p.161 of Geroch's general relativity notes http://home.uchicago.edu/~geroch/Links_to_Notes.html [Broken] and p.164 of said notes for the solution.

Coming back to the photon radius in Schwarzschild space-time, it is clear from the above prescription that Fermi-transport of the gyroscope implies the gyroscope always points tangent to the circle because a photon emitted from my photon gun will travel along this circle, and so will the reflected photon, irrespective of my angular velocity. I have two questions with regards to the above:

(1) I don't quite understand what constraints there are on the worldline of the mirror in the above operational procedure, which is basically Pirani's "bouncing photon" method. Certainly it cannot follow an arbitrary worldline in my infinitesimal neighborhood, so what kind of restrictions are there on said mirror's worldline? Consider for example an observer and a mirror at rest on a rigidly rotating disk in flat space-time that are separated by an infinitesimal radial amount. Then there exist both future and past directed radial null geodesics given by $\frac{dr}{dt} = \pm (1 - \omega^2 r^2)^{1/2}$ from my worldline to that of the mirror's so a photon emitted by myself towards the mirror will come back to me in the same radial direction. Will not then the gyroscope remain fixed in the radial direction? But this does not constitute Fermi-Walker transport as the polar axes of the local frame fixed to the disk rotate relative to the momentarily comoving local inertial frame due to Thomas precession. So where in the above is my incorrect application of Pirani's "bouncing photon" method?

(2) Let $(M,g_{\mu\nu})$ be a static axisymmetric space-time having a time-like and hypersurface orthogonal Killing field $\xi^{\mu}$ and an axial Killing field $\psi^{\nu}$; it is easy to see that $M$ being static is equivalent to $\psi_{\mu}\xi^{\mu} = 0$. Now consider a congruence of circular orbits in the rest space of $\xi^{\mu}$ given covariantly by the time-like Killing field $\eta^{\mu} = \xi^{\mu} + \omega \psi^{\mu}$. Then, given the above conditions, it can be shown that $\eta_{[\alpha}\nabla_{\beta}\eta_{\gamma]} = 0$ for all permissible values of $\omega$ on a time-like integral 2-manifold of $\eta^{\mu}$ if and only if the null vector field $k^{\mu} = \xi^{\mu} + \Omega \psi^{\mu}$ is a geodesic on this 2-manifold, where $\Omega^2 = -\xi_{\mu}\xi^{\mu}/\psi_{\nu}\psi^{\nu}$. This is a generalization of the situation at the photon radius in Schwarzschild space-time due to $\eta_{[\alpha}\nabla_{\beta}\eta_{\gamma]} = 0$ being equivalent to Fermi-Walker transport.

However intuitively I do not understand why the space-time must be static as opposed to just stationary. Consider a freely falling circular photon orbit in some stationary (but not static) axisymmetric space-time. If an observer were to be in circular orbit at this photon radius then why couldn't he just use Pirani's "bouncing photon" procedure from above to conclude that Fermi-Walker transport of a comoving gyroscope implies that its axis remains tangent to the circle? Intuitively speaking, under what circumstances would this fail in a stationary but non-static space-time, and why?

Last edited by a moderator: May 6, 2017
2. Jul 8, 2014

### Staff: Mentor

The only constraint I can see is that the worldlines of the observer, the two bouncing photons, and the two mirrors, must all lie in the same 2-surface. Equations (2) and (3) of the Lammerzal-Neugebauer paper express this for the photons and the observer, but I don't see any equation in that paper that expresses it for the mirrors.

But, relative to a momentarily comoving inertial frame, this "radial direction" is changing. So I don't think the constraint I described above is fulfilled by this setup; i.e., I don't think the worldlines of the observer, the bouncing photon, and the mirror will all lie in the same 2-surface in spacetime.

It's not strictly relevant to the rest of this discussion, but it's worth reminding ourselves here that even without Thomas precession, a gyroscope that is forced to remain fixed in the radial direction, relative to the rotating disk, is not Fermi-Walker transported. Fermi-Walker transport, minus Thomas precession, would make the gyroscope rotate, relative to the rotating disk, in the opposite sense to the disk's rotation--i.e., the gyroscope would always point in the same direction relative to infinity (e.g., it might always point at the same fixed star far away). Thomas precession adds a retrograde drift of the gyroscope relative to some fixed object at infinity.

I think the key property that holds in the static case, but not in the stationary/non-static case, is that the prograde and retrograde photon orbits are at the same radius. In the stationary/non-static case, the prograde photon orbit is at a different radius than the retrograde one. I'm not sure exactly what difference this makes in the details, but I suspect that it plays a role.

3. Jul 12, 2014

### WannabeNewton

Thank you very much for the reply Peter.

I think my main issue is I don't fully understand the relationship between the mathematical constraints and physics of the construction in that paper. Geroch's construction makes somewhat more sense to me physically so I figured it would be easier for us to discuss his construction instead; I've attached the relevant parts.

To start with, it isn't clear to me from the diagrams but in order to make sense of quantities like $\mathcal{L}_{\xi}k^a$ and $\mathcal{L}_{\xi}l^a$ we need $k^a, l^a$ to be (future and past directed respectively) null vector fields which means that they must be defined at every point on the worldline $\gamma$ so presumably this would only make sense if at each instant of proper time at $\gamma$, there is a light ray being emitted and a light ray being received, right? Then the conditions $\mathcal{L}_{\xi}k^a = \mathcal{L}_{\xi}l^a = 0$ simply mean that at each event on $\gamma$ there exists a light ray going from $\gamma$ to the worldline of the infinitesimally neighboring mirror and there exists a light ray received at $\gamma$ from the worldline of said mirror.

I was initially confused by the physical significance of the condition that $\xi^a$ be linearly dependent on $k^a$ and $l^a$ i.e. that $\xi^a$ be a linear combination $\xi^a = \alpha k^a - \beta l^a$ where I've chosen $\alpha,\beta > 0$ since $\xi^ak_a < 0$ and $\xi^a l_a > 0$. But I think I've shown that this condition holds if and only if the spatial direction of $k^a$ relative to $\xi^a$ is the same as the spatial direction of $l^a$ relative to $\xi^a$ at any given event; this is in fact the hypothesis of the problem in the first attachment below. In order not to clutter this reply, I've attached at the bottom my verification of this equivalence.

But then Geroch says to "normalize" things so that $\alpha = \beta = 1$ on $\gamma$ itself but not necessarily anywhere else. I do not understand how one achieves this in principle. Could you explain this if possible?

If we follow Geroch's calculations, the conditions that must be satisfied are $\xi^{a} = \alpha k^{a} - \beta l^{a}$ and $\mathcal{L}_{\xi}k^{a} = \mathcal{L}_{\xi}l^a = 0$. As noted above, the first condition i.e. that of $\xi^a$ lying in the 2-plane spanned by $k^a$ and $-l^a$ is equivalent to that of the emitted light ray at any instant being in the same direction as that of the light ray received at that instant. The second condition is equivalent to the statement that the null vector fields $k^a,l^a$ correspond respectively to future-directed and past-directed null vectors at each event on $\gamma$ pointing to the infinitesimally neighboring mirror.

Coming back to the rigidly rotating disk, if an observer $O$ at rest on the disk at $r$ emits a light ray in the forward radial direction towards $r - dr$ then, in the background global inertial frame, this light ray will just keep going straight in the radial direction until it reaches $r - dr$ (since this is flat space-time) but because the observers on $r - dr$ are rotating around with the disk, this light ray will not reach the observer $O'$ separated from $O$ in the instantaneous radial direction $e_r$ fixed to the disk (so that $e_r$ rotates relative to the global inertial frame) but rather it will reach the observer $O''$ to the left of $O'$ by some amount $d\varphi$ as viewed by $O$.

If $O''$ wants to reflect the light ray back to $O$, then, as viewed in the background global inertial frame wherein $O$ and $O''$ are rigidly rotating around at the same angular velocity $\omega$ but are separated by $d\tilde{\varphi}$, $O''$ must reflect the light at an angle to the instantaneous radial direction $e_r$ fixed to the disk so that the light travels, in the background global inertial frame, in a straight line back to $O$ but because it was emitted at an angle it will not come back to $O$ in the direction $e_r$ at the location of $O$. Does that sound right to you? And is this what you were talking about above since this implies that $\xi^a \neq \alpha k^a - \beta l^a$?

That makes perfect sense, thanks. Indeed there is no reason in general to expect the prograde and retrograde light orbits to exist at the same radius since there is only stationary symmetry as opposed to static symmetry i.e. $\psi_a \xi^a \neq 0$. If so Pirani's "bouncing photon" method clearly cannot be applied to conclude that $\eta_{[a}\nabla_{b}\eta_{c]} = 0$ for all $\omega$ at the location of any circular null geodesic since at this location in general only the prograde or retrograde light orbit will exist but not both.

Just for fun I'll try to write up a more rigorous calculation in order to see if $\psi_{a}\xi^a = 0$ is a necessary condition in order for $k^b \nabla_b k^a = 0 \Leftrightarrow \eta_{[a}\nabla_{b}\eta_{c]} = 0$ for all $\omega$ on top of being a sufficient condition which we already know to be true.

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• ###### Null vectors linear combination.pdf
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4. Jul 12, 2014

### Staff: Mentor

It seems that way to me, yes.

I'm not sure; I'll have to look at Geroch's presentation in more detail. Possibly he's relying on the fact that a Lorentz boost doesn't change the direction of null vectors, so he's thinking that one can always change the relative magnitudes of $\alpha$ and $\beta$ by a boost while keeping the directions of the null vectors the same.

Yes.

5. Jul 13, 2014

### WannabeNewton

Yes that is probably what he is doing. We have that $0 = \mathcal{L}_{\xi}\xi^a = k^a \mathcal{L}_{\xi}\alpha + l^a \mathcal{L}_{\xi}\beta$ since $\mathcal{L}_{\xi}k^a = \mathcal{L}_{\xi}l^a = 0$. Then if we contract both sides once with $k^a$ and once with $l^a$, and use the fact that $k^al_a > 0$, we get $\xi^a \nabla_a \alpha = \xi^a \nabla_a \beta = 0$.
Let $\alpha_0 = \alpha|_{\gamma}, \beta_0 = \beta|_{\gamma}$ where as above $\gamma$ is the worldline of the observer emitting to and receiving light rays from the neighboring swarm of mirrors.

Since $\xi^a \nabla_a \alpha = \xi^a \nabla_a \beta = 0$, $\alpha_0,\beta_0$ are just constants so we can define the new functions $f(x^{\mu}) = \alpha_0, g(x^{\mu}) = \beta_0$ on space-time so that $\nabla_a f = \nabla_a g = 0$. Now $f, g \neq 0$ so define then the null vector fields $\tilde{k}^a = f k^a$ and $\tilde{l}^a = g l^a$. These are still null geodesics of course because $\tilde{k}^a \nabla_a \tilde{k}^b = fk^a \nabla_a (f k^b) = f^2 k^a \nabla_a k^b = 0$, and similarly for $\tilde{l}^a$. Furthermore $\mathcal{L}_{\xi}\tilde{k}^a = \mathcal{L}_{\xi}\tilde{l}^a = 0$ still obviously holds so we have the exact same construction as we did with $k^a$ and $l^a$ except now $\xi^a|_{\gamma}= \alpha_0 k^a - \beta_0 l^a = f k^a - g l^a = \tilde{k}^a - \tilde{l}^a$.

Does that seem about right? Thanks.

6. Jul 13, 2014

### Staff: Mentor

Looks right to me, yes.