sorry for the "generic" question, just that the last month I have been studying GR from 0 and having problems understanding some of the concepts. Anyways, to the question I wanted to ask:

From my understandings, ZAMO (Zero Angular Momentum Observer) Co-Rotates with the same Angular speed around the black hole, right?

Does this means that he HAS Angular momentum for an Euclidean/Lorentzian Observer far far away from the Black hole?

ZAMO holds the equation $r=C_0$, but $\phi \neq C_1$ right?

  • $\begingroup$ Wikipedia says yes, the ZAMO observer is co-rotating. $\endgroup$ – PM 2Ring Oct 21 at 18:26
  • $\begingroup$ @PM2Ring Then how is he "Zero Angular Momentum" ? $\endgroup$ – billy Oct 21 at 18:40
  • $\begingroup$ Because he's at rest with respect to the local spacetime. But that spacetime itself is rotating because of frame dragging. $\endgroup$ – PM 2Ring Oct 21 at 18:46
  • $\begingroup$ @PM2Ring Yeah but the Angular velocity then should depend on $r$. Also, since its stattc ($\partial t = 0$) then $\phi $ does change for ZAMO, no? If you move with an Angular Velocity, the line $\phi =C_1$ is not chaning with time. $\endgroup$ – billy Oct 21 at 19:35

ZAMOs are basically the observers having zero angular momentum at infinity. So what does this mean?

In General Relativity (GR), angular momentum of a test particle is defined as $$l=g_{t\phi}\frac{dt}{d\tau}+g_{\phi\phi}\frac{d\phi}{d\tau}$$ where $t$ is the coordinate time and $\tau$ is the proper time. Now far from a compact object (i.e. at infinity), we have $g_{t\phi}\rightarrow0$ and $g_{\phi\phi}\rightarrow1$. So at large distances, we have $$l=\frac{d\phi}{d\tau}$$ This is exactly the definition of angular momentum defined in the proper frame. Now when we say that the angular momentum is zero, we mean that $l=0$.

Q. What would happen when the particle approach a compact object (e.g., Kerr black hole)?

A. Put $l=0$ in the first equation to obtain $$\frac{d\phi}{dt}=\frac{d\phi/d\tau}{dt/d\tau}=\frac{\dot{\phi}}{\dot{t}}=-\frac{g_{t\phi}}{g_{\phi\phi}}$$

Recall that this is nothing but the frame dragging angular velocity which one can obtain from the Kerr metric, so that we obtain $$\omega(r,\theta)=-\frac{g_{t\phi}}{g_{\phi\phi}}$$


ZAMOs are the observers who have zero angular momentum ($l=0$) in their proper frame at large distances. But as they approach any compact object, frame dragging pulls it along with the geometry so that they attain the angular velocity $\omega(r,\theta)$, the frame dragging angular velocity.

  • $\begingroup$ First of all, REALLY thank you for the answer. I have some questions in case you are able to answer :) - although you have already helped me- Firstly, I thought the system of a Kerr Black Hole is stationary , meaning that $\dfrac{\partial}{\partial t} \cdot A = 0 $ for any Physical Quantity A. What's that Frame Dragging velocity you refer too. Secondly, I guess you also assume that $r=C1$ , $\phi = C2$ in ZAMO. Do both of these equations hold at an observer at infinity? Lastly, about the angular momentum definition. Do you have any source to recommend me to conclude that myself? $\endgroup$ – billy Oct 28 at 15:37
  • $\begingroup$ maybe you could help a little bit more? :( $\endgroup$ – billy Nov 10 at 1:43
  • $\begingroup$ @billy Frame dragging means that all inertial frames starts to co-rotate with the geometry as they approach a Kerr BH. However, the ZAMOs are at rest in their local rest frame (i.e. proper frame), but the fact is that the proper frame is itself dragged to rotate along the direction in which the Kerr BH is rotating. This frame dragging is true for all rotating objects, e.g., the induced frame dragging angular velocity for the Earth is $\omega=\frac{GM_E}{R_Ec^2}\Omega_E\sim 0.3''/$year which is too small to detect. $\endgroup$ – Richard 2 days ago
  • $\begingroup$ @billy Note that the expression for $\omega$ depends on the compactness ratio $M/R$ indicating that frame dragging is significant for compact objects when this ratio becomes large. $\endgroup$ – Richard 2 days ago

In general questions like

Does this means that he HAS Angular momentum for an Euclidean/Lorentzian Observer far far away from the Black hole?

are meaningless in general relativity. In general, a distant observer would have no meaningful definition of the angular momentum of an object at a different location.

More to the point. In the case of a Kerr black hole a ZAMO is defined to be an observer with zero angular momentum defined as the conserved quantity related to the rotational symmetry of the Kerr spacetime.

As it happens, because Kerr is asymptotically flat, it also has a notion of the total angular momentum of the spacetime, in the form of the so-called ADM angular momentum. One can prove (but this is far from trivial) that a ZAMO will not contribute to the ADM angular momentum.


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