Assuming a particle is at rest in a certain frame of reference around a kerr spacetime and hence the 4-velocity (in spherical polar coordinates) of the particle is given by

$$ u=(u^{t},0,0,0)$$

Now, the invariant quantity $u_{\alpha}u^{\alpha}$ (for the signature $-,+,+,+$ of kerr metric) can be calculated as follows

$$ u_{\alpha}u^{\alpha} = g_{\alpha\beta}u^{\alpha}u^{\beta} = g_{tt}u^{t}u^{t}=-1$$

from the kerr solution we put in the value for $g_{tt}$ we get

$$-\left(1- \dfrac{2GMr}{\rho^{2}} \right)\left(u^{t}\right)^{2} = -1 $$

which leads me to conclude that for a real solution

$$1-\dfrac{2GMr}{\rho^2} > 0 $$

since $\rho = \sqrt{r^2+a^2cos^2\theta}$ hence the range of $r$ for which no real solution exist is $$r<GM+\sqrt{G^2M^2 - a^2cos^2\theta}$$ but the outer event horizon $r_{+}$ of the kerr metric is given by

$$ r_{+} = GM + \sqrt{G^2M^2 - a^2}$$

which means depending on the value of $\theta$, $r$ can be greater than $r_+$. Hence in a finite region outside the event horizon the particle can never stay at rest around the black hole. This is because we initially assumed the particle is at rest and got a contradiction for a certain range of $r$.

Now, my doubt is that if the particle is inside the radius given by the inequality above the particle then cannot be at rest with respect to what? The origin of the coordinate system?

But, it can always (?) be assumed that the reference frame is attached to the particle itself and 4-velocity equation above remains the same hence it leads to an illogical statement that the particle is at not at rest with respect to itself.

  • $\begingroup$ Isn't this just frame-dragging? $\endgroup$ Jun 19, 2018 at 1:41
  • $\begingroup$ Yes. The motion is due to frame dragging but what I was saying is that I had attached the reference frame to the particle itself. Hence it must always be at rest w.r.t itself (which is absurd) . But simultaneously it is being shown that it cannot be at rest. So how is my assumption that the frame is attached to the particle is wrong? It's the only place it could be wrong. $\endgroup$ Jun 19, 2018 at 2:04
  • $\begingroup$ in the above comment "which is absurd" will come after the just next sentence. $\endgroup$ Jun 19, 2018 at 3:30
  • $\begingroup$ The particle isn't at rest wrt the comoving inertial frame. $\endgroup$
    – PM 2Ring
    Jun 19, 2018 at 6:42
  • $\begingroup$ what is the co-moving frame? can you explain in detail? $\endgroup$ Jun 19, 2018 at 6:43

1 Answer 1


You have shown that in a particular coordinate system, the one from which you used $g_{tt}$
&c, the particle does not remain at rest in a certain region. This coordinate system is not the coordinate system based on the particle: you can't just pick the metric components from another coordinate system and use them in a coordinate system based on the particle.


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