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I'm currently studying the Bonnor solution starting with this paper on Black Diholes. The metric is given by :

$$ ds^2 = \left(1-\frac{2Mr}\Sigma\right)^2 \left[-dt^2 + \frac{\Sigma^4}{(\Delta + (M^2 +a^2)\sin^2 \theta)^3}\left(\frac{dr^2}{\Delta} + d\theta^2\right)\right] + \frac{\Delta \sin^2\theta}{\left(1-\frac{2Mr}\Sigma\right)^2}d\phi^2 \tag{1}$$

where $\Delta = r^2 + 2 M r - a^2 = (r-r_+)(r-r_-)$ and $\Sigma = r^2 - a^2 \cos^2\theta$.

I'm getting the "picture" behind the conical defect on a manifold: take a piece of paper, cut off a "wedge" and then glue the two sides together -- a cone is obtained (see the figure for clarity).

enter image description here

In more concrete terms, any circumference of center O in the original manifold doesn't close anymore (it spans an angle $ \theta < 2\pi$ because of the cut), while identifying the two sides it can still close. The presence of a deficit angle is hence signaled if the ratio between the circumference and the radius is not $2\pi$.

What I would do to compute the deficit angle in the context of pseudo-Riemannian manifolds: take some cylindrical-like coordinates with z coordinate coinciding with the axis of the "cone", then the radius is the proper distance of a radial path starting from the axis ($r=0$) to $r=R$; as a circumference one uses the proper length of a curve around the given axis (i.e. in the identified direction), taken $r=R$. Take the ratio of such quantities, then send $R$ to zero.

But is finding such an exact change of coordinates always an efficient strategy? Since eq.(1) is given in Boyer-Lindquist-like coordinates, I guess there is a more general approach to the problem, which doesn't require any particular choice of coordinates.

Luckily the paper offers two examples, each for a different axis of simmetry, from which we can learn :

$$ \delta_0 = 2\pi - \left|\frac{\Delta \phi \space d \sqrt{ g_{\phi \phi}}}{\sqrt{ g_{\theta\theta}}\ d\theta}\right| \tag{4}$$

$$ \delta_{r_+} = 2\pi - \left|\frac{\Delta \phi \space d \sqrt{ g_{\phi \phi}}}{\sqrt{ g_{rr}}\ dr}\right| \tag{5}$$

Where it is said that the $\phi$ coordinate is identified with period $\Delta \phi$.

Apparently both numerators look like what I was thinking about. But I can't understand the factors in the denominators. Moreover I don't get whether the "d" in both formulae is a true derivative or not.

Can anyone provide an insight on these formulae?


UPDATE: I tried moving to cylindrical coordinates. Since eq.(1), as said, is expressed in Boyer-Lindquist coordinates, the change of coordinates involves of course just the ($r$,$\theta$) coordinates:

\begin{cases} \rho = \sin{\theta} \sqrt{r^2 - 2 M r + a^2} \\ z = (r-M) \cos{ \theta} \end{cases}

Let's compute the deficit as explained, in these coordinates: I'm taking an infinitesimal circumference $\gamma_{\epsilon}$ around the z-axis ($\theta = 0$). The proper length of such thing would be (remember that $ \phi \in (0, \Delta \phi) $)

$$ \int_{\gamma_{\epsilon}} \sqrt{g_{\mu \nu} dx^\mu dx^\nu} = \int_0^{\Delta \phi} \sqrt{g_{\phi \phi}}\ d\phi \vert_{\rho = \epsilon} $$

and the proper radius

$$ \int_0^\epsilon \sqrt{g_{\rho \rho}}\ d\rho$$

Hence the defect of the angle would be something like :

$$ \delta_0 = 2\pi - \lim_{\epsilon \to 0} \frac{\int_0^{\Delta \phi} \sqrt{g_{\phi \phi}}\ d\phi \vert_{\rho = \epsilon}}{\int_0^\epsilon \sqrt{g_{\rho \rho}}\ d\rho} $$

Now for sloppiness we drop the integration sign, substituting the limit with the infinitesimal version of the integrand. So I'm interpreting the "d"s in the paper as just a compact way to express this limit.

However my result looks more like eq.(5) rather than eq.(4). Any ideas of what's going on?

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1 Answer 1

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I figured out what happened: I was trying to "see" the cone as a 3D structure, "emerging" in the $\theta = 0$ direction, hence forgetting it is actually a 2D plane, which has undergone an angular identification; in this way I couldn't even generalize the procedure for the conical singularity in the "axis" $r = r_+$, since it was lacking an intuitive geometrical interpretation.

I think the following procedure could still be possible in Weyl/Cylindrical coordinates, however this would imply a mixing of both $(r,\theta)$ coordinates, introducing unncessary complications in the limits. Using Boyer-Lindquist coordinates is actually a simple way to get things done.

Since we're looking for a 2D manifold, let's take the induced metric for $t, r = const.$ :

$$ ds^2 = \left(1-\frac{2Mr}\Sigma\right)^2 \left[\frac{\Sigma^4}{(\Delta + (M^2 +a^2)\sin^2 \theta)^3} d\theta^2\right] + \frac{\Delta \sin^2\theta}{\left(1-\frac{2Mr}\Sigma\right)^2}d\phi^2 $$

and since one seeks $g_{\phi \phi} = 0$, from eq.(1) one has to let $\theta \to 0 \lor \pi$. In both limits, one sees that the metric simply becomes :

$$ ds^2 = C \left( d\theta^2 + \sin^2 \theta \space d\phi^2 \right)$$

where C is a constant. This has the more familiar face of a 2D-Euclidean manifold with polar-like coordinates (just use the "$\theta$" as a radius). One can finally take the proper length of circumference as said before :

$$ L = \int_0^{\Delta \phi} \sqrt{g_{\phi \phi}}\ d\phi \vert_{\theta = \epsilon} = C \Delta \phi \sin \epsilon $$

and the radius :

$$ R = \int_0^{\epsilon} \sqrt{g_{\theta \theta}} \space d \theta = C \epsilon $$

Now taking the limit, one recovers the result of the paper:

$$ \lim_{\epsilon \to 0} \frac{L}{R} = \lim_{\epsilon \to 0} \frac{C \Delta \phi \sin\epsilon}{C \epsilon} = \Delta \phi$$

In analogue fashion, one can look at the other piece of the axis of symmetry, i.e. $r = r_+$. Let's consider the induced metric for $t, \theta = const.$ and take the first leading order of the metric in the limit $r \to r_+$. What you get is:

$$ ds^2 = C \left( \frac{f(r) dr^2}{\left (a^2 + M^2 \right)^3} + \frac{d\phi^2}{f(r)} \right) $$

where C is a constant (that depends on $\theta$, which is fixed) and $f(r) = \frac{a^4}{2 \sqrt{a^2 + M^2} \left(r-r_+ \right)}$.

Now it's simply a matter of redifining R to make the metric look like this:

$$ ds^2 = C \left( \frac{dR^2}{\left (a^2 + M^2 \right)^3} + \frac{R^2 d\phi^2}{4A^2} \right) $$

Since the 2D metric in polar coordinates is recovered just by finding $R = R(r)$, the OP's eq.(5) is still valid - by taking integrals in the place of differentials as mentioned in the "UPDATE" section.

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