Solving for g_φφ=0 in charged/rotating BHs

[stevebd1]

Based on http://casa.colorado.edu/~ajsh/phys5770_08/bh.pdf" (page 37), closed timelike curves exist within a boundary dictated by $g_{\phi \phi}=0$ which occurs where

$$\frac{R^4}{\Delta}=a^2 \sin^2 \theta$$

where $R=\sqrt(r^2+a^2)$ and $\Delta=R^2-2Mr+Q^2$

I'd appreciate it if someone could give some indication of how to solve for r in the following equation-

$$\frac{(r^2+a^2)^2}{(r^2+a^2-2Mr+Q^2)}=a^2\sin^2\theta$$

 

[tiny-tim]

hi stevebd1!

written as R⁴ = a²sin²θ(R² + A), isn't it a straightforward quadratic equation in R² ?

(btw, is that a² in the denominator correct?)

 

[stevebd1]

Hi tiny-tim

Thanks for the response. I'm trying to solve for lower case r (radius) for $g_{\phi \phi}=0$ as in the case for establishing radii for the ergospheres where $g_{tt}=0$ where

$$\Delta=a^2 \sin^2 \theta$$

which becomes

$$(r^2+a^2-2Mr+Q^2)=a^2\sin^2\theta$$

and produces the following equation relative to r-

$$r_{e\pm}=M\pm\sqrt{M^2-Q^2-a^2\cos^2\theta}$$

As far as I can tell, the a² in the denominator is correct.

 

[stevebd1]

I tried a couple of online equation solvers but the results I got were a choice of four very long equations-

http://www.numberempire.com/equationsolver.php

http://www.solvemymath.com/online_math_calculator/algebra_combinatorics/equations/equation_solver.php

even when I reduced the equation down to being relative to just the equator-

(r^2+a^2)^2/(r^2-2*M*r+a^2+Q^2)=a^2

I also tried the following-

$$g_{\phi\phi}=(r^2+a^2+[(2Mr-Q^2)a^2 \sin^2\theta]/\rho^2) \sin^2\theta$$

where $\rho^2=r^2+a^2 \cos^2\theta$

reduced to-

0=r^2+a^2+((2*M*r-Q^2)*a^2)/r^2

and got the same results.

I also tried the following online program-

http://www.myalgebra.com/algebra_solver.aspx

which is normally pretty good. I used the option 'solve the equation' then chose r as the 'variable to solve'. This produced much shorter equations but which were unfortunately incorrect, though they beared some resemblance to the quadratic equations shown in the wiki page for http://en.wikipedia.org/wiki/Kerr_metric#Important_surfaces"

You say the equation might be quadratic tiny-tim, is there anyway of rewriting the equation so that it is in the form Ar²+Br+C=0 which should be more straightforward to solve?

 

[tiny-tim]

hi stevebd1!

(forget what i said about quadratic, i got confused about R and r )

perhaps drawing a different coordinate grid would help?

if you draw lines of equal (r² + a²)/sinθ,

ie r² + a² = aksinθ for various k (i don't know what they look like)

then those lines are crossed by the solutions of r² - 2mr + Q² + a² - k² = 0

does that help?

 

[stevebd1]

Hi tiny-tim

Thanks again for the input. Could you elaborate a little on what you're suggesting and/or recommend an online software that could be used for this. I'm guessing that you're suggesting that I plot two lines, each one relating to some part of the original equation and where they cross (probably in two places) will provide the solutions.

 

[tiny-tim]

hi stevebd1!

sorry, I've never used any graphing software

i'm hoping that the (r² + a²)/sinθ contour lines will be fairly easy to plot, then you can mark on each contour line the two values of r where that quadratic equation is satisfied, and then join-the-dots

(or, to put it another way, plot r² - 2mr + Q² + a² - k² as if k was a standard y-coordinate, and then squash the result by replacing the horizontal constant-k lines by the constant-(r² + a²)/sinθ lines)

 

[stevebd1]

In http://philosophyfaculty.ucsd.edu/faculty/wuthrich/pub/WuthrichChristian1999MScThesis.pdf" page 58, there is a similar attempt to establish a radius for $g_{\phi \phi}=0$ including for charge using the equation in post #1, they state the following-

'..we cannot reduce the polynomial to a cubic expression since we have an "inhomogeneity" due to the charge. Though this complicates matters, we can find an analytical solution of (the) equation. In case of equality, two real and two complex roots may be found. The lengthy form of the solution is confusing and therefore not worth being printed. For this reason, we are satised with a purely numerical analysis. This analysis will at least have a pedagogical impact.'

In the end, considering that M,Q and a are constants, I simply rewrote the equation including for these quantities, leaving r as the variable (Q=0.8M, a=0.56M)-

(r^2+2481.11^2)^2/(r^2-2*4430.55*r+2481.11^2+3544.44^2)=2481.11^2*sin(45)^2

and using the following software-

http://www.wolframalpha.com/

simply repeated the process with different theta.

this produced two sets of real numbers (mainly one positive, one negative) between a range of theta 34-90. When converted to elliptical coordinates and plotted, the image matched exactly figure 7, page 35 from http://casa.colorado.edu/~ajsh/phys5770_08/bh.pdf" .

It's not a definitive final equation but it's a solution that produces exact results.

 

[tiny-tim]

hi stevebd1!

When converted to elliptical coordinates and plotted, the image matched exactly figure 7, page 35 from http://casa.colorado.edu/~ajsh/phys5770_08/bh.pdf" .

that shows the "turnaround radius", g_tφ = 0 …

you're saying that g_φφ = 0 is similar?

btw, have i got this right? … the "turnaround radius" (inside the inner event horizon) is the junkyard of a charged rotating black hole, in that all freefalling bodies approach it at zero velocity, from either side ?

 

[stevebd1]

Hi tiny-tim

$g_{\phi \phi}=0$ is the blue dotted line marking the boundary of the CTCs. The dotted blue curve above the ring singularity is the plotting of the positive numbers (which ideally would be the same below the RS), the larger dotted curve below the ring singularity is the plotting of the negative numbers where the CTCs are in negative space (i.e. within the singularity). In the case of uncharged Kerr metric, all the CTCs are within the ring singularity.

Regarding the turnaround radius in RN metric-

'..Outside the outer horizon r₊ space falls at less than the speed of light, at the horizon space falls at the speed of light, and inside the horizon space falls faster than light. But the gravitational repulsion produced by the tension of the radial electric field starts to slow down the inflow of space, so that the infall velocity reaches a maximum at $r=Q^2/M$. The infall slows back down to the speed of light at the inner horizon r_-. Inside the inner horizon, the flow of space slows all the way to zero velocity, $\beta=0$, at the turnaround radius $r_0=Q^2/2M$..'

While I don't entirely agree with the river model (the idea of space itself flowing into the black hole), the river concept is synonymous with infall velocity so the above should still apply when considering infalling matter, if so, it would appear that matter would collect within the inner horizon, unless the matter approaches with such velocity, entering the black hole as a hail frame, that it overcomes the repulsion. In the case of Kerr-Newman metric, I suppose the idea is is that if you have the velocity to overcome the repulsion, you'll pass through the ring singularity, encounter another turnaround radius which will push you away, any other matter where the velocity isn't high enough will be pulled back to the original turnaround radius where it will collect (though maybe the frame dragging keeps this matter predominately at the equator).

 

[tiny-tim]

hi stevebd1!

$g_{\phi \phi}=0$ is the blue dotted line marking the boundary of the CTCs. The dotted blue curve above the ring singularity is the plotting of the positive numbers …

isn't the blue dotted curve above the ring the boundary of the inner ergosphere?

 

[stevebd1]

Hi tiny-tim

The purple dotted line that makes contact with the inner horizon at the pole is the inner ergosphere.

 

[tiny-tim]

oh i see now!

i was confusing the blue and the purple

 

[stevebd1]

The RN metric can also use something described as Misner-Sharp mass M(r)-

$$M(r)=M-\frac{Q^2}{2r}$$

http://arxiv.org/PS_cache/gr-qc/pdf/0411/0411060v2.pdf" in post #1, page 20, makes RN metric resemble Schwarzschild metric when replacing M for M(r). M(r)=M at infinity and M(r)=0 at Q²/2M (turnaround radius).


On a side note, radial infall velocity (β) is defined using the following equation-

$$\beta=\frac{\sqrt{2Mr-Q^2}}{\sqrt{r^2+a^2}}$$

if we just consider Q (charge) then max. infall velocity is at r=Q²/M and the turnaround radius is at r=Q²/2M. If we just consider a (spin) then max. infall velocity is at r=a and the turnaround radius is at r=0. For a Kerr-Newman black hole (with both charge and spin) the turnaround radius is still at r=Q²/2M but the equation for radius of max. infall velocity doesn't seem apparent. Any suggestions as to what this might be?

 

[tiny-tim]

turnaround radius

On a side note, radial infall velocity (β) is defined using the following equation-

$$\beta=\frac{\sqrt{2Mr-Q^2}}{\sqrt{r^2+a^2}}$$

if we just consider Q (charge) then max. infall velocity is at r=Q²/M and the turnaround radius is at r=Q²/2M. If we just consider a (spin) then max. infall velocity is at r=a and the turnaround radius is at r=0.

in that case, I think that my interpretation was wrong …

btw, have i got this right? … the "turnaround radius" (inside the inner event horizon) is the junkyard of a charged rotating black hole, in that all freefalling bodies approach it at zero velocity, from either side ?

i think the turnaround radius is a purely imaginary line which just happens to be the perihelion of freefalling bodies whose speed "at infinity" is zero (because √(2Mr - Q²) < 0 is inadmissible) …

in two dimensions, such a body would reach perihelion (in "conjunction" rather than "opposition"!) with zero radial but non-zero tangential speed, and would return towards the inner horizon (with constant "angular momentum" parameter) …

if it had more energy, its personal turnaround radius would be further in

 

[stevebd1]

Hi tiny-tim

You mention the possibility that the turnaround radius is a perihelion of sorts where infalling objects have zero radial but non-zero tangential velocity, this might be the case for the Kerr-Newman solution where the black hole is rotating but in the Reissner-Nordstrom, where the black hole has charge but no rotation, there is still predicted to be a turnaround radius and that even infalling objects on a strict radial path will have zero velocity at the turnaround radius, suggesting some kind of electrostatic repulsion, this effect is used in some papers to explain mass-inflation at the Cauchy horizon where outgoing light meets infalling light (the process being slightly different for a rotating black hole).

 

[tiny-tim]

hi steve!

i wasn't suggesting it had anything to do with the rotation of the black hole …

i just find a 2D orbit easier to visualise than a 1D orbit!

(but it can't be electrostatic, since it applies to bodies of any charge or none)

 

[stevebd1]

I've just reread your post and caught the '..and would return towards the inner horizon..' part, I guess it was the talk of zero radial and non-zero tangential speed that made me think you were talking conventional orbits. What exactly do you mean when you talk about a 2D orbit, are you plotting the objects 1D infall radius against time?

 

[tiny-tim]

yes, i am talking about conventional orbits (if anything's conventional in there! )

i'm assuming (i haven't done any calculations) coming in on the equatorial plane with a non-zero angular momentum …

it seemed to me that g_tφ = 0 gives 1D "orbits" with perihelion there for freefalling objects with energy parameter corresponding to zero speed "at infinity", and so similarly would give 2D orbits with perihelion either there or nearby

 

[stevebd1]

I can imagine a 2D orbit but what is a 1D orbit? Also, are you suggesting that matter that falls in on a strict radial path (for the RN solution) will be pulled into orbit at the turnaround radius, regardless of having no initial AM, working its way back to the inner horizon? (For the record, g_φφ=0 denotes the boundary for CTCs, the turnaround radius is at g_tφ=0 for Kerr-Newman metric and r₀=Q²/2M for RN metric.

 

[tiny-tim]

by "1D" orbit i meant coming radially in, stopping, and going radially out again (infinite eccentricity!)

(oh, yes i should have written g_tφ = 0 … i got carried away by the thread title! … i'll edit it now )

 

[stevebd1]

I suppose that begs the question, if the turnaround radius implies that falling matter is simply reversed and returned to the inner horizon (in the river model, this implies a white hole which there is currently no evidence of), how is anything supposed to reach the CTCs in the Kerr-Newman solution? In http://arxiv.org/PS_cache/arxiv/pdf/0708/0708.2324v2.pdf" , the r₀ is recognised but is referred to as the counter-rotation radius where space begins to rotate in the opposite direction of the ring singularity but there is no mention of there being any turnaround.

Also from the same paper that talks about the turnaround radius (http://casa.colorado.edu/~ajsh/phys5770_08/bh.pdf" page 23), the following is stated-

'..What happens inward of the turnaround radius r₀, equation r₀=Q²/2M? Inside this radius the interior mass M(r), equation M(r)=Q²/2r, is negative, and the velocity β is imaginary. The interior mass M(r) diverges to negative infinity towards the central singularity at r→0. The singularity is timelike, and infinitely gravitationally repulsive, unlike the central singularity of the Schwarzschild geometry. Is it physically realistic to have a singularity that has infinite negative mass and is infinitely gravitationally repulsive? Undoubtedly not...'

While it's open to interpretation what happens within a black hole, it would good to find a model that recognises the turnaround radius but still allows you to pass it and reach the singularity in the RN solution or ring singularity and CTCs in the KN solution.On a side note, at the turnaround radius (r₀), g_tφ=0 (which means the frame dragging rate is zero), g_tt=-1 and g_φφ=r₀²+a² '..This is almost the same as in Minkowski spacetime, the only difference being that g_φφ is not r², but a bit larger. Hence the light cones at r₀ are "standing", but they are a bit narrower than at infinity (the bigger a is, the narrower the light cones)..'

 

[tiny-tim]

I suppose that begs the question, if the turnaround radius implies that falling matter is simply reversed and returned to the inner horizon (in the river model, this implies a white hole which there is currently no evidence of), how is anything supposed to reach the CTCs in the Kerr-Newman solution?

perhaps it only needs an extra push … there's (i assume) a different turnaround radius for different initial velocities

or perhaps it can't … is there any reason why it should … if nothing escapes from a CTC, how does anything get into one? (i'm just rambling: I've never looked at the CTC equations )

'..What happens inward of the turnaround radius r₀, equation r₀=Q²/2M? Inside this radius the interior mass M(r), equation M(r)=Q²/2r, is negative, and the velocity β is imaginary.

isn't imaginary velocity exactly the condition which shows that you're beyond the perihelion (or aphelion)? … if there's no real solution, the orbit doesn't go there!

The interior mass M(r) diverges to negative infinity towards the central singularity at r→0. The singularity is timelike, and infinitely gravitationally repulsive, unlike the central singularity of the Schwarzschild geometry. Is it physically realistic to have a singularity that has infinite negative mass and is infinitely gravitationally repulsive? Undoubtedly not...'

i don't understand "infinitely gravitationally repulsive" … the repulsion is very local, and unnoticeable until you get near the turnaround radius

While it's open to interpretation what happens within a black hole, it would good to find a model that recognises the turnaround radius but still allows you to pass it and reach the singularity in the RN solution or ring singularity and CTCs in the KN solution.

why? i quite like the idea of a black hole "pulling up the ladder behind it" … after the initial enthusiasm of being formed, not allowing any further mass into the singularity (but still allowing it in through the event horizon, to orbit forever), or at least not until there's enough for some sort of "pressure" to overcome the repulsion (i'm still rambling, of course! )

 

[stevebd1]

It's a slightly amusing idea that after finding a spinning SMBH with some degree of charge, braving the outer region and event horizon, surviving the inner horizon, to then approach the CTCs near the ring singularity, only to find yourself suddenly on your way back out towards the inner horizon of (presumably) a theoretical white hole! I suppose the idea of the turnaround radius only applies if white holes exist (which there is currently no evidence of) as the river model paper proposes whereas the other paper, which only seems to recognise counter-rotation, doesn't seem to recognise the notion of white holes. There's still the issue of infall velocity becoming zero within the inner horizon so it would be interesting to see the alternatives as to what happens here if white holes is not an option.