# I Black hole orbits

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1. Jun 19, 2016

### edguy99

Are videos displayed in coordinate time or proper time? From the video, I had assumed they we are in coordinate time (ie. what you would see from earth looking at the black hole in the center of the milky way), but now I see they must be proper time (from the particles point of view that is falling into the black hole). Still a little confused.

I agree.

2. Jun 19, 2016

### m4r35n357

All my simulations are done in particle proper time, but can be plotted in coordinate time (well, against any of the coordinates really if you want to . . . ). The main difference when plotting against coordinate time is that particles travel slower near to the mass and faster far away. I think it's more dramatic to have them "whoosh" by, and requires no post processing ;)

Of course, if I were to simulate n test particles, it would only make sense to use coordinate time to show where they "are".

3. Jun 19, 2016

### Yukterez

If I had to simulate more than 1 particle free falling I would also use coordinate time steps:

But i was also wondering where the red particle would be in terms of the blue particles system after the blue particle has already disappeared behind the horizon. I'll see if I can figure that out too, but right now I have no idea how that would look like.

Last edited: Jun 19, 2016
4. Jun 19, 2016

### m4r35n357

Off the top of my head, you might want to look at "rain coordinates" (Gullstrand-Painleve), but to be honest I'm still wondering precisely what the question means ;)

5. Jun 19, 2016

### Yukterez

In the above example I can say
in coordinate time, when the blue particle is at r=4.277, the red particle is ar r=11.61
when the blue particle is at r=0.1, at which r is the red particle at that time in the blue particle's plane of simultaneity?
I think I really have to switch to another coordinate system than Schwarzschild, but maybe I also just don't see the obvious. I'll have to take a nap over that.

6. Jun 19, 2016

### Staff: Mentor

Yes, you do. The Schwarzschild coordinate lines of simultaneity outside the horizon do not extend inside the horizon. So the question you are asking is meaningless in Schwarzschild coordinates.

7. Jun 25, 2016

### Yukterez

I finished the simulator for rotating black holes using Kerr metric.

It takes intial positions in terms of {r0, θ0, ψ0} for radial distance and polar and spin angle, and initial velocities in terms of {v0, δ0, φ0} for total velcity and horizontal and vertical launch angle, or alternatively {vr, vθ, vψ} for the seperate components.

Initial conditions:
Spin parameter: a=0.998, r0=4GM/c², initial position of the test projectile: θ0=90° (equatorial plane), horizontal launching angle: δ0=60° in south pole direction, vertical launch angle: φ0=0° (horizontal), local inital velocity: v0=0.6c

Top, side and front view relative to the spin axis of the black hole:

http://yukterez.net/org/kerr4.gif

It still needs some updates to have the same displays as the Schwarzschild-simulator which I will do when I have the time (hopefully soon).

Last edited: Jun 25, 2016
8. Jun 25, 2016

### m4r35n357

Nice. If you were to keep track of and display the max and min r values, and the maximum latitude, we could compare notes ;)

You can also compare against http://staff.science.nus.edu.sg/~phylyk/downloads/reports/sp2172_report.pdf [Broken] for particle orbits, and this for light orbits.

Last edited by a moderator: May 8, 2017
9. Jun 25, 2016

### Yukterez

That's next on the list, and also the local and delayed velocities as well as the time dilation factor.

The latitude goes over all values, as seen in the right image (as well as the longitude, which is seen in the left image)

Last edited: Jun 25, 2016
10. Jun 25, 2016

### m4r35n357

You might want to reconsider your last statement, in the light of the top view on the left ;)
Anyway, this will become more explicit when you implement the latitude detection . . .

BTW this and this is what I mean by a maximum latitude/polar orbit.

11. Jun 25, 2016

### Yukterez

That's why I went online again, after thinking about it it became clear that this was an error.

12. Jun 25, 2016

### m4r35n357

Heh, I realize you know your stuff, just proving that I am awake!

13. Jun 26, 2016

### Yukterez

The most important values are now displayed below the animation (I call it "the butterfly", because with a little fantasy you can see one in the right image with the polar perspective):

http://i.imgur.com/5LuDiKl.gif

Unfortunately I did the index in german, but the latitude is the "Breitengrad θ" with 0° for the north and 180° for the south pole.
Edit: the picture embedding does not seem to work (maybe the gif is too large, the running numbers cost an extra megabyte) but if you click on the icon you should see it.

Last edited: Jun 26, 2016
14. Jun 27, 2016

### m4r35n357

Just pulled in the Python implementation and made the algorithms identical. The Vala version is ~3.5 times as fast as the PyPy version, and at least 40 times the speed of bare Python. I scaled the number of timesteps in the run for Vala and PyPy (I didn't have the patience to wait for basic Python), and the execution times scaled by 10, so I think I've eliminated any VM startup delays from the calculation.

The difference is very significant when running on a Raspberry Pi.

15. Jun 28, 2016

### edguy99

Great animations, best I have seen. I do step by step animations using Newtons law calculation an any number of particles between each step (ie. force in 3d calculated, force added to velocity, velocity added to position, then re-display). Would love to convert to Kerr metric but I have a little trouble following your linked equations. Do you have any suggestions to assist in understanding how to calculate the x/y/z force on each particle for each frame using your linked equations?

Sample: http://www.animatedphysics.com/planets/planet_edit.htm

16. Jun 28, 2016

### Yukterez

If you use Mathematica you can automatically differentiate r'[t], θ'[t] and ψ'[t] (which are defined in the differential equation) with

Evaluate[r''[t] /. sol][[1]]
Evaluate[θ''[t] /. sol][[1]]
Evaluate[ψ''[t] /. sol][[1]]

to get the second (proper) time derivative of the motion. The transformation from r, θ, ψ components to x, y, z components is

x = r Sin[θ] Cos[ψ])
y = r Sin[θ] Sin[ψ]
z = r Cos[θ]

17. Jun 28, 2016

### pervect

Staff Emeritus
I assume r, theta, and psi are Schwarzschild coordinates? The difficulty with constructing x,y, and z in the manner you did above from r, theta, and psi is that the speed of light is not isotropic in Schwarzschild coordinates, i.e. it depends on direction. Thus with these relationships, you have the speed of light being different in the x direction, the y direction, and the z direction. This means that the distance scales are not the same - if you consider the distance light travels in one "tick" of proper time (or, if you prefer, one "tick" of coordinate time), the distance the light moves depends on the direction, i.e. the distance is different in the x, y, and z directions.

This is rather incompatible with the standard SI definition of the meter if you attempt to interpret your coordinate values as having physical significance (i.e. representing distances). The proportionality between a change in the coordinate value and the change in distance depends on which coordinate you pick.

The usual solution to this is to use an alternate coordinate system, the isotropic coordinate system, which you can find the metric for in The "Alternate coordinate" section of https://en.wikipedia.org/wiki/Schwarzschild_metric. It's rather painful to compute things in isotropic coordinates, though it can be done, the equations are are lot more complicated. It would probably be easiest to compute the answer in Schwarzschild coordinates, and convert the answer into isotropic coordinates. Then you could convert that answer into x,y, and z, and the results would be much more physically significant.

I don't recall the transformation equations offhand, however, and I didn't notice them in the wiki article I quoted.

18. Jun 28, 2016

### Yukterez

In Schwarzschild the factor for the transversal direction is √(1-rs/r) and for the radial direction (1-rs/r) without the square (because you have time dilation and radial length expansion of the same magnitude, also see this link). In Kerr you replace the Schwarzschild radius rs=2 with rk=1+√(1-a²Cos²[θ]) where a is the spin parameter.

19. Jun 28, 2016

### Yukterez

I see I have a bug in the .txt-File code and mixed up some r0 and r(t) in the formula for the digit velocity display (the simulation is ok, but the digit output below the animation needs a repair). I'll fix that tomorrow, until then you can use the equations of motion from my animation code, but if you want to display the numbers transform from r', θ' & ψ' to vr, vθ & vψ like in my last posting above, where all r are r(t).

20. Jun 29, 2016

### Yukterez

The bug in the code is fixed now, the transformation from proper to coordinate velocity should give the right numers now. If you use the kerr.txt force refresh (ctrl f5) to make sure to get the updated version.

I'll put the transformation in Latex (with G, M and c set to 1):
to transform from the derivatives in the differential equation to the local velocity components use

$\dot{r} = \frac{v_r (1-r_k/r)^{(3/2)}}{\sqrt{1-v^2}}$

$\dot{\theta} = \frac{v_{\theta}}{r \sqrt{1-v^2}}$

$\dot{\psi} = \frac{v_{\psi} }{r \sqrt{1-v^2}}$

The $v_{\psi}$ component is observed faster by

$d_{\psi} = \pm \frac{2 \alpha r^2 }{ \Sigma}$

with

$r_k = \sqrt{1-a^2 \cos^2 (\theta )}+1$

because of frame dragging, where

$\Sigma =\left(\alpha ^2+r^2\right)^2-\alpha ^2 \sin ^2(\theta ) \Delta$

and

$\Delta =r^{2}-2r+\alpha ^{2}$

After solving for $v_r$, $v_{\theta}$, $v_{\psi}$ and $d_{v_{\psi}}$ you can use Pythagoras for the total velocity $v$.

Last edited: Jun 29, 2016