Ask Your Question
0

I want to calculate the Riemann tensor and Kretschmann scalar for Kerr- De Sitter spacetime but it is taking too long and not giving any answer please help

asked 2023-11-25 14:37:26 +0100

I want to calculate the Riemann tensor and Kretschmann scalar for Kerr- De Sitter spacetime but it is taking too long and not giving any answer, it is giving Christoffel symbols,

edit retag flag offensive close merge delete

Comments

As general advice, try to give your metric in a closed form as much as you can. For example, instead of (r^2 + a^2*(cosθ)^2), use rho(r,theta), Delta_r(r), etc. Then perform the calculations. After everything is set, put the actual functions and simplify. If this cannot help, please share your code and we will see what we can do.

tolga gravatar imagetolga ( 2023-11-25 18:43:24 +0100 )edit

@tolga thanks for the comment, I have given my metric in closed form, but it is still not working. I want to give you the code please tell me how I can give it

Techeye gravatar imageTecheye ( 2023-11-27 13:33:22 +0100 )edit

Please check the answer of @eric_g below. If it does not work out, you can gather your code in one cell and paste it here, or upload it to your GitHub as a public file and share the link.

tolga gravatar imagetolga ( 2023-11-27 17:58:19 +0100 )edit

@tolga here is the link http://localhost:8888/notebooks/Untitled51.ipynb (http://localhost:8888/notebooks/Untit...)

Techeye gravatar imageTecheye ( 2023-11-28 07:42:10 +0100 )edit

@tolga here is the actual one I tried http://localhost:8889/notebooks/Untitled50.ipynb (http://localhost:8889/notebooks/Untit...)

Techeye gravatar imageTecheye ( 2023-11-28 07:44:01 +0100 )edit

2 Answers

Sort by » oldest newest most voted
0

answered 2023-11-28 11:33:55 +0100

tolga gravatar image

You actually put the functions Delta, rho2, and Sigma in the metric at the beginning. You should use them in a closed form as in the following code. Please try this (but first check if the metric is correct).

(If it succeeds, then you can substitute the actual functions in the results):

reset()
M = Manifold(4, 'M', latex_name=r'\mathcal{M}', structure='Lorentzian')
BL.<t,r,th,ph> = M.chart(r"t r th:(0,pi):\theta ph:(0,2*pi):\phi")

var('m a H E')

g = M.metric()

rho2=function('rho2')(r,th)
Delta=function('Delta')(r)
Sigma=function('Sigma')(th)

g[0,0] = -(Delta - a^2*sin(th)^2*Sigma)/rho2
g[0,3] = -(a*sin(th)^2)*((r^2 + a^2)*Sigma - Delta)/(rho2*E)
g[1,1], g[2,2] = rho2/Delta, rho2/Sigma
g[3,3] = (sin(th)^2/(rho2*E))*((r^2 + a^2)^2*Sigma - Delta*a^2*sin(th)^2)
show(g.display())

Riem = g.riemann()
R_up = Riem.up(g)
R_down = Riem.down(g)

Kretschmann_scalar = R_up['^{abcd}']*R_down['_{abcd}']
Kretschmann_scalar.display()
edit flag offensive delete link more

Comments

@tolga and @eric_g really thanks for your answers. @tolga I was using this material for the metric https://arxiv.org/abs/1710.00997#:~:text=The%20Kerr-de%20Sitter%20spacetime%20has%20two%20ergospheres%20--%20associated,cosmological%20ergoregion%20is%20never%20possible (https://arxiv.org/abs/1710.00997#:~:t...). and there is a mismatch between your and my material.

Techeye gravatar imageTecheye ( 2023-11-28 18:48:20 +0100 )edit

I think that the metric given by Eq. (1) of your reference (https://arxiv.org/abs/1710.00997) is not correct (compare with Table 1 of https://arxiv.org/abs/1805.00243, which I think is correct): a $\Xi^2$ factor is missing in $g_{tt}$ and $\Xi$ should be $\Xi^2$ in $g_{t\phi}$. As a proof of the above assertion, the metric from https://arxiv.org/abs/1805.00243 fulfills the Einstein equation, as I've checked with SageMath (cf. my answer), so this one is truly the Kerr-de Sitter metric.

eric_g gravatar imageeric_g ( 2023-11-28 20:07:57 +0100 )edit

@tolga and eric_g maybe I got the wrong metric, I will look into it. @tolga I didn't understand the simplification part can you send me the code in detail.

Techeye gravatar imageTecheye ( 2023-11-29 06:38:24 +0100 )edit

We can discuss the substitution and simplification if @eric_g 's solution does not fit your needs.

tolga gravatar imagetolga ( 2023-11-29 07:12:59 +0100 )edit

Actually, since $\Xi := 1 + H_0^2 a^2$ is a constant, one can rescale the $t$ of https://arxiv.org/abs/1805.00243 as $t' = t/\Xi$ to get Eq. (1) of https://arxiv.org/abs/1710.00997, so that both metrics are equivalent. So contrary to what I said in the above comment, the metric of https://arxiv.org/abs/1710.00997 is correct and represents the Kerr-dS solution as well. The original issue with your notebook was then a mistake in the definition of g[3,3]: the factor E (which denotes $\Xi$) must be replaced by E^2:

# g[3,3] = (sin(th)^2/(rho2*E))*((r^2 + a^2)^2*Sigma - Delta*a^2*sin(th)^2)  # original version
g[3,3] = (sin(th)^2/(rho2*E^2))*((r^2 + a^2)^2*Sigma - Delta*a^2*sin(th)^2)  # correct version
eric_g gravatar imageeric_g ( 2023-11-29 08:54:27 +0100 )edit
0

answered 2023-11-27 16:56:18 +0100

eric_g gravatar image

updated 2023-11-28 15:26:45 +0100

The 5-dimensional Kerr-AdS spacetime has been treated there. The Riemann tensor was computed in a reasonable time (it is not shown in the notebook but its computation was triggered by g.ricci() in cell 12). Since the 4-dimensional case should a priori be simpler, you could get some inspiration from this notebook (note the use of $\mu = \cos\theta$ instead of $\theta$).

Update (28 Nov 2023): Thanks for sharing the code in the above comment. There were actually some errors in the entries for the metric components $g_{00}$ (a factor E ^2 was missing), $g_{03}$ (E should be replaced by E^2) and $g_{33}$ (idem). Compare with Table 1 of https://arxiv.org/abs/1805.00243. Here are the corrected components (the commented lines are the original ones):

g = M.metric()
rho2 = r^2 + (a*cos(th))^2
Delta = (r^2 + a^2)*(1 - H^2*r^2) -2*m*r
Sigma = 1 + H^2*a^2*cos(th)^2
E = 1 + H^2*a^2
#g[0,0] = -(Delta - a^2*sin(th)^2*Sigma)/rho2
g[0,0] = -(Delta - a^2*sin(th)^2*Sigma)/(rho2*E^2)
#g[0,3] = -(a*sin(th)^2)*((r^2 + a^2)*Sigma - Delta)/(rho2*E)
g[0,3] = -(a*sin(th)^2)*((r^2 + a^2)*Sigma - Delta)/(rho2*E^2)
g[1,1], g[2,2] = rho2/Delta, rho2/Sigma
#g[3,3] = (sin(th)^2/(rho2*E))*((r^2 + a^2)^2*Sigma - Delta*a^2*sin(th)^2)
g[3,3] = (sin(th)^2/(rho2*E^2))*((r^2 + a^2)^2*Sigma - Delta*a^2*sin(th)^2)

As a test that this is the correct Kerr-de Sitter metric, one can check the Einstein equation:

g.ricci() == 3*H^2*g

returns True.

With the corrected metric, the Riemann tensor is computed in 1 min 8 s on my Ubuntu 22.04 lntel Xeon E5-2623 computer running Sage 10.1 with parallization on 8 cores (switched on via Parallelism().set(nproc=8)) and the Kretschmann scalar is computed in 3 min 55 s. Here is the result (outcome of Kr_scalar.expr().factor()):

$$ \frac{24 \, {\left(H^{4} a^{12} \cos\left({\theta}\right)^{12} + 6 \, H^{4} a^{10} r^{2} \cos\left({\theta}\right)^{10} + 15 \, H^{4} a^{8} r^{4} \cos\left({\theta}\right)^{8} + 20 \, H^{4} a^{6} r^{6} \cos\left({\theta}\right)^{6} + 15 \, H^{4} a^{4} r^{8} \cos\left({\theta}\right)^{4} + 6 \, H^{4} a^{2} r^{10} \cos\left({\theta}\right)^{2} + H^{4} r^{12} - 2 \, a^{6} m^{2} \cos\left({\theta}\right)^{6} + 30 \, a^{4} m^{2} r^{2} \cos\left({\theta}\right)^{4} - 30 \, a^{2} m^{2} r^{4} \cos\left({\theta}\right)^{2} + 2 \, m^{2} r^{6}\right)}}{{\left(a^{2} \cos\left({\theta}\right)^{2} + r^{2}\right)}^{6}}$$

Note on the computational time: the rather long computational time (a few minutes for the Kretschmann scalar) is due to the default simplification chain used by SageMath for calculus on manifolds. The computational time can be significantly reduced by customizing the simplification chain via the method set_simplify_function and using $\mu := \cos\theta$ as a coordinate instead of $\theta$. For instance, introducing $\mu$ and setting

M.set_simplify_function(factor)

just after the definition of the chart $(t,r,\mu,\phi)$ yields a computational time of only 4 seconds for the Riemann tensor and 6 seconds for the Kretschmann scalar.

edit flag offensive delete link more

Comments

Thanks for the answer but what I want is the Kretschmann scalar for Kerr- De Sitter spacetime, not Kerr- Anti De Sitter spacetime. And I have also used these simplifications

Techeye gravatar imageTecheye ( 2023-11-28 07:36:11 +0100 )edit

Sorry I read your post too fast. I have updated my answer to the Kerr-de Sitter case.

eric_g gravatar imageeric_g ( 2023-11-28 12:15:58 +0100 )edit

Your Answer

Please start posting anonymously - your entry will be published after you log in or create a new account.

Add Answer

Question Tools

Stats

Asked: 2023-11-25 14:37:26 +0100

Seen: 154 times

Last updated: Nov 28 '23