Revision history [back]
 | 1 | initial version |
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 $\theta$).
| | 2 | No.2 Revision |
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 was actually an error 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()):
$$ \frac{24 \, {\left(H^{4} a^{12} \cos\left({\theta}\right)^{12} + 6 \, H^{4} a^{10} r^{2} \cos\left({\theta}\right)^{10} + 15 \, H^{4} a^{8} r^{4} \cos\left({\theta}\right)^{8} + 20 \, H^{4} a^{6} r^{6} \cos\left({\theta}\right)^{6} + 15 \, H^{4} a^{4} r^{8} \cos\left({\theta}\right)^{4} + 6 \, H^{4} a^{2} r^{10} \cos\left({\theta}\right)^{2} + H^{4} r^{12} - 2 \, a^{6} m^{2} \cos\left({\theta}\right)^{6} + 30 \, a^{4} m^{2} r^{2} \cos\left({\theta}\right)^{4} - 30 \, a^{2} m^{2} r^{4} \cos\left({\theta}\right)^{2} + 2 \, m^{2} r^{6}\right)}}{{\left(a^{2} \cos\left({\theta}\right)^{2} + r^{2}\right)}^{6}}$$
| | 3 | No.3 Revision |
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 was were actually an error 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()):
$$ \frac{24 \, {\left(H^{4} a^{12} \cos\left({\theta}\right)^{12} + 6 \, H^{4} a^{10} r^{2} \cos\left({\theta}\right)^{10} + 15 \, H^{4} a^{8} r^{4} \cos\left({\theta}\right)^{8} + 20 \, H^{4} a^{6} r^{6} \cos\left({\theta}\right)^{6} + 15 \, H^{4} a^{4} r^{8} \cos\left({\theta}\right)^{4} + 6 \, H^{4} a^{2} r^{10} \cos\left({\theta}\right)^{2} + H^{4} r^{12} - 2 \, a^{6} m^{2} \cos\left({\theta}\right)^{6} + 30 \, a^{4} m^{2} r^{2} \cos\left({\theta}\right)^{4} - 30 \, a^{2} m^{2} r^{4} \cos\left({\theta}\right)^{2} + 2 \, m^{2} r^{6}\right)}}{{\left(a^{2} \cos\left({\theta}\right)^{2} + r^{2}\right)}^{6}}$$
| | 4 | No.4 Revision |
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.
