Kerr-Newman spacetime

This notebook demonstrates a few capabilities of SageMath in computations regarding Kerr-Newman spacetime. The corresponding tools have been developed within the SageManifolds project.

NB: a version of SageMath at least equal to 9.2 is required to run this notebook:

In [1]:
version()
Out[1]:
'SageMath version 9.5.beta2, Release Date: 2021-09-26'

First we set up the notebook to display mathematical objects using LaTeX rendering:

In [2]:
%display latex

and we initialize a time counter for benchmarking:

In [3]:
import time
comput_time0 = time.perf_counter()

Since some computations are quite heavy, we ask for running them in parallel on 8 threads:

In [4]:
Parallelism().set(nproc=8)

Spacetime manifold

We declare the Kerr-Newman spacetime (or more precisely the part of it covered by Boyer-Lindquist coordinates) as a 4-dimensional Lorentzian manifold $\mathcal{M}$:

In [5]:
M = Manifold(4, 'M', latex_name=r'\mathcal{M}', structure='Lorentzian')

We then introduce the standard Boyer-Lindquist coordinates as a chart BL (for Boyer-Lindquist) on $\mathcal{M}$, via the method chart(), the argument of which is a string (delimited by r"..." because of the backslash symbols) expressing the coordinates names, their ranges (the default is $(-\infty,+\infty)$) and their LaTeX symbols:

In [6]:
BL.<t,r,th,ph> = M.chart(r't r th:(0,pi):\theta ph:(0,2*pi):\phi') 
print(BL); BL
Chart (M, (t, r, th, ph))
Out[6]:
\[\newcommand{\Bold}[1]{\mathbf{#1}}\left(\mathcal{M},(t, r, {\theta}, {\phi})\right)\]

Metric tensor

The 3 parameters $m$, $a$ and $q$ of the Kerr-Newman spacetime are declared as symbolic variables:

In [7]:
var('m a q')
Out[7]:
\[\newcommand{\Bold}[1]{\mathbf{#1}}\left(m, a, q\right)\]

We get the (yet undefined) spacetime metric:

In [8]:
g = M.metric()

The metric is defined by its components in the coordinate frame associated with Boyer-Lindquist coordinates, which is the current manifold's default frame:

In [9]:
rho2 = r^2 + (a*cos(th))^2
Delta = r^2 -2*m*r + a^2 + q^2
g[0,0] = -1 + (2*m*r-q^2)/rho2
g[0,3] = -a*sin(th)^2*(2*m*r-q^2)/rho2
g[1,1], g[2,2] = rho2/Delta, rho2
g[3,3] = (r^2 + a^2 + (2*m*r-q^2)*(a*sin(th))^2/rho2)*sin(th)^2
g.display()
Out[9]:
\[\newcommand{\Bold}[1]{\mathbf{#1}}g = \left( -\frac{q^{2} - 2 \, m r}{a^{2} \cos\left({\theta}\right)^{2} + r^{2}} - 1 \right) \mathrm{d} t\otimes \mathrm{d} t + \left( \frac{{\left(q^{2} - 2 \, m r\right)} a \sin\left({\theta}\right)^{2}}{a^{2} \cos\left({\theta}\right)^{2} + r^{2}} \right) \mathrm{d} t\otimes \mathrm{d} {\phi} + \left( \frac{a^{2} \cos\left({\theta}\right)^{2} + r^{2}}{a^{2} + q^{2} - 2 \, m r + r^{2}} \right) \mathrm{d} r\otimes \mathrm{d} r + \left( a^{2} \cos\left({\theta}\right)^{2} + r^{2} \right) \mathrm{d} {\theta}\otimes \mathrm{d} {\theta} + \left( \frac{{\left(q^{2} - 2 \, m r\right)} a \sin\left({\theta}\right)^{2}}{a^{2} \cos\left({\theta}\right)^{2} + r^{2}} \right) \mathrm{d} {\phi}\otimes \mathrm{d} t -{\left(\frac{{\left(q^{2} - 2 \, m r\right)} a^{2} \sin\left({\theta}\right)^{2}}{a^{2} \cos\left({\theta}\right)^{2} + r^{2}} - a^{2} - r^{2}\right)} \sin\left({\theta}\right)^{2} \mathrm{d} {\phi}\otimes \mathrm{d} {\phi}\]

The list of the non-vanishing components:

In [10]:
g.display_comp()
Out[10]:
\[\newcommand{\Bold}[1]{\mathbf{#1}}\begin{array}{lcl} g_{ \, t \, t }^{ \phantom{\, t}\phantom{\, t} } & = & -\frac{q^{2} - 2 \, m r}{a^{2} \cos\left({\theta}\right)^{2} + r^{2}} - 1 \\ g_{ \, t \, {\phi} }^{ \phantom{\, t}\phantom{\, {\phi}} } & = & \frac{{\left(q^{2} - 2 \, m r\right)} a \sin\left({\theta}\right)^{2}}{a^{2} \cos\left({\theta}\right)^{2} + r^{2}} \\ g_{ \, r \, r }^{ \phantom{\, r}\phantom{\, r} } & = & \frac{a^{2} \cos\left({\theta}\right)^{2} + r^{2}}{a^{2} + q^{2} - 2 \, m r + r^{2}} \\ g_{ \, {\theta} \, {\theta} }^{ \phantom{\, {\theta}}\phantom{\, {\theta}} } & = & a^{2} \cos\left({\theta}\right)^{2} + r^{2} \\ g_{ \, {\phi} \, t }^{ \phantom{\, {\phi}}\phantom{\, t} } & = & \frac{{\left(q^{2} - 2 \, m r\right)} a \sin\left({\theta}\right)^{2}}{a^{2} \cos\left({\theta}\right)^{2} + r^{2}} \\ g_{ \, {\phi} \, {\phi} }^{ \phantom{\, {\phi}}\phantom{\, {\phi}} } & = & -{\left(\frac{{\left(q^{2} - 2 \, m r\right)} a^{2} \sin\left({\theta}\right)^{2}}{a^{2} \cos\left({\theta}\right)^{2} + r^{2}} - a^{2} - r^{2}\right)} \sin\left({\theta}\right)^{2} \end{array}\]

The component $g^{tt}$ of the inverse metric:

In [11]:
g.inverse()[0,0]
Out[11]:
\[\newcommand{\Bold}[1]{\mathbf{#1}}\frac{a^{4} + 2 \, a^{2} r^{2} + r^{4} - {\left(a^{4} + a^{2} q^{2} - 2 \, a^{2} m r + a^{2} r^{2}\right)} \sin\left({\theta}\right)^{2}}{2 \, m r^{3} - r^{4} - {\left(a^{2} + q^{2}\right)} r^{2} - {\left(a^{4} + a^{2} q^{2} - 2 \, a^{2} m r + a^{2} r^{2}\right)} \cos\left({\theta}\right)^{2}}\]

The lapse function:

In [12]:
N = 1/sqrt(-(g.inverse()[[0,0]])); N
Out[12]:
\[\newcommand{\Bold}[1]{\mathbf{#1}}\mbox{Scalar field on the 4-dimensional Lorentzian manifold M}\]
In [13]:
N.display()
Out[13]:
\[\newcommand{\Bold}[1]{\mathbf{#1}}\begin{array}{llcl} & \mathcal{M} & \longrightarrow & \mathbb{R} \\ & \left(t, r, {\theta}, {\phi}\right) & \longmapsto & \frac{\sqrt{a^{2} \cos\left({\theta}\right)^{2} + r^{2}} \sqrt{{\left| a^{2} + q^{2} - 2 \, m r + r^{2} \right|}}}{\sqrt{a^{4} + 2 \, a^{2} r^{2} + r^{4} - {\left(a^{4} + a^{2} q^{2} - 2 \, a^{2} m r + a^{2} r^{2}\right)} \sin\left({\theta}\right)^{2}}} \end{array}\]

Electromagnetic field tensor

Let us first introduce the 1-form basis associated with Boyer-Lindquist coordinates:

In [14]:
dBL = BL.coframe()
dBL
Out[14]:
\[\newcommand{\Bold}[1]{\mathbf{#1}}\left(\mathcal{M}, \left(\mathrm{d} t,\mathrm{d} r,\mathrm{d} {\theta},\mathrm{d} {\phi}\right)\right)\]

The electromagnetic field tensor $F$ is formed as [cf. e.g. Eq. (33.5) of Misner, Thorne & Wheeler (1973)]

In [15]:
F = q/rho2^2 * (r^2-a^2*cos(th)^2)* dBL[1].wedge( dBL[0] - a*sin(th)^2* dBL[3] ) + \
    2*q/rho2^2 * a*r*cos(th)*sin(th)* dBL[2].wedge( (r^2+a^2)* dBL[3] - a* dBL[0] ) 
F.set_name('F')
F.display()
Out[15]:
\[\newcommand{\Bold}[1]{\mathbf{#1}}F = \left( \frac{a^{2} q \cos\left({\theta}\right)^{2} - q r^{2}}{a^{4} \cos\left({\theta}\right)^{4} + 2 \, a^{2} r^{2} \cos\left({\theta}\right)^{2} + r^{4}} \right) \mathrm{d} t\wedge \mathrm{d} r + \left( \frac{2 \, a^{2} q r \cos\left({\theta}\right) \sin\left({\theta}\right)}{a^{4} \cos\left({\theta}\right)^{4} + 2 \, a^{2} r^{2} \cos\left({\theta}\right)^{2} + r^{4}} \right) \mathrm{d} t\wedge \mathrm{d} {\theta} + \left( \frac{{\left(a^{3} q \cos\left({\theta}\right)^{2} - a q r^{2}\right)} \sin\left({\theta}\right)^{2}}{a^{4} \cos\left({\theta}\right)^{4} + 2 \, a^{2} r^{2} \cos\left({\theta}\right)^{2} + r^{4}} \right) \mathrm{d} r\wedge \mathrm{d} {\phi} + \left( \frac{2 \, {\left(a^{3} q r + a q r^{3}\right)} \cos\left({\theta}\right) \sin\left({\theta}\right)}{a^{4} \cos\left({\theta}\right)^{4} + 2 \, a^{2} r^{2} \cos\left({\theta}\right)^{2} + r^{4}} \right) \mathrm{d} {\theta}\wedge \mathrm{d} {\phi}\]

The list of non-vanishing components:

In [16]:
F.display_comp()
Out[16]:
\[\newcommand{\Bold}[1]{\mathbf{#1}}\begin{array}{lcl} F_{ \, t \, r }^{ \phantom{\, t}\phantom{\, r} } & = & \frac{a^{2} q \cos\left({\theta}\right)^{2} - q r^{2}}{a^{4} \cos\left({\theta}\right)^{4} + 2 \, a^{2} r^{2} \cos\left({\theta}\right)^{2} + r^{4}} \\ F_{ \, t \, {\theta} }^{ \phantom{\, t}\phantom{\, {\theta}} } & = & \frac{2 \, a^{2} q r \cos\left({\theta}\right) \sin\left({\theta}\right)}{a^{4} \cos\left({\theta}\right)^{4} + 2 \, a^{2} r^{2} \cos\left({\theta}\right)^{2} + r^{4}} \\ F_{ \, r \, t }^{ \phantom{\, r}\phantom{\, t} } & = & -\frac{a^{2} q \cos\left({\theta}\right)^{2} - q r^{2}}{a^{4} \cos\left({\theta}\right)^{4} + 2 \, a^{2} r^{2} \cos\left({\theta}\right)^{2} + r^{4}} \\ F_{ \, r \, {\phi} }^{ \phantom{\, r}\phantom{\, {\phi}} } & = & \frac{{\left(a^{3} q \cos\left({\theta}\right)^{2} - a q r^{2}\right)} \sin\left({\theta}\right)^{2}}{a^{4} \cos\left({\theta}\right)^{4} + 2 \, a^{2} r^{2} \cos\left({\theta}\right)^{2} + r^{4}} \\ F_{ \, {\theta} \, t }^{ \phantom{\, {\theta}}\phantom{\, t} } & = & -\frac{2 \, a^{2} q r \cos\left({\theta}\right) \sin\left({\theta}\right)}{a^{4} \cos\left({\theta}\right)^{4} + 2 \, a^{2} r^{2} \cos\left({\theta}\right)^{2} + r^{4}} \\ F_{ \, {\theta} \, {\phi} }^{ \phantom{\, {\theta}}\phantom{\, {\phi}} } & = & \frac{2 \, {\left(a^{3} q r + a q r^{3}\right)} \cos\left({\theta}\right) \sin\left({\theta}\right)}{a^{4} \cos\left({\theta}\right)^{4} + 2 \, a^{2} r^{2} \cos\left({\theta}\right)^{2} + r^{4}} \\ F_{ \, {\phi} \, r }^{ \phantom{\, {\phi}}\phantom{\, r} } & = & -\frac{{\left(a^{3} q \cos\left({\theta}\right)^{2} - a q r^{2}\right)} \sin\left({\theta}\right)^{2}}{a^{4} \cos\left({\theta}\right)^{4} + 2 \, a^{2} r^{2} \cos\left({\theta}\right)^{2} + r^{4}} \\ F_{ \, {\phi} \, {\theta} }^{ \phantom{\, {\phi}}\phantom{\, {\theta}} } & = & -\frac{2 \, {\left(a^{3} q r + a q r^{3}\right)} \cos\left({\theta}\right) \sin\left({\theta}\right)}{a^{4} \cos\left({\theta}\right)^{4} + 2 \, a^{2} r^{2} \cos\left({\theta}\right)^{2} + r^{4}} \end{array}\]

The Hodge dual of $F$:

In [17]:
star_F = F.hodge_dual(g)
star_F.display()
Out[17]:
\[\newcommand{\Bold}[1]{\mathbf{#1}}\star F = \left( \frac{2 \, a q r \sqrt{\cos\left({\theta}\right) + 1} \sqrt{-\cos\left({\theta}\right) + 1} \cos\left({\theta}\right)}{{\left(a^{4} \cos\left({\theta}\right)^{4} + 2 \, a^{2} r^{2} \cos\left({\theta}\right)^{2} + r^{4}\right)} \sin\left({\theta}\right)} \right) \mathrm{d} t\wedge \mathrm{d} r + \left( -\frac{{\left(a^{3} q \cos\left({\theta}\right)^{2} - a q r^{2}\right)} \sqrt{\cos\left({\theta}\right) + 1} \sqrt{-\cos\left({\theta}\right) + 1}}{a^{4} \cos\left({\theta}\right)^{4} + 2 \, a^{2} r^{2} \cos\left({\theta}\right)^{2} + r^{4}} \right) \mathrm{d} t\wedge \mathrm{d} {\theta} + \left( \frac{2 \, a^{2} q r \sqrt{\cos\left({\theta}\right) + 1} \sqrt{-\cos\left({\theta}\right) + 1} \cos\left({\theta}\right) \sin\left({\theta}\right)}{a^{4} \cos\left({\theta}\right)^{4} + 2 \, a^{2} r^{2} \cos\left({\theta}\right)^{2} + r^{4}} \right) \mathrm{d} r\wedge \mathrm{d} {\phi} + \left( -\frac{{\left(a^{4} q - q r^{4} - {\left(a^{4} q + a^{2} q r^{2}\right)} \sin\left({\theta}\right)^{2}\right)} \sqrt{\cos\left({\theta}\right) + 1} \sqrt{-\cos\left({\theta}\right) + 1}}{a^{4} \cos\left({\theta}\right)^{4} + 2 \, a^{2} r^{2} \cos\left({\theta}\right)^{2} + r^{4}} \right) \mathrm{d} {\theta}\wedge \mathrm{d} {\phi}\]

Maxwell equations

Let us check that $F$ obeys the two (source-free) Maxwell equations:

In [18]:
diff(F).display()
Out[18]:
\[\newcommand{\Bold}[1]{\mathbf{#1}}\mathrm{d}F = 0\]
In [19]:
diff(star_F).display()
Out[19]:
\[\newcommand{\Bold}[1]{\mathbf{#1}}\mathrm{d}\star F = 0\]

Levi-Civita Connection

The Levi-Civita connection $\nabla$ associated with $g$:

In [20]:
nabla = g.connection()
print(nabla)
Levi-Civita connection nabla_g associated with the Lorentzian metric g on the 4-dimensional Lorentzian manifold M

Let us verify that the covariant derivative of $g$ with respect to $\nabla$ vanishes identically:

In [21]:
nabla(g) == 0
Out[21]:
\[\newcommand{\Bold}[1]{\mathbf{#1}}\mathrm{True}\]

Another view of the above property:

In [22]:
nabla(g).display()
Out[22]:
\[\newcommand{\Bold}[1]{\mathbf{#1}}\nabla_{g} g = 0\]

The nonzero Christoffel symbols (skipping those that can be deduced by symmetry of the last two indices):

In [23]:
g.christoffel_symbols_display()
Out[23]:
\[\newcommand{\Bold}[1]{\mathbf{#1}}\begin{array}{lcl} \Gamma_{ \phantom{\, t} \, t \, r }^{ \, t \phantom{\, t} \phantom{\, r} } & = & \frac{a^{4} m + a^{2} q^{2} r + q^{2} r^{3} - m r^{4} - {\left(a^{4} m + a^{2} m r^{2}\right)} \sin\left({\theta}\right)^{2}}{2 \, m r^{5} - r^{6} - {\left(a^{2} + q^{2}\right)} r^{4} - {\left(a^{6} + a^{4} q^{2} - 2 \, a^{4} m r + a^{4} r^{2}\right)} \cos\left({\theta}\right)^{4} + 2 \, {\left(2 \, a^{2} m r^{3} - a^{2} r^{4} - {\left(a^{4} + a^{2} q^{2}\right)} r^{2}\right)} \cos\left({\theta}\right)^{2}} \\ \Gamma_{ \phantom{\, t} \, t \, {\theta} }^{ \, t \phantom{\, t} \phantom{\, {\theta}} } & = & \frac{{\left(a^{2} q^{2} - 2 \, a^{2} m r\right)} \cos\left({\theta}\right) \sin\left({\theta}\right)}{a^{4} \cos\left({\theta}\right)^{4} + 2 \, a^{2} r^{2} \cos\left({\theta}\right)^{2} + r^{4}} \\ \Gamma_{ \phantom{\, t} \, r \, {\phi} }^{ \, t \phantom{\, r} \phantom{\, {\phi}} } & = & -\frac{a^{3} q^{2} r - a^{3} m r^{2} + 2 \, a q^{2} r^{3} - 3 \, a m r^{4} - {\left(a^{5} m + a^{3} q^{2} r - a^{3} m r^{2}\right)} \cos\left({\theta}\right)^{4} + {\left(a^{5} m - 2 \, a q^{2} r^{3} + 3 \, a m r^{4}\right)} \cos\left({\theta}\right)^{2}}{2 \, m r^{5} - r^{6} - {\left(a^{2} + q^{2}\right)} r^{4} - {\left(a^{6} + a^{4} q^{2} - 2 \, a^{4} m r + a^{4} r^{2}\right)} \cos\left({\theta}\right)^{4} + 2 \, {\left(2 \, a^{2} m r^{3} - a^{2} r^{4} - {\left(a^{4} + a^{2} q^{2}\right)} r^{2}\right)} \cos\left({\theta}\right)^{2}} \\ \Gamma_{ \phantom{\, t} \, {\theta} \, {\phi} }^{ \, t \phantom{\, {\theta}} \phantom{\, {\phi}} } & = & \frac{{\left(a^{5} q^{2} - 2 \, a^{5} m r\right)} \cos\left({\theta}\right) \sin\left({\theta}\right)^{5} - {\left(a^{5} q^{2} - 2 \, a^{5} m r + a^{3} q^{2} r^{2} - 2 \, a^{3} m r^{3}\right)} \cos\left({\theta}\right) \sin\left({\theta}\right)^{3}}{a^{6} \cos\left({\theta}\right)^{6} + 3 \, a^{4} r^{2} \cos\left({\theta}\right)^{4} + 3 \, a^{2} r^{4} \cos\left({\theta}\right)^{2} + r^{6}} \\ \Gamma_{ \phantom{\, r} \, t \, t }^{ \, r \phantom{\, t} \phantom{\, t} } & = & \frac{m r^{4} - {\left(2 \, m^{2} + q^{2}\right)} r^{3} + {\left(a^{2} m + 3 \, m q^{2}\right)} r^{2} - {\left(a^{4} m + a^{2} m q^{2} - 2 \, a^{2} m^{2} r + a^{2} m r^{2}\right)} \cos\left({\theta}\right)^{2} - {\left(a^{2} q^{2} + q^{4}\right)} r}{a^{6} \cos\left({\theta}\right)^{6} + 3 \, a^{4} r^{2} \cos\left({\theta}\right)^{4} + 3 \, a^{2} r^{4} \cos\left({\theta}\right)^{2} + r^{6}} \\ \Gamma_{ \phantom{\, r} \, t \, {\phi} }^{ \, r \phantom{\, t} \phantom{\, {\phi}} } & = & -\frac{{\left(a m r^{4} - {\left(2 \, a m^{2} + a q^{2}\right)} r^{3} + {\left(a^{3} m + 3 \, a m q^{2}\right)} r^{2} - {\left(a^{5} m + a^{3} m q^{2} - 2 \, a^{3} m^{2} r + a^{3} m r^{2}\right)} \cos\left({\theta}\right)^{2} - {\left(a^{3} q^{2} + a q^{4}\right)} r\right)} \sin\left({\theta}\right)^{2}}{a^{6} \cos\left({\theta}\right)^{6} + 3 \, a^{4} r^{2} \cos\left({\theta}\right)^{4} + 3 \, a^{2} r^{4} \cos\left({\theta}\right)^{2} + r^{6}} \\ \Gamma_{ \phantom{\, r} \, r \, r }^{ \, r \phantom{\, r} \phantom{\, r} } & = & -\frac{a^{2} m + q^{2} r - m r^{2} - {\left(a^{2} m - a^{2} r\right)} \sin\left({\theta}\right)^{2}}{2 \, m r^{3} - r^{4} - {\left(a^{2} + q^{2}\right)} r^{2} - {\left(a^{4} + a^{2} q^{2} - 2 \, a^{2} m r + a^{2} r^{2}\right)} \cos\left({\theta}\right)^{2}} \\ \Gamma_{ \phantom{\, r} \, r \, {\theta} }^{ \, r \phantom{\, r} \phantom{\, {\theta}} } & = & -\frac{a^{2} \cos\left({\theta}\right) \sin\left({\theta}\right)}{a^{2} \cos\left({\theta}\right)^{2} + r^{2}} \\ \Gamma_{ \phantom{\, r} \, {\theta} \, {\theta} }^{ \, r \phantom{\, {\theta}} \phantom{\, {\theta}} } & = & \frac{2 \, m r^{2} - r^{3} - {\left(a^{2} + q^{2}\right)} r}{a^{2} \cos\left({\theta}\right)^{2} + r^{2}} \\ \Gamma_{ \phantom{\, r} \, {\phi} \, {\phi} }^{ \, r \phantom{\, {\phi}} \phantom{\, {\phi}} } & = & \frac{{\left(a^{2} m r^{4} - {\left(2 \, a^{2} m^{2} + a^{2} q^{2}\right)} r^{3} + {\left(a^{4} m + 3 \, a^{2} m q^{2}\right)} r^{2} - {\left(a^{6} m + a^{4} m q^{2} - 2 \, a^{4} m^{2} r + a^{4} m r^{2}\right)} \cos\left({\theta}\right)^{2} - {\left(a^{4} q^{2} + a^{2} q^{4}\right)} r\right)} \sin\left({\theta}\right)^{4} + {\left(2 \, m r^{6} - r^{7} - {\left(a^{2} + q^{2}\right)} r^{5} + {\left(2 \, a^{4} m r^{2} - a^{4} r^{3} - {\left(a^{6} + a^{4} q^{2}\right)} r\right)} \cos\left({\theta}\right)^{4} + 2 \, {\left(2 \, a^{2} m r^{4} - a^{2} r^{5} - {\left(a^{4} + a^{2} q^{2}\right)} r^{3}\right)} \cos\left({\theta}\right)^{2}\right)} \sin\left({\theta}\right)^{2}}{a^{6} \cos\left({\theta}\right)^{6} + 3 \, a^{4} r^{2} \cos\left({\theta}\right)^{4} + 3 \, a^{2} r^{4} \cos\left({\theta}\right)^{2} + r^{6}} \\ \Gamma_{ \phantom{\, {\theta}} \, t \, t }^{ \, {\theta} \phantom{\, t} \phantom{\, t} } & = & \frac{{\left(a^{2} q^{2} - 2 \, a^{2} m r\right)} \cos\left({\theta}\right) \sin\left({\theta}\right)}{a^{6} \cos\left({\theta}\right)^{6} + 3 \, a^{4} r^{2} \cos\left({\theta}\right)^{4} + 3 \, a^{2} r^{4} \cos\left({\theta}\right)^{2} + r^{6}} \\ \Gamma_{ \phantom{\, {\theta}} \, t \, {\phi} }^{ \, {\theta} \phantom{\, t} \phantom{\, {\phi}} } & = & -\frac{{\left(a^{3} q^{2} - 2 \, a^{3} m r + a q^{2} r^{2} - 2 \, a m r^{3}\right)} \cos\left({\theta}\right) \sin\left({\theta}\right)}{a^{6} \cos\left({\theta}\right)^{6} + 3 \, a^{4} r^{2} \cos\left({\theta}\right)^{4} + 3 \, a^{2} r^{4} \cos\left({\theta}\right)^{2} + r^{6}} \\ \Gamma_{ \phantom{\, {\theta}} \, r \, r }^{ \, {\theta} \phantom{\, r} \phantom{\, r} } & = & -\frac{a^{2} \cos\left({\theta}\right) \sin\left({\theta}\right)}{2 \, m r^{3} - r^{4} - {\left(a^{2} + q^{2}\right)} r^{2} - {\left(a^{4} + a^{2} q^{2} - 2 \, a^{2} m r + a^{2} r^{2}\right)} \cos\left({\theta}\right)^{2}} \\ \Gamma_{ \phantom{\, {\theta}} \, r \, {\theta} }^{ \, {\theta} \phantom{\, r} \phantom{\, {\theta}} } & = & \frac{r}{a^{2} \cos\left({\theta}\right)^{2} + r^{2}} \\ \Gamma_{ \phantom{\, {\theta}} \, {\theta} \, {\theta} }^{ \, {\theta} \phantom{\, {\theta}} \phantom{\, {\theta}} } & = & -\frac{a^{2} \cos\left({\theta}\right) \sin\left({\theta}\right)}{a^{2} \cos\left({\theta}\right)^{2} + r^{2}} \\ \Gamma_{ \phantom{\, {\theta}} \, {\phi} \, {\phi} }^{ \, {\theta} \phantom{\, {\phi}} \phantom{\, {\phi}} } & = & -\frac{{\left({\left(a^{6} + a^{4} q^{2} - 2 \, a^{4} m r + a^{4} r^{2}\right)} \cos\left({\theta}\right)^{5} - 2 \, {\left(2 \, a^{2} m r^{3} - a^{2} r^{4} - {\left(a^{4} + a^{2} q^{2}\right)} r^{2}\right)} \cos\left({\theta}\right)^{3} - {\left(a^{4} q^{2} - 2 \, a^{4} m r + 2 \, a^{2} q^{2} r^{2} - 4 \, a^{2} m r^{3} - a^{2} r^{4} - r^{6}\right)} \cos\left({\theta}\right)\right)} \sin\left({\theta}\right)}{a^{6} \cos\left({\theta}\right)^{6} + 3 \, a^{4} r^{2} \cos\left({\theta}\right)^{4} + 3 \, a^{2} r^{4} \cos\left({\theta}\right)^{2} + r^{6}} \\ \Gamma_{ \phantom{\, {\phi}} \, t \, r }^{ \, {\phi} \phantom{\, t} \phantom{\, r} } & = & \frac{a^{3} m \cos\left({\theta}\right)^{2} + a q^{2} r - a m r^{2}}{2 \, m r^{5} - r^{6} - {\left(a^{2} + q^{2}\right)} r^{4} - {\left(a^{6} + a^{4} q^{2} - 2 \, a^{4} m r + a^{4} r^{2}\right)} \cos\left({\theta}\right)^{4} + 2 \, {\left(2 \, a^{2} m r^{3} - a^{2} r^{4} - {\left(a^{4} + a^{2} q^{2}\right)} r^{2}\right)} \cos\left({\theta}\right)^{2}} \\ \Gamma_{ \phantom{\, {\phi}} \, t \, {\theta} }^{ \, {\phi} \phantom{\, t} \phantom{\, {\theta}} } & = & \frac{{\left(a q^{2} - 2 \, a m r\right)} \cos\left({\theta}\right)}{{\left(a^{4} \cos\left({\theta}\right)^{4} + 2 \, a^{2} r^{2} \cos\left({\theta}\right)^{2} + r^{4}\right)} \sin\left({\theta}\right)} \\ \Gamma_{ \phantom{\, {\phi}} \, r \, {\phi} }^{ \, {\phi} \phantom{\, r} \phantom{\, {\phi}} } & = & -\frac{a^{2} q^{2} r - a^{2} m r^{2} + q^{2} r^{3} - 2 \, m r^{4} + r^{5} - {\left(a^{4} m - a^{4} r\right)} \cos\left({\theta}\right)^{4} + {\left(a^{4} m - a^{2} m r^{2} + 2 \, a^{2} r^{3}\right)} \cos\left({\theta}\right)^{2}}{2 \, m r^{5} - r^{6} - {\left(a^{2} + q^{2}\right)} r^{4} - {\left(a^{6} + a^{4} q^{2} - 2 \, a^{4} m r + a^{4} r^{2}\right)} \cos\left({\theta}\right)^{4} + 2 \, {\left(2 \, a^{2} m r^{3} - a^{2} r^{4} - {\left(a^{4} + a^{2} q^{2}\right)} r^{2}\right)} \cos\left({\theta}\right)^{2}} \\ \Gamma_{ \phantom{\, {\phi}} \, {\theta} \, {\phi} }^{ \, {\phi} \phantom{\, {\theta}} \phantom{\, {\phi}} } & = & \frac{a^{4} \cos\left({\theta}\right) \sin\left({\theta}\right)^{4} - {\left(2 \, a^{4} + a^{2} q^{2} - 2 \, a^{2} m r + 2 \, a^{2} r^{2}\right)} \cos\left({\theta}\right) \sin\left({\theta}\right)^{2} + {\left(a^{4} + 2 \, a^{2} r^{2} + r^{4}\right)} \cos\left({\theta}\right)}{{\left(a^{4} \cos\left({\theta}\right)^{4} + 2 \, a^{2} r^{2} \cos\left({\theta}\right)^{2} + r^{4}\right)} \sin\left({\theta}\right)} \end{array}\]

Killing vectors

The default vector frame on the spacetime manifold is the coordinate basis associated with Boyer-Lindquist coordinates:

In [24]:
M.default_frame() is BL.frame()
Out[24]:
\[\newcommand{\Bold}[1]{\mathbf{#1}}\mathrm{True}\]
In [25]:
BL.frame()
Out[25]:
\[\newcommand{\Bold}[1]{\mathbf{#1}}\left(\mathcal{M}, \left(\frac{\partial}{\partial t },\frac{\partial}{\partial r },\frac{\partial}{\partial {\theta} },\frac{\partial}{\partial {\phi} }\right)\right)\]

Let us consider the first vector field of this frame:

In [26]:
xi = BL.frame()[0] ; xi
Out[26]:
\[\newcommand{\Bold}[1]{\mathbf{#1}}\frac{\partial}{\partial t }\]
In [27]:
print(xi)
Vector field ∂/∂t on the 4-dimensional Lorentzian manifold M

The 1-form associated to it by metric duality is

In [28]:
xi_form = xi.down(g)
xi_form.display()
Out[28]:
\[\newcommand{\Bold}[1]{\mathbf{#1}}\left( -\frac{q^{2} - 2 \, m r}{a^{2} \cos\left({\theta}\right)^{2} + r^{2}} - 1 \right) \mathrm{d} t + \left( \frac{{\left(q^{2} - 2 \, m r\right)} a \sin\left({\theta}\right)^{2}}{a^{2} \cos\left({\theta}\right)^{2} + r^{2}} \right) \mathrm{d} {\phi}\]

Its covariant derivative is

In [29]:
nab_xi = nabla(xi_form)
print(nab_xi)
nab_xi.display()
Tensor field of type (0,2) on the 4-dimensional Lorentzian manifold M
Out[29]:
\[\newcommand{\Bold}[1]{\mathbf{#1}}\left( \frac{m r^{4} - {\left(2 \, m^{2} + q^{2}\right)} r^{3} + {\left(a^{2} m + 3 \, m q^{2}\right)} r^{2} - {\left(a^{4} m + a^{2} m q^{2} - 2 \, a^{2} m^{2} r + a^{2} m r^{2}\right)} \cos\left({\theta}\right)^{2} - {\left(a^{2} q^{2} + q^{4}\right)} r}{2 \, m r^{5} - r^{6} - {\left(a^{2} + q^{2}\right)} r^{4} - {\left(a^{6} + a^{4} q^{2} - 2 \, a^{4} m r + a^{4} r^{2}\right)} \cos\left({\theta}\right)^{4} + 2 \, {\left(2 \, a^{2} m r^{3} - a^{2} r^{4} - {\left(a^{4} + a^{2} q^{2}\right)} r^{2}\right)} \cos\left({\theta}\right)^{2}} \right) \mathrm{d} t\otimes \mathrm{d} r + \left( -\frac{{\left(a^{2} q^{2} - 2 \, a^{2} m r\right)} \cos\left({\theta}\right) \sin\left({\theta}\right)}{a^{4} \cos\left({\theta}\right)^{4} + 2 \, a^{2} r^{2} \cos\left({\theta}\right)^{2} + r^{4}} \right) \mathrm{d} t\otimes \mathrm{d} {\theta} + \left( -\frac{m r^{4} - {\left(2 \, m^{2} + q^{2}\right)} r^{3} + {\left(a^{2} m + 3 \, m q^{2}\right)} r^{2} - {\left(a^{4} m + a^{2} m q^{2} - 2 \, a^{2} m^{2} r + a^{2} m r^{2}\right)} \cos\left({\theta}\right)^{2} - {\left(a^{2} q^{2} + q^{4}\right)} r}{2 \, m r^{5} - r^{6} - {\left(a^{2} + q^{2}\right)} r^{4} - {\left(a^{6} + a^{4} q^{2} - 2 \, a^{4} m r + a^{4} r^{2}\right)} \cos\left({\theta}\right)^{4} + 2 \, {\left(2 \, a^{2} m r^{3} - a^{2} r^{4} - {\left(a^{4} + a^{2} q^{2}\right)} r^{2}\right)} \cos\left({\theta}\right)^{2}} \right) \mathrm{d} r\otimes \mathrm{d} t + \left( -\frac{a^{3} m \cos\left({\theta}\right)^{4} - a q^{2} r + a m r^{2} - {\left(a^{3} m - a q^{2} r + a m r^{2}\right)} \cos\left({\theta}\right)^{2}}{a^{4} \cos\left({\theta}\right)^{4} + 2 \, a^{2} r^{2} \cos\left({\theta}\right)^{2} + r^{4}} \right) \mathrm{d} r\otimes \mathrm{d} {\phi} + \left( \frac{{\left(a^{2} q^{2} - 2 \, a^{2} m r\right)} \cos\left({\theta}\right) \sin\left({\theta}\right)}{a^{4} \cos\left({\theta}\right)^{4} + 2 \, a^{2} r^{2} \cos\left({\theta}\right)^{2} + r^{4}} \right) \mathrm{d} {\theta}\otimes \mathrm{d} t + \left( -\frac{{\left(a^{3} q^{2} - 2 \, a^{3} m r + a q^{2} r^{2} - 2 \, a m r^{3}\right)} \cos\left({\theta}\right) \sin\left({\theta}\right)}{a^{4} \cos\left({\theta}\right)^{4} + 2 \, a^{2} r^{2} \cos\left({\theta}\right)^{2} + r^{4}} \right) \mathrm{d} {\theta}\otimes \mathrm{d} {\phi} + \left( \frac{a^{3} m \cos\left({\theta}\right)^{4} - a q^{2} r + a m r^{2} - {\left(a^{3} m - a q^{2} r + a m r^{2}\right)} \cos\left({\theta}\right)^{2}}{a^{4} \cos\left({\theta}\right)^{4} + 2 \, a^{2} r^{2} \cos\left({\theta}\right)^{2} + r^{4}} \right) \mathrm{d} {\phi}\otimes \mathrm{d} r + \left( \frac{{\left(a^{3} q^{2} - 2 \, a^{3} m r + a q^{2} r^{2} - 2 \, a m r^{3}\right)} \cos\left({\theta}\right) \sin\left({\theta}\right)}{a^{4} \cos\left({\theta}\right)^{4} + 2 \, a^{2} r^{2} \cos\left({\theta}\right)^{2} + r^{4}} \right) \mathrm{d} {\phi}\otimes \mathrm{d} {\theta}\]

Let us check that the vector field $\xi=\frac{\partial}{\partial t}$ obeys Killing equation:

In [30]:
nab_xi.symmetrize() == 0
Out[30]:
\[\newcommand{\Bold}[1]{\mathbf{#1}}\mathrm{True}\]

Similarly, let us check that $\chi := \frac{\partial}{\partial\phi}$ is a Killing vector:

In [31]:
chi = BL.frame()[3] ; chi
Out[31]:
\[\newcommand{\Bold}[1]{\mathbf{#1}}\frac{\partial}{\partial {\phi} }\]
In [32]:
nabla(chi.down(g)).symmetrize() == 0
Out[32]:
\[\newcommand{\Bold}[1]{\mathbf{#1}}\mathrm{True}\]

Another way to check that $\xi$ and $\chi$ are Killing vectors is the vanishing of the Lie derivative of the metric tensor along them:

In [33]:
g.lie_derivative(xi) == 0
Out[33]:
\[\newcommand{\Bold}[1]{\mathbf{#1}}\mathrm{True}\]
In [34]:
g.lie_derivative(chi) == 0
Out[34]:
\[\newcommand{\Bold}[1]{\mathbf{#1}}\mathrm{True}\]

Curvature

The Ricci tensor associated with $g$:

In [35]:
Ric = g.ricci()
print(Ric)
Field of symmetric bilinear forms Ric(g) on the 4-dimensional Lorentzian manifold M
In [36]:
Ric.display()
Out[36]:
\[\newcommand{\Bold}[1]{\mathbf{#1}}\mathrm{Ric}\left(g\right) = \left( -\frac{a^{2} q^{2} \cos\left({\theta}\right)^{2} - 2 \, a^{2} q^{2} - q^{4} + 2 \, m q^{2} r - q^{2} r^{2}}{a^{6} \cos\left({\theta}\right)^{6} + 3 \, a^{4} r^{2} \cos\left({\theta}\right)^{4} + 3 \, a^{2} r^{4} \cos\left({\theta}\right)^{2} + r^{6}} \right) \mathrm{d} t\otimes \mathrm{d} t + \left( \frac{{\left(2 \, a^{5} q^{2} + a^{3} q^{4} - 2 \, a^{3} m q^{2} r + 2 \, a^{3} q^{2} r^{2}\right)} \sin\left({\theta}\right)^{4} - {\left(2 \, a^{5} q^{2} + a^{3} q^{4} - 2 \, a^{3} m q^{2} r - 2 \, a m q^{2} r^{3} + 2 \, a q^{2} r^{4} + {\left(4 \, a^{3} q^{2} + a q^{4}\right)} r^{2}\right)} \sin\left({\theta}\right)^{2}}{a^{8} \cos\left({\theta}\right)^{8} + 4 \, a^{6} r^{2} \cos\left({\theta}\right)^{6} + 6 \, a^{4} r^{4} \cos\left({\theta}\right)^{4} + 4 \, a^{2} r^{6} \cos\left({\theta}\right)^{2} + r^{8}} \right) \mathrm{d} t\otimes \mathrm{d} {\phi} + \left( \frac{q^{2}}{2 \, m r^{3} - r^{4} - {\left(a^{2} + q^{2}\right)} r^{2} - {\left(a^{4} + a^{2} q^{2} - 2 \, a^{2} m r + a^{2} r^{2}\right)} \cos\left({\theta}\right)^{2}} \right) \mathrm{d} r\otimes \mathrm{d} r + \left( \frac{q^{2}}{a^{2} \cos\left({\theta}\right)^{2} + r^{2}} \right) \mathrm{d} {\theta}\otimes \mathrm{d} {\theta} + \left( \frac{{\left(2 \, a^{5} q^{2} + a^{3} q^{4} - 2 \, a^{3} m q^{2} r + 2 \, a^{3} q^{2} r^{2}\right)} \sin\left({\theta}\right)^{4} - {\left(2 \, a^{5} q^{2} + a^{3} q^{4} - 2 \, a^{3} m q^{2} r - 2 \, a m q^{2} r^{3} + 2 \, a q^{2} r^{4} + {\left(4 \, a^{3} q^{2} + a q^{4}\right)} r^{2}\right)} \sin\left({\theta}\right)^{2}}{a^{8} \cos\left({\theta}\right)^{8} + 4 \, a^{6} r^{2} \cos\left({\theta}\right)^{6} + 6 \, a^{4} r^{4} \cos\left({\theta}\right)^{4} + 4 \, a^{2} r^{6} \cos\left({\theta}\right)^{2} + r^{8}} \right) \mathrm{d} {\phi}\otimes \mathrm{d} t + \left( -\frac{{\left(a^{6} q^{2} + a^{4} q^{4} - 2 \, a^{4} m q^{2} r + a^{4} q^{2} r^{2}\right)} \sin\left({\theta}\right)^{6} - {\left(a^{4} q^{4} - 2 \, a^{4} m q^{2} r + a^{2} q^{4} r^{2} - 2 \, a^{2} m q^{2} r^{3}\right)} \sin\left({\theta}\right)^{4} - {\left(a^{6} q^{2} + 3 \, a^{4} q^{2} r^{2} + 3 \, a^{2} q^{2} r^{4} + q^{2} r^{6}\right)} \sin\left({\theta}\right)^{2}}{a^{8} \cos\left({\theta}\right)^{8} + 4 \, a^{6} r^{2} \cos\left({\theta}\right)^{6} + 6 \, a^{4} r^{4} \cos\left({\theta}\right)^{4} + 4 \, a^{2} r^{6} \cos\left({\theta}\right)^{2} + r^{8}} \right) \mathrm{d} {\phi}\otimes \mathrm{d} {\phi}\]

A matric view of the components:

In [37]:
Ric[:]
Out[37]:
\[\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rrrr} -\frac{a^{2} q^{2} \cos\left({\theta}\right)^{2} - 2 \, a^{2} q^{2} - q^{4} + 2 \, m q^{2} r - q^{2} r^{2}}{a^{6} \cos\left({\theta}\right)^{6} + 3 \, a^{4} r^{2} \cos\left({\theta}\right)^{4} + 3 \, a^{2} r^{4} \cos\left({\theta}\right)^{2} + r^{6}} & 0 & 0 & \frac{{\left(2 \, a^{5} q^{2} + a^{3} q^{4} - 2 \, a^{3} m q^{2} r + 2 \, a^{3} q^{2} r^{2}\right)} \sin\left({\theta}\right)^{4} - {\left(2 \, a^{5} q^{2} + a^{3} q^{4} - 2 \, a^{3} m q^{2} r - 2 \, a m q^{2} r^{3} + 2 \, a q^{2} r^{4} + {\left(4 \, a^{3} q^{2} + a q^{4}\right)} r^{2}\right)} \sin\left({\theta}\right)^{2}}{a^{8} \cos\left({\theta}\right)^{8} + 4 \, a^{6} r^{2} \cos\left({\theta}\right)^{6} + 6 \, a^{4} r^{4} \cos\left({\theta}\right)^{4} + 4 \, a^{2} r^{6} \cos\left({\theta}\right)^{2} + r^{8}} \\ 0 & \frac{q^{2}}{2 \, m r^{3} - r^{4} - {\left(a^{2} + q^{2}\right)} r^{2} - {\left(a^{4} + a^{2} q^{2} - 2 \, a^{2} m r + a^{2} r^{2}\right)} \cos\left({\theta}\right)^{2}} & 0 & 0 \\ 0 & 0 & \frac{q^{2}}{a^{2} \cos\left({\theta}\right)^{2} + r^{2}} & 0 \\ \frac{{\left(2 \, a^{5} q^{2} + a^{3} q^{4} - 2 \, a^{3} m q^{2} r + 2 \, a^{3} q^{2} r^{2}\right)} \sin\left({\theta}\right)^{4} - {\left(2 \, a^{5} q^{2} + a^{3} q^{4} - 2 \, a^{3} m q^{2} r - 2 \, a m q^{2} r^{3} + 2 \, a q^{2} r^{4} + {\left(4 \, a^{3} q^{2} + a q^{4}\right)} r^{2}\right)} \sin\left({\theta}\right)^{2}}{a^{8} \cos\left({\theta}\right)^{8} + 4 \, a^{6} r^{2} \cos\left({\theta}\right)^{6} + 6 \, a^{4} r^{4} \cos\left({\theta}\right)^{4} + 4 \, a^{2} r^{6} \cos\left({\theta}\right)^{2} + r^{8}} & 0 & 0 & -\frac{{\left(a^{6} q^{2} + a^{4} q^{4} - 2 \, a^{4} m q^{2} r + a^{4} q^{2} r^{2}\right)} \sin\left({\theta}\right)^{6} - {\left(a^{4} q^{4} - 2 \, a^{4} m q^{2} r + a^{2} q^{4} r^{2} - 2 \, a^{2} m q^{2} r^{3}\right)} \sin\left({\theta}\right)^{4} - {\left(a^{6} q^{2} + 3 \, a^{4} q^{2} r^{2} + 3 \, a^{2} q^{2} r^{4} + q^{2} r^{6}\right)} \sin\left({\theta}\right)^{2}}{a^{8} \cos\left({\theta}\right)^{8} + 4 \, a^{6} r^{2} \cos\left({\theta}\right)^{6} + 6 \, a^{4} r^{4} \cos\left({\theta}\right)^{4} + 4 \, a^{2} r^{6} \cos\left({\theta}\right)^{2} + r^{8}} \end{array}\right)\]

Let us check that in the Kerr case, i.e. when $q=0$, the Ricci tensor is zero:

In [38]:
Ric_Kerr = Ric.copy()
Ric_Kerr.apply_map(lambda f: f.subs({q: 0}))
Ric_Kerr[:]
Out[38]:
\[\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rrrr} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right)\]

The Riemann curvature tensor associated with $g$:

In [39]:
R = g.riemann()
print(R)
Tensor field Riem(g) of type (1,3) on the 4-dimensional Lorentzian manifold M

The component $R^0_{\ \, 101}$ of the Riemann tensor is

In [40]:
R[0,1,0,1]
Out[40]:
\[\newcommand{\Bold}[1]{\mathbf{#1}}\frac{4 \, a^{2} q^{2} r^{2} - 3 \, a^{2} m r^{3} + 3 \, q^{2} r^{4} - 2 \, m r^{5} + {\left(a^{4} q^{2} - 3 \, a^{4} m r\right)} \cos\left({\theta}\right)^{4} - {\left(2 \, a^{4} q^{2} - 9 \, a^{4} m r + 2 \, a^{2} q^{2} r^{2} - 7 \, a^{2} m r^{3}\right)} \cos\left({\theta}\right)^{2}}{2 \, m r^{7} - r^{8} - {\left(a^{2} + q^{2}\right)} r^{6} - {\left(a^{8} + a^{6} q^{2} - 2 \, a^{6} m r + a^{6} r^{2}\right)} \cos\left({\theta}\right)^{6} + 3 \, {\left(2 \, a^{4} m r^{3} - a^{4} r^{4} - {\left(a^{6} + a^{4} q^{2}\right)} r^{2}\right)} \cos\left({\theta}\right)^{4} + 3 \, {\left(2 \, a^{2} m r^{5} - a^{2} r^{6} - {\left(a^{4} + a^{2} q^{2}\right)} r^{4}\right)} \cos\left({\theta}\right)^{2}}\]

The expression in the uncharged limit (Kerr spacetime) is

In [41]:
R[0,1,0,1].expr().subs(q=0).simplify_rational()
Out[41]:
\[\newcommand{\Bold}[1]{\mathbf{#1}}\frac{3 \, a^{4} m r \cos\left({\theta}\right)^{4} + 3 \, a^{2} m r^{3} + 2 \, m r^{5} - {\left(9 \, a^{4} m r + 7 \, a^{2} m r^{3}\right)} \cos\left({\theta}\right)^{2}}{a^{2} r^{6} - 2 \, m r^{7} + r^{8} + {\left(a^{8} - 2 \, a^{6} m r + a^{6} r^{2}\right)} \cos\left({\theta}\right)^{6} + 3 \, {\left(a^{6} r^{2} - 2 \, a^{4} m r^{3} + a^{4} r^{4}\right)} \cos\left({\theta}\right)^{4} + 3 \, {\left(a^{4} r^{4} - 2 \, a^{2} m r^{5} + a^{2} r^{6}\right)} \cos\left({\theta}\right)^{2}}\]

while in the non-rotating limit (Reissner-Nordström spacetime), it is

In [42]:
R[0,1,0,1].expr().subs(a=0).simplify_rational()
Out[42]:
\[\newcommand{\Bold}[1]{\mathbf{#1}}-\frac{3 \, q^{2} - 2 \, m r}{q^{2} r^{2} - 2 \, m r^{3} + r^{4}}\]

In the Schwarzschild limit, it reduces to

In [43]:
R[0,1,0,1].expr().subs(a=0, q=0).simplify_rational()
Out[43]:
\[\newcommand{\Bold}[1]{\mathbf{#1}}-\frac{2 \, m}{2 \, m r^{2} - r^{3}}\]

Obviously, it vanishes in the flat space limit:

In [44]:
R[0,1,0,1].expr().subs(m=0, a=0, q=0)
Out[44]:
\[\newcommand{\Bold}[1]{\mathbf{#1}}0\]

Bianchi identity

Let us check the Bianchi identity $\nabla_p R^i_{\ \, j kl} + \nabla_k R^i_{\ \, jlp} + \nabla_l R^i_{\ \, jpk} = 0$:

In [45]:
DR = nabla(R)   # long (takes a while)
print(DR)  
Tensor field nabla_g(Riem(g)) of type (1,4) on the 4-dimensional Lorentzian manifold M
In [46]:
for i in M.irange():
    for j in M.irange():
        for k in M.irange():
            for l in M.irange():
                for p in M.irange():
                    print(DR[i,j,k,l,p] + DR[i,j,l,p,k] + DR[i,j,p,k,l], end=' ')
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 

If the last sign in the Bianchi identity is changed to minus, the identity does no longer hold:

In [47]:
DR[0,1,2,3,1] + DR[0,1,3,1,2] + DR[0,1,1,2,3]  # should be zero (Bianchi identity)
Out[47]:
\[\newcommand{\Bold}[1]{\mathbf{#1}}0\]
In [48]:
DR[0,1,2,3,1] + DR[0,1,3,1,2] - DR[0,1,1,2,3]  # note the change of the second + to -
Out[48]:
\[\newcommand{\Bold}[1]{\mathbf{#1}}-\frac{4 \, {\left({\left(a^{5} q^{2} - 6 \, a^{5} m r + a^{3} q^{2} r^{2} - 6 \, a^{3} m r^{3}\right)} \cos\left({\theta}\right)^{3} - {\left(5 \, a^{3} q^{2} r^{2} - 6 \, a^{3} m r^{3} + 5 \, a q^{2} r^{4} - 6 \, a m r^{5}\right)} \cos\left({\theta}\right)\right)} \sin\left({\theta}\right)}{2 \, m r^{7} - r^{8} - {\left(a^{2} + q^{2}\right)} r^{6} - {\left(a^{8} + a^{6} q^{2} - 2 \, a^{6} m r + a^{6} r^{2}\right)} \cos\left({\theta}\right)^{6} + 3 \, {\left(2 \, a^{4} m r^{3} - a^{4} r^{4} - {\left(a^{6} + a^{4} q^{2}\right)} r^{2}\right)} \cos\left({\theta}\right)^{4} + 3 \, {\left(2 \, a^{2} m r^{5} - a^{2} r^{6} - {\left(a^{4} + a^{2} q^{2}\right)} r^{4}\right)} \cos\left({\theta}\right)^{2}}\]

Ricci scalar

The Ricci scalar $R = g^{ab} R_{ab}$ of the Kerr-Newman spacetime vanishes identically:

In [49]:
g.ricci_scalar().display()
Out[49]:
\[\newcommand{\Bold}[1]{\mathbf{#1}}\begin{array}{llcl} \mathrm{r}\left(g\right):& \mathcal{M} & \longrightarrow & \mathbb{R} \\ & \left(t, r, {\theta}, {\phi}\right) & \longmapsto & 0 \end{array}\]

Einstein equation

The Einstein tensor is

In [50]:
G = Ric - 1/2*g.ricci_scalar()*g
print(G)
Field of symmetric bilinear forms Ric(g)-unnamed metric on the 4-dimensional Lorentzian manifold M

Since the Ricci scalar is zero, the Einstein tensor reduces to the Ricci tensor:

In [51]:
G == Ric
Out[51]:
\[\newcommand{\Bold}[1]{\mathbf{#1}}\mathrm{True}\]

The invariant $F_{ab} F^{ab}$ of the electromagnetic field:

In [52]:
Fuu = F.up(g)
F2 = F['_ab']*Fuu['^ab']
print(F2)
Scalar field on the 4-dimensional Lorentzian manifold M
In [53]:
F2.display()
Out[53]:
\[\newcommand{\Bold}[1]{\mathbf{#1}}\begin{array}{llcl} & \mathcal{M} & \longrightarrow & \mathbb{R} \\ & \left(t, r, {\theta}, {\phi}\right) & \longmapsto & -\frac{2 \, {\left(a^{4} q^{2} \cos\left({\theta}\right)^{4} - 6 \, a^{2} q^{2} r^{2} \cos\left({\theta}\right)^{2} + q^{2} r^{4}\right)}}{a^{8} \cos\left({\theta}\right)^{8} + 4 \, a^{6} r^{2} \cos\left({\theta}\right)^{6} + 6 \, a^{4} r^{4} \cos\left({\theta}\right)^{4} + 4 \, a^{2} r^{6} \cos\left({\theta}\right)^{2} + r^{8}} \end{array}\]

The energy-momentum tensor of the electromagnetic field:

In [54]:
Fud = F.up(g,0)
T = 1/(4*pi)*( F['_k.']*Fud['^k_.'] - 1/4*F2 * g )
print(T)
Tensor field of type (0,2) on the 4-dimensional Lorentzian manifold M
In [55]:
T[:]
Out[55]:
\[\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rrrr} -\frac{a^{2} q^{2} \cos\left({\theta}\right)^{2} - 2 \, a^{2} q^{2} - q^{4} + 2 \, m q^{2} r - q^{2} r^{2}}{8 \, {\left(\pi a^{6} \cos\left({\theta}\right)^{6} + 3 \, \pi a^{4} r^{2} \cos\left({\theta}\right)^{4} + 3 \, \pi a^{2} r^{4} \cos\left({\theta}\right)^{2} + \pi r^{6}\right)}} & 0 & 0 & -\frac{{\left(2 \, a^{3} q^{2} + a q^{4} - 2 \, a m q^{2} r + 2 \, a q^{2} r^{2}\right)} \sin\left({\theta}\right)^{2}}{8 \, {\left(\pi a^{6} \cos\left({\theta}\right)^{6} + 3 \, \pi a^{4} r^{2} \cos\left({\theta}\right)^{4} + 3 \, \pi a^{2} r^{4} \cos\left({\theta}\right)^{2} + \pi r^{6}\right)}} \\ 0 & \frac{q^{2}}{8 \, {\left(2 \, \pi m r^{3} - \pi r^{4} - {\left(\pi a^{2} + \pi q^{2}\right)} r^{2} - {\left(\pi a^{4} + \pi a^{2} q^{2} - 2 \, \pi a^{2} m r + \pi a^{2} r^{2}\right)} \cos\left({\theta}\right)^{2}\right)}} & 0 & 0 \\ 0 & 0 & \frac{q^{2}}{8 \, {\left(\pi a^{2} \cos\left({\theta}\right)^{2} + \pi r^{2}\right)}} & 0 \\ -\frac{{\left(2 \, a^{3} q^{2} + a q^{4} - 2 \, a m q^{2} r + 2 \, a q^{2} r^{2}\right)} \sin\left({\theta}\right)^{2}}{8 \, {\left(\pi a^{6} \cos\left({\theta}\right)^{6} + 3 \, \pi a^{4} r^{2} \cos\left({\theta}\right)^{4} + 3 \, \pi a^{2} r^{4} \cos\left({\theta}\right)^{2} + \pi r^{6}\right)}} & 0 & 0 & -\frac{{\left(2 \, a^{2} m q^{2} r^{5} - 2 \, a^{2} q^{2} r^{6} - {\left(2 \, a^{4} q^{2} + a^{2} q^{4}\right)} r^{4} - {\left(2 \, a^{8} q^{2} + a^{6} q^{4} - 2 \, a^{6} m q^{2} r + 2 \, a^{6} q^{2} r^{2}\right)} \cos\left({\theta}\right)^{4} + 2 \, {\left(2 \, a^{4} m q^{2} r^{3} + 2 \, a^{4} q^{2} r^{4} + {\left(2 \, a^{6} q^{2} - a^{4} q^{4}\right)} r^{2}\right)} \cos\left({\theta}\right)^{2}\right)} \sin\left({\theta}\right)^{4} - {\left(a^{2} q^{2} r^{6} + q^{2} r^{8} + {\left(a^{8} q^{2} + a^{6} q^{2} r^{2}\right)} \cos\left({\theta}\right)^{6} - 5 \, {\left(a^{6} q^{2} r^{2} + a^{4} q^{2} r^{4}\right)} \cos\left({\theta}\right)^{4} + {\left(8 \, a^{6} q^{2} r^{2} + 11 \, a^{4} q^{2} r^{4} + 3 \, a^{2} q^{2} r^{6}\right)} \cos\left({\theta}\right)^{2}\right)} \sin\left({\theta}\right)^{2}}{8 \, {\left(\pi a^{10} \cos\left({\theta}\right)^{10} + 5 \, \pi a^{8} r^{2} \cos\left({\theta}\right)^{8} + 10 \, \pi a^{6} r^{4} \cos\left({\theta}\right)^{6} + 10 \, \pi a^{4} r^{6} \cos\left({\theta}\right)^{4} + 5 \, \pi a^{2} r^{8} \cos\left({\theta}\right)^{2} + \pi r^{10}\right)}} \end{array}\right)\]

Check that the Einstein equation is satisfied:

In [56]:
G == 8*pi*T
Out[56]:
\[\newcommand{\Bold}[1]{\mathbf{#1}}\mathrm{True}\]

Kretschmann scalar

The tensor $R^\flat$, of components $R_{abcd} = g_{am} R^m_{\ \, bcd}$:

In [57]:
dR = R.down(g)
print(dR)
Tensor field of type (0,4) on the 4-dimensional Lorentzian manifold M

The tensor $R^\sharp$, of components $R^{abcd} = g^{bp} g^{cq} g^{dr} R^a_{\ \, pqr}$:

In [58]:
uR = R.up(g)
print(uR)
Tensor field of type (4,0) on the 4-dimensional Lorentzian manifold M

The Kretschmann scalar $K := R^{abcd} R_{abcd}$:

In [59]:
Kr_scalar = uR['^abcd']*dR['_abcd']
Kr_scalar.display()
Out[59]:
\[\newcommand{\Bold}[1]{\mathbf{#1}}\begin{array}{llcl} & \mathcal{M} & \longrightarrow & \mathbb{R} \\ & \left(t, r, {\theta}, {\phi}\right) & \longmapsto & -\frac{8 \, {\left(6 \, m^{2} r^{8} - 12 \, {\left(m^{3} + m q^{2}\right)} r^{7} + {\left(6 \, a^{2} m^{2} + 30 \, m^{2} q^{2} + 7 \, q^{4}\right)} r^{6} - 6 \, {\left(a^{8} m^{2} + a^{6} m^{2} q^{2} - 2 \, a^{6} m^{3} r + a^{6} m^{2} r^{2}\right)} \cos\left({\theta}\right)^{6} - 2 \, {\left(6 \, a^{2} m q^{2} + 13 \, m q^{4}\right)} r^{5} + 7 \, {\left(a^{2} q^{4} + q^{6}\right)} r^{4} + {\left(7 \, a^{6} q^{4} + 7 \, a^{4} q^{6} + 90 \, a^{4} m^{2} r^{4} - 60 \, {\left(3 \, a^{4} m^{3} + a^{4} m q^{2}\right)} r^{3} + {\left(90 \, a^{6} m^{2} + 210 \, a^{4} m^{2} q^{2} + 7 \, a^{4} q^{4}\right)} r^{2} - 2 \, {\left(30 \, a^{6} m q^{2} + 37 \, a^{4} m q^{4}\right)} r\right)} \cos\left({\theta}\right)^{4} - 2 \, {\left(45 \, a^{2} m^{2} r^{6} - 30 \, {\left(3 \, a^{2} m^{3} + 2 \, a^{2} m q^{2}\right)} r^{5} + {\left(45 \, a^{4} m^{2} + 165 \, a^{2} m^{2} q^{2} + 17 \, a^{2} q^{4}\right)} r^{4} - 2 \, {\left(30 \, a^{4} m q^{2} + 47 \, a^{2} m q^{4}\right)} r^{3} + 17 \, {\left(a^{4} q^{4} + a^{2} q^{6}\right)} r^{2}\right)} \cos\left({\theta}\right)^{2}\right)}}{2 \, m r^{13} - r^{14} - {\left(a^{2} + q^{2}\right)} r^{12} - {\left(a^{14} + a^{12} q^{2} - 2 \, a^{12} m r + a^{12} r^{2}\right)} \cos\left({\theta}\right)^{12} + 6 \, {\left(2 \, a^{10} m r^{3} - a^{10} r^{4} - {\left(a^{12} + a^{10} q^{2}\right)} r^{2}\right)} \cos\left({\theta}\right)^{10} + 15 \, {\left(2 \, a^{8} m r^{5} - a^{8} r^{6} - {\left(a^{10} + a^{8} q^{2}\right)} r^{4}\right)} \cos\left({\theta}\right)^{8} + 20 \, {\left(2 \, a^{6} m r^{7} - a^{6} r^{8} - {\left(a^{8} + a^{6} q^{2}\right)} r^{6}\right)} \cos\left({\theta}\right)^{6} + 15 \, {\left(2 \, a^{4} m r^{9} - a^{4} r^{10} - {\left(a^{6} + a^{4} q^{2}\right)} r^{8}\right)} \cos\left({\theta}\right)^{4} + 6 \, {\left(2 \, a^{2} m r^{11} - a^{2} r^{12} - {\left(a^{4} + a^{2} q^{2}\right)} r^{10}\right)} \cos\left({\theta}\right)^{2}} \end{array}\]

A variant of this expression can be obtained by invoking the factor() method on the coordinate function representing the scalar field in the manifold's default chart:

In [60]:
Kr = Kr_scalar.coord_function()
Kr.factor()
Out[60]:
\[\newcommand{\Bold}[1]{\mathbf{#1}}-\frac{8 \, {\left(6 \, a^{6} m^{2} \cos\left({\theta}\right)^{6} - 7 \, a^{4} q^{4} \cos\left({\theta}\right)^{4} + 60 \, a^{4} m q^{2} r \cos\left({\theta}\right)^{4} - 90 \, a^{4} m^{2} r^{2} \cos\left({\theta}\right)^{4} + 34 \, a^{2} q^{4} r^{2} \cos\left({\theta}\right)^{2} - 120 \, a^{2} m q^{2} r^{3} \cos\left({\theta}\right)^{2} + 90 \, a^{2} m^{2} r^{4} \cos\left({\theta}\right)^{2} - 7 \, q^{4} r^{4} + 12 \, m q^{2} r^{5} - 6 \, m^{2} r^{6}\right)}}{{\left(a^{2} \cos\left({\theta}\right)^{2} + r^{2}\right)}^{6}}\]

As a check, we can compare Kr to the formula given by R. Conn Henry, Astrophys. J. 535, 350 (2000):

In [61]:
Kr == 8/(r^2+(a*cos(th))^2)^6 *( 
          6*m^2*(r^6 - 15*r^4*(a*cos(th))^2 + 15*r^2*(a*cos(th))^4 - (a*cos(th))^6) 
        - 12*m*q^2*r*(r^4 - 10*(a*r*cos(th))^2 + 5*(a*cos(th))^4) 
        + q^4*(7*r^4 - 34*(a*r*cos(th))^2 + 7*(a*cos(th))^4) )
Out[61]:
\[\newcommand{\Bold}[1]{\mathbf{#1}}\mathrm{True}\]

The Schwarzschild value of the Kretschmann scalar is recovered by setting $a=0$ and $q=0$:

In [62]:
Kr.expr().subs(a=0, q=0)
Out[62]:
\[\newcommand{\Bold}[1]{\mathbf{#1}}\frac{48 \, m^{2}}{r^{6}}\]

Let us plot the Kretschmann scalar for $m=1$, $a=0.9$ and $q=0.5$:

In [63]:
K1 = Kr.expr().subs(m=1, a=0.9, q=0.5)
plot3d(K1, (r,1,3), (th, 0, pi), axes_labels=['r', 'theta', 'Kr'])
Out[63]:
In [64]:
print("Total elapsed time: {} s".format(time.perf_counter() - comput_time0))
Total elapsed time: 776.7964824969999 s