This Jupyter/SageMath worksheet is relative to the lectures Introduction to black hole physics
These computations are based on SageManifolds (version 1.0, as included in SageMath 7.5)
Click here to download the worksheet file (ipynb format). To run it, you must start SageMath with the Jupyter notebook, with the command sage -n jupyter
NB: a version of SageMath at least equal to 7.5 is required to run this worksheet:
version()
First we set up the notebook to display mathematical objects using LaTeX rendering:
%display latex
We declare the spacetime manifold $M$:
M = Manifold(4, 'M')
print(M)
and declare the chart of Lemaître synchronous coordinates on it:
X.<t,x,th,ph> = M.chart(r't:\tau x:(0,+oo):\chi th:(0,pi):\theta ph:(0,2*pi):\phi')
X
The most general metric tensor, assuming spherical symmetry and synchronous coordinates:
g = M.lorentzian_metric('g')
a = function('a')
r = function('r')
g[0,0] = -1
g[1,1] = a(t,x)^2
g[2,2] = r(t,x)^2
g[3,3] = (r(t,x)*sin(th))^2
g.display()
The cosmological constant:
var('Lamb', latex_name='\Lambda')
The Ricci tensor:
Ric = g.ricci()
print(Ric)
Ric.display()
The Einstein tensor:
G = Ric - 1/2*g.ricci_scalar() * g
G.set_name('G')
print(G)
G.display_comp()
Let us consider a pressureless fluid ("dust"). Moreover, we assume that the coordinates $(\tau,\chi,\theta,\phi)$ are comoving, i.e. that the fluid 4-velocity is equal to $\partial_\tau$:
u = M.vector_field('u')
u[0] = 1
u.display()
Since $(\tau,\chi,\theta,\chi)$ are synchronous, the above does define a unit timelike vector:
g(u,u).display()
The 1-form associated to the fluid 4-velocity by metric duality:
u_form = u.down(g)
print(u_form)
u_form.display()
The pressureless energy-momentum tensor:
rho = function('rho')
T = rho(t,x)*(u_form * u_form)
T.set_name('T')
print(T)
T.display()
The Einstein equation:
E = G + Lamb*g - 8*pi*T
E.set_name('E')
print(E)
E.display_comp()
Let us first consider the $01 = \tau\chi$ component of the Einstein equation:
E[0,1]
A slight rearrangement of the equation:
eq = E[0,1]*r(t,x)/(-2*a(t,x))
eq
We see that this equation is equivalent to $$ \frac{\partial}{\partial\tau} \left( \frac{1}{a}\frac{\partial r}{\partial\chi} \right) = 0 $$ since
drdx = diff(r(t,x), x)
eq - diff(drdx/a(t,x), t)
Hence there exists a function of $\chi$ only, $f(\chi)$ say, such that
$\frac{1}{a}\frac{\partial r}{\partial\chi} = f(\chi)$.
We disregard the case $f(\chi)=0$, which would imply $\frac{\partial r}{\partial\chi}=0$
and would lead to the so-called Datt model (1938). Accordingly, we may write
$$
a(\tau,\chi) = \frac{1}{f(\chi)}\frac{\partial r}{\partial\chi}
$$
Let us call af
this expression of $a$:
f = function('f')
af(t,x) = drdx / f(x)
af(t,x)
We check that if we substitute $a$ by af
in the $\tau\chi$ component of the Einstein equation, we get identically zero:
E[0,1].expr().substitute_function(a, af)
NB: expr()
returns a Sage symbolic expression from the coordinate function E[0,1]
, so that we may apply substitute_function
Hence the first Lemaitre-Tolman equation is
LT1 = a(t,x) == af(t,x)
LT1
The $11 = \chi\chi$ component of Einstein equation is
E[1,1]
It is equivalent to
eq = (- E[1,1] * r(t,x)^2).expr() == 0
eq
Let us substitute for $a(\tau,\chi)$ the value found above when solving the $\tau\chi$ component:
eq1 = eq.substitute_function(a, af)
eq1
Some slight rearrangement and simplification:
eq2 = (eq1 * f(x)^2 / diff(r(t,x), x)^2).simplify_full()
eq2
eq3 = (eq2 * diff(r(t,x),t)).simplify_full()
eq3
We notice that the left-hand side of this equation is nothing but the partial derivative w.r.t. $\tau$ of the following quantity:
A = (diff(r(t,x),t)^2 + 1 - f(x)^2 - (Lamb/3)*r(t,x)^2) * r(t,x)
A
bool(eq3.lhs() == diff(A, t))
Hence eq3
tells that $A$ is independent of $\tau$, i.e. is a function of $\chi$ only, which we call $2 m(\chi)$:
m = function('m')
eq4 = A - 2*m(x) == 0
eq4
Let us solve extract $(\partial r/\partial\tau)^2$ from this equation:
drdt2_sol = solve(eq4, diff(r(t,x),t)^2)
drdt2_sol
We thus obtain the second Lemaitre-Tolman equation:
LT2 = drdt2_sol[0].expand()
LT2
drdt2 = LT2.rhs()
drdt2
The $00 = \tau\tau$ component of Einstein equation is
E[0,0]
It is equivalent to
eq = (- E[0,0] * a(t,x)^3 * r(t,x)^2).expr() == 0
eq
As above, we substitute for $a(\tau,\chi)$ the value found when solving the $\tau\chi$ component:
eq1 = eq.substitute_function(a, af)
eq1
eq2 = (eq1 * f(x)^3).simplify_full()
eq2
Let us substitute for $\partial r/\partial \tau$ the positive square root of the value of $(\partial r/\partial \tau)^2$ found when solving the $\chi\chi$ component:
drdt = sqrt(drdt2)
drdt
eq3 = eq2.subs({diff(r(t,x),t): drdt, diff(r(t,x),t,x): diff(drdt, x)}).simplify_full()
eq3
If we use the negative square root of $(\partial r/\partial \tau)^2$ instead, we get the same result:
drdt = - sqrt(drdt2)
drdt
eq3_minus = eq2.subs({diff(r(t,x),t): drdt, diff(r(t,x),t,x): diff(drdt, x)}).simplify_full()
eq3_minus
eq3_minus == eq3
Thus we continue with eq3
and rearrange it to get the third Lemaitre-Tolman equation:
eq4 = (eq3 / (2*diff(r(t,x),x)^2)).simplify_full()
eq4
dmdx_sol = solve(eq4, diff(m(x),x))
dmdx_sol
LT3 = dmdx_sol[0]
LT3
First we notice that the $\theta\theta$ and $\phi\phi$ components of the Einstein equation are equivalent:
E[3,3] == E[2,2] * sin(th)^2
Let us thus consider only the $22 = \theta\theta$ component:
E[2,2]
It is equivalent to
eq = (- E[2,2] * a(t,x)^3).expr() == 0
eq
We substitute for $a(\tau,\chi)$ the value found when solving the $\tau\chi$ component:
eq1 = eq.substitute_function(a, af).simplify_full()
eq1
eq2 = (eq1 * f(x)^3).simplify_full()
eq2
Then we substitute for $\partial r/\partial\tau$ the value obtained when solving the $\tau\tau$ component:
eq3 = eq2.subs({diff(r(t,x),t,t,x): diff(drdt,t,x), diff(r(t,x),t,t): diff(drdt,t)}).simplify_full()
eq4 = eq3.subs({diff(r(t,x),t): drdt, diff(r(t,x),t,x): diff(drdt,x)}).simplify_full()
eq4
We conclude that the $\theta\theta$ component of Einstein equation does add any independent equation.
Let us collect the three independent equations obtained from the Einstein equation, constituting the Lemaître-Tolman system:
for eq in [LT1, LT2, LT3]:
show(eq)
The first equation is the unnumbered one just above Eq. (8.1) in Lemaître's article L'univers en expansion, Annales de la Société Scientifique de Bruxelles A 53, 51 (1933), translated in English in Gen. Relativ. Gravit. 29, 641 (1997). The second equation is Eq. (8.2) in Lemaître's article, while the third one is Eq. (8.3).