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Author: Eric Gourgoulhon
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Lemaître-Tolman solutions

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:

In [1]:
version()
'SageMath version 8.0.beta6, Release Date: 2017-05-12'

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

In [2]:
%display latex

Spacetime

We declare the spacetime manifold MM:

In [3]:
M = Manifold(4, 'M') print(M)
4-dimensional differentiable manifold M

and declare the chart of Lemaître synchronous coordinates on it:

In [4]:
X.<t,x,th,ph> = M.chart(r't:\tau x:(0,+oo):\chi th:(0,pi):\theta ph:(0,2*pi):\phi') X
(M,(τ,χ,θ,ϕ))\left(M,({\tau}, {\chi}, {\theta}, {\phi})\right)

The most general metric tensor, assuming spherical symmetry and synchronous coordinates:

In [5]:
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()
g=dτdτ+a(τ,χ)2dχdχ+r(τ,χ)2dθdθ+r(τ,χ)2sin(θ)2dϕdϕg = -\mathrm{d} {\tau}\otimes \mathrm{d} {\tau} + a\left({\tau}, {\chi}\right)^{2} \mathrm{d} {\chi}\otimes \mathrm{d} {\chi} + r\left({\tau}, {\chi}\right)^{2} \mathrm{d} {\theta}\otimes \mathrm{d} {\theta} + r\left({\tau}, {\chi}\right)^{2} \sin\left({\theta}\right)^{2} \mathrm{d} {\phi}\otimes \mathrm{d} {\phi}

Einstein equation

The cosmological constant:

In [6]:
var('Lamb', latex_name='\Lambda')
Λ{\Lambda}

The Ricci tensor:

In [7]:
Ric = g.ricci() print(Ric)
Field of symmetric bilinear forms Ric(g) on the 4-dimensional differentiable manifold M
In [8]:
Ric.display()
Ric(g)=(r(τ,χ)2aτ2+2a(τ,χ)2rτ2a(τ,χ)r(τ,χ))dτdτ2(a(τ,χ)2rτχaτrχ)a(τ,χ)r(τ,χ)dτdχ2(a(τ,χ)2rτχaτrχ)a(τ,χ)r(τ,χ)dχdτ+(a(τ,χ)2r(τ,χ)2aτ2+2a(τ,χ)2aτrτ+2aχrχ2a(τ,χ)2rχ2a(τ,χ)r(τ,χ))dχdχ+(a(τ,χ)2r(τ,χ)aτrτ+a(τ,χ)3rτ2+a(τ,χ)3r(τ,χ)2rτ2+a(τ,χ)3+r(τ,χ)aχrχa(τ,χ)rχ2a(τ,χ)r(τ,χ)2rχ2a(τ,χ)3)dθdθ+(a(τ,χ)2r(τ,χ)aτrτ+a(τ,χ)3rτ2+a(τ,χ)3r(τ,χ)2rτ2+a(τ,χ)3+r(τ,χ)aχrχa(τ,χ)rχ2a(τ,χ)r(τ,χ)2rχ2)sin(θ)2a(τ,χ)3dϕdϕ\mathrm{Ric}\left(g\right) = \left( -\frac{r\left({\tau}, {\chi}\right) \frac{\partial^2\,a}{\partial {\tau} ^ 2} + 2 \, a\left({\tau}, {\chi}\right) \frac{\partial^2\,r}{\partial {\tau} ^ 2}}{a\left({\tau}, {\chi}\right) r\left({\tau}, {\chi}\right)} \right) \mathrm{d} {\tau}\otimes \mathrm{d} {\tau} -\frac{2 \, {\left(a\left({\tau}, {\chi}\right) \frac{\partial^2\,r}{\partial {\tau}\partial {\chi}} - \frac{\partial\,a}{\partial {\tau}} \frac{\partial\,r}{\partial {\chi}}\right)}}{a\left({\tau}, {\chi}\right) r\left({\tau}, {\chi}\right)} \mathrm{d} {\tau}\otimes \mathrm{d} {\chi} -\frac{2 \, {\left(a\left({\tau}, {\chi}\right) \frac{\partial^2\,r}{\partial {\tau}\partial {\chi}} - \frac{\partial\,a}{\partial {\tau}} \frac{\partial\,r}{\partial {\chi}}\right)}}{a\left({\tau}, {\chi}\right) r\left({\tau}, {\chi}\right)} \mathrm{d} {\chi}\otimes \mathrm{d} {\tau} + \left( \frac{a\left({\tau}, {\chi}\right)^{2} r\left({\tau}, {\chi}\right) \frac{\partial^2\,a}{\partial {\tau} ^ 2} + 2 \, a\left({\tau}, {\chi}\right)^{2} \frac{\partial\,a}{\partial {\tau}} \frac{\partial\,r}{\partial {\tau}} + 2 \, \frac{\partial\,a}{\partial {\chi}} \frac{\partial\,r}{\partial {\chi}} - 2 \, a\left({\tau}, {\chi}\right) \frac{\partial^2\,r}{\partial {\chi} ^ 2}}{a\left({\tau}, {\chi}\right) r\left({\tau}, {\chi}\right)} \right) \mathrm{d} {\chi}\otimes \mathrm{d} {\chi} + \left( \frac{a\left({\tau}, {\chi}\right)^{2} r\left({\tau}, {\chi}\right) \frac{\partial\,a}{\partial {\tau}} \frac{\partial\,r}{\partial {\tau}} + a\left({\tau}, {\chi}\right)^{3} \frac{\partial\,r}{\partial {\tau}}^{2} + a\left({\tau}, {\chi}\right)^{3} r\left({\tau}, {\chi}\right) \frac{\partial^2\,r}{\partial {\tau} ^ 2} + a\left({\tau}, {\chi}\right)^{3} + r\left({\tau}, {\chi}\right) \frac{\partial\,a}{\partial {\chi}} \frac{\partial\,r}{\partial {\chi}} - a\left({\tau}, {\chi}\right) \frac{\partial\,r}{\partial {\chi}}^{2} - a\left({\tau}, {\chi}\right) r\left({\tau}, {\chi}\right) \frac{\partial^2\,r}{\partial {\chi} ^ 2}}{a\left({\tau}, {\chi}\right)^{3}} \right) \mathrm{d} {\theta}\otimes \mathrm{d} {\theta} + \frac{{\left(a\left({\tau}, {\chi}\right)^{2} r\left({\tau}, {\chi}\right) \frac{\partial\,a}{\partial {\tau}} \frac{\partial\,r}{\partial {\tau}} + a\left({\tau}, {\chi}\right)^{3} \frac{\partial\,r}{\partial {\tau}}^{2} + a\left({\tau}, {\chi}\right)^{3} r\left({\tau}, {\chi}\right) \frac{\partial^2\,r}{\partial {\tau} ^ 2} + a\left({\tau}, {\chi}\right)^{3} + r\left({\tau}, {\chi}\right) \frac{\partial\,a}{\partial {\chi}} \frac{\partial\,r}{\partial {\chi}} - a\left({\tau}, {\chi}\right) \frac{\partial\,r}{\partial {\chi}}^{2} - a\left({\tau}, {\chi}\right) r\left({\tau}, {\chi}\right) \frac{\partial^2\,r}{\partial {\chi} ^ 2}\right)} \sin\left({\theta}\right)^{2}}{a\left({\tau}, {\chi}\right)^{3}} \mathrm{d} {\phi}\otimes \mathrm{d} {\phi}

The Einstein tensor:

In [9]:
G = Ric - 1/2*g.ricci_scalar() * g G.set_name('G') print(G)
Field of symmetric bilinear forms G on the 4-dimensional differentiable manifold M
In [10]:
G.display_comp()
Gττττ=2a(τ,χ)2r(τ,χ)aτrτ+a(τ,χ)3rτ2+a(τ,χ)3+2r(τ,χ)aχrχa(τ,χ)rχ22a(τ,χ)r(τ,χ)2rχ2a(τ,χ)3r(τ,χ)2Gτχτχ=2(a(τ,χ)2rτχaτrχ)a(τ,χ)r(τ,χ)Gχτχτ=2(a(τ,χ)2rτχaτrχ)a(τ,χ)r(τ,χ)Gχχχχ=a(τ,χ)2(rτ)2+2a(τ,χ)2r(τ,χ)2rτ2+a(τ,χ)2(rχ)2r(τ,χ)2Gθθθθ=a(τ,χ)2r(τ,χ)22aτ2+a(τ,χ)2r(τ,χ)aτrτ+a(τ,χ)3r(τ,χ)2rτ2+r(τ,χ)aχrχa(τ,χ)r(τ,χ)2rχ2a(τ,χ)3Gϕϕϕϕ=(a(τ,χ)2r(τ,χ)22aτ2+a(τ,χ)2r(τ,χ)aτrτ+a(τ,χ)3r(τ,χ)2rτ2+r(τ,χ)aχrχa(τ,χ)r(τ,χ)2rχ2)sin(θ)2a(τ,χ)3\begin{array}{lcl} G_{ \, {\tau} \, {\tau} }^{ \phantom{\, {\tau}}\phantom{\, {\tau}} } & = & \frac{2 \, a\left({\tau}, {\chi}\right)^{2} r\left({\tau}, {\chi}\right) \frac{\partial\,a}{\partial {\tau}} \frac{\partial\,r}{\partial {\tau}} + a\left({\tau}, {\chi}\right)^{3} \frac{\partial\,r}{\partial {\tau}}^{2} + a\left({\tau}, {\chi}\right)^{3} + 2 \, r\left({\tau}, {\chi}\right) \frac{\partial\,a}{\partial {\chi}} \frac{\partial\,r}{\partial {\chi}} - a\left({\tau}, {\chi}\right) \frac{\partial\,r}{\partial {\chi}}^{2} - 2 \, a\left({\tau}, {\chi}\right) r\left({\tau}, {\chi}\right) \frac{\partial^2\,r}{\partial {\chi} ^ 2}}{a\left({\tau}, {\chi}\right)^{3} r\left({\tau}, {\chi}\right)^{2}} \\ G_{ \, {\tau} \, {\chi} }^{ \phantom{\, {\tau}}\phantom{\, {\chi}} } & = & -\frac{2 \, {\left(a\left({\tau}, {\chi}\right) \frac{\partial^2\,r}{\partial {\tau}\partial {\chi}} - \frac{\partial\,a}{\partial {\tau}} \frac{\partial\,r}{\partial {\chi}}\right)}}{a\left({\tau}, {\chi}\right) r\left({\tau}, {\chi}\right)} \\ G_{ \, {\chi} \, {\tau} }^{ \phantom{\, {\chi}}\phantom{\, {\tau}} } & = & -\frac{2 \, {\left(a\left({\tau}, {\chi}\right) \frac{\partial^2\,r}{\partial {\tau}\partial {\chi}} - \frac{\partial\,a}{\partial {\tau}} \frac{\partial\,r}{\partial {\chi}}\right)}}{a\left({\tau}, {\chi}\right) r\left({\tau}, {\chi}\right)} \\ G_{ \, {\chi} \, {\chi} }^{ \phantom{\, {\chi}}\phantom{\, {\chi}} } & = & -\frac{a\left({\tau}, {\chi}\right)^{2} \left(\frac{\partial\,r}{\partial {\tau}}\right)^{2} + 2 \, a\left({\tau}, {\chi}\right)^{2} r\left({\tau}, {\chi}\right) \frac{\partial^2\,r}{\partial {\tau} ^ 2} + a\left({\tau}, {\chi}\right)^{2} - \left(\frac{\partial\,r}{\partial {\chi}}\right)^{2}}{r\left({\tau}, {\chi}\right)^{2}} \\ G_{ \, {\theta} \, {\theta} }^{ \phantom{\, {\theta}}\phantom{\, {\theta}} } & = & -\frac{a\left({\tau}, {\chi}\right)^{2} r\left({\tau}, {\chi}\right)^{2} \frac{\partial^2\,a}{\partial {\tau} ^ 2} + a\left({\tau}, {\chi}\right)^{2} r\left({\tau}, {\chi}\right) \frac{\partial\,a}{\partial {\tau}} \frac{\partial\,r}{\partial {\tau}} + a\left({\tau}, {\chi}\right)^{3} r\left({\tau}, {\chi}\right) \frac{\partial^2\,r}{\partial {\tau} ^ 2} + r\left({\tau}, {\chi}\right) \frac{\partial\,a}{\partial {\chi}} \frac{\partial\,r}{\partial {\chi}} - a\left({\tau}, {\chi}\right) r\left({\tau}, {\chi}\right) \frac{\partial^2\,r}{\partial {\chi} ^ 2}}{a\left({\tau}, {\chi}\right)^{3}} \\ G_{ \, {\phi} \, {\phi} }^{ \phantom{\, {\phi}}\phantom{\, {\phi}} } & = & -\frac{{\left(a\left({\tau}, {\chi}\right)^{2} r\left({\tau}, {\chi}\right)^{2} \frac{\partial^2\,a}{\partial {\tau} ^ 2} + a\left({\tau}, {\chi}\right)^{2} r\left({\tau}, {\chi}\right) \frac{\partial\,a}{\partial {\tau}} \frac{\partial\,r}{\partial {\tau}} + a\left({\tau}, {\chi}\right)^{3} r\left({\tau}, {\chi}\right) \frac{\partial^2\,r}{\partial {\tau} ^ 2} + r\left({\tau}, {\chi}\right) \frac{\partial\,a}{\partial {\chi}} \frac{\partial\,r}{\partial {\chi}} - a\left({\tau}, {\chi}\right) r\left({\tau}, {\chi}\right) \frac{\partial^2\,r}{\partial {\chi} ^ 2}\right)} \sin\left({\theta}\right)^{2}}{a\left({\tau}, {\chi}\right)^{3}} \end{array}

Dust matter model

Let us consider a pressureless fluid ("dust"). Moreover, we assume that the coordinates (τ,χ,θ,ϕ)(\tau,\chi,\theta,\phi) are comoving, i.e. that the fluid 4-velocity is equal to τ\partial_\tau:

In [11]:
u = M.vector_field('u') u[0] = 1 u.display()
u=τu = \frac{\partial}{\partial {\tau} }

Since (τ,χ,θ,χ)(\tau,\chi,\theta,\chi) are synchronous, the above does define a unit timelike vector:

In [12]:
g(u,u).display()
g(u,u):MR(τ,χ,θ,ϕ)1\begin{array}{llcl} g\left(u,u\right):& M & \longrightarrow & \mathbb{R} \\ & \left({\tau}, {\chi}, {\theta}, {\phi}\right) & \longmapsto & -1 \end{array}

The 1-form associated to the fluid 4-velocity by metric duality:

In [13]:
u_form = u.down(g) print(u_form)
1-form on the 4-dimensional differentiable manifold M
In [14]:
u_form.display()
dτ-\mathrm{d} {\tau}

The pressureless energy-momentum tensor:

In [15]:
rho = function('rho') T = rho(t,x)*(u_form * u_form) T.set_name('T') print(T)
Field of symmetric bilinear forms T on the 4-dimensional differentiable manifold M
In [16]:
T.display()
T=ρ(τ,χ)dτdτT = \rho\left({\tau}, {\chi}\right) \mathrm{d} {\tau}\otimes \mathrm{d} {\tau}

The Einstein equation:

In [17]:
E = G + Lamb*g - 8*pi*T E.set_name('E') print(E)
Field of symmetric bilinear forms E on the 4-dimensional differentiable manifold M
In [18]:
E.display_comp()
Eττττ=8πa(τ,χ)3r(τ,χ)2ρ(τ,χ)+Λa(τ,χ)3r(τ,χ)22a(τ,χ)2r(τ,χ)aτrτa(τ,χ)3rτ2a(τ,χ)32r(τ,χ)aχrχ+a(τ,χ)rχ2+2a(τ,χ)r(τ,χ)2rχ2a(τ,χ)3r(τ,χ)2Eτχτχ=2(a(τ,χ)2rτχaτrχ)a(τ,χ)r(τ,χ)Eχτχτ=2(a(τ,χ)2rτχaτrχ)a(τ,χ)r(τ,χ)Eχχχχ=Λa(τ,χ)2r(τ,χ)2a(τ,χ)2(rτ)22a(τ,χ)2r(τ,χ)2rτ2a(τ,χ)2+(rχ)2r(τ,χ)2Eθθθθ=Λa(τ,χ)3r(τ,χ)2a(τ,χ)2r(τ,χ)22aτ2a(τ,χ)2r(τ,χ)aτrτa(τ,χ)3r(τ,χ)2rτ2r(τ,χ)aχrχ+a(τ,χ)r(τ,χ)2rχ2a(τ,χ)3Eϕϕϕϕ=(Λa(τ,χ)3r(τ,χ)2a(τ,χ)2r(τ,χ)22aτ2a(τ,χ)2r(τ,χ)aτrτa(τ,χ)3r(τ,χ)2rτ2r(τ,χ)aχrχ+a(τ,χ)r(τ,χ)2rχ2)sin(θ)2a(τ,χ)3\begin{array}{lcl} E_{ \, {\tau} \, {\tau} }^{ \phantom{\, {\tau}}\phantom{\, {\tau}} } & = & -\frac{8 \, \pi a\left({\tau}, {\chi}\right)^{3} r\left({\tau}, {\chi}\right)^{2} \rho\left({\tau}, {\chi}\right) + {\Lambda} a\left({\tau}, {\chi}\right)^{3} r\left({\tau}, {\chi}\right)^{2} - 2 \, a\left({\tau}, {\chi}\right)^{2} r\left({\tau}, {\chi}\right) \frac{\partial\,a}{\partial {\tau}} \frac{\partial\,r}{\partial {\tau}} - a\left({\tau}, {\chi}\right)^{3} \frac{\partial\,r}{\partial {\tau}}^{2} - a\left({\tau}, {\chi}\right)^{3} - 2 \, r\left({\tau}, {\chi}\right) \frac{\partial\,a}{\partial {\chi}} \frac{\partial\,r}{\partial {\chi}} + a\left({\tau}, {\chi}\right) \frac{\partial\,r}{\partial {\chi}}^{2} + 2 \, a\left({\tau}, {\chi}\right) r\left({\tau}, {\chi}\right) \frac{\partial^2\,r}{\partial {\chi} ^ 2}}{a\left({\tau}, {\chi}\right)^{3} r\left({\tau}, {\chi}\right)^{2}} \\ E_{ \, {\tau} \, {\chi} }^{ \phantom{\, {\tau}}\phantom{\, {\chi}} } & = & -\frac{2 \, {\left(a\left({\tau}, {\chi}\right) \frac{\partial^2\,r}{\partial {\tau}\partial {\chi}} - \frac{\partial\,a}{\partial {\tau}} \frac{\partial\,r}{\partial {\chi}}\right)}}{a\left({\tau}, {\chi}\right) r\left({\tau}, {\chi}\right)} \\ E_{ \, {\chi} \, {\tau} }^{ \phantom{\, {\chi}}\phantom{\, {\tau}} } & = & -\frac{2 \, {\left(a\left({\tau}, {\chi}\right) \frac{\partial^2\,r}{\partial {\tau}\partial {\chi}} - \frac{\partial\,a}{\partial {\tau}} \frac{\partial\,r}{\partial {\chi}}\right)}}{a\left({\tau}, {\chi}\right) r\left({\tau}, {\chi}\right)} \\ E_{ \, {\chi} \, {\chi} }^{ \phantom{\, {\chi}}\phantom{\, {\chi}} } & = & \frac{{\Lambda} a\left({\tau}, {\chi}\right)^{2} r\left({\tau}, {\chi}\right)^{2} - a\left({\tau}, {\chi}\right)^{2} \left(\frac{\partial\,r}{\partial {\tau}}\right)^{2} - 2 \, a\left({\tau}, {\chi}\right)^{2} r\left({\tau}, {\chi}\right) \frac{\partial^2\,r}{\partial {\tau} ^ 2} - a\left({\tau}, {\chi}\right)^{2} + \left(\frac{\partial\,r}{\partial {\chi}}\right)^{2}}{r\left({\tau}, {\chi}\right)^{2}} \\ E_{ \, {\theta} \, {\theta} }^{ \phantom{\, {\theta}}\phantom{\, {\theta}} } & = & \frac{{\Lambda} a\left({\tau}, {\chi}\right)^{3} r\left({\tau}, {\chi}\right)^{2} - a\left({\tau}, {\chi}\right)^{2} r\left({\tau}, {\chi}\right)^{2} \frac{\partial^2\,a}{\partial {\tau} ^ 2} - a\left({\tau}, {\chi}\right)^{2} r\left({\tau}, {\chi}\right) \frac{\partial\,a}{\partial {\tau}} \frac{\partial\,r}{\partial {\tau}} - a\left({\tau}, {\chi}\right)^{3} r\left({\tau}, {\chi}\right) \frac{\partial^2\,r}{\partial {\tau} ^ 2} - r\left({\tau}, {\chi}\right) \frac{\partial\,a}{\partial {\chi}} \frac{\partial\,r}{\partial {\chi}} + a\left({\tau}, {\chi}\right) r\left({\tau}, {\chi}\right) \frac{\partial^2\,r}{\partial {\chi} ^ 2}}{a\left({\tau}, {\chi}\right)^{3}} \\ E_{ \, {\phi} \, {\phi} }^{ \phantom{\, {\phi}}\phantom{\, {\phi}} } & = & \frac{{\left({\Lambda} a\left({\tau}, {\chi}\right)^{3} r\left({\tau}, {\chi}\right)^{2} - a\left({\tau}, {\chi}\right)^{2} r\left({\tau}, {\chi}\right)^{2} \frac{\partial^2\,a}{\partial {\tau} ^ 2} - a\left({\tau}, {\chi}\right)^{2} r\left({\tau}, {\chi}\right) \frac{\partial\,a}{\partial {\tau}} \frac{\partial\,r}{\partial {\tau}} - a\left({\tau}, {\chi}\right)^{3} r\left({\tau}, {\chi}\right) \frac{\partial^2\,r}{\partial {\tau} ^ 2} - r\left({\tau}, {\chi}\right) \frac{\partial\,a}{\partial {\chi}} \frac{\partial\,r}{\partial {\chi}} + a\left({\tau}, {\chi}\right) r\left({\tau}, {\chi}\right) \frac{\partial^2\,r}{\partial {\chi} ^ 2}\right)} \sin\left({\theta}\right)^{2}}{a\left({\tau}, {\chi}\right)^{3}} \end{array}

Solving the Einstein equation

τχ\tau\chi component

Let us first consider the 01=τχ01 = \tau\chi component of the Einstein equation:

In [19]:
E[0,1]
2(a(τ,χ)2rτχaτrχ)a(τ,χ)r(τ,χ)-\frac{2 \, {\left(a\left({\tau}, {\chi}\right) \frac{\partial^2\,r}{\partial {\tau}\partial {\chi}} - \frac{\partial\,a}{\partial {\tau}} \frac{\partial\,r}{\partial {\chi}}\right)}}{a\left({\tau}, {\chi}\right) r\left({\tau}, {\chi}\right)}

A slight rearrangement of the equation:

In [20]:
eq = E[0,1]*r(t,x)/(-2*a(t,x)) eq
a(τ,χ)2rτχaτrχa(τ,χ)2\frac{a\left({\tau}, {\chi}\right) \frac{\partial^2\,r}{\partial {\tau}\partial {\chi}} - \frac{\partial\,a}{\partial {\tau}} \frac{\partial\,r}{\partial {\chi}}}{a\left({\tau}, {\chi}\right)^{2}}

We see that this equation is equivalent to τ(1arχ)=0 \frac{\partial}{\partial\tau} \left( \frac{1}{a}\frac{\partial r}{\partial\chi} \right) = 0 since

In [21]:
drdx = diff(r(t,x), x) eq - diff(drdx/a(t,x), t)
00

Hence there exists a function of χ\chi only, f(χ)f(\chi) say, such that 1arχ=f(χ)\frac{1}{a}\frac{\partial r}{\partial\chi} = f(\chi). We disregard the case f(χ)=0f(\chi)=0, which would imply rχ=0\frac{\partial r}{\partial\chi}=0 and would lead to the so-called Datt model (1938). Accordingly, we may write a(τ,χ)=1f(χ)rχ a(\tau,\chi) = \frac{1}{f(\chi)}\frac{\partial r}{\partial\chi} Let us call af this expression of aa:

In [22]:
f = function('f') af(t,x) = drdx / f(x) af(t,x)
χr(τ,χ)f(χ)\frac{\frac{\partial}{\partial {\chi}}r\left({\tau}, {\chi}\right)}{f\left({\chi}\right)}

We check that if we substitute aa by af in the τχ\tau\chi component of the Einstein equation, we get identically zero:

In [23]:
E[0,1].expr().substitute_function(a, af)
00

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

In [24]:
LT1 = a(t,x) == af(t,x) LT1
a(τ,χ)=χr(τ,χ)f(χ)a\left({\tau}, {\chi}\right) = \frac{\frac{\partial}{\partial {\chi}}r\left({\tau}, {\chi}\right)}{f\left({\chi}\right)}

χχ\chi\chi component

The 11=χχ11 = \chi\chi component of Einstein equation is

In [25]:
E[1,1]
Λa(τ,χ)2r(τ,χ)2a(τ,χ)2(rτ)22a(τ,χ)2r(τ,χ)2rτ2a(τ,χ)2+(rχ)2r(τ,χ)2\frac{{\Lambda} a\left({\tau}, {\chi}\right)^{2} r\left({\tau}, {\chi}\right)^{2} - a\left({\tau}, {\chi}\right)^{2} \left(\frac{\partial\,r}{\partial {\tau}}\right)^{2} - 2 \, a\left({\tau}, {\chi}\right)^{2} r\left({\tau}, {\chi}\right) \frac{\partial^2\,r}{\partial {\tau} ^ 2} - a\left({\tau}, {\chi}\right)^{2} + \left(\frac{\partial\,r}{\partial {\chi}}\right)^{2}}{r\left({\tau}, {\chi}\right)^{2}}

It is equivalent to

In [26]:
eq = (- E[1,1] * r(t,x)^2).expr() == 0 eq
Λa(τ,χ)2r(τ,χ)2+a(τ,χ)2τr(τ,χ)2+2a(τ,χ)2r(τ,χ)2(τ)2r(τ,χ)+a(τ,χ)2χr(τ,χ)2=0-{\Lambda} a\left({\tau}, {\chi}\right)^{2} r\left({\tau}, {\chi}\right)^{2} + a\left({\tau}, {\chi}\right)^{2} \frac{\partial}{\partial {\tau}}r\left({\tau}, {\chi}\right)^{2} + 2 \, a\left({\tau}, {\chi}\right)^{2} r\left({\tau}, {\chi}\right) \frac{\partial^{2}}{(\partial {\tau})^{2}}r\left({\tau}, {\chi}\right) + a\left({\tau}, {\chi}\right)^{2} - \frac{\partial}{\partial {\chi}}r\left({\tau}, {\chi}\right)^{2} = 0

Let us substitute for a(τ,χ)a(\tau,\chi) the value found above when solving the τχ\tau\chi component:

In [27]:
eq1 = eq.substitute_function(a, af) eq1
Λr(τ,χ)2χr(τ,χ)2f(χ)2+τr(τ,χ)2χr(τ,χ)2f(χ)2+2r(τ,χ)2(τ)2r(τ,χ)χr(τ,χ)2f(χ)2χr(τ,χ)2+χr(τ,χ)2f(χ)2=0-\frac{{\Lambda} r\left({\tau}, {\chi}\right)^{2} \frac{\partial}{\partial {\chi}}r\left({\tau}, {\chi}\right)^{2}}{f\left({\chi}\right)^{2}} + \frac{\frac{\partial}{\partial {\tau}}r\left({\tau}, {\chi}\right)^{2} \frac{\partial}{\partial {\chi}}r\left({\tau}, {\chi}\right)^{2}}{f\left({\chi}\right)^{2}} + \frac{2 \, r\left({\tau}, {\chi}\right) \frac{\partial^{2}}{(\partial {\tau})^{2}}r\left({\tau}, {\chi}\right) \frac{\partial}{\partial {\chi}}r\left({\tau}, {\chi}\right)^{2}}{f\left({\chi}\right)^{2}} - \frac{\partial}{\partial {\chi}}r\left({\tau}, {\chi}\right)^{2} + \frac{\frac{\partial}{\partial {\chi}}r\left({\tau}, {\chi}\right)^{2}}{f\left({\chi}\right)^{2}} = 0

Some slight rearrangement and simplification:

In [28]:
eq2 = (eq1 * f(x)^2 / diff(r(t,x), x)^2).simplify_full() eq2
Λr(τ,χ)2f(χ)2+τr(τ,χ)2+2r(τ,χ)2(τ)2r(τ,χ)+1=0-{\Lambda} r\left({\tau}, {\chi}\right)^{2} - f\left({\chi}\right)^{2} + \frac{\partial}{\partial {\tau}}r\left({\tau}, {\chi}\right)^{2} + 2 \, r\left({\tau}, {\chi}\right) \frac{\partial^{2}}{(\partial {\tau})^{2}}r\left({\tau}, {\chi}\right) + 1 = 0
In [29]:
eq3 = (eq2 * diff(r(t,x),t)).simplify_full() eq3
f(χ)2τr(τ,χ)+τr(τ,χ)3+2r(τ,χ)τr(τ,χ)2(τ)2r(τ,χ)(Λr(τ,χ)21)τr(τ,χ)=0-f\left({\chi}\right)^{2} \frac{\partial}{\partial {\tau}}r\left({\tau}, {\chi}\right) + \frac{\partial}{\partial {\tau}}r\left({\tau}, {\chi}\right)^{3} + 2 \, r\left({\tau}, {\chi}\right) \frac{\partial}{\partial {\tau}}r\left({\tau}, {\chi}\right) \frac{\partial^{2}}{(\partial {\tau})^{2}}r\left({\tau}, {\chi}\right) - {\left({\Lambda} r\left({\tau}, {\chi}\right)^{2} - 1\right)} \frac{\partial}{\partial {\tau}}r\left({\tau}, {\chi}\right) = 0

We notice that the left-hand side of this equation is nothing but the partial derivative w.r.t. τ\tau of the following quantity:

In [30]:
A = (diff(r(t,x),t)^2 + 1 - f(x)^2 - (Lamb/3)*r(t,x)^2) * r(t,x) A
13(Λr(τ,χ)2+3f(χ)23τr(τ,χ)23)r(τ,χ)-\frac{1}{3} \, {\left({\Lambda} r\left({\tau}, {\chi}\right)^{2} + 3 \, f\left({\chi}\right)^{2} - 3 \, \frac{\partial}{\partial {\tau}}r\left({\tau}, {\chi}\right)^{2} - 3\right)} r\left({\tau}, {\chi}\right)
In [31]:
bool(eq3.lhs() == diff(A, t))
True\mathrm{True}

Hence eq3 tells that AA is independent of τ\tau, i.e. is a function of χ\chi only, which we call 2m(χ)2 m(\chi):

In [32]:
m = function('m') eq4 = A - 2*m(x) == 0 eq4
13(Λr(τ,χ)2+3f(χ)23τr(τ,χ)23)r(τ,χ)2m(χ)=0-\frac{1}{3} \, {\left({\Lambda} r\left({\tau}, {\chi}\right)^{2} + 3 \, f\left({\chi}\right)^{2} - 3 \, \frac{\partial}{\partial {\tau}}r\left({\tau}, {\chi}\right)^{2} - 3\right)} r\left({\tau}, {\chi}\right) - 2 \, m\left({\chi}\right) = 0

Let us solve extract (r/τ)2(\partial r/\partial\tau)^2 from this equation:

In [33]:
drdt2_sol = solve(eq4, diff(r(t,x),t)^2) drdt2_sol
[τr(τ,χ)2=Λr(τ,χ)3+3f(χ)2r(τ,χ)+6m(χ)3r(τ,χ)3r(τ,χ)]\left[\frac{\partial}{\partial {\tau}}r\left({\tau}, {\chi}\right)^{2} = \frac{{\Lambda} r\left({\tau}, {\chi}\right)^{3} + 3 \, f\left({\chi}\right)^{2} r\left({\tau}, {\chi}\right) + 6 \, m\left({\chi}\right) - 3 \, r\left({\tau}, {\chi}\right)}{3 \, r\left({\tau}, {\chi}\right)}\right]

We thus obtain the second Lemaitre-Tolman equation:

In [34]:
LT2 = drdt2_sol[0].expand() LT2
τr(τ,χ)2=13Λr(τ,χ)2+f(χ)2+2m(χ)r(τ,χ)1\frac{\partial}{\partial {\tau}}r\left({\tau}, {\chi}\right)^{2} = \frac{1}{3} \, {\Lambda} r\left({\tau}, {\chi}\right)^{2} + f\left({\chi}\right)^{2} + \frac{2 \, m\left({\chi}\right)}{r\left({\tau}, {\chi}\right)} - 1
In [35]:
drdt2 = LT2.rhs() drdt2
13Λr(τ,χ)2+f(χ)2+2m(χ)r(τ,χ)1\frac{1}{3} \, {\Lambda} r\left({\tau}, {\chi}\right)^{2} + f\left({\chi}\right)^{2} + \frac{2 \, m\left({\chi}\right)}{r\left({\tau}, {\chi}\right)} - 1

ττ\tau\tau component

The 00=ττ00 = \tau\tau component of Einstein equation is

In [36]:
E[0,0]
8πa(τ,χ)3r(τ,χ)2ρ(τ,χ)+Λa(τ,χ)3r(τ,χ)22a(τ,χ)2r(τ,χ)aτrτa(τ,χ)3rτ2a(τ,χ)32r(τ,χ)aχrχ+a(τ,χ)rχ2+2a(τ,χ)r(τ,χ)2rχ2a(τ,χ)3r(τ,χ)2-\frac{8 \, \pi a\left({\tau}, {\chi}\right)^{3} r\left({\tau}, {\chi}\right)^{2} \rho\left({\tau}, {\chi}\right) + {\Lambda} a\left({\tau}, {\chi}\right)^{3} r\left({\tau}, {\chi}\right)^{2} - 2 \, a\left({\tau}, {\chi}\right)^{2} r\left({\tau}, {\chi}\right) \frac{\partial\,a}{\partial {\tau}} \frac{\partial\,r}{\partial {\tau}} - a\left({\tau}, {\chi}\right)^{3} \frac{\partial\,r}{\partial {\tau}}^{2} - a\left({\tau}, {\chi}\right)^{3} - 2 \, r\left({\tau}, {\chi}\right) \frac{\partial\,a}{\partial {\chi}} \frac{\partial\,r}{\partial {\chi}} + a\left({\tau}, {\chi}\right) \frac{\partial\,r}{\partial {\chi}}^{2} + 2 \, a\left({\tau}, {\chi}\right) r\left({\tau}, {\chi}\right) \frac{\partial^2\,r}{\partial {\chi} ^ 2}}{a\left({\tau}, {\chi}\right)^{3} r\left({\tau}, {\chi}\right)^{2}}

It is equivalent to

In [37]:
eq = (- E[0,0] * a(t,x)^3 * r(t,x)^2).expr() == 0 eq
8πa(τ,χ)3r(τ,χ)2ρ(τ,χ)+Λa(τ,χ)3r(τ,χ)22a(τ,χ)2r(τ,χ)τa(τ,χ)τr(τ,χ)a(τ,χ)3τr(τ,χ)2a(τ,χ)32r(τ,χ)χa(τ,χ)χr(τ,χ)+a(τ,χ)χr(τ,χ)2+2a(τ,χ)r(τ,χ)2(χ)2r(τ,χ)=08 \, \pi a\left({\tau}, {\chi}\right)^{3} r\left({\tau}, {\chi}\right)^{2} \rho\left({\tau}, {\chi}\right) + {\Lambda} a\left({\tau}, {\chi}\right)^{3} r\left({\tau}, {\chi}\right)^{2} - 2 \, a\left({\tau}, {\chi}\right)^{2} r\left({\tau}, {\chi}\right) \frac{\partial}{\partial {\tau}}a\left({\tau}, {\chi}\right) \frac{\partial}{\partial {\tau}}r\left({\tau}, {\chi}\right) - a\left({\tau}, {\chi}\right)^{3} \frac{\partial}{\partial {\tau}}r\left({\tau}, {\chi}\right)^{2} - a\left({\tau}, {\chi}\right)^{3} - 2 \, r\left({\tau}, {\chi}\right) \frac{\partial}{\partial {\chi}}a\left({\tau}, {\chi}\right) \frac{\partial}{\partial {\chi}}r\left({\tau}, {\chi}\right) + a\left({\tau}, {\chi}\right) \frac{\partial}{\partial {\chi}}r\left({\tau}, {\chi}\right)^{2} + 2 \, a\left({\tau}, {\chi}\right) r\left({\tau}, {\chi}\right) \frac{\partial^{2}}{(\partial {\chi})^{2}}r\left({\tau}, {\chi}\right) = 0

As above, we substitute for a(τ,χ)a(\tau,\chi) the value found when solving the τχ\tau\chi component:

In [38]:
eq1 = eq.substitute_function(a, af) eq1
8πr(τ,χ)2ρ(τ,χ)χr(τ,χ)3f(χ)3+2(χf(χ)χr(τ,χ)f(χ)22(χ)2r(τ,χ)f(χ))r(τ,χ)χr(τ,χ)+Λr(τ,χ)2χr(τ,χ)3f(χ)32r(τ,χ)τr(τ,χ)2τχr(τ,χ)χr(τ,χ)2f(χ)3+χr(τ,χ)3f(χ)τr(τ,χ)2χr(τ,χ)3f(χ)3+2r(τ,χ)χr(τ,χ)2(χ)2r(τ,χ)f(χ)χr(τ,χ)3f(χ)3=0\frac{8 \, \pi r\left({\tau}, {\chi}\right)^{2} \rho\left({\tau}, {\chi}\right) \frac{\partial}{\partial {\chi}}r\left({\tau}, {\chi}\right)^{3}}{f\left({\chi}\right)^{3}} + 2 \, {\left(\frac{\frac{\partial}{\partial {\chi}}f\left({\chi}\right) \frac{\partial}{\partial {\chi}}r\left({\tau}, {\chi}\right)}{f\left({\chi}\right)^{2}} - \frac{\frac{\partial^{2}}{(\partial {\chi})^{2}}r\left({\tau}, {\chi}\right)}{f\left({\chi}\right)}\right)} r\left({\tau}, {\chi}\right) \frac{\partial}{\partial {\chi}}r\left({\tau}, {\chi}\right) + \frac{{\Lambda} r\left({\tau}, {\chi}\right)^{2} \frac{\partial}{\partial {\chi}}r\left({\tau}, {\chi}\right)^{3}}{f\left({\chi}\right)^{3}} - \frac{2 \, r\left({\tau}, {\chi}\right) \frac{\partial}{\partial {\tau}}r\left({\tau}, {\chi}\right) \frac{\partial^{2}}{\partial {\tau}\partial {\chi}}r\left({\tau}, {\chi}\right) \frac{\partial}{\partial {\chi}}r\left({\tau}, {\chi}\right)^{2}}{f\left({\chi}\right)^{3}} + \frac{\frac{\partial}{\partial {\chi}}r\left({\tau}, {\chi}\right)^{3}}{f\left({\chi}\right)} - \frac{\frac{\partial}{\partial {\tau}}r\left({\tau}, {\chi}\right)^{2} \frac{\partial}{\partial {\chi}}r\left({\tau}, {\chi}\right)^{3}}{f\left({\chi}\right)^{3}} + \frac{2 \, r\left({\tau}, {\chi}\right) \frac{\partial}{\partial {\chi}}r\left({\tau}, {\chi}\right) \frac{\partial^{2}}{(\partial {\chi})^{2}}r\left({\tau}, {\chi}\right)}{f\left({\chi}\right)} - \frac{\frac{\partial}{\partial {\chi}}r\left({\tau}, {\chi}\right)^{3}}{f\left({\chi}\right)^{3}} = 0
In [39]:
eq2 = (eq1 * f(x)^3).simplify_full() eq2
2f(χ)r(τ,χ)χf(χ)χr(τ,χ)22r(τ,χ)τr(τ,χ)2τχr(τ,χ)χr(τ,χ)2+f(χ)2χr(τ,χ)3+(8πr(τ,χ)2ρ(τ,χ)+Λr(τ,χ)2τr(τ,χ)21)χr(τ,χ)3=02 \, f\left({\chi}\right) r\left({\tau}, {\chi}\right) \frac{\partial}{\partial {\chi}}f\left({\chi}\right) \frac{\partial}{\partial {\chi}}r\left({\tau}, {\chi}\right)^{2} - 2 \, r\left({\tau}, {\chi}\right) \frac{\partial}{\partial {\tau}}r\left({\tau}, {\chi}\right) \frac{\partial^{2}}{\partial {\tau}\partial {\chi}}r\left({\tau}, {\chi}\right) \frac{\partial}{\partial {\chi}}r\left({\tau}, {\chi}\right)^{2} + f\left({\chi}\right)^{2} \frac{\partial}{\partial {\chi}}r\left({\tau}, {\chi}\right)^{3} + {\left(8 \, \pi r\left({\tau}, {\chi}\right)^{2} \rho\left({\tau}, {\chi}\right) + {\Lambda} r\left({\tau}, {\chi}\right)^{2} - \frac{\partial}{\partial {\tau}}r\left({\tau}, {\chi}\right)^{2} - 1\right)} \frac{\partial}{\partial {\chi}}r\left({\tau}, {\chi}\right)^{3} = 0

Let us substitute for r/τ\partial r/\partial \tau the positive square root of the value of (r/τ)2(\partial r/\partial \tau)^2 found when solving the χχ\chi\chi component:

In [40]:
drdt = sqrt(drdt2) drdt
13Λr(τ,χ)2+f(χ)2+2m(χ)r(τ,χ)1\sqrt{\frac{1}{3} \, {\Lambda} r\left({\tau}, {\chi}\right)^{2} + f\left({\chi}\right)^{2} + \frac{2 \, m\left({\chi}\right)}{r\left({\tau}, {\chi}\right)} - 1}
In [41]:
eq3 = eq2.subs({diff(r(t,x),t): drdt, diff(r(t,x),t,x): diff(drdt, x)}).simplify_full() eq3
8πr(τ,χ)2ρ(τ,χ)χr(τ,χ)32χm(χ)χr(τ,χ)2=08 \, \pi r\left({\tau}, {\chi}\right)^{2} \rho\left({\tau}, {\chi}\right) \frac{\partial}{\partial {\chi}}r\left({\tau}, {\chi}\right)^{3} - 2 \, \frac{\partial}{\partial {\chi}}m\left({\chi}\right) \frac{\partial}{\partial {\chi}}r\left({\tau}, {\chi}\right)^{2} = 0

If we use the negative square root of (r/τ)2(\partial r/\partial \tau)^2 instead, we get the same result:

In [42]:
drdt = - sqrt(drdt2) drdt
13Λr(τ,χ)2+f(χ)2+2m(χ)r(τ,χ)1-\sqrt{\frac{1}{3} \, {\Lambda} r\left({\tau}, {\chi}\right)^{2} + f\left({\chi}\right)^{2} + \frac{2 \, m\left({\chi}\right)}{r\left({\tau}, {\chi}\right)} - 1}
In [43]:
eq3_minus = eq2.subs({diff(r(t,x),t): drdt, diff(r(t,x),t,x): diff(drdt, x)}).simplify_full() eq3_minus
8πr(τ,χ)2ρ(τ,χ)χr(τ,χ)32χm(χ)χr(τ,χ)2=08 \, \pi r\left({\tau}, {\chi}\right)^{2} \rho\left({\tau}, {\chi}\right) \frac{\partial}{\partial {\chi}}r\left({\tau}, {\chi}\right)^{3} - 2 \, \frac{\partial}{\partial {\chi}}m\left({\chi}\right) \frac{\partial}{\partial {\chi}}r\left({\tau}, {\chi}\right)^{2} = 0
In [44]:
eq3_minus == eq3
True\mathrm{True}

Thus we continue with eq3 and rearrange it to get the third Lemaitre-Tolman equation:

In [45]:
eq4 = (eq3 / (2*diff(r(t,x),x)^2)).simplify_full() eq4
4πr(τ,χ)2ρ(τ,χ)χr(τ,χ)χm(χ)=04 \, \pi r\left({\tau}, {\chi}\right)^{2} \rho\left({\tau}, {\chi}\right) \frac{\partial}{\partial {\chi}}r\left({\tau}, {\chi}\right) - \frac{\partial}{\partial {\chi}}m\left({\chi}\right) = 0
In [46]:
dmdx_sol = solve(eq4, diff(m(x),x)) dmdx_sol
[χm(χ)=4πr(τ,χ)2ρ(τ,χ)χr(τ,χ)]\left[\frac{\partial}{\partial {\chi}}m\left({\chi}\right) = 4 \, \pi r\left({\tau}, {\chi}\right)^{2} \rho\left({\tau}, {\chi}\right) \frac{\partial}{\partial {\chi}}r\left({\tau}, {\chi}\right)\right]
In [47]:
LT3 = dmdx_sol[0] LT3
χm(χ)=4πr(τ,χ)2ρ(τ,χ)χr(τ,χ)\frac{\partial}{\partial {\chi}}m\left({\chi}\right) = 4 \, \pi r\left({\tau}, {\chi}\right)^{2} \rho\left({\tau}, {\chi}\right) \frac{\partial}{\partial {\chi}}r\left({\tau}, {\chi}\right)

θθ\theta\theta and ϕϕ\phi\phi components

First we notice that the θθ\theta\theta and ϕϕ\phi\phi components of the Einstein equation are equivalent:

In [48]:
E[3,3] == E[2,2] * sin(th)^2
True\mathrm{True}

Let us thus consider only the 22=θθ22 = \theta\theta component:

In [49]:
E[2,2]
Λa(τ,χ)3r(τ,χ)2a(τ,χ)2r(τ,χ)22aτ2a(τ,χ)2r(τ,χ)aτrτa(τ,χ)3r(τ,χ)2rτ2r(τ,χ)aχrχ+a(τ,χ)r(τ,χ)2rχ2a(τ,χ)3\frac{{\Lambda} a\left({\tau}, {\chi}\right)^{3} r\left({\tau}, {\chi}\right)^{2} - a\left({\tau}, {\chi}\right)^{2} r\left({\tau}, {\chi}\right)^{2} \frac{\partial^2\,a}{\partial {\tau} ^ 2} - a\left({\tau}, {\chi}\right)^{2} r\left({\tau}, {\chi}\right) \frac{\partial\,a}{\partial {\tau}} \frac{\partial\,r}{\partial {\tau}} - a\left({\tau}, {\chi}\right)^{3} r\left({\tau}, {\chi}\right) \frac{\partial^2\,r}{\partial {\tau} ^ 2} - r\left({\tau}, {\chi}\right) \frac{\partial\,a}{\partial {\chi}} \frac{\partial\,r}{\partial {\chi}} + a\left({\tau}, {\chi}\right) r\left({\tau}, {\chi}\right) \frac{\partial^2\,r}{\partial {\chi} ^ 2}}{a\left({\tau}, {\chi}\right)^{3}}

It is equivalent to

In [50]:
eq = (- E[2,2] * a(t,x)^3).expr() == 0 eq
Λa(τ,χ)3r(τ,χ)2+a(τ,χ)2r(τ,χ)22(τ)2a(τ,χ)+a(τ,χ)2r(τ,χ)τa(τ,χ)τr(τ,χ)+a(τ,χ)3r(τ,χ)2(τ)2r(τ,χ)+r(τ,χ)χa(τ,χ)χr(τ,χ)a(τ,χ)r(τ,χ)2(χ)2r(τ,χ)=0-{\Lambda} a\left({\tau}, {\chi}\right)^{3} r\left({\tau}, {\chi}\right)^{2} + a\left({\tau}, {\chi}\right)^{2} r\left({\tau}, {\chi}\right)^{2} \frac{\partial^{2}}{(\partial {\tau})^{2}}a\left({\tau}, {\chi}\right) + a\left({\tau}, {\chi}\right)^{2} r\left({\tau}, {\chi}\right) \frac{\partial}{\partial {\tau}}a\left({\tau}, {\chi}\right) \frac{\partial}{\partial {\tau}}r\left({\tau}, {\chi}\right) + a\left({\tau}, {\chi}\right)^{3} r\left({\tau}, {\chi}\right) \frac{\partial^{2}}{(\partial {\tau})^{2}}r\left({\tau}, {\chi}\right) + r\left({\tau}, {\chi}\right) \frac{\partial}{\partial {\chi}}a\left({\tau}, {\chi}\right) \frac{\partial}{\partial {\chi}}r\left({\tau}, {\chi}\right) - a\left({\tau}, {\chi}\right) r\left({\tau}, {\chi}\right) \frac{\partial^{2}}{(\partial {\chi})^{2}}r\left({\tau}, {\chi}\right) = 0

We substitute for a(τ,χ)a(\tau,\chi) the value found when solving the τχ\tau\chi component:

In [51]:
eq1 = eq.substitute_function(a, af).simplify_full() eq1
f(χ)r(τ,χ)χf(χ)χr(τ,χ)2+(Λr(τ,χ)2r(τ,χ)2(τ)2r(τ,χ))χr(τ,χ)3(r(τ,χ)23(τ)2χr(τ,χ)+r(τ,χ)τr(τ,χ)2τχr(τ,χ))χr(τ,χ)2f(χ)3=0-\frac{f\left({\chi}\right) r\left({\tau}, {\chi}\right) \frac{\partial}{\partial {\chi}}f\left({\chi}\right) \frac{\partial}{\partial {\chi}}r\left({\tau}, {\chi}\right)^{2} + {\left({\Lambda} r\left({\tau}, {\chi}\right)^{2} - r\left({\tau}, {\chi}\right) \frac{\partial^{2}}{(\partial {\tau})^{2}}r\left({\tau}, {\chi}\right)\right)} \frac{\partial}{\partial {\chi}}r\left({\tau}, {\chi}\right)^{3} - {\left(r\left({\tau}, {\chi}\right)^{2} \frac{\partial^{3}}{(\partial {\tau})^{2}\partial {\chi}}r\left({\tau}, {\chi}\right) + r\left({\tau}, {\chi}\right) \frac{\partial}{\partial {\tau}}r\left({\tau}, {\chi}\right) \frac{\partial^{2}}{\partial {\tau}\partial {\chi}}r\left({\tau}, {\chi}\right)\right)} \frac{\partial}{\partial {\chi}}r\left({\tau}, {\chi}\right)^{2}}{f\left({\chi}\right)^{3}} = 0
In [52]:
eq2 = (eq1 * f(x)^3).simplify_full() eq2
f(χ)r(τ,χ)χf(χ)χr(τ,χ)2(Λr(τ,χ)2r(τ,χ)2(τ)2r(τ,χ))χr(τ,χ)3+(r(τ,χ)23(τ)2χr(τ,χ)+r(τ,χ)τr(τ,χ)2τχr(τ,χ))χr(τ,χ)2=0-f\left({\chi}\right) r\left({\tau}, {\chi}\right) \frac{\partial}{\partial {\chi}}f\left({\chi}\right) \frac{\partial}{\partial {\chi}}r\left({\tau}, {\chi}\right)^{2} - {\left({\Lambda} r\left({\tau}, {\chi}\right)^{2} - r\left({\tau}, {\chi}\right) \frac{\partial^{2}}{(\partial {\tau})^{2}}r\left({\tau}, {\chi}\right)\right)} \frac{\partial}{\partial {\chi}}r\left({\tau}, {\chi}\right)^{3} + {\left(r\left({\tau}, {\chi}\right)^{2} \frac{\partial^{3}}{(\partial {\tau})^{2}\partial {\chi}}r\left({\tau}, {\chi}\right) + r\left({\tau}, {\chi}\right) \frac{\partial}{\partial {\tau}}r\left({\tau}, {\chi}\right) \frac{\partial^{2}}{\partial {\tau}\partial {\chi}}r\left({\tau}, {\chi}\right)\right)} \frac{\partial}{\partial {\chi}}r\left({\tau}, {\chi}\right)^{2} = 0

Then we substitute for r/τ\partial r/\partial\tau the value obtained when solving the ττ\tau\tau component:

In [53]:
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
0=00 = 0

We conclude that the θθ\theta\theta component of Einstein equation does add any independent equation.