The expanding Universe - Astroparticles and Cosmology

In the previous chapter, through simple geometrical considerations, we managed to write down the form of the metric solution to Einstein’s equation for a homogeneous, isotropic Universe. From an unknown tensor with 10 components (because the metric is a symmetrical tensor), through symmetry arguments we have arrived at the FLRW metric, which contains just one unknown function of time $a(t)$ . To describe the dynamics of the Universe, rather than its geometry, we need to solve Einstein’s equation to understand how matter and energy content affect the expansion of the Universe via the scale factor $a(t)$ .

1The energy-momentum tensor¶

The energy-momentum tensor represents the flux of quadri-momentum p^\mu and the energy density \epsilon in a local volume of space-time. If the physical system studied in this local volume is not subject to any working force apart from gravitation, then we have the conservation equation T^{\mu\nu}_{\;\;\;;\mu}=0. — The energy-momentum tensor represents the flux of quadri-momentum $p^\mu$ and the energy density ε in a local volume of space-time. If the physical system studied in this local volume is not subject to any working force apart from gravitation, then we have the conservation equation $T^{\mu\nu}_{\;\;\;;\mu}=0$ .

The energy-momentum tensor $T^{\mu\nu}$ in Einstein’s equation describes the energy density and momentum fluxes in relativistic mechanics. It is a second-order tensor, constructed from the 4-momentum vector, and takes the following form:

T^{\mu\nu}=\begin{pmatrix} T^{00}= \text{energy density}\,\,\,\, & T^{01}= \text{energy/c flux through }x_1\,\,\,\,\, & T^{02}=\text{energy/c flux through }x_2\,\,\,\,\, & T^{03}=\text{energy/c flux through }x_3 \\ T^{10}=c\times \text{density of }p_1,\,\,\,\,\,\, & T^{11}= \text{flux of }p_1\text{ through }x_1\,\,\,\,\,\,\, & T^{12}= \text{flux of }p_1\text{ through }x_2\,\,\,\,\,\,\, & T^{13}= \text{flux of }p_1\text{ through }x_3 \\ T^{20}= c\times\text{density of }p_2\,\,\,\,\,\,\,\, & T^{21}= \text{flux of }p_2\text{ through }x_1\,\,\,\,\,\,\, & T^{22}= \text{flux of }p_2\text{ through }x_2\,\,\,\,\,\,\, & T^{31}= \text{flux of }p_2\text{ through }x_3 \\ T^{30}= c\times\text{density of }p_3\,\,\,\,\,\,\,\, & T^{31}= \text{flux of }p_3\text{ through }x_1\,\,\,\,\,\,\, & T^{32}= \text{flux of }p_3\text{ through }x_2\,\,\,\,\,\,\,\, & T^{33}= \text{flux of }p_3\text{ through }x_3 \end{pmatrix}

(1)

If the physical system studied in this local volume is not subject to any working force other than gravitation, then we have the conservation equation $T^{\mu\nu}_{\;\;\;;\mu}=0$ , a set of four equations representing the local conservation equation for energy and momentum.

A few remarks on the components of this tensor:

$T^{00}$ is the local ε energy density, generally the dominant component of the energy-momentum tensor;
$T^{ii}$ represent the flux of momentum through a surface, and hence the kinetic pressure $P$ exerted by the physical system in the $\vec e_i$ direction;
$T^{ij},\ i \neq j$ represent momentum flows lateral to displacements, i.e. viscosity or shear phenomena.

Now, in our hypothesis of a Universe of maximum symmetry, let’s first recall that we can define a cosmic, universal time, using the physical evolution of the Universe as a clock (matter density, CMB temperature...). The hypersurfaces of space-time parametrized by this universal time are then themselves subspaces of maximum symmetry. The $\mathcal{T}$ tensors representing the cosmological observables of such maximally symmetric subspaces must then be of form invariant, i.e. they remain the same functions of the spatial coordinates at a date $t$ whatever the chosen coordinate system: if we go from a $x^\rho$ system to $x'^\rho$ , we must have $\mathcal{T}'_{\mu\nu\ldots}(x'^\rho) = \mathcal{T}_{\mu\nu\ldots}(x'^\rho)$ . Intuitively, if $\mathcal{T}$ is the energy-momentum tensor, this means, among other things, that the energy density must be identical at all points for any choice of coordinate system Weinberg, 1972[p. 409]. We can then demonstrate an important property concerning the form that the tensors of these subspaces Weinberg, 1972[p. 392] must take.

Therefore, mathematically we can introduce $\epsilon(t)$ and $P(t)$ two functions of time such that the energy-momentum tensor simplifies into :

\begin{align} T^{00} & = \epsilon(t) &\quad \text{(scalar)} \\ T^{i0} & = T^{0i} = 0 & \quad \text{(vector)} \\ T^{ij} & = P(t) \gamma^{ij}& \quad \text{(second-order tensor)} \end{align}

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More elegantly, we can introduce the quadri-vector $U^\mu$ defined by :

U^0 = 1, \quad U^i = 0

(12)

and obtain a compact expression for the energy-momentum tensor of a homogeneous, isotropic Universe:

T^{\mu\nu} = (\epsilon + P) U^\mu U^\nu + P g^{\mu\nu}

(13)

Using the FLRW metric, solution of a homogeneous and isotropic universe as well, the energy-momentum tensor is written :

T^{\mu\nu} = (\epsilon + P) U^\mu U^\nu + P g^{\mu\nu} = \begin{pmatrix} -\epsilon g^{00} & 0 & 0 & 0 \\ 0 & P g^{11} & 0 & 0 \\ 0 & 0 & P g^{22} & 0 \\ 0 & 0 & 0 & P g^{33} \\ \end{pmatrix}

(14)

In a Cartesian basis and flat space, the energy-momentum tensor takes the simple form:

T^{mu\nu} = \begin{pmatrix} \epsilon & 0 & 0 & 0 \\ 0 & P/a^2(t) & 0 & 0 \\ 0 & 0 & P/a^2(t) & 0 \\ 0 & 0 & 0 & P/a^2(t) \\ \end{pmatrix}.

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How should we interpret these mathematical considerations? Firstly, if we compare equation (14) with (1) then we identify ε with energy density and $P$ with kinetic pressure (flow of momentum across a surface)^[1]. Next, the energy-momentum tensor $T^{\mu\nu}$ is identified with that of a perfect fluid. This means that in a homogeneous, isotropic Universe, matter can be described as a continuous medium, whose evolution can be described without taking into account viscosity and thermal conduction effects. The thermodynamic evolution of the Universe is therefore adiabatic. Finally, $U^\mu$ is then identified with the comobile quadri-velocity of the fluid, so the fact that $U^i = 0$ shows that the physical system under study is at rest in the comobile coordinates, as expected.

Energy-momentum tensor of a perfect fluid Weinberg, 1972[p. 48]

Let’s study a perfect fluid, i.e. a set of particles whose mean free path is small compared with the distances at which it is studied, and without viscosity. Given the definition of an energy-momentum tensor, in the $\mathcal{R}'$ frame of reference where the perfect fluid is at rest, we can write:

T'^{ij} = P \delta^{ij}, \quad T'^{i0} = T'^{0i} = 0, \quad T'^{00} = \rho c^2

(16)

where ρ is explicitly the fluid’s own massic density and $P$ its kinetic pressure (i.e. a flow of momentum across a surface). In another reference frame, that of a flow observer for example, this energy-momentum tensor is rewritten:

T^{\mu\nu} = \Lambda^{\mu}_{\;\alpha} \Lambda^{\nu}_{\;\beta} T^{\alpha\beta}

(17)

with $\Lambda^\mu_{\;\alpha}$ the Lorentz transformation defined by equation (7). More explicitly:

T^{ij} = P \delta^{ij} + (P + \rho c^2) \frac{v^i v^j}{c^2- v^2}, \quad T^{i0} = (P + \rho c^2) \frac{c v ^i}{c^2 - v^2}, \quad T^{00} = \frac{\rho c^4 + P v^2}{c^2 - v^2}

(18)

Let’s define the quadri-velocity in comobile coordinates as follows:

\vec U = \frac{ \dd \vec x }{c \dd \tau} = \frac{ \vec v / c }{ \sqrt{1-v^2}}, \quad U^0 = \frac{ \dd t }{ \dd \tau} = \frac{1 }{ \sqrt{1-v^2}}, \quad U^\mu U^\mu = -c^2

(19)

then the tensor is written:

T^{\mu\nu} = (\rho c^2 + P) U^\mu U^\nu + P \eta^{\mu\nu}

(20)

Equation (14) therefore corresponds well to the definition of an energy-momentum tensor for a perfect fluid in the relativistic framework.

Note that in a flat space-time, the conservation of the energy-momentum tensor of a perfect fluid allows us to recover the Navier-Stokes equation without viscosity or external forces, and the conservation of matter equation. For simplicity’s sake, let’s return to the case of an incompressible fluid, so $\dd \rho / \dd t = 0$ and non-relativistic, so $P / \rho c^2 \propto (v/c)^2 \ll 1$ . Then:

0 = \frac{\partial T^{\mu\nu}}{\partial x^\nu} \Rightarrow \left\lbrace\begin{align*} & \mu = i:\ & 0 = & \vec\nabla P + \rho \frac{\partial \vec v}{\partial t} + \rho (\vec v \cdot \vec\nabla)(\vec v)\\ \mu = 0:\ & 0 = & \rho \vec \nabla \vec v + \frac{\partial \rho}{\partial t} \end{align*}\right.

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2Friedmann equations¶

Solving Einstein’s equation (42) involves finding a solution metric, given the distribution of matter and energy encoded in $T^{\mu\nu}$ . Assuming the principles of homogeneity and isotropy for this tensor, the metric is the Friedmann-Lemaître-Robertson-Walker (FLRW) metric, using the usual set of spherical comobile coordinates $(ct, \sigma, \theta, \phi)$ :

\begin{aligned} \displaystyle g_{mu\nu} = \begin{pmatrix} -1 & 0 & 0 & 0 \\ 0 & \dfrac{a^2(t)}{1-k\sigma^2} & 0 & 0 \\ 0 & 0 & a^2(t)\sigma^2 & 0 \\ 0 & 0 & 0 & a^2(t) \sigma^2 \sin^2 \theta \\ \end{pmatrix},\end{aligned}

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where $a(t)$ is an unknown function. The scale parameter $a(t)$ can be obtained by solving the Einstein equation knowing the content of the Universe’s energy-momentum tensor $T^{\mu\nu}$ and the value of $k$ . For the FLRW metric, its inverse is simply:

g^{\mu\nu} = \begin{pmatrix} -1 & 0 & 0 & 0 \\ 0 & \dfrac{1-k\sigma^2}{a^2(t)} & 0 & 0 \\ 0 & 0 & \dfrac{1}{a^2(t)\sigma^2} & 0 \\ 0 & 0 & 0 & \dfrac{1}{a^2(t) \sigma^2 \sin^2 \theta} \\ \end{pmatrix}.

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Using the FLRW metric (22), let’s calculate the following affine connection from equation (20):

\begin{aligned} \Gamma^1_{\ 01} & = \frac{1}{2} g^{1 \mu} \left( \partial_0 g_{1\mu} + \partial_1 g_{0 \mu} - \partial_\mu g_{01} \right) \\ & = \frac{1}{2} g^{1 1} \left(\frac{\partial g_{11}}{c\partial t} + \partial_\sigma g_{01} - 0 \right) \text{ because }\forall \mu \neq 1, g^{1\mu}=0\\ & = \frac{1}{2} \frac{1-k\sigma^2}{a^2} \left( \frac{2 \dot{a} a}{c(1-k\sigma^2)} + 0 \right) \\ & = \frac{\dot a}{ca} = \frac{H}{c}. \end{aligned}

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In the same way, we obtain the other affine connections, then the Riemann and Ricci tensors. In the end, the Einstein tensor is diagonal and is worth:

\begin{aligned} G_{00} & = - 3 \left( \frac{\dot{a}^2}{c^2 a^2}+ \frac{k}{a^2} \right), \\ G_{ij} & = \frac{2\ddot{a}a + \dot{a}^2 + c^2 k}{c^2 a^2}g_{ij} \text{ for } i=j\neq 0. \end{aligned}

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From Einstein’s equation (42) and the energy-momentum tensor (15), we obtain for the coordinate 00 and for the spatial coordinates $ij$ :

\begin{aligned} G_{\mu\nu}-\Lambda g_{\mu\nu} & = -8\pi \GN T_{\mu\nu}/c^4 \GN \Leftrightarrow & \left\lbrace \begin{array}{rl} \text{00: } & \displaystyle{3 \left( \frac{\dot{a}^2}{a^2}+ \frac{c^2 k}{a^2} \right) = 8\pi \GN \rho + c^2 \Lambda} \\ ij\text{: } & \displaystyle{\frac{2\ddot{a}a + \dot{a}^2 + c^2 k}{a^2} = - \frac{8\pi \GN}{c^2 } P + c^2 \Lambda } \end{array} \right.\end{aligned}

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These are the two Friedmann equations. Here they are expressed in terms of the Hubble parameter $H=\dot{a}/a$ :

\left\lbrace \begin{array}{rl} \text{00: } & \displaystyle{H^2 = \frac{8\pi \GN \rho}{3} + \frac{c^2 \Lambda}{3} - \frac{c^2 k}{a^2}}\\ ij\text{: } & \displaystyle{2\dot{H} + 3H^2 = - \frac{8\pi \GN}{c^2 } P + c^2 \Lambda - \frac{c^2 k}{a^2}} \end{array} \right.

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The first Friedmann equation explicitly relates the evolution of the scale factor $a(t)$ to the energy content of the Universe. Moreover, by subtracting these two equations and combining the result with the time derivative of the first, we can obtain the energy conservation equation that would also be obtained directly by calculating $T^{\mu\nu}_{\;\;\;;\mu}=0$ in the FLRW metric:

\boxed{\dot{\epsilon} = -3 H( \epsilon + P )}

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Exercise 1

In this exercise, we use the FLRW metric in cartesian coordinates, and the corresponding stress energy tensor for a perfect fluid:

g_{\mu\nu} = \begin{pmatrix} -1 & 0 & 0 & 0 \\ 0 & a^2(t) & 0 & 0 \\ 0 & 0& a^2(t) & 0 \\ 0 & 0& 0& a^2(t) \\ \end{pmatrix}\qquad T_{\mu\nu}=\begin{pmatrix} \rho c^2 & 0 & 0 & 0 \\ 0 &a^2 P & 0 & 0 \\ 0&0&a^2 P &0 \\ 0&0&0&a^2 P \end{pmatrix}

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We recall the following formula for the Christoffel and Riemann tensors:

\begin{align} \Gamma^\kappa_{\ \mu\nu} &= \frac{1}{2}g^{\kappa\lambda}\left( g_{\mu\lambda,\nu}+g_{\lambda\nu,\mu}-g_{\mu\nu,\lambda} \right) \\ R^{\mu}_{\ \nu\alpha\beta} &= -\Gamma^\mu_{\ \nu\beta,\alpha} + \Gamma^\mu_{\ \nu\alpha,\beta} - \Gamma^\lambda_{\ \nu\beta}\Gamma^\mu_{\ \lambda\alpha} + \Gamma^\lambda_{\ \nu\alpha}\Gamma^\mu_{\ \lambda\beta} \\ R_{\mu\nu}& = R^\rho_{\ \mu\rho\nu} = -\Gamma^\rho_{\ \mu\nu,\rho} + \Gamma^\rho_{\ \mu\rho,\nu} - \Gamma^\rho_{\ \rho\lambda}\Gamma^\lambda_{\ \mu\nu} + \Gamma^\rho_{\ \nu\lambda}\Gamma^\lambda_{\ \mu\rho} \\ R &= g^{\mu\nu}R_{\mu\nu} \end{align}

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with $_{,\mu}$ the derivative $\partial / \partial x^\mu$ .

Show that $\Gamma^0_{\ ij} = \dot{a}a/c \delta_{ij}$ and $\Gamma^j_{\ 0i}=\dot{a}/(ac)\delta^j_i$ , with the latin index standing for the spatial coordinates $i=1,2,3$ . The other Christoffel symbols are null.
Show that $R_{00}=3\ddot{a}/(ac^2)$ and $R_{ij}=-(\ddot{a}a+2\dot{a}^2)\delta_{ij}/c^2$ .
Show that $R=-6\left(\ddot{a}/a+(\dot{a}/a)^2\right)/c^2$ and finally that $G_{\mu\nu}=R_{\mu\nu}-g_{\mu\nu}R/2$ is:

G_{\mu\nu}={1 \over c^2}\begin{pmatrix} 3{\dot{a}^2\over a^2} & 0 & 0 & 0 \\ 0 &-2\ddot{a}a-\dot{a}^2 & 0 & 0 \\ 0&0& -2\ddot{a}a-\dot{a}^2 &0 \\ 0&0&0&-2\ddot{a}a-\dot{a}^2 \end{pmatrix}

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Write the two Friedmann equations in term of the Hubble parameter $H=\dot{a}/a$ instead of $a$ .
From the two Friedmann equations, find the energy conservation equation law $\dot{\rho} c^2 = -3 H(\rho c^2 + P)$ .

Solution to Exercise 1

Let’s compute the Christoffel symbols:

\begin{align} \Gamma^{0}_{\mu\nu} & = \frac{1}{2}g^{0\lambda}\left( g_{\mu\lambda,\nu}+g_{\lambda\nu,\mu}-g_{\mu\nu,\lambda} \right) \\ & = \frac{1}{2}g^{00}\left( g_{\mu 0,\nu}+g_{0\nu,\mu}-g_{\mu\nu,0} \right) \\ & = \frac{1}{2}g^{00}\left( g_{0 0,\nu}+g_{00,\mu}-g_{\mu\nu,0} \right) \\ & = \frac{1}{2}(-1)\times\left( 0+ 0 -g_{\mu\nu,0} \right) \\ & = \frac{1}{2}g_{\mu\nu,0} \end{align}

(32)

We find:

\Gamma^{0}_{ij} = -\frac{1}{2}g_{ij,0} = -\frac{1}{2}\frac{\partial (-a^2(t))}{c\partial t} \delta_{ij}= \frac{\dot{a}a}{c} \delta_{ij}

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with $\delta_{ij}$ the Kronecker symbol such that $\delta_{ij}=1$ only if $j=i$ . It is a number (not a matrix). Then:

\begin{align} \Gamma^{j}_{0i} & = \frac{1}{2}g^{j\lambda}\left( g_{0\lambda,i}+g_{\lambda i,0}-g_{0i,\lambda} \right) \\ & = \frac{1}{2}g^{jk}\left( g_{0k,i}+g_{ki,0}-g_{0i,k} \right) \\ & = \frac{1}{2}g^{jk}\left( 0+g_{ki,0}-0 \right) \\ & = \frac{1}{2}g^{jk}\times\left( \frac{-2a\dot{a}}{c}\delta_{ik} \right) \\ & = g^{jk}\times\left( +\frac{\dot{a}}{ac}g_{ik} \right) \\ & = \frac{\dot{a}}{ac}\delta^j_i \end{align}

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with $\delta^j_i$ the identity matrix ( $\delta^i_i = 3$ , but $\delta_{ii}=1$ ).

Now the Ricci tensors :

\begin{align} R_{00} & = -\Gamma^\rho_{\ 00,\rho} + \Gamma^\rho_{\ 0\rho,0} - \Gamma^\rho_{\ \rho\lambda}\Gamma^\lambda_{\ 00} + \Gamma^\rho_{\ 0\lambda}\Gamma^\lambda_{\ 0\rho} \\ & = 0 + \Gamma^\rho_{\ 0\rho,0} + 0 + \Gamma^\rho_{\ 0\lambda}\Gamma^\lambda_{\ 0\rho} \\ & = + \Gamma^i_{\ 0i,0} + 0 + \Gamma^i_{\ 0j}\Gamma^j_{\ 0i} \\ & = 3\frac{\ddot{a}a-\dot{a}^2}{a^2 c^2} + \left(\frac{\dot{a}}{ac}\right)^2 \delta^i_j \delta^j_i\\ & = 3\frac{\ddot{a}a-\dot{a}^2}{a^2 c^2} + 3 \left(\frac{\dot{a}}{ac}\right)^2\\ & = 3 \frac{\ddot{a}}{ac^2} \end{align}

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\begin{align} R_{ij} & = -\Gamma^\rho_{\ ij,\rho} + \Gamma^\rho_{\ i\rho,j} - \Gamma^\rho_{\ \rho\lambda}\Gamma^\lambda_{\ ij} + \Gamma^\rho_{\ i\lambda}\Gamma^\lambda_{\ j\rho} \\ & = -\Gamma^0_{\ ij,0} + 0 -\Gamma^k_{\ k0}\Gamma^0_{\ ij}+\Gamma^0_{\ ik}\Gamma^k_{\ j0} + \Gamma^k_{\ i0} \Gamma^0_{\ jk} \\ & = -\frac{\ddot{a}a+\dot{a}^2}{c^2}\delta_{ij} -\frac{\dot{a}}{ac}\frac{\dot{a}a}{c}\delta^k_k \delta_{ij} + \frac{\dot{a}}{ac}\frac{\dot{a}a}{c} \delta_{ik}\delta^k_j + \frac{\dot{a}}{ac}\frac{\dot{a}a}{c} \delta_{jk}\delta^k_i\\ & = -\frac{\ddot{a}a+\dot{a}^2}{c^2}\delta_{ij} -3\frac{\dot{a}}{ac}\frac{\dot{a}a}{c} \delta_{ij} + 2 \frac{\dot{a}}{ac}\frac{\dot{a}a}{c} \delta_{ij}\\ & = -\frac{a^2}{c^2}\left(\frac{\ddot{a}}{a}+2\frac{\dot{a}^2}{a^2}\right)\delta_{ij} \end{align}

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with $\delta_{ik}\delta^k_j=\delta_{ij}$ and $\delta^k_k=3$ .

Now the Einstein tensor : From the Einstein equation $G_{\mu\nu} - \Lambda g_{\mu\nu} = -8\pi G T_{\mu\nu}/c^4$ write the Friedmann equations:

\left\lbrace \begin{array}{l} \displaystyle{3 \left( \frac{\dot{a}^2}{a^2}+ \frac{c^2k}{a^2} \right) = 8\pi \GN \rho + c^2 \Lambda} \\ \displaystyle{\frac{2\ddot{a}a + \dot{a}^2 + c^2 k}{a^2} = - 8\pi \GN P/c^2 + c^2 \Lambda } \end{array} \right.

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(here we admit the Friedmann equations with an arbitrary curvature $k$ ).

\begin{align} R &= g^{\mu\nu}R_{\mu\nu} \\ & = g^{00}R_{00} + g^{ij}R_{ij}\\ & = -1\times\left( 3 \frac{\ddot{a}}{ac^2}\right) - g^{ij}\delta_{ij} \frac{a^2}{c^2}\left(\frac{\ddot{a}}{a}+2\frac{\dot{a}^2}{a^2}\right) \\ & = -1\times\left( - 3 \frac{\ddot{a}}{ac^2}\right) -\frac{\delta^i_i}{a^2} \frac{a^2}{c^2}\left(\frac{\ddot{a}}{a}+2\frac{\dot{a}^2}{a^2}\right) \\ & = -6 \left(\frac{\ddot{a}}{ac^2} + \frac{\dot{a}^2}{a^2c^2}\right) \\ \end{align}

(38)

with $g^{ij}=\delta^{ij}/a^2$ and $\delta^i_i=3$ .

\begin{align} G_{00} &= R_{00}-g_{00}R/2 \\ & = 3 \frac{\ddot{a}}{ac^2} - \frac{6}{2}\left(\frac{\ddot{a}}{ac^2} + \frac{\dot{a}^2}{a^2c^2}\right) \\ & =- 3 \frac{\dot{a}^2}{a^2c^2} \end{align}

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\begin{align} G_{ij} &= R_{ij}-g_{ij}R/2 \\ & =-\frac{a^2}{c^2}\left(\frac{\ddot{a}}{a}+2\frac{\dot{a}^2}{a^2}\right)\delta_{ij} -(a^2\delta_{ij})\times \frac{-6}{2}\left(\frac{\ddot{a}}{ac^2} + \frac{\dot{a}^2}{a^2c^2}\right) \\ & = a^2\left(2\frac{\ddot{a}}{ac^2}+ \frac{\dot{a}^2}{a^2c^2}\right) \delta_{ij} \end{align}

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Using the definition of $T_{\mu\nu}$ , we obtain directly the two Friedmann equations:

G_{\mu\nu} - \Lambda g_{\mu\nu} = -\frac{8\pi \GN}{c^4}T_{\mu\nu}

(41)

\begin{align} 3{\dot{a}^2 \over a^2}&=& 8 \pi \GN \rho + c^2 \Lambda\\ 2{\ddot{a} \over a} + {\dot{a}^2 \over a^2}& =& - {8 \pi \GN \over c^2}P + c^2 \Lambda \end{align}

(42)

Using $H=\dot a / a$ :

\left\lbrace \begin{array}{l} \displaystyle{3 H^2+ \frac{3kc^2}{a^2} = 8\pi \GN \rho + c^2\Lambda} \\ \displaystyle{2\dot{H} + 3H^2 + \frac{c^2k}{a^2} = - 8\pi \GN P/c^2 + c^2\Lambda } \end{array} \right.

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using $\ddot{a}/a=\dot{H}+H^2$ .

And finally the energy-momentum tensor conservation:

\begin{align} 8\pi G \dot{\rho} & = 6\dot{H}H - \frac{6c^2 k \dot{a}}{a^3} \\ & = 3H \left[-8\pi \GN P / c^2 + c^2 \Lambda - 3 H^2 - \frac{kc^2}{a^2}\right] - 6 H \frac{kc^2}{a^2} \\ & = 3H \left[-8\pi \GN P / c^2 -8 \pi \GN \rho + 2\frac{kc^2}{a^2}\right] - 6 H \frac{kc^2}{a^2} \\ & = 3H \left[-8\pi \GN P / c^2 -8 \pi \GN \rho \right] \end{align}

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and $\dot{\rho} c^2 = -3 H(\rho c^2 + P)$ . This is the same result as in the $T^{\beta \alpha}_{\,\,\,\,\,\,;\alpha}=0$ computation (as expected).

Solution to Exercise 2

\dd U = T \dd S - P \dd V

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U = a^3 \epsilon, \quad V = a^3

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\dd(a^3 \epsilon) = - P \dd (a^3) + T \dd S \Rightarrow 3 \dot{a} a^2 \epsilon + a^3 \dot{\epsilon} = - 3 P \dot{a} a^2 + T \frac{\dd S}{\dd t}\Rightarrow \dot{\epsilon} = -3\frac{\dot{a}}{a}(P+\epsilon) +T \frac{\dd S}{\dd t}

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\frac{\dd S}{\dd t} = 0

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and the expansion is isentropic. This is to be expected, given that for a homogeneous, isotropic Universe, the energy-momentum tensor is that of a perfect fluid, i.e. without viscosity or heat transfer. Evolution is therefore adiabatic ( $\delta Q=0$ ).

3Cosmological inventory¶

The energy-momentum tensor includes non-relativistic and relativistic matter. Relativistic matter is generally referred to as radiation, since CMB photon radiation now dominates this component.

3.1Matter¶

Non-relativistic matter exerts no pressure, so

P_m=0,

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then :

\dot{\rho}_m = -3 H\rho_m \Rightarrow \rho_m = \rho_m^0 \left(\frac{a_0}{a}\right)^{3}.

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This last relationship reflects the fact that if a box of side $a$ containing a certain quantity of matter sees the length of its sides doubled, then the density of matter is indeed divided by 2³.

3.2Photons and neutrinos¶

For relativistic matter (photons, neutrinos),

P_r = \frac{1}{3} \epsilon_r,

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therefore :

\dot{\epsilon}_r = -4 H\epsilon_r \Rightarrow \epsilon_r = \epsilon_r^0 \left(\frac{a_0}{a}\right)^{4}.

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The reasoning with a cubic box of side $a$ also applies here, but if all the lengths double, so does the wavelength of the radiation, so its energy is divided by 2. The radiation energy density decreases in 2⁴.

3.3Cosmological constant¶

In Friedmann’s equations (27), the cosmological constant Λ and curvature $k$ can be interpreted in terms of energy densities in the same way as the energy-momentum tensor ρ.

The energy density associated with the cosmological constant is sometimes called the dark energy density, because of the strange properties associated with it:

\epsilon_\Lambda(t) = \rho_\Lambda c^2 = \frac{c^4 \Lambda}{8\pi \GN} = \text{ constant }.

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As we can see, the energy density associated with the cosmological constant is constant over time, so its behavior is quite unusual: whatever the size of the Universe, there is always as much energy per unit volume. It is therefore not diluted like ordinary energy when the Universe is expanding. What’s more, thanks to Friedmann’s second equation, we can see that the pressure associated with the cosmological constant would be :

P_\Lambda = - \epsilon_\Lambda,

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a negative pressure! In ordinary physics, one of the rare phenomena where negative pressures are involved is cavitation (<wiki:Pressure#Negative_pressures). By positing $\epsilon_{\mathrm{tot}}=\epsilon + \epsilon_\Lambda$ (and $P_{\mathrm{tot}}=P + P_\Lambda$ ) then combining the two Friedmann equations (27) so as to eliminate the curvature term, we obtain:

2\dot{H} + 2H^2 = \frac{2\ddot{a}}{a} = -\frac{8\pi \GN}{3}\left( \epsilon _{\mathrm{tot}} + 3P_{\mathrm{tot}}\right).

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We see that the expansion of the Universe accelerates ( $\ddot{a}>0$ ) if $P_{\mathrm{tot}}<-\epsilon_{\mathrm{tot}}/3$ . Since the Universe consists essentially of non-relativistic matter and radiation, the previous condition becomes equivalent to :

\ddot{a} > 0 \Leftrightarrow \epsilon_\Lambda > \epsilon_r + \epsilon_m/2

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. In conclusion, if the cosmological constant dominates the energy content of the Universe, then it generates such negative pressure that the Universe enters accelerated expansion.

3.4Curvature¶

The energy density associated with curvature energy can be identified as :

\epsilon_k(t) =\rho_k(t) c^2 = - \frac{3 c^4 k }{8\pi \GN a^2(t)}.

(63)

Similarly, its effect in terms of pressure is :

P_k = \frac{c^4 k}{8\pi \GN a^2(t)}.

(64)

4Cosmological parameters¶

4.1Equation-of-state parameters¶

The equation of state $w$ associated with a component of the Universe is defined by the ratio of its pressure to its energy density:

\boxed{w=P/\epsilon}

(65)

Non-relativistic cold matter exerts no pressure on its external environment from which $P_m=0$ hence $w_m=0$ .
Relativistic matter, on the other hand, exerts a pressure on its medium relativistic matter exerts a pressure on its environment of $P_r = \epsilon_r / 3$ , hence $w_r=1/3$ .
For the cosmological constant, we have $P_\Lambda = - \epsilon_\Lambda$ , so its equation of state is constant and negative $w_\Lambda = -1$ .
Curvature assimilated to a perfect fluid would have an equation-of-state parameter $w_k=1/3$ .

4.2Energy density parameters¶

We can define a critical density, corresponding to the density we should have in a homogeneous, isotropic, expanding universe with zero spatial curvature (cf equation (27)):

\rho_c(t) = \frac{3H^2(t)}{8\pi \GN}.

(66)

It’s also convenient to define its current value:

\rho_{c}^0 = \frac{3H^2_0}{8\pi \GN} = 1.1 \times 10^{-29} \left( \frac{H_0}{75\text{ km/s/Mpc}}\right)^2\text{ g/cm}^3 \approx 6 \text{ protons/m}^3.

(67)

where $H_0$ is the Hubble constant.

The density parameters (dimensionless) are defined by normalizing the energy densities by the critical density, i.e. :

\Omega_m(t) = \frac{\rho_m(t)}{\rho_c(t)},\quad \Omega_\Lambda(t) = \frac{c^2 \Lambda}{3H^2(t)}, \quad \Omega_k(t) = -\frac{c^2 k}{a^2(t)H^2(t)}

(68)

\Omega_m^0 = \frac{\rho_m^0}{\rho_c^0},\quad \Omega_\Lambda^0 = \frac{c^2 \Lambda}{3H^2_0}, \quad \Omega_k^0 = -\frac{c^2 k}{a_0^2 H^2_0}.

(69)

The first Friedmann equation is then simply written:

1 = \Omega_m(t) + \Omega_r(t) + \Omega_\Lambda(t) + \Omega_k(t)

(70)

\bar H^2 (t) \equiv \frac{H^2(t)}{H_0^2} = \Omega_m^0 \left(\frac{a_0}{a(t)}\right)^{3} + \Omega_r^0 \left(\frac{a_0}{a(t)}\right)^{4} + \Omega_\Lambda^0 + \Omega_k^0 \left(\frac{a_0}{a(t)}\right)^{2}.

(71)

This model of the Universe, linking the prediction of its expansion $\bar H(z)$ to its contents of cosmological constant, matter and radiation, is called the ΛCDM model (Λ for the cosmological constant and CDM for Cold Dark Matter) in the case $k=0$ (flat Universe). This is the standard model of cosmology.

4.3Dark energy models¶

What is the true nature of dark energy? Is it the manifestation of vacuum energy? A second fundamental constant of gravitation? Or a new fifth force? The manifestation of additional spatial dimensions? These questions about the nature of dark energy remain unanswered for the moment, but since the discovery of accelerated expansion in 1998 Riess et al., 1998Perlmutter et al., 1999 new cosmological surveys are underway to precisely measure dark energy’s equation of state $w_{DE}$ : as long as we measure $w_{DE} = w_\Lambda=-1$ then the accelerating expansion can be explained with a single parameter, which is the value of Λ. If the measurements deviate significantly from -1, then more complex models will have to be tested.

This is why today, in addition to the standard ΛCDM model, cosmologists are testing empirical models that look for deviations from the standard model:

Flat $w$ CDM: flat Universe model with free parameters $\Omega_m^0$ , $\Omega_r^0$ and $w_{DE}$ ;
$w$ CDM: any curvature model with free parameters $\Omega_m^0$ , $\Omega_r^0$ , $\Omega_\Lambda^0$ and $w_{DE}$ ;
$w_0w_a$ CDM: model where the equation-of-state parameter for dark energy is given by two free parameters:
$w_{DE}(a) = w_0 + \left(1 - \frac{a}{a_0}\right)w_a$
(72)

The major challenge for current and future cosmological surveys is to measure $w_a$ , in order to measure variations in the acceleration of the expansion of the Universe.

5Cosmological distances¶

Cosmology is an observational science. We need to infer the properties of the Universe without being able to move around or repeat the Big Bang experience, but from our observations alone. Cosmological parameters are linked to the expansion rate of the Universe $H(z)$ . So, to estimate them, we need to be able to measure $H(z)$ . This expansion rate is present in the defined proper distance and comobile distance Sec. 6, but these are not measurable. Telescopes, on the other hand, can measure luminous fluxes and angles: if we know the luminosity of the observed object or its physical size, we can define its distance and link it to the expansion rate $H(z)$ .

5.1Hubble distance¶

With the parameters $c$ and $H_0$ , it’s possible to construct a quantity homogeneous to a length. This distance, typical in cosmology, is called the Hubble distance and is equal to :

D_H = \frac{c}{H_0} = 3000\,\text{Mpc/}h

(73)

where $h$ is usually defined by :

H_0 = 100\,h\,\text{km/s/Mpc}

(74)

So for $h=0.7$ , we find $D_H \approx 4.3 \,\text{Gpc} \approx 14 \,\text{Gly}$ . This value will appear for all (non-moving) distances defined below.

5.2Luminosity distance¶

In a static, flat space, the apparent luminosity of a source at rest at distance $D_L$ would be $L_E/4\pi D_L^2$ . We therefore propose to define the luminosity distance of a source $D_L(z)$ in cosmology as:

\Phi_0 \equiv \frac{L_E}{4 \pi D_L^2(z)}

(75)

Let’s consider a source located in $\sigma_E$ , emitting $\delta N_E$ photons of mean frequency $\nu_E$ at time $t_E$ during $\delta t_E$ (refer again to the Figure 6). Its luminosity is :

L_E = h\nu_E \frac{\delta N_E }{\delta t_E}.

(76)

So the flux density received by an observer with a telescope of aperture size $A$ is :

\Phi_0 = h \nu_0\frac{\delta N_0 }{A \delta t_0}.

(77)

The surface over which the emitted flux is distributed at time $t_0$ is:

S = \int_0^{2\pi} \int_0^\pi \sqrt{-g} \dd\theta \dd\phi = \int_0^{2\pi} \int_0^\pi a^2(t_0)\sigma^2(t_0)\sin^2\theta \dd\theta \dd\phi = 4 \pi a^2_0 \sigma^2_E.

(78)

with $\sigma(t_0)=\sigma_E$ . The number of emitted photons $\delta N_E$ intercepted by the collecting surface of size $A$ is therefore :

\delta N_0 = \delta N_E \frac{A}{4 \pi a^2_0 \sigma^2_E}.

(79)

From the equation (41), we have:

\nu_E = \nu_0 a_0/a(t_E) = \nu_0 (1+z)

(80)

and also:

\delta t_E = \delta t_0/(1+z).

(81)

Hence the received flux:

\Phi_0 = h \nu_0\frac{\delta N_0 }{A \delta t_0 } = h \nu_0 \frac{\delta N_E}{4 \pi a^2_0 \sigma^2_E \delta t_0 } = \frac{L_E}{4 \pi a^2_0 \sigma^2_E(1+z)^2}.

(82)

We deduce the expression for the luminosity distance in a curved, expanding universe, a function of cosmological parameters and redshift:

\Rightarrow D_L(z) = a_0 \sigma_E (1+z) = a_0 (1+z) \left\lbrace \begin{array}{cl} \sin \chi(z) & \text{ si } k=+1 \\ \chi(z) & \text{ si } k=0 \\ \text{sh } \chi(z) & \text{ si } k=-1 \end{array} \right.

(83)

Today, the scale factor $a_0$ is not accessible via Friedmann’s equations, which only give the expansion rate. However, it can be expressed as a function of cosmological parameters and $H_0$ :

\Omega_k^0 = - \frac{kc^2}{H_0^2 a_0^2} \Rightarrow a_0 = \left\lbrace\begin{array}{l} \displaystyle{\frac{c}{H_0\sqrt{-\Omega_k^0}}} \text{ if } k=+1 \\ \text{indeterminate but usually worth } 1 \text{ if } k=0 \\ \displaystyle{\frac{c}{H_0\sqrt{\Omega_k^0}}} \text{ if } k=-1 \end{array} \right.

(84)

Thus:

\displaystyle{\chi(z) = \left\lbrace\begin{array}{cl} \displaystyle{H_0\sqrt{-\Omega_k^0}\int_0^z\frac{dz}{H(z)}} & \text{ if } k=+1 \\\ \displaystyle{\frac{1}{a_0}\int_0^z\frac{cdz}{H(z)}} & \text{ if } k=0 \\\\ \displaystyle{H_0\sqrt{\Omega_k^0}\int_0^z\frac{dz}{H(z)} } & \text{ if } k=-1 \end{array} \right.}

(85)

D_L(z) = (1+z) \left\lbrace \begin{array}{cl} \displaystyle \frac{c}{H_0 \sqrt{-\Omega_k^0}} \sin\left[ H_0 \sqrt{-\Omega_k^0} \int_0^z \frac{\dd z}{H(z)} \right] & \text{ if } k=+1 \\ \displaystyle \int_0^z \frac{c\dd z}{H(z)} & \text{ if } k=0 \\ \displaystyle \frac{c}{H_0 \sqrt{\Omega_k^0}} \sh\left[ H_0 \sqrt{\Omega_k^0} \int_0^z \frac{\dd z}{H(z)} \right] & \text{ if } k=-1 \ \end{array} \right. .

(86)

We have thus obtained a link between a distance measurement obtained by measuring the $\Phi_0$ flux of a star, and a cosmological model based on parameters to be determined. By measuring the fluxes of objects of known intrinsic luminosity $L_E$ , cosmological parameters can be estimated.

5.3Angular distances¶

Angular distance of an object of transverse physical size l. — Figure 2:Angular distance of an object of transverse physical size $l$ .

The last important distance in cosmology is the angular distance of an object $D_A(z)$ . In a static, flat space, the apparent angle δ of an object of physical size $l$ at rest at distance $D_A$ would be $l/D_A$ . We therefore propose to define the angular distance $D_A(z)$ in cosmology as:

\delta = \frac{l}{D_A(z)}

(87)

How is this distance modelled in the FLRW metric? Let’s consider an object of transverse physical size $l$ located in $\sigma=\sigma_E,t=t_E$ and observed today in $\sigma=0,t=t_0$ .

In physical space, the angle δ is the same as in comobile space (we pass from one to the other by a homothety), but also the same at reception and transmission. The angle under which the object is seen is therefore, in all cases, and for any curvature (see Figure 8) :

\delta = \frac{l}{a_E \sigma_E} = \frac{l (a_0/a_E)}{a_0 \sigma_E} = \frac{l_c}{\sigma_E}

(88)

with $l_c = l / a_E$ the comoving size of the object at emission $t_E$ .We propose to define the comoving angular distance or comoving transverse distance simply as:

d_A(z) = \frac{l_c}{\delta} = \sigma_E = \left\lbrace\begin{array}{cl} \sin \chi(z) & \text{ if } k=+1 \\ \chi(z) & \text{ if } k=0 \\ \sinh \chi(z) & \text{ if } k=-1 \end{array} \right.

(89)

We can also deduce the expression for the angular distance in a curved, expanding universe, as a function of cosmological parameters and redshift:

\Rightarrow D_A(z) \equiv\frac{l}{\delta} = a(t_E) \sigma_E=\frac{a_0 \sigma_E}{1+z} = \frac{a_0}{1+z}d_A(z)=\frac{D_L(z)}{(1+z)^2}

(90)

D_A(z) = \frac{a_0}{1+z} \left\lbrace\begin{array}{cl} \sin \chi(z) & \text{ if } k=+1 \\ \chi(z) & \text{ if } k=0 \\ \sinh \chi(z) & \text{ if } k=-1 \end{array} \right.

(91)

D_A(z) = \frac{1}{1+z} \left\lbrace \begin{array}{cl} \displaystyle \dfrac{c}{H_0 \sqrt{-\Omega_k^0}} \sin\left[ H_0 \sqrt{-\Omega_k^0} \int_0^z \dfrac{\dd z}{H(z)} \right] & \text{ if } k=+1 \\ \displaystyle \int_0^z \dfrac{c \dd z}{H(z)} & \text{ if } k=0 \\ \displaystyle \dfrac{c}{H_0 \sqrt{\Omega_k^0}} \sh\left[ H_0 \sqrt{\Omega_k^0} \int_0^z \dfrac{\dd z}{H(z)} \right] & \text{ if } k=-1 \\ \end{array} \right. .

(92)

From exercise Exercise 2, we can see that the use of σ instead of χ is well suited to the three types of Universe curvatures in these distance definitions.

Cosmological parameter measurements

In the context of a flat Universe, angular measurements made by the Planck satellite on the cosmological diffuse background, combined with flux measurements of type Ia supernovae, and angular distances of baryon acoustic oscillations allow us to infer the matter and dark energy content for a flat Universe today (Aghanim et al. (2020)) :

\Omega_m^0 = 0.3111 \pm 0.0056, \quad \Omega_\Lambda^0 = 0.6889 \pm 0.0056

(94)

If curvature is a free parameter of the model (we speak of an extension to the standard ΛCDM model), we measure :

\Omega_k^0 = 0.0007\pm 0.0037

(95)

If we propose a two-parameter model for dark energy, we obtain (Aghanim et al. (2020)) :

w_0 = -0.957 \pm 0.080, \quad w_a = -0.29^{+0.32}_{ -0.26}

(96)

Solution to Exercise 3

Suppose that density of the dark energy as cosmological constant is equal to the present critical density, $\rho_\Lambda=\rho_c$ . What is then the total amount of dark energy inside the Solar System? Compare this number with $M_\odot c^2$ .

\rho_c\approx 10^{-29}\,\text{g/cm}^3

(97)

R\approx 50\,\text{A.U.}

(98)

$$1,\text{A.U.}approx 1.5\times 1011m;EDESS/c2≃0.2⋅1014 kg;M⊙≃2⋅1030 kg;EDESSM⊙c2≃10-17.

Transform Lambda into a length: Length = sqrt(1/Lambda) = ....

Footnotes¶

The choice of notations for these mathematical functions was not made by chance...
↩

References¶

Weinberg, S. (1972). Gravitation and cosmology: principles and applications of the general theory of relativity. http://www.lavoisier.fr/livre/notice.asp?ouvrage=1382255
Riess, A. G., Filippenko, A. V., Challis, P., Clocchiatti, A., Diercks, A., Garnavich, P. M., Gilliland, R. L., Hogan, C. J., Jha, S., Kirshner, R. P., Leibundgut, B., Phillips, M. M., Reiss, D., Schmidt, B. P., Schommer, R. A., Smith, R. C., Spyromilio, J., Stubbs, C., Suntzeff, N. B., & Tonry, J. (1998). Observational Evidence from Supernovae for an Accelerating Universe and a Cosmological Constant. The Astronomical Journal, 116(3), 1009–1038. 10.1086/300499
Perlmutter, S., Aldering, G., Goldhaber, G., Knop, R. A., Nugent, P., Castro, P. G., Deustua, S., Fabbro, S., Goobar, A., Groom, D. E., Hook, I. M., Kim, A. G., Kim, M. Y., Lee, J. C., Nunes, N. J., Pain, R., Pennypacker, C. R., Quimby, R., Lidman, C., … Project, T. S. C. (1999). Measurements of Ω and Λ from 42 High-Redshift Supernovae. The Astrophysical Journal, 517(2), 565–586. 10.1086/307221
Aghanim, N., Akrami, Y., Ashdown, M., Aumont, J., Baccigalupi, C., Ballardini, M., Banday, A. J., Barreiro, R. B., Bartolo, N., Basak, S., Battye, R., Benabed, K., Bernard, J.-P., Bersanelli, M., Bielewicz, P., Bock, J. J., Bond, J. R., Borrill, J., Bouchet, F. R., … Zonca, A. (2020). Planck2018 results: VI. Cosmological parameters. Astronomy & Astrophysics, 641, A6. 10.1051/0004-6361/201833910