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In cosmology, it is usually assumed that the cosmological fluid made of galaxies could be described as a gas of "particles" without any pressure (the dust gas), of density $\rho_{\text{matter}} \propto a^{-3}$ (where $a(t)$ is the scale factor).

At small scales, galaxies tend to clump together because of their mutual gravity, thus building galactic clusters and mega clusters. At a very large scale, this "fluid" appears to be in tension, so it should have a negative pressure instead of the usual $p = 0$ assumption. The dust model appears to ignore any small scale effect.

I'm not talking of any kind of Dark Matter or Dark Energy here, just simple galaxies-particles that are attracting themselves at a small scale. Their local attraction appears to me like a kind of van der Waals interaction. So would it be better to use something else, instead of the dust model, to describe "matter" (i.e. galaxies) at the large - homogeneous - scale?

Even at relatively short times after the Big Bang, when there wasn't any galaxies yet (just an uniform gas of hot particles), the global matter fluid should behave differently than "dust". I never understood the constant use of the equation of state $p = 0$ in modern cosmology (except that it's a simple model that allows easy calculations!).

What could be the equation of state of a "galactic van der Waals" gas describing the cosmological matter fluid at a very large scale?


Could the cosmological matter fluid be described as a kind of "foam"? About the properties of ordinary foam: https://en.wikipedia.org/wiki/Foam. The discussion in the Structure section is interesting. I'm wondering if the cosmological fluid could better be described as a kind of foam (instead of dust), as seen on a very large scale (much larger than the foam bubbles).


The van der Waals equation of state is the following: \begin{equation}\tag{1} p = \frac{n R T}{V - n b} - \frac{a n^2}{V^2} \sim \frac{\alpha \rho}{1 - \beta \rho} - \kappa \, \rho^2, \end{equation} where $a$, $b$, $\beta$ and $\kappa$ are constants. Since the galaxies are stationnary (on average) in the comoving frame, they are defining a non-relativistic gas of "particles" at rest. Their temperature should then be $T \approx 0$, so that $\alpha \approx 0$ (this is the "dust" scenario, for which $w = 0$ in the linear relation $p = w \, \rho$). In that case, the first term of (1) is negligible and we get \begin{equation}\tag{2} p \approx -\, \kappa \, \rho^2. \end{equation} Is that making any sense for the cosmological fluid of matter?

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  • $\begingroup$ I think that the assumption is done at and fior large scale. If you think like this a vdw gas is just dust if is density is low enough. The question is interesting but it seems to me that as soon as we assume homogeneity and isotropy than small scale interaction are excluded a priori $\endgroup$ – Alchimista Jan 4 at 10:29
  • $\begingroup$ @Alchimista, even an homegeneous and isotropic substance could be under tension (negative pressure). Think of soap foam on a sufficiently large scale, where it looks smooth. What is the "pressure" of that foam on a large scale? Is it really $0$? I think it should exert some tension on the walls of the container. $\endgroup$ – Cham Jan 4 at 13:55
  • $\begingroup$ İf the equation of state is described by $P=w \epsilon$ and then the energy density can be written as, $ \epsilon=a(t)^{-3(1+w)}$ To find the $w$ for foam we just need to find how it behaves with respect ot expension of the universe (or in general how the density behaves if we increase the volume of the box) For matter its $w=0$ and $ \epsilon_m=a(t)^{-3}$ for radiation $w=1/3$ hence $ \epsilon_r=a(t)^{-4}$ etc. $\endgroup$ – Reign Jan 6 at 9:22
  • $\begingroup$ @Reign, for foam, I'm expecting $w < 0$. When you say $w = 0$ for matter, don't forget it's actually a simplistic model of "dust". I don't think we could just transpose and apply what we know about matter in a laboratory to "matter" on a very large cosmological scale. I believe that the standard relation $p = w \, \rho$ is too simple (when the state parameter $w$ is independant of $\rho$). I'm expecting something non-linear (i.e. $w$ is a function of $\rho$). The "galactic matter" fluid isn't ordinary matter. $\endgroup$ – Cham Jan 6 at 14:52
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It is a very lengthy answer deriving an energy equation to describe a galaxy as an ideal, non-relativistic plasma with infinite conductivity to prove that $p=0$ is no necessity.


I don't think that $p=0$ is always applied. I will derive an energy equation for a plasma that could describe e.g. a galaxy, if you'd like to, that includes a non-zero pressure term. I'll use an magnetohydrodynamics (MHD) approach. In a formula describing energy conservation in MHD, we expect terms describing the energy of the fluid in the absence of any magnetic fields, such as kinetic energy or fluid pressure, but also terms describing the interaction between the fluid and the magnetic field, such as the work done by the Lorentz force.

If we stay in the Lagrangian frame (a volume element co-moving with the fluid), the energy flow across the surface (It is negative because the energy is flowing out of the volume element.) $-\nabla\cdot\boldsymbol{q}$ plus the volumetric heating rate $R_V$ must equal the heat input rate per unit volume $\rho\frac{\mathrm{d}Q}{\mathrm{d}t}$: \begin{equation}\label{eq:HeatInputRate} \rho\frac{\mathrm{d}Q}{\mathrm{d}t}=-\nabla\cdot\boldsymbol{q}+R_V, \end{equation} with $Q$ the heat unit per mass and $\boldsymbol{q}$ the heat flux through the boundary. The volumetric heating rate $R_V$ can be broken down into the viscous heating rate $\boldsymbol{\Pi}\left(\nabla\boldsymbol{V}\right)$ and the Ohmic heating rate $\boldsymbol{J}\cdot\boldsymbol{E}=\frac{1}{\sigma}\boldsymbol{J}^2$, which is zero under the assumption of infinite conductivity. These quantities can be derived as follows:

Volumetric Viscous Heating Rate The viscous stress tensor $\boldsymbol{\Pi}$, is of the form \begin{equation} \boldsymbol{\Pi}= \begin{pmatrix} 0 & \alpha & \beta \\ \alpha & 0 & \gamma\\ \beta & \gamma & 0 \end{pmatrix}, \end{equation} with, $\alpha$, $\beta$, and $\gamma$ scalars. Herewith, we can calculate the work done on a volume element, $W_V=\boldsymbol{V}\cdot\boldsymbol{F}_V$, by the viscous force $\boldsymbol{F}_V=\nabla\cdot\boldsymbol{\Pi}$: \begin{align} W_F&=\boldsymbol{V}\cdot\boldsymbol{F}_V\\ &=\boldsymbol{V}\cdot\left(\nabla\cdot\boldsymbol{\Pi}\right)\\ &=\nabla\cdot\left(\boldsymbol{\Pi}\cdot\boldsymbol{V}\right)-\boldsymbol{\Pi}\left(\nabla\boldsymbol{V}\right)\\ &=\partial_j\left(v_i\Pi_{ji}\right)-\Pi_{ji}\partial_jv_i, \end{align} where $\nabla\boldsymbol{V}$ refers to the covariant derivative. If we are now interested in the corresponding energies, we have to integrate over the whole volume. Then, we find \begin{align} \int_V \!W_F\,\mathrm{d}V&=\int_V\!\nabla\cdot\left(\boldsymbol{\Pi}\cdot\boldsymbol{V}\right)\,\mathrm{d}V-\int_V\!\boldsymbol{\Pi}\left(\nabla\boldsymbol{V}\right)\,\mathrm{d}V\\ &=\oint_S\!\mathrm{d}\boldsymbol{S}\cdot\left(\boldsymbol{\Pi}\cdot\boldsymbol{V}\right)-\int_V\!\boldsymbol{\Pi}\left(\nabla\boldsymbol{V}\right)\,\mathrm{d}V. \end{align} The first term on the RHS describes the energy on the surface, or more precisely, the work done on the surface; the second the energy stored within the volume. Kinetic energy is lost from the fluid if $\boldsymbol{\Pi}\left(\nabla\boldsymbol{V}\right)>0$ and the surface term is absent. This energy must therefore show up as internal energy! $\boldsymbol{\Pi}\left(\nabla\boldsymbol{V}\right)$ is therefore the volumetric viscous heating rate.

Volumetric Ohmic Heating Rate Analogous to mechanical friction, electric friction$\,—\,$resistivity$\,—\,$also causes heat. $\boldsymbol{J}\cdot\boldsymbol{E}$ gives the rate of work done by the fields in the volume element. Again, taking the integral over the volume would give us the energy, but $\boldsymbol{J}\cdot\boldsymbol{E}$ can be simplified under the assumptions we already made. In the Lagrangian frame we find $\boldsymbol{J}=\frac{1}{\sigma}\boldsymbol{E}$, which is, under the assumption of infinite conductivity, zero.

So far, we have not taken into consideration that we are dealing with a plasma$\,—\,$a statistical ensemble describable by thermodynamics (TD). In TD, we assume that we have a huge, not necessarily known, number of particles acting in a way such that their collective behaviour can be statistically described. One of the most important properties of such an ensemble is its entropy $S$. The total differential of entropy is given by \begin{align} \mathrm{d}S&= \left(\frac{\partial S}{\partial \mathcal{E}}\right)_{V,N}\mathrm{d}\mathcal{E}+ \left(\frac{\partial S}{\partial V}\right)_{\mathcal{E},N}\mathrm{d}V- \left(\frac{\partial S}{\partial N}\right)_{\mathcal{E},V}\mathrm{d}N\\ &=\frac{1}{T}\mathrm{d}\mathcal{E}+\frac{p}{T}\mathrm{d}V-\frac{\mu}{T}\mathrm{d}N\label{eq:dSentropy} \end{align} with $\mathcal{E}$ the energy, $V$ the volume, $p$ the pressure, $\mu$ the chemical potential, $N$ the number of particles, and $T$ the temperature. Energy conservation is a fundamental rule in physics. A galaxy as a whole loses energy by radiation to the surrounding intergalactic space. This is represented by d$\mathcal{E}$.

Let us proceed with the second term of the entropy equation. One could say that the pressure in the galaxy is very low$\,—\,$it is mostly empty space, a vacuum. But the temperature in these areas is also close to $0$ K, so the term $\frac{p}{T}$ doesn't necessarily vanish, it can just as well be unity, leaving the change in volume d$V$ non-negligible. The chemical potential as well as the number of particles changes constantly in a galaxy. Stars are fusing lighter elements to heavier elements, increasing the chemical potential and decreasing the number of particles. But how does this compare to the total number of particles in a galaxy? If we assume that a galaxy starts from a pure hydrogen cloud and we consider the fact that today about 75% of all baryonic mass are still hydrogen and 24% are helium both, the number of particles as well as the chemical potential, change very slowly. As we observe galactic magnetic fields already in young galaxies (high-redshift galaxies), we do not have to consider the whole lifetime of, say, our Milky Way, when we try to understand the origin of magnetic fields. Only a fraction of the age of our Milky Way should be enough to establish a magnetic field. Therefore, the last term from the entropy equation can be neglected and we only have to consider \begin{equation}\label{eq:SentropyReduced} \mathrm{d}S=\frac{1}{T}\mathrm{d}\mathcal{E}+\frac{p}{T}\mathrm{d}V. \end{equation} The entropy is connected to the heat by d$S=\delta Q_{rev}/T$, where $\delta Q_{rev}$ is the change in heat in a reversible process.

We started our considerations about energy with volume elements. The reduced entropy equation describes the galactic magnetic field as a whole. To consider its inner structure, we must also consider the interaction between volume elements. Small packets within the galaxy can exchange energy. Therefore, when considering the inner behaviour, this change in energy has to be considered. For the change in heat for smaller volume elements, we can restate the reduced entropy equation in agreement with the laws of thermodynamics: \begin{equation} \mathrm{d}Q=p\mathrm{d}\left(\frac{1}{\rho}\right)+\mathrm{d}\epsilon, \end{equation} with $p\mathrm{d}\left(\frac{1}{\rho}\right)$ the PV-work per unit mass and $\mathrm{d}\epsilon$ the change in energy per unit mass. Substituting this into the equation describing the heat input rate from the beginning and using the fact that there is no Ohmic heating under infinite conductivity, we find \begin{equation} p\rho\frac{\mathrm{d}\left(\frac{1}{\rho}\right)}{\mathrm{d}t} +\rho\frac{\mathrm{d}\epsilon}{\mathrm{d}t}=-\nabla\cdot\boldsymbol{q}+\boldsymbol{\Pi}\left(\nabla\boldsymbol{V}\right). \end{equation}

The factor $\frac{\mathrm{d}\left(\frac{1}{\rho}\right)}{\mathrm{d}t}$ can be rewritten, using the continuity equation: \begin{align} \frac{\mathrm{d}\left(\frac{1}{\rho}\right)}{\mathrm{d}t}&=\frac{-1}{\rho^2}\frac{\mathrm{d}\rho}{\mathrm{d}t}\\ &=\frac{-1}{\rho^2}\left(-\rho\nabla\cdot\boldsymbol{V}\right)\\ &=\frac{1}{\rho}\nabla\cdot\boldsymbol{V}. \end{align} With this we can express the rate of change of energy per unit volume as \begin{equation} \rho\frac{\mathrm{d}\epsilon}{\mathrm{d}t}=-p\nabla\cdot\boldsymbol{V}-\nabla\cdot\boldsymbol{q}+\boldsymbol{\Pi}\left(\nabla\boldsymbol{V}\right). \end{equation} Again, we can use the continuity equation, but this time to rewrite \begin{equation} \rho\frac{\mathrm{d}\epsilon}{\mathrm{d}t}=\frac{\mathrm{d}}{\mathrm{d}t}\left(\rho\epsilon\right)+\rho\epsilon\nabla\cdot\boldsymbol{V}, \end{equation} which leads to the energy equation

$\begin{equation}\label{eq:energy} \frac{\mathrm{d}}{\mathrm{d}t}\left(\rho\epsilon\right)=-\rho \epsilon\nabla\cdot\boldsymbol{V}-p\nabla\cdot\boldsymbol{V}-\nabla\cdot\boldsymbol{q}+\boldsymbol{\Pi}\left(\nabla\boldsymbol{V}\right). \end{equation}$

If we now approximate the plasma as an ideal gas, we can rewrite this equation. Densities in the interstellar medium are very low, which means few interactions between particles, which basically describes an ideal gas. In TD the energy of an ideal gas depends only on the pressure applied to it: \begin{equation} \rho\epsilon= \frac{p}{\Gamma-1}, \end{equation} with $\Gamma=5/3$ the adiabatic index for plasma. This yields: \begin{align} \frac{\mathrm{d}}{\mathrm{d}t}\left(p\right) &=\left(\Gamma-1\right)\left( -\frac{p}{\Gamma-1}\nabla\cdot\boldsymbol{V}-p\nabla\cdot\boldsymbol{V}-\nabla\cdot\boldsymbol{q}+\boldsymbol{\Pi}\left(\nabla\boldsymbol{V}\right) \right)\\ &=-\Gamma p\nabla\cdot\boldsymbol{V}-\left(\Gamma-1\right)\left( \nabla\cdot\boldsymbol{q}-\boldsymbol{\Pi}\left(\nabla\boldsymbol{V}\right) \right).\label{eq:idealEnergy} \end{align} The first term on the RHS represents the reversible PV-work, the second term the irreversible heating process. The pressure does not have to be zero. q.e.d.

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  • $\begingroup$ There's a minor sign error for $dS$. You guys should really add numbers to equations, using the command \tag{number}. $\endgroup$ – Cham Jan 10 at 15:25
  • $\begingroup$ Thanks a lot! I wasn't aware of the possibility to include equation numbering here in SE, will use it henceforth! $\endgroup$ – kalle Jan 10 at 15:37
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In the Einstein equations

$$G_{\mu\nu} = 8\pi G T_{\mu\nu}$$

$T_{\mu\nu}$ is the stress energy tensor, which includes all properties of matter other than gravity. After all, the Einstein equations let us solve for the gravitational field; you can't know the effects of gravity on matter before you solve them. We don't know a priori that galaxies attract each other; it is a consequence of the gravitational field they generate.

Now, we haven't forgotten that stuff tends to clump up in small scales. This is handled through perturbation theory: you have a fluid that is, on average, homogeneous, but can have deviations from homogeneity, and the same for the gravitational field. You then solve for these deviations up to whatever order you want, and the correct behavior of galaxies will pop up from the solution. You're not supposed to put it in by hand.

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  • $\begingroup$ You fall in the same fallacy trap as most cosmologists. The perfect fluid tensor on the right side of Einstein's equation already includes the metric, so gravity is there: \begin{equation} T_{\mu \nu} = (\rho + p) \, u_{\mu} \, u_{\nu} - g_{\mu \nu}\,p. \end{equation} The pressure $p$ that appears in the same tensor could also get an indirect influence from the metric. Even the scale factor $a(t)$ (which comes from the metric) is included in the energy density ($\rho(t) \propto a^{-3}$, for example). So why not consider that the pressure $p$ may also get more contributions from the metric? $\endgroup$ – Cham Jan 8 at 18:49
  • $\begingroup$ And this is why we still need to impose an equation of state, even when we don't know the effects of gravity on matter. Gravity may still have an - unknown - effect on that equation of state. This doesn't prevent us to try various models. The dust case $p = 0$ is the simplest model and I believe it is unrealistic because it doesn't take any effect of gravity into consideration. $\endgroup$ – Cham Jan 8 at 18:51

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