Relativistic speed of sound in an inhomogeneous medium

Question

What is the expression for the speed of sound in a medium if the relativistic effects are important when the sound travels there?

In the non-relativistic limit the sound speed is

$$v = \sqrt{\left(\dfrac{\partial P}{\partial \rho}\right)_s}$$

where $P$ is the pressure and $\rho$ is the density of the medium, and the derivative is taken isentropically, that is, at constant entropy $s$. This is because a sound wave travels so fast that its propagation can be approximated as an adiabatic process, according to Wikipedia. It then says,

If relativistic effects are important, the speed of sound is calculated from the relativistic Euler equations.

Question: how this can be done for a general case? (No reference is given in the aforementioned Wikipedia article!)

NOTE: I've checked this question in SE Physics, but found almost nothing relevant for my question.

My Attempt: As said in this question: How to compute the speed of sound in relativistic hydrodynamics?, a formula for relativistic speed of sound has been derived in Weinberg: Gravitation and Cosmology chapter $2.10$ (Relativistic Hydrodynamics), equation $2.10.31$ and that is quite similar to the non-relativistic expression: $v = \sqrt{\left(\dfrac{\partial P}{\partial \rho}\right)_s}$ . But while deriving this, author assumed that

1. the fluid (in which the sound wave is propagating) is homogeneous
2. number of particles in the fluid is conserved (which need not to be necessarily held in case of a relativistic fluid)

So what if these two assumptions are not hold? What will be formula for the speed of sound in a inhomogeneous relativistic medium (fluid), where particle number may not be conserved?

Answer 1

The relativistic Euler equation reads $$a^\mu = \frac{-P_{,\nu}}{e + P}(\eta^{\mu\nu} + u^\mu u^\nu)$$ where $a^\mu = d^2 x^\mu/d\tau^2$ is the proper acceleration of the fluid element, $u^\mu = d x^\mu/d \tau$ its four-velocity, $P$ the pressure, and $e$ its total energy density (including rest mass density $\rho$). The comma denotes a partial derivative. The Euler equation has to be coupled to the relativistic continuity equation for the rest mass density $\rho$: $$(\rho u^\mu)_{,\mu} = 0$$

Now we assume that at zeroth order the fluid is homogeneous and static, we are in its rest frame ($\to t=\tau=x^0$), and that the fluid is subject to a small periodic perturbation, $$u^\mu = \delta^\mu_0 + \delta u^\mu \exp(ik_\nu x^\nu), a^\mu = 0 + \delta u^\mu i k_0 \exp(i k_\nu x^\nu),$$ $$P = P_{(0)} + \delta P \exp(i k_\nu x^\nu), e = e_{(0)} + \delta e \exp(i k_\nu x^\nu), \rho = \rho_{(0)} + \delta \rho \exp(i k_\nu x^\nu),$$ where $\delta u^\mu, P_{(0)}, e_{(0)},\rho_{(0)}, \delta e, \delta P,\delta \rho, k_\mu$ are constant. Furthermore, we assume that the perturbation is adiabatic and that the pressure and energy perturbations can be related to mass density by a state equation, $\delta e = (\partial e/\partial \rho) \delta \rho$, $\delta P = (\partial P/\partial \rho) \delta \rho$. Now we plug this all into the continuity equation and get to leading order in the perturbations $$\delta \rho k_0 + \rho_{(0)} k_\mu \delta u^\mu = 0\,.$$ From the Euler equation we get to leading order $$k_0 \delta u^\mu = -\frac{\delta \rho}{e_{(0)}+ P_{(0)}} \frac{\partial P}{\partial \rho} k_\nu \left(\eta^{\mu\nu} + \delta^\mu_0 \delta^\nu_0\right)\,.$$ Now by contracting the Euler equation with $k_\mu$, substituting into the relation above, and examining the meaning of the dispersion relation you obtain, you should be able to derive the speed of sound in relativistic media.

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As for your edits, the speed of sound is only meaningful when a wave on a background can be meaningfully identified. That implies that the wavelength of the wave has to be much smaller than the variability length of the background and the period is much shorter than the time variability scale of the "background". Then the derivation works the same to leading order. At sub-leading order, one can examine corrections by expanding in $k_\nu \ll P_{(0),\nu}/P_{(0)}, \rho_{(0),\nu}/\rho_{(0)},...$ but of course, the waves will then have a sound speed dependent on frequency (wavelength).

Answer 2

Short Answer: The answer depends on what exactly you mean. If you mean that the wave motion is close to the speed of light, then there is no single speed of sound. This case is highly nonlinear, and it turns out that the wave speed depends on the instantaneous amplitude of the signal. The usual reference wave speed is the "small-signal sound speed", which is just the wave speed in the case of infinitesimal amplitudes, which is the usual non-relativistic sound speed. If, on the other hand, you mean a more moderate wave motion within a relativistic flow, a constant sound speed may be obtained, and an expression is given below.

Longer Answer:

I am not an expert in relativity, so I will make the up-front assumption that the non-relativistic continuity equation is still valid, and so we may write $$ \frac{\partial\rho}{\partial t} + \nabla\cdot(\rho\vec v) = 0, $$ where $\rho$ is the mass density, $t$ is time time, $\nabla$ is the gradient operator, and $\vec v$ is the fluid particle velocity. For simplicity, I will assume one-dimensional sound propagation in the $x$ direction (which will also be the direction of any potential flow), and so we obtain $$ \frac{\partial\rho}{\partial t} + \frac{\partial}{\partial x}(\rho v) = 0. $$ From Wikipedia we may then write the relativistic Euler equation as $$ (e + P)\frac{\gamma}{c}\left( \frac{\partial v}{\partial t} + v\frac{\partial v}{\partial x} \right) = - \frac{\partial P}{\partial x} - \frac{\gamma^2}{c^2}v\left( \frac{\partial P}{\partial t} + v\frac{\partial P}{\partial x} \right), $$ where $e=\gamma\rho c^2 + \rho\epsilon$ is the energy density of the system, $\gamma=\sqrt{1-v^2/c^2}$, $c$ is the speed of light, $\epsilon$ is the internal energy, $P$ is the pressure, and I have expanded the material derivatives. These two equations have three unknowns (particle velocity, mass density, and pressure), and so at least one supplemental equation is necessary. I have no idea if it is a reasonable assumption, but I will assume an adiabatic fluid, such that we may write an equation of state formally as $$ P = P(\rho). $$

At this point, we need more information about what we are modelling. The two options I am imagining are (1) the motion of the wave itself is very high amlitude in a quiet environment, and (2) the wave motion is small in a very high-speed environment. Let's look at them seperately.

High-Amplitude Wave

In this case we are assuming that the particle velocity of the wave is a non-negligble fraction of the speed of light. Since acoustics is usually considered a perturbation analysis, this limit doesn't even make sense at some level. However, we will use the field of nonlinear acoustics as a guide and see what we find. In non-relativistic nonlinear acoustics we find that there is an exact solution for the sound speed, which I will write $a$ here since $c$ is already the speed of light, given by (Nonlinear Acoustics, edited by Hamilton and Blackstock, 2008) $$ a = a_0 + \frac{1}{2}(\gamma_{sp} - 1)v, $$ where $a_0$ is the sound speed in the limit of small signals and $\gamma_{sp}$ is the ratio of specific heats. Thus, the speed of sound depends on the instantaneous particle velocity. (The methods used to obtain this expression are convoluted and unique to the non-relativistic case, and so discussing them here would not be helpful.) Usually in nonlinear acoustics we just assume that the small-signal sound speed, $a_0$, is the sound speed, and we use perturbations to account for the amplitude dependence, leading to such equations as the Burgers' equation. If we follow this line of reasoning, then the relativistic speed of sound is identical to the non-relativistic speed of sound.

Relativistic Flow

Now let us assume that the acoustic perturbation is indeed small, but that there is a uniform but relativistic flow, $v_0$, in the background. Then we may linearize the equations to become \begin{gather} p = \left(\frac{\partial P}{\partial\rho}\right)_s\rho' \equiv b^2\rho' \tag{state}, \\ \frac{\partial\rho'}{\partial t} + \rho_0\frac{\partial v'}{\partial x} + v_0\frac{\partial\rho'}{\partial x} = 0 \tag{continuity}, \\ (e_0 + p_0)\frac{\gamma_0}{c}\left( \frac{\partial v'}{\partial t} + v_0\frac{\partial v'}{\partial x} \right) = - \frac{\partial p}{\partial x} - \frac{\gamma^2_0}{c^2}v_0\left( \frac{\partial p}{\partial t} + v_0\frac{\partial p}{\partial x} \right) \tag{Euler}, \end{gather} where $\rho'=\rho-\rho_0$, $\rho_0$ is the ambient density, $p=P-p_0$, $p_0$ is the ambient pressure, $v'=v-v_0$, and the subscript 0 on $\gamma$ and $e$ denote that they are evaluated at $v'=0$. Combining these equations and eliminating $\rho'$ and $v'$ we then obtain the wave-like equation \begin{equation} -(e_0 + p_0)\frac{\gamma_0}{\rho_0 b^2c}\left( \frac{\partial^2p}{\partial t^2} + 2v_0\frac{\partial^2p}{\partial t\partial x} + v_0^2\frac{\partial^2p}{\partial x^2} \right) = - \frac{\partial^2 p}{\partial x^2} - \frac{\gamma^2_0}{c^2}v_0\left( \frac{\partial^2 p}{\partial t\partial x} + v_0\frac{\partial^2 p}{\partial x^2} \right). \end{equation} This equation is rather distinct from the usual wave equation, so it is challenging to pull out the wave speed directly. However, if we assume a time-harmonic wave we may write the dispersion relationship by assuming $p=Ae^{-i\omega t + ikx}$: \begin{equation} 0 = k^2 - \frac{\gamma^2_0}{c^2}v_0k\left( \omega - v_0k \right) - (e_0 + p_0)\frac{\gamma_0}{\rho_0 b^2c}\left( \omega - v_0 k \right)^2. \end{equation} Solving for the angular frequency $\omega$ divided by the wavenumber $k$ (the phase speed of the wave, $a$) then yields \begin{equation} a = \frac{\omega}{k} = 2v_0\frac{\frac{\gamma^2_0}{c^2}v_0 - (e_0 + p_0)\frac{\gamma_0}{\rho_0 b^2c}v_0 + 1}{\frac{\gamma^2_0}{c^2}v_0 - 2(e_0 + p_0)\frac{\gamma_0}{\rho_0 b^2c}v_0 \pm \sqrt{(\frac{\gamma^2_0}{c^2}v_0)^2+4(e_0 + p_0)\frac{\gamma_0}{\rho_0 b^2c}}}. \end{equation} This expression can be simplified, but I need to get back to work. I would expect it is relatively close to simply $v_0 + b$.

A Final Caution: I am an acoustician, not a specialist in relativity. There are likely a number of issues with my derivation (such as assuming non-relativistic continuity), so take my conclusions with a grain of salt.