Speed of sound in a relativistic fluid

[etotheipi]

Homework Statement

Derive the speed of sound in an $\mathrm{\mathbf{isolated}}$ and $\mathrm{\mathbf{homogeneous}}$ simple fluid, by considering small first-order adiabatic perturbations to the fluid.

Assume an equation of state $p = p(\epsilon, S)$ where $\epsilon$ is the $\mathrm{\mathbf{proper \, energy \, density}}:= \mathbf{T}(\mathbf{\vec{u}}, \mathbf{\vec{u}})$ of the fluid [with $\mathbf{\vec{u}}$ the 4-velocity of a co-moving observer $\mathscr{C}$] and $S:= s/n$ is the $\mathrm{\mathbf{entropy\, per\, baryon}}$.

Relevant Equations

Fluid energy equation:$$\partial_t E + \boldsymbol{\nabla} \cdot ([E+p]\mathbf{\vec{V}}) = P_{\text{ext}}$$Relativistic Euler equation:$$\partial_t \mathbf{\vec{V}} + \boldsymbol{\nabla}_{\mathbf{\vec{V}}} \mathbf{\vec{V}} = - \frac{c^2}{E+p} \left( \tilde{\nabla} p + \frac{1}{c^2} \left( \partial_t p + P_{\text{ext}} \right) \mathbf{\vec{V}} \right) + \frac{c^2}{E+p} \mathbf{\vec{F}}_{\text{ext}}$$where $\tilde{\nabla}$ denotes the purely spatial gradient operator. Also, $\mathbf{\vec{V}}$ [the "fluid velocity with respect to $\mathscr{O}$" 4-vector] is defined by the orthogonal decomposition:$$\mathbf{\vec{u}}_{\mathscr{C}} = \gamma \left( \mathbf{\vec{u}}_{\mathscr{O}} + \frac{1}{c} \mathbf{\vec{V}} \right)$$with $\mathscr{C}$ being a co-moving observer and $\mathscr{O}$ a general observer.

Let us consider the co-moving observer $\mathscr{C}$ for whom $E = \epsilon$ and $\mathbf{\vec{V}} = \mathbf{\vec{0}}$. Doing the perturbation stuff to the first of the relevant equations gives$$\partial_t \delta \epsilon + \boldsymbol{\nabla} \cdot ([\epsilon + p] \delta \mathbf{\vec{V}}) = \delta P_{\text{ext}} = 0$$since $\delta [(\epsilon + p) \mathbf{\vec{V}}] = (\delta [\epsilon + p]) \mathbf{\vec{0}} + (\epsilon + p) \delta \mathbf{\vec{V}}$. To the second relevant equation the perturbation is similarly$$
\begin{align*}
\partial_t \delta \mathbf{\vec{V}} + \boldsymbol{\nabla}_{\mathbf{\vec{V}}} \delta \mathbf{\vec{V}} &= - \delta \left[\frac{c^2}{\epsilon+p} \left( \tilde{\nabla} p + \frac{1}{c^2} \left( \partial_t p + P_{\text{ext}} \right) \mathbf{\vec{V}} \right) \right] + \delta \left[ \frac{c^2}{\epsilon+p} \mathbf{\vec{F}}_{\text{ext}} \right] \\

\end{align*}$$Because the fluid is homogenous, $\tilde{\nabla} p = 0$ and given also that $\mathbf{\vec{V}} = \mathbf{\vec{0}}$, I think that the first term will reduce to:$$
\begin{align*}
\delta \left[\frac{c^2}{\epsilon+p} \left( \tilde{\nabla} p + \frac{1}{c^2} \left( \partial_t p + P_{\text{ext}} \right) \mathbf{\vec{V}} \right) \right] &= \frac{c^2}{\epsilon + p} \left( \tilde{\nabla} \delta p + \frac{1}{c^2} \mathbf{\vec{V}} \partial_t \delta p+ \frac{1}{c^2} (\partial_t p + P_{\text{ext}}) \delta \mathbf{\vec{V}} \right) \\

&= \frac{c^2}{\epsilon + p} \left( \tilde{\nabla} \delta p + \frac{1}{c^2} (\partial_t p + P_{\text{ext}}) \delta \mathbf{\vec{V}} \right)

\end{align*}$$whilst since $\delta \mathbf{\vec{F}}_{\text{ext}} = \mathbf{\vec{0}}$, the second term is$$\delta \left[ \frac{c^2}{\epsilon+p} \mathbf{\vec{F}}_{\text{ext}} \right] = \frac{-c^2}{(\epsilon + p)^2} \mathbf{\vec{F}}_{\text{ext}} \delta (\epsilon + p)$$I'm not really sure how to clean this up. I don't know what $\boldsymbol{\nabla}_{\mathbf{\vec{V}}} \delta \mathbf{\vec{V}}$ reduces to, and I don't know how to get rid of the 4-force $\mathbf{\vec{F}}_{\text{ext}}$ and external power density $P_{\text{ext}}$. Also, given the change is adiabatic I can write down from the equation of state:$$\delta p = \frac{\partial p}{\partial \epsilon} \big{|}_S \delta \epsilon + \frac{\partial p}{\partial S} \big{|}_{\epsilon} \delta S = \frac{\partial p}{\partial \epsilon} \big{|}_S \delta \epsilon$$How do I tidy up the perturbation, and then somehow extract a wave equation from that? Thanks!

 

[etotheipi]

Oh wait; it's isolated and so both $\mathbf{F}_{\text{ext}}$ and $P_{\text{ext}} = 0$, right, and furthermore due to the homogeneity $\epsilon$ and $p$ are constants thus $\delta(\epsilon + p) = 0$ too? Also, presumably $\partial_t p = 0$ also due to time-independence! Okay, I think I can clean this up! We have:$$
\begin{align*}

\partial_t \delta \mathbf{\vec{V}} + \boldsymbol{\nabla}_{\mathbf{\vec{V}}} \delta \mathbf{\vec{V}} &=

\frac{-c^2}{\epsilon + p} \left( \tilde{\nabla} \delta p \right) = \frac{-c^2}{\epsilon + p} \frac{\partial p}{\partial \epsilon} \big{|}_{S} \tilde{\nabla} \delta \epsilon
\end{align*}
$$and for the other equation, again since $\epsilon$ and $p$ are independent of position,$$\partial_t \delta \epsilon + (\epsilon + p)\boldsymbol{\nabla} \cdot \delta \mathbf{\vec{V}}= 0$$Hmm, it's a bit simpler, but still not that simple! I can substitute in my $\delta p$ from the thermodynamic equation of state, but I still don't know how to deal with the $\boldsymbol{\nabla}_{\mathbf{\vec{V}}} \delta \mathbf{\vec{V}}$ term!

 

[vanhees71]

To derive the sound waves you work in the rest frame of the unpertubed fluid, assuming it's in thermal equilibrium. Then $\vec{v}=\delta \vec{v}$. Since thus $\nabla_{\vec{v}} \vec{V}=\delta \vec{v} \cdot \delta \vec{v}=\mathcal{O}(\delta^2)=\simeq 0$. Further you can also use $\epsilon+p=\epsilon_0+p_0$ and $(\partial p/\partial \epsilon)_S=(\partial p/\partial \epsilon)_{S0}$, because these expressions multiply quantities which are already in 1st order of the perturbations. Now eliminate $\delta \vec{v}$ from your 2nd equation, using the first (taking into account the said approximations). Then interpret the resulting equation to get the speed of sound.

 

[etotheipi]

Thanks @vanhees71, yes I think I can do it now!

Although how I defined it above $\mathbf{\vec{V}}$ is a 4-vector, it's an orthogonal projection onto the observer's local rest space $\mathscr{E}_{\mathbf{\vec{u}}_{\mathscr{O}}}$ and hence has zero time component w.r.t. his basis vectors, i.e. $\mathbf{\vec{V}} \equiv (0, \vec{V})$. Thus reformulating in terms of 3-vectors and then acting $\tilde{\nabla} \cdot$ on the first equation gives$$
\partial_t \tilde{\nabla} \cdot \delta \vec{V} = \frac{-c^2}{\epsilon + p} \frac{\partial p}{\partial \epsilon} \big{|}_{S} \tilde{\nabla}^2 \delta \epsilon
$$whilst taking the time derivative of the second equation gives$$\partial_t \delta \epsilon + (\epsilon + p) \partial_t \tilde{\nabla} \cdot \delta \vec{V}= 0$$Hence eliminating $\partial_t \tilde{\nabla} \cdot \delta \vec{V}$ results in the equation$$\partial_t^2 \delta \epsilon = c^2 \frac{\partial p}{\partial \epsilon} \big{|}_{S} \tilde{\nabla}^2 \delta \epsilon$$from which it follows that$$c_{\text{sound}} = c \sqrt{\frac{\partial p}{\partial \epsilon} \big{|}_{S}}$$

 

[vanhees71]

Argh. I should have read your posting more carefully. For me everything with an arrow is a three-vector, which can be simply the three spatial components of a four-vector wrt. a Minkowski-orthonormal basis or one of the two three-vectors making up an antisymmetric Minkowski tensor (as $\vec{E}$ and $\vec{B}$ making up the Faraday tensor components $F_{\mu \nu}$ wrt. a Minkowski-orthonormal basis).