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Starting from the following definition of stress-energy tensor for a perfect fluid in special relativity :

$${\displaystyle T^{\mu \nu }=\left(\rho+{\frac {p}{c^{2}}}\right)\,v^{\mu }v^{\nu }-p\,\eta ^{\mu \nu }\,}\quad(1)$$

with $$v^{\nu}=\dfrac{\text{d}x^{\nu}}{\text{d}\tau}$$ and

$$V^{\nu}=\dfrac{\text{d}x^{\nu}}{\text{d}t}$$ (we have $v^{\nu}=\gamma\,V^{\nu}$)

So, finally, I have to get the following relation :

$$\dfrac{\partial \vec{V}}{\partial t} + (\vec{V}.\vec{grad})\vec{V} = -\dfrac{1}{\gamma^2(\rho+\dfrac{p}{c^2})} \bigg(\vec{grad}\,p+\dfrac{\vec{V}}{c^2}\dfrac{\partial p}{\partial t}\bigg)\quad(2)$$

To get this relation, I must use the conservation of energy for $\nu=i$ and $\nu=0$ with :

$$\partial_{\mu}T^{\mu\nu}=0\quad(3)$$

If someone could help me to find the equation $(2)$ from $(1)$ and $(3)$, this would be nice to indicate the tricks to apply.

EDIT 1 :

For the moment, below where I am :

I recognize in the left member of wanted relation $(2)$ the Lagrangian derivative :

$$\dfrac{\text{D}\,\vec{V}}{\text{d}t}=\dfrac{\partial \vec{V}}{\partial t} + (\vec{V}.\vec{\nabla})\vec{V}\quad(4)$$

and I can rewrite $(1)$ with the $V^{\mu}$ components like :

$$T^{\mu\nu}=\left(\rho+\dfrac{p}{c^{2}}\right)\,\gamma^2\,V^{\mu}V^{\nu }-p\,\eta^{\mu\nu}\,\quad(5)$$

But from this point, I don't know how to make the link between $(4)$, $(5)$, $(3)$ (the divergence of stress-energy equal to zero), and $(1)$ ...

Any help is welcome

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  • $\begingroup$ You may take a look there : physics.stackexchange.com/q/321982 $\endgroup$ – Cham Dec 13 '18 at 2:15
  • $\begingroup$ -@Cham . Sorry, this was a mistake, it is the time derivated of pression $p$. I have corrected it. $\endgroup$ – youpilat13 Dec 13 '18 at 2:43
  • $\begingroup$ The equation I'm getting is this (I use $c = 1$ to simplify things) : \begin{equation}\gamma \, (\rho + p) \, \frac{d \gamma \, \vec{v}}{dt} = - \vec{\nabla} p - \gamma^2 \vec{v} \, \frac{dp}{dt}.\end{equation} The $\gamma$ factors aren't placed at the same place as in your equation (2), and I get a total time derivative of pressure, not a partial one. $\endgroup$ – Cham Dec 13 '18 at 3:41
  • $\begingroup$ Ok, after contracting with $\vec{v}$ and playing a bit with my previous equation, I get yours : \begin{equation}\gamma^2 (\rho + p) \frac{d\vec{v}}{dt} = - \vec{\nabla} p - \vec{v} \, \frac{\partial p}{\partial t}.\end{equation} Writing a complete answer would be pretty long, though. $\endgroup$ – Cham Dec 13 '18 at 3:53
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Here are the main steps. I use $c = 1$ as usual in relativity, and $a, b = 0, 1, 2, 3$ are flat spacetime indices : \begin{gather} T^{ab} = (\rho + p) \, u^a \, u^b - \eta^{ab} p,\tag{1} \\[12pt] \partial_a T^{ab} = u^b u^a \, \partial_a \, \rho + u^b u^a \, \partial_a \, p + (\rho + p)\big( (\partial_a \, u^a) \, u^b + u^a \, \partial_a \, u^b \,\big) - \partial^b p = 0. \tag{2} \end{gather} Contract (2) on $u_b$ and use properties $u_b \, u^b = 1$ and $u_b \, \partial_a \, u^b \equiv 0$. You should get the continuity equation : \begin{equation}\tag{3} (\rho + p) \, \partial_a \, u^a = -\, u^a \, \partial_a \, \rho. \end{equation} Subsitute this constraint into equation (2). You should get this : \begin{equation}\tag{4} (\rho + p) \, u^a \, \partial_a \, u^b = -\, u^b u^a \, \partial_a \, p + \partial^b \, p. \end{equation} Write these for simplicity (total proper-time derivative) : \begin{align}\tag{5} u^a \, \partial_a \, u^b &\equiv \frac{d u^b}{d\tau}, & u^a \, \partial_a \, p &\equiv \frac{d p}{d\tau}. \end{align} Then you get this, for index $b = i = 1, 2, 3$ : \begin{align}\tag{6} (\rho + p) \frac{d u^i}{d\tau} = -\, u^i \, \frac{dp}{d\tau} + \partial^i \, p. \end{align} Use $u^i = \gamma \, v^i$ and $\partial^i p = -\, \partial_i \, p$ and $d\tau = dt / \gamma$, so (6) becomes a vectorial equation : \begin{align}\tag{7} \gamma \, (\rho + p) \frac{d \gamma \, \vec{v}}{dt} = -\, \gamma^2 \, \vec{v} \, \frac{dp}{dt} - \vec{\nabla} \, p. \end{align} Now scalar-contract this with vector $\vec{v}$ and use $\dot{\gamma} = \gamma^3 \, \vec{v} \cdot \dot{\vec{v}}$ and the identity $1 + \gamma^2 \, v^2 \equiv \gamma^2$ : \begin{gather} \gamma^2 (\rho + p) \frac{d \vec{v}}{dt} + \gamma \, (\rho + p) \, \gamma^3 (\vec{v} \cdot \dot{\vec{v}}) \, \vec{v} = -\, \vec{\nabla} \, p - \gamma^2 \, \vec{v} \, \frac{dp}{dt} \tag{8} \\[12pt] \gamma^4 (\rho + p)(\vec{v} \cdot \dot{\vec{v}}) = -\, \vec{v} \cdot \vec{\nabla} \, p - \gamma^2 \, v^2 \, \frac{d p}{dt}. \tag{9} \end{gather} Subsitute (9) into the second term of left part of equation (8). After some simplification algebra and using \begin{equation}\tag{10} \frac{d p}{dt} = \frac{\partial p}{\partial t} + \vec{v} \cdot \vec{\nabla} \, p, \end{equation} you should get your equation : \begin{equation}\tag{11} \gamma^2 (\rho + p) \frac{d \vec{v}}{dt} = -\, \vec{\nabla} \, p - \vec{v} \, \frac{\partial p}{\partial t}. \end{equation} That's the relativistic Euler equation for a perfect fluid. Usually : $p \ll \rho$ and $\gamma \approx 1$ for a slowly moving fluid. The last term should be negligible.

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  • $\begingroup$ -@Cham thanks. I would like to know how you get the relation $u_b \, \partial_a \, u^b \equiv 0$ ? $\endgroup$ – youpilat13 Dec 13 '18 at 11:30
  • $\begingroup$ Since $u_b u^b = 1$, just derive it : $\partial_a (u_b u^b) = 2 \, u_b \, \partial_a \, u^b = 0$. $\endgroup$ – Cham Dec 13 '18 at 11:33
  • $\begingroup$ -@Cham. Sorry, I didn't understand all the demo : how do you get the continuity equations,(3) i.e : $(\rho + p) \, \partial_a \, u^a = -\, u^a \, \partial_a \, p$ from (1) ? $\endgroup$ – youpilat13 Dec 13 '18 at 12:27
  • $\begingroup$ You get (3) from (2). Contract (2) with the components $u_b$, and use $u_b u^b = 1$ and $u_b \, \partial_a \, u^b = 0$. $\endgroup$ – Cham Dec 13 '18 at 12:32
  • $\begingroup$ Sorry, there was a small error in (3) : $p \Rightarrow \rho$. I'm fixing it. $\endgroup$ – Cham Dec 13 '18 at 12:37

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