Fluid Dynamics¶

Stress-Energy Tensor¶

In general, the stress energy tensor is the flux of momentum $p^\mu$ over the surface $x^\nu$ . It is a machine that contains a knowledge of the energy density, momentum density and stress as measured by any observer of the event.

Imagine a (small) box in the spacetime. Then the observer with a 4-velocity $u^\mu$ measures the density of 4-momentum $\d p^\alpha\over\d V$ in his frame as:

${\d p^\alpha\over\d V} = -T^\alpha{}_\beta u^\beta$

and the energy density that he measures is:

$\rho = {E\over V} = -{u^\alpha p_\alpha \over V} = - u^\alpha {\d p_\alpha\over\d V} = u^\alpha T_{\alpha\beta} u^\beta$

One can also obtain the stress energy tensor from the Lagrangian $\L=\L(\eta_\rho, \partial_\nu \eta_\rho, x^\nu)$ by combining the Euler-Lagrange equations

${ \partial \L\over\partial \eta_\rho} - \partial_\nu\left( { \partial \L\over\partial (\partial_\nu \eta_\rho)} \right) =0$

with the total derivative ${\d \L\over \d x^\mu}$ :

${\d \L\over \d x^\mu} = {\partial\L\over\partial\eta_\rho} \partial_\mu \eta_\rho + { \partial \L\over\partial (\partial_\nu \eta_\rho)} \partial_\mu\partial_\nu\eta_\rho + \partial_\mu\L = = \partial_\nu\left( { \partial \L\over\partial (\partial_\nu \eta_\rho)} \right) \partial_\mu \eta_\rho + { \partial \L\over\partial (\partial_\nu \eta_\rho)} \partial_\nu\partial_\mu\eta_\rho + \partial_\mu\L = = \partial_\nu\left( { \partial \L\over\partial (\partial_\nu \eta_\rho)} \partial_\mu \eta_\rho \right) + \partial_\mu\L$

or

$\partial_\nu\left( { \partial \L\over\partial (\partial_\nu \eta_\rho)} \partial_\mu \eta_\rho -\L \delta_\mu{}^\nu \right) + \partial_\mu\L =0$

This can be written as:

$\partial_\nu T_\mu{}^\nu + f_\mu = 0$

where

$T_\mu{}^\nu = { \partial \L\over\partial (\partial_\nu \eta_\rho)} \partial_\mu \eta_\rho -\L \delta_\mu{}^\nu f_\mu = \partial_\mu\L$

The Navier-Stokes equations can be derived from the conservation law:

$\partial_\nu T^{\mu\nu} + f^\mu = 0$

To obtain some Lagrangian (and action) for the perfect fluid, so that we can derive the stress energy tensor $T^{\mu\nu}$ from that, is not trivial, see for example arXiv:gr-qc/9304026. One has to take into account the equation of state and incorporate the particle number conservation $\nabla_\mu(nu^\mu)=0$ and no entropy exchange $\nabla_\mu(nsu^\mu)=0$ constraints.

The equation of continuity follows from the conservation of the baryon number — the volume $V$ that contains certain number of baryons can change, but the total number of baryons $nV$ must remain constant:

${\d (nV)\over\d\tau} = 0 {\d n\over\d\tau}V + n{\d V\over\d\tau} = 0 u^\alpha (\partial_\alpha n)V + n(\partial_\alpha u^\alpha) V = 0 \partial_\alpha (n u^\alpha) = 0$

Perfect Fluids¶

Perfect fluids have no heat conduction ( $T^{i0} = T^{0i} = 0$ ) and no viscosity ( $T^{ij} = p\one$ ), so in the comoving frame:

$T^{\alpha\beta} = \diag(\rho c^2, p, p, p) = \left(\rho+{p\over c^2}\right)u^\alpha u^\beta + p g^{\alpha\beta}$

where in the comoving frame we have $g^{\mu\nu} = \diag(-1, 1, 1, 1)$ , $u^0=c$ and $u^i=0$ , but $\partial_\alpha U^i\neq0$ . $p$ is the pressure with units $[p] =\rm N\,m^{-2}=kg\,m^{-1}\,s^{-2}$ (then $[{p\over c^2}] =\rm kg\,m^{-3}$ ), $\rho$ is the rest mass density with units $[\rho] =\rm kg\,m^{-3}$ , and $\rho c^2$ is the energy density with units $[\rho c^2] =\rm kg\,m^{-1}\,s^{-2}$ .

The last equation is a tensor equation so it holds in any frame. Let’s write the components explicitly:

$T^{00} = \left(\rho+{p\over c^2}\right)u^0u^0 - p = \left(\rho+{p\over c^2}\right)c^2 \gamma^2 - p = \left(\rho c^2+p\left(1-{1\over\gamma^2}\right)\right) \gamma^2 = \left(\rho c^2+p {v^2\over c^2}\right) \gamma^2 T^{0i} = T^{i0} = \left(\rho+{p\over c^2}\right)u^0u^i = \left(\rho+{p\over c^2}\right) c v^i \gamma^2 = {1\over c}\left(\rho c^2+p\right) v^i \gamma^2 T^{ij} = \left(\rho+{p\over c^2}\right) u^iu^j + p \delta^{ij} = \left(\rho+{p\over c^2}\right) v^iv^j\gamma^2 + p \delta^{ij}$

We now use the conservation of the stress energy tensor and the conservation of the number of particles:

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

(2) $\partial_\mu(nu^\mu) = 0$

The equation (2) gives:

$\partial_t (n\gamma) + \partial_i(n v^i \gamma) = 0$

(3) $\partial_t (n m\gamma) + \partial_i(n m v^i \gamma) = 0$

(4) $\partial_t (n m c^2\gamma) + \partial_i(n m c^2 v^i \gamma) = 0$

The equation (1) gives for $\mu=0$ :

$\partial_\nu T^{0\nu} = 0 \partial_0 T^{00} + \partial_i T^{0i} = 0 \partial_t\left({1\over c}\left(\rho c^2 + p {v^2\over c^2}\right) \gamma^2\right) + \partial_i\left({1\over c}\left(\rho c^2 + p\right) v^i \gamma^2\right) = 0$

(5) $\partial_t\left(\left(\rho c^2 + p {v^2\over c^2}\right) \gamma^2\right) + \partial_i\left(\left(\rho c^2 + p\right) v^i \gamma^2\right) = 0$

We now substract the equation (4) from (5):

$\partial_t\left(\left(\rho c^2\gamma - n m c^2 + p {v^2\over c^2} \gamma\right) \gamma\right) + \partial_i\left(\left(\rho c^2\gamma -n m c^2 + p \gamma\right) v^i \gamma\right) = 0$

We define the nonrelativistic energy as:

$E = \rho c^2\gamma -n m c^2 = \half \rho v^2 + (\rho - nm)c^2 + O\left(v^4\over c^2\right)$

so it contains the kinetic plus internal energies. We substitute back into (5):

(6) $\partial_t\left(\left(E + p {v^2\over c^2} \gamma\right) \gamma\right) + \partial_i\left(\left(E + p \gamma\right) v^i \gamma\right) = 0$

This is the relativistic equation for the energy. Substituting $nm = \rho\gamma - {E\over c^2}$ into (3):

(7) $\partial_t\left(\rho\gamma^2 - {E\gamma\over c^2}\right) + \partial_i\left(\left(\rho\gamma^2 - {E\gamma\over c^2}\right) v^i \right) = 0$

For $\mu=i$ we get:

$\partial_\nu T^{i\nu} = 0 \partial_0 T^{i0} + \partial_j T^{ij} = 0 \partial_t \left({1\over c^2}\left(\rho c^2 + p \right) v^i\gamma^2\right) + \partial_j \left( \left(\rho+{p\over c^2}\right) v^iv^j\gamma^2 + p \delta^{ij} \right) = 0$

(8) $\partial_t \left(\left(\rho + {p\over c^2} \right) v^i\gamma^2\right) + \partial_j \left( \left(\rho+{p\over c^2}\right) v^iv^j\gamma^2 + p \delta^{ij} \right) = 0$

This is the momentum equation. The equations (7), (8) and (6) are the correct relativistic equations for the perfect fluid (no approximations were done). We can take either (7) or (5) as the equation of continuity (both give the same nonrelativistic equation of continuity). Their Newtonian limit is:

$\partial_t \rho + \partial_i(\rho v^i) = 0 \partial_t \left(\rho v^i\right) + \partial_j \left( \rho v^iv^j + p \delta^{ij} \right) = 0 \partial_t E + \partial_j\left(v^j\left(E + p \right)\right) = 0$

those are the Euler equations, also sometimes written as:

${\partial \rho\over \partial t} + \nabla\cdot(\rho{\bf v}) = 0 {\partial (\rho{\bf v})\over\partial t} + \nabla \cdot (\rho {\bf v}{\bf v}^T) + \nabla p = 0 {\partial E\over\partial t} + \nabla\cdot\left({\bf v}\left(E + p \right)\right) = 0$

Energy Equation¶

The energy equation can also be derived from thermodynamic and the other two Euler equations. We have the following two Euler equations:

$\partial_t\rho + \partial_i(\rho u^i) = 0 \rho\partial_t u^i + \rho u^j\partial_j u^i + \delta^{ij}\partial_j p = 0$

We’ll need the following formulas:

$\partial_t (u_i u^i) = (\partial_t u_i) u^i + u_i \partial_t u^i = (\partial_t u_i)\delta^{ij} u_j + u_i \partial_t u^i = = (\partial_t u_i\delta^{ij}) u_j + u_i \partial_t u^i = (\partial_t u^j) u_j + u_i \partial_t u^i = 2 u_i \partial_t u^i \partial_j (u_i u^i) = 2 u_i \partial_j u^i \partial_t\rho =- \partial_i(\rho u^i) \partial_t u^i =- u^j\partial_j u^i - {\delta^{ij}\over\rho}\partial_j p - u^j\partial_j p + \partial_t(\rho U) = = - {\d p \over\d t} +\partial_t p + \partial_t(\rho U) = = - {\d p \over\d t} +\partial_t (\rho U + p) = = - {\d p \over\d t} +{\d\over\d t} (\rho U + p) -u^j\partial_j (\rho U + p)= = - {\d p \over\d t} +{\d\rho\over\d t} \left(U + {p\over\rho}\right) +\rho{\d\over\d t} \left(U + {p\over\rho}\right) -u^j\partial_j (\rho U + p)= = - {\d p \over\d t} +{\d\rho\over\d t} \left(U + {p\over\rho}\right) +\rho{\d\over\d t} \left(U + {p\over\rho}\right) + (\rho U + p)\partial_j u^j -\partial_j (\rho U u^j + p u^j) = = \left[\rho {\d\over\d t}\left(U + {p\over\rho}\right) - {\d p\over\d t} \right] + \left(U + {p\over\rho}\right)\left[ {\d\rho\over\d t} + \rho \partial_j u^j \right] -\partial_j (\rho U u^j + p u^j) = = - \partial_j(\rho U u^j + p u^j) 0 = \d Q = T\d S = \d U + p\d V = \d (U + pV) - V\d p = \d\left(U+{p\over\rho}\right) - {1\over \rho}\d p = \d H - {1\over \rho}\d p$

where $V = {1\over\rho}$ is the specific volume and $H = U+{p\over\rho}$ is entalphy (heat content).

Then:

$\partial_t E = = \partial_t (\half \rho u_i u^i + \rho U) = = \half u_i u^i \partial_t \rho +\half\rho\partial_t(u_i u^i) + \partial_t(\rho U) = = -\half u_i u^i \partial_j(\rho u^j) +\rho u_i\partial_t u^i + \partial_t(\rho U) = = -\half u_i u^i \partial_j(\rho u^j) - \rho u_i u^j\partial_j u^i - u_i\delta^{ij}\partial_j p + \partial_t(\rho U) = = -\half u_i u^i \partial_j(\rho u^j) - \half\rho u^j\partial_j (u_i u^i) - u_i\delta^{ij}\partial_j p + \partial_t(\rho U) = = -\half\partial_j(\rho u_i u^i u^j) - u^j\partial_j p + \partial_t(\rho U) = = -\half\partial_j(\rho u_i u^i u^j) - \partial_j(\rho U u^j + p u^j) = = -\partial_j\left(u^j\left(\half\rho u_i u^i+\rho U + p \right)\right) = = -\partial_j\left(u^j\left(E + p \right)\right)$

so:

$\partial_t E + \partial_j\left(u^j\left(E + p \right)\right) = 0 {\partial E\over\partial t} + \nabla\cdot\left({\bf u}\left(E + p \right)\right) = 0$

Navier-Stokes Equations¶

When we write the relativistic conservation law in a nonrelativistic limit (for a general fluid), we get the Cauchy momentum equation:

$\rho\left({\partial {\bf v}\over\partial t} +{\bf v}\cdot\nabla{\bf v} \right) = \nabla \cdot \mathds{\sigma} + {\bf f}$

where the stress tensor $\sigma$ can be written as:

$\sigma=-p\mathds{1} + \mathds{T}$

and we get the Navier-Stokes equations:

$\rho\left({\partial {\bf v}\over\partial t} +{\bf v}\cdot\nabla{\bf v} \right) = -\nabla p + \nabla \cdot \mathds{T} + {\bf f}$

Those are the most general equations. If we assume some more things about the fluid, they can be further simplified.

For Newtonean fluids, we want $\mathds{T}$ to be isotropic, linear in strain rates and it’s divergence zero for fluid at rest. It follows that the only way to write the tensor under these conditions is:

$T_{ij} = 2\mu\epsilon_{ij} + \delta_{ij} \lambda \nabla\cdot{\bf v}$

where the strain rate is:

$\epsilon_{ij}={1\over 2}\left(\partial_j v_i+\partial_i v_j\right)$

The divergence of the tensor is:

$\partial_j T_{ij} =2\mu\partial_j\epsilon_{ij} + \partial_j\delta_{ij} \lambda \nabla\cdot{\bf v} =\mu\partial_j\partial_j v_i+\mu\partial_i \nabla\cdot{\bf v} + \lambda \partial_i \nabla\cdot{\bf v} =\mu\partial_j\partial_j v_i+(\mu+\lambda)\partial_i \nabla\cdot{\bf v}$

or in vector form (these are usually called the compressible Navier-Stokes equations):

$\nabla \cdot \mathds{T} =\mu\nabla^2{\bf v}+(\mu+\lambda)\nabla \nabla\cdot{\bf v}$

For incompressible fluid we have $\nabla\cdot\bf v=0$ , so we get the incompressible Navier-Stokes equations:

$\nabla \cdot \mathds{T} =\mu\nabla^2{\bf v}$

and for a perfect fluid we have no viscosity, e.g. $\mu=0$ , then we get the Euler equations (for perfect fluid):

$\nabla \cdot \mathds{T}=0$

Bernoulli’s Principle¶

Bernoulli’s principle works for a perfect fluid, so we take the Euler equations:

$\rho\left({\partial {\bf v}\over\partial t} +{\bf v}\cdot\nabla{\bf v} \right) = -\nabla p + {\bf f}$

and put it into a vertical gravitational field ${\bf f} = (0, 0, -\rho g)=-\rho g\nabla z$ , so:

$\rho\left({\partial {\bf v}\over\partial t} +{\bf v}\cdot\nabla{\bf v} \right) = -\nabla p - \rho g\nabla z$

we divide by $\rho$ :

${\partial {\bf v}\over\partial t} +{\bf v}\cdot\nabla{\bf v} = -\nabla \left({p\over\rho} + g z\right)$

and use the identity ${\bf v}\cdot\nabla{\bf v}={1\over 2}\nabla v^2 + (\nabla \times {\bf v})\times{\bf v}$ :

${\partial {\bf v}\over\partial t} +{1\over 2}\nabla v^2+(\nabla \times {\bf v})\times{\bf v} +\nabla \left({p\over\rho} + g z\right)=0$

so:

${\partial {\bf v}\over\partial t} +(\nabla \times {\bf v})\times{\bf v} +\nabla \left({v^2\over 2} + gz + {p\over\rho} \right)=0$

If the fluid is moving, we integrate this along a streamline from the point $A$ to $B$ :

$\int {\partial {\bf v}\over\partial t} \cdot \d {\bf l} +\left[{v^2\over 2} + gz + {p\over\rho} \right]_A^B=0$

So far we didn’t do any approximation (besides having a perfect fluid in a vertical gravitation field). Now we assume a steady flow, so ${\partial {\bf v}\over\partial t}=0$ and since points $A$ and $B$ are arbitrary, we get:

${v^2\over 2} + gz + {p\over\rho}={\rm const.}$

along the streamline. This is called the Bernoulli’s principle. If the fluid is not moving, we set ${\bf v}=0$ in the equations above and immediately get:

${v^2\over 2} + gz + {p\over\rho}={\rm const.}$

The last equation then holds everywhere in the (nonmoving) fluid (as opposed to the previous equation that only holds along the streamline).

Hydrostatic Pressure¶

Let $p_1$ be the pressure on the water surface and $p_2$ the pressure $h$ meters below the surface. From the Bernoulli’s principle:

${p_1\over\rho} = g\cdot (-h) + {p_2\over \rho}$

so

$p_1 + h\rho g = p_2$

and we can see, that the pressure $h$ meters below the surface is $h\rho g$ plus the (atmospheric) pressure $p_1$ on the surface.

Torricelli’s Law¶

We want to find the speed $v$ of the water flowing out of the tank (of the height $h$ ) through a small hole at the bottom. The (atmospheric) pressure at the water surface and also near the small hole is $p_1$ . From the Bernoulli’s principle:

${p_1\over\rho} = {v^2\over 2} + g\cdot (-h) + {p_1\over \rho}$

so:

$v=\sqrt{2g h}$

This is called the Torricelli’s law.

Venturi Effect¶

A pipe with a cross section $A_1$ , pressure $p_1$ and the speed of a perfect liquid $v_1$ changes it’s cross section to $A_2$ , so the pressure changes to $p_2$ and the speed to $v_2$ . Given $\Delta p = p_1-p_2$ , $A_1$ and $A_2$ , calculate $v_1$ and $v_2$ .

We use the continuity equation:

$A_1 v_1 = A_2 v_2$

and the Bernoulli’s principle:

${v_1^2\over 2} + {p_1\over\rho} = {v_2^2\over 2} + {p_2\over\rho}$

so we have two equations for two unknowns $v_1$ and $v_2$ , after solving it we get:

$v_1 = A_2\sqrt{2\Delta p\over \rho(A_1^2-A_2^2)}$

$v_2 = A_1\sqrt{2\Delta p\over \rho(A_1^2-A_2^2)}$

Hagen-Poiseuille Law¶

We assume incompressible (but viscuous) Newtonean fluid (in no external force field):

$\rho\left({\partial {\bf v}\over\partial t} +{\bf v}\cdot\nabla{\bf v} \right) = -\nabla p + \mu\nabla^2{\bf v}$

flowing in the vertical pipe of radius $R$ and we further assume steady flow ${\partial {\bf v}\over\partial t}=0$ , axis symmetry $v_r=v_\theta=\partial_\theta(\cdots)=0$ and a fully developed flow $\partial_z v_z=0$ . We write the Navier-Stokes equations above in the cylindrical coordinates and using the stated assumptions, the only nonzero equations are:

$0=-\partial_r p$

$0=-\partial_z p+\mu{1\over r}\partial_r(r\partial_r v_z)$

from the first one we can see the $p=p(z)$ is a function of $z$ only and we can solve the second one for $v_z=v_z(r)$ :

$v_z(r) = {1\over 4\mu}(\partial_z p) r^2 + C_1\log r + C_2$

We want $v_z(r=0)$ to be finite, so $C_1=0$ , next we assume the no slip boundary conditions $v_z(r=R)=0$ , so $C_2 = -{1\over 4\mu}(\partial_z p) R^2$ and we get the parabolic velocity profile: