# 7.1. Fluid Dynamics¶

## 7.1.1. Stress-Energy Tensor¶

In general, the stress energy tensor is the flux of momentum over the surface . 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 measures the density of 4-momentum in his frame as:

and the energy density that he measures is:

One can also obtain the stress energy tensor from the Lagrangian by combining the Euler-Lagrange equations

with the total derivative :

or

This can be written as:

where

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

To obtain some Lagrangian (and action) for the perfect fluid, so that we can derive the stress energy tensor 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 and no entropy exchange constraints.

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

### Perfect Fluids¶

Perfect fluids have no heat conduction () and no viscosity (), so in the comoving frame:

where in the comoving frame we have , and , but . is the pressure with units (then ), is the rest mass density with units , and is the energy density with units .

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

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

(7.1.1.1)¶

(7.1.1.2)¶

The equation (7.1.1.2) gives:

(7.1.1.3)¶

(7.1.1.4)¶

The equation (7.1.1.1) gives for :

(7.1.1.5)¶

We now substract the equation (7.1.1.4) from (7.1.1.5):

We define the nonrelativistic energy as:

so it contains the kinetic plus internal energies. We substitute back into (7.1.1.5):

(7.1.1.6)¶

This is the relativistic equation for the energy. Substituting into (7.1.1.3):

(7.1.1.7)¶

The equation (7.1.1.1) for gives:

(7.1.1.8)¶

This is the momentum equation. The equations (7.1.1.7), (7.1.1.8) and (7.1.1.6) are the correct relativistic equations for the perfect fluid (no approximations were done). We can take either (7.1.1.7) or (7.1.1.5) as the equation of continuity (both give the same nonrelativistic equation of continuity). Their Newtonian limit is obtained by (which implies ):

those are the Euler equations, also sometimes written as:

(7.1.1.9)¶

(7.1.1.10)¶

(7.1.1.11)¶

The momentum equation can be further simplified by expanding the parentheses and using the continuity equation:

(7.1.1.12)¶

Where we used:

#### Alternative Derivation¶

We can also take the non-relativistic limit in the stress energy tensor:

and plug it into the equation (7.1.1.1). For we get the equation of continuity:

and for we get the momentum equation:

However, in order to derive the equation for energy , one needs to take into account the full relativistic stress energy tensor, see the previous section for details.

### Energy Equation¶

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

We’ll need the following formulas:

where is the specific volume and is entalphy (heat content).

Then:

so:

## 7.1.3. Incompressible Equations¶

Incompressible flow means that the material derivative of density is zero:

(7.1.3.1)¶

Putting this into the equation of continuity (7.1.1.9) one obtains or equivalently:

(7.1.3.2)¶

But also (7.1.3.2) implies (7.1.3.1), so these two equations are equivalent: the divergence of the velocity field is zero if and only if the material derivative of the density is zero.

Using the condition in (7.1.1.9) and (7.1.1.12) we obtain:

In addition to incompressibility, we can also assume a constant density , then we obtain the incompressible Navier-Stokes equations:

(7.1.3.3)¶

(7.1.3.4)¶

For they become the incompressible Euler equations. At the given time step with known and , the equation (7.1.3.4) is solved for at the new time step. Then we solve for new as follows. Apply divergence to (7.1.3.4):

now we use the following identities:

to get:

Finally we use the equation (7.1.3.3) to simplify:

(7.1.3.5)¶

which is a Poisson equation for . Note again that . The equation (7.1.3.5) is then used to solve for at the new time step.

### Divergence Free Velocity¶

Typically by propagating (7.1.3.4), we obtain a velocity that is not divergence free. To make it so, we want to find such a divergence free that is closest to in the norm , in other words we want to find the projection onto the divergence free subspace, so we have to minimize the following functional:

where we used a Langrange multiplier in the second term to impose the zero divergence on for all points (that is why is a function of and not a constant) and in the first term we ensure that is as close as possible to the original field in the sense. Let’s calculate the variation:

From the condition and assuming the surface integral vanishes (i.e. either or everywhere on the boundary) we obtain the two Euler-Lagrange equations:

(7.1.3.6)¶

(7.1.3.7)¶

Applying divergence to (7.1.3.6) and using (7.1.3.7) we obtain:

(7.1.3.8)¶

After solving this Poisson equation for we can calculate the divergence free from (7.1.3.6):

(7.1.3.9)¶

### Time Discretization¶

The incompressible Euler equations are:

(7.1.3.10)¶

We use first order time discretization:

(7.1.3.11)¶

(7.1.3.12)¶

(7.1.3.13)¶

The velocity at time steps and must be divergence free, per (7.1.3.11) and (7.1.3.12). The simplest discretization of (7.1.3.10) is to use an explicit scheme, so we evaluate the term at the time step . Regarding the pressure term , if we evaluated it at the time step , then from (7.1.3.13) we could calculate that would not be divergence free, per (7.1.3.12). So we are led to evaluate the pressure term at the time step , then all the equations (7.1.3.11), (7.1.3.12) and (7.1.3.13) can be satisfied.

To solve this system of equations, we use an operator splitting on (7.1.3.13), the most natural is probably the following:

(7.1.3.14)¶

(7.1.3.15)¶

The first equation (7.1.3.14) is just like (7.1.3.13), except that the pressure term is evaluated at the time step , which forces us to change into , which is not divergence free. The second equation (7.1.3.15) is then uniquely given by the condition that the sum of (7.1.3.14) and (7.1.3.15) is equal to (7.1.3.13).

The equation (7.1.3.15) is equivalent to (7.1.3.9), with , so this is an projection of onto the divergence free subspace to obtain , also sometimes called a pressure projection. We use the same method as was used to obtain the Poisson equation (7.1.3.8) for , i.e. take a divergence and rearrange:

(7.1.3.16)¶

One solves (7.1.3.14) for , then the Poisson equation (7.1.3.16) for (i.e. the pressure update ), and then one computes using (7.1.3.15) (or equivalently (7.1.3.9)).

These equations are derived from Euler equations (7.1.3.11) and (7.1.3.12) using a time discretization and an operator splitting technique. The theory of the projection onto the divergence free subspace is not needed to derive these equations, but it helps with understanding of what is going on.

Note 1: the operator splitting of (7.1.3.13) into (7.1.3.14) and (7.1.3.15) is not unique. Another option is:

(7.1.3.17)¶

(7.1.3.18)¶

The sum of (7.1.3.17) and (7.1.3.18) is still (7.1.3.13) and the equation (7.1.3.18) is still equivalent to (7.1.3.9), only this time with . The Poisson equation then becomes:

(7.1.3.19)¶

The only difference to the previous scheme is that now the norm of is larger, because now depends on the full pressure instead of the pressure difference, so is not as close to as in the previous scheme.

Note 2: By applying divergence to (7.1.3.17) we obtain:

and substituting into (7.1.3.19) we obtain:

or

(7.1.3.20)¶

which is the discrete analog of the equation (7.1.3.5). The same result is obtained by applying a divergence to (7.1.3.14) and substituting into (7.1.3.16):

Which simplifies to (7.1.3.20).

## 7.1.4. Bernoulli’s Principle¶

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

and put it into a vertical gravitational field , so:

we divide by :

and use the identity :

so:

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

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 and since points and are arbitrary, we get:

along the streamline. This is called the Bernoulli’s principle. If the fluid is not moving, we set in the equations above and immediately get:

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 be the pressure on the water surface and the pressure meters below the surface. From the Bernoulli’s principle:

so

and we can see, that the pressure meters below the surface is plus the (atmospheric) pressure on the surface.

### Torricelli’s Law¶

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

so:

This is called the Torricelli’s law.

### Venturi Effect¶

A pipe with a cross section , pressure and the speed of a perfect liquid changes it’s cross section to , so the pressure changes to and the speed to . Given , and , calculate and .

We use the continuity equation:

and the Bernoulli’s principle:

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

### Hagen-Poiseuille Law¶

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

flowing in the vertical pipe of radius and we further assume steady flow , axis symmetry and a fully developed flow . We write the Navier-Stokes equations above in the cylindrical coordinates and using the stated assumptions, the only nonzero equations are:

from the first one we can see the is a function of only and we can solve the second one for :

We want to be finite, so , next we assume the no slip boundary conditions , so and we get the parabolic velocity profile:

Assuming that the pressure decreases linearly across the length of the pipe, we have and we get:

We can now calculate the volumetric flow rate:

so we can see that it depends on the 4th power of . This is called the Hagen-Poiseuille law.