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 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:

(1)

(2)

The equation (2) gives:

(3)

(4)

The equation (1) gives for :

(5)

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

We define the nonrelativistic energy as:

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

(6)

This is the relativistic equation for the energy. Substituting into (3):

(7)

For we get:

(8)

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:

those are the Euler equations, also sometimes written as:

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:

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).

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.

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.

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:

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.