The Navier-Stokes equation is in terms
of the velocity field of the fluid,
but we'll actually be solving for what's
called the vorticity field of the fluid.
I need to manipulate the Navier-Stokes equation to
get the equation for the vorticity equation.
We define the vorticity,
we call it omega,
which is a vector.
It has the curl of the velocity field.
Del cross the velocity field.
In order to find the equation for the vorticity,
we need to take the curl of the Navier-Stokes equations.
We're going to take the curl
of the left-hand side of the Navier-Stokes equation,
which is du/dt plus u.del u.
We can put a parenthesis here,
but it's not absolutely necessary.
That's going to be equal to
the curl of the left-hand side,
which is minus 1 over the density
rho times the gradient of
the pressure plus the viscous term
nu times del squared u.
We do this term by term.
Start with the first term.
The curl of the time derivative of u,
these two derivatives commute,
so this is equal to the time derivative of the curl of u,
which is just the time derivative of the vorticity field.
This is a complicated term.
I'll do that last.
Let's do the viscous term next.
The curl of the viscous term,
the curl of nu del squared u.
Again, these two derivatives commute.
This is the constant nu times
del squared of the curl of u,
which is nu times del squared, the vorticity field.
Also a very simple result.
Now, the pressure term,
the curl of the pressure term minus 1 over the density,
which is a constant,
times the gradient of the pressure.
Well, the curl of a gradient is always equal to zero,
so this term goes to zero,
which is very nice.
The pressure then disappears from the equation.
We've got all the terms except
this non-linear term, u.del u.
Let's have a look at that one.
We have the curl of u.del u.
This is fairly complicated.
It was actually a homework problem
for you in the vector calculus course that I taught,
but I asked you to do it again in this course.
You can manipulate this equation
using the Levi-Civita symbol.
What you'll find is that this is equal to
the curl manipulating u.del u.
This is 1/2 the gradient of u.u,
which is a scalar plus the vorticity cross u.
Again, the curl of a gradient is zero.
This is just del cross omega cross u.
Now we have the Navier-Stokes equation
putting it all together.
We have d omega dt plus the non-linear term,
which is plus del cross
omega cross u equal to the right-hand side.
The pressure went away.
This is nu del squared omega.
That's a form of the vorticity equation.
There's one more manipulation you can do with this.
You can expand this del cross omega cross u term.
If you do that,
we get a second form of the vorticity equation,
which is d omega dt plus u.del omega
equals omega.del u plus
nu del squared omega.
That's a second form of the vorticity equation.
Here, just del cross omega cross u is equal to
u.del omega minus omega.del u.
You can prove that as part of your homework assignment.
This extra term, this looks
like the Navier-Stokes equations,
except that you replace u by the vorticity,
but there's this extra term on the right-hand side.
In fluid mechanics, this is called
the vortex stretching term because
this is the term when you study turbulence that we say is
responsible for the energy cascade in turbulence when
large eddies transfer their energy to smaller eddies.
That's our vorticity equation in three dimensions.
We're still working in
three dimensions even though
our problem will be in two dimensions.
But this is the general equation in three dimensions.
Let me summarize.
We're going to be solving for
the vorticity rather than for the velocity of the fluid.
We've defined the vorticity as the curl of the velocity.
We derive the vorticity equation by taking
the curl of the Navier-Stokes equations.
The nice thing about taking
the curl of the Navier-Stokes equation is
that the curl of the gradient of the pressure is zero,
so the pressure term is eliminated from the equation.
Then after some manipulation,
we get two forms of the vorticity equation.
We'll be working with this second form,
which we'll simplify further when we
go to a two-dimensional problem.
I'm Jeff Chasnov.
Thanks for watching,
and I'll see you in the next video.