OK, in this video we're going to conclude our discussion of coordinate transformations and hopefully this is going to bring together what
we talked about in the previous videos. So you may remember that we've gone to a lot of trouble to establish relationships between
the position coordinates in different coordinate systems. Particularly we focused on Cartesian coordinates where position is
located in terms of the components x, y, and z. And cylindrical coordinates where they are expressed in terms of components r, theta, z.
And we also looked at the velocity vectors and how those components are expressed, or can be related.
And how derivatives associated with those velocity components can be expressed.
The reason for that is that you may remember when we talked about conservation of mass, we ended up with a differential equation
that involves the position and velocity, and derivatives of these quantities.
So in order to transform between coordinate systems, we need to be able to express
these relationships between them. So this is complicated if we want to take this expression of the conservation of mass.
In Cartesian coordinates where you have
[Reading equation]
This is the form that the equation takes assuming incompressible flow.
If we want to express the corresponding equation in cylindrical coordinates, it's not so straightforward. We can't just say it's
partial of v,r with respect to r, plus partial of v,theta with respect to theta, plus partial of v,z with respect to z.
And the reason is, if you remember when we talked about the velocity vectors in the Cartesian system.
The x and y axes remain fixed in space. Whereas in the cylindrical system,
the r and theta unit vectors can change their orientation depending on the orientation of the velocity vector.
So since the relative orientation between the two coordinate systems can be constantly changing, then that introduces some
additional of things that we need to consider in this coordinate transformation.
So choosing these coordinates and then establishing relationships between them is actually not a trivial thing.
And as you get into more complex flows and deformations of non-Newtonian fluids, those issues become
even more important. But we're not going to worry about that for the purposes of this class. So with the background that we've
established in the previous videos, now we can start to look at that these components. So let's start with the first component.
The partial derivative of u with respect to x.
So this is the partial derivative with respect x of u. So we obtained an expression for the velocity component u, in terms of v,r and v,theta.
So it is [Reading Equation].
We want to take the partial derivative with respect to x of this quantity.
So when we take the derivative of a function involving products, we can apply the chain rule.
So I'm going to do that.
I can express this derivative with respect to x as the partial derivative of r with respect to x
times the partial derivative with respect to r of [Reading Equation]
plus [Reading Equation]
So remember, because x can depend on r and theta, this gives me a complete mapping of the derivatives
in terms of the interdependence of these variables.
So now I can go through and evaluate the derivative with respect to r and theta of these terms.
So that's what I'm going to do here in the next step. Also, we have the same thing for the partial derivative with respect to z,
but this is a wash because these components v,r and v,theta don't have a component in the z direction.
Because the z direction is orthogonal to the r and theta directions.
OK, so then we just work through this. The partial derivative of r with respect to x is cos(theta). We found that previously.
And then we can work through here. [Reading Equation]
cos(theta) doesn't have a derivative with respect to R, so I left it out here.
And then same thing. We can apply the chain rule here.
[Reading Equation]
And again, sin(theta), because r and theta are orthogonal, there is no derivative with respect to r, so it's zero.
So we're left with these two terms when we evaluate
their partial derivative with respect to r.
Similarly we can do the same thing for the theta direction.
So previously we found that the partial derivative of theta with respect to x is
minus sin(theta) over r.
And then we can evaluate the partial derivative with respect to theta of these components.
And so here, both v,r and v,theta can have theta dependence. So we get all the terms when we use the chain rule.
[Reading Equation]
So those two come from this term. And then
[Reading Equation]
So we have that term
[Reading Equation]
And then, like I said before, we get nothing for the partial derivative with respect to z
because there is no z dependence of these quantities.
OK, so now we can collect terms and see what we end up with.
[Reading Equations]
Similarly, we can go through the same analysis with the other components.
[Reading Equations]
And then for the z component, we get that the partial derivative with respect to z is the same in both coordinate systems.
Cartesian and cylindrical.
OK, so now we can notice a few things to combine these. Because remember the conservation of mass is the sum of these.
Or the continuity equation is the sum of these partial derivatives.
So notice that these terms cancel. [Reading Equations].
And there are some other terms that cancel also. So all the terms that have sin(theta) * cos(theta) cancel.
Also, some of the terms add together. [Reading Equations]
So remember the trig identity, sin^2(theta) + cos^2(theta) = 1.
So when I add these together, the sin^2(theta) + cos^2(theta) just equals to 1.
So these simplifications allow me to reduce the continuity equation as follows for cylindrical coordinates.
[Reading Equations]
So this is an expression of the continuity equation, or the differential form of the conservation of mass, in cylindrical coordinates.
So as you can see, it's not a one to one mapping.
You don't just substitue u, v, and w with v,r, v,theta, and Vz.
[Reading Equation]
We also talked about a more compact way to express this equation in terms of the gradient or the del operator.
So we can see that in cylindrical coordinates that the del of something is.
[Reading Equation]
So now, using basically the geometry that we examined in previous videos,
we found that if we're careful and we keep track of what we're doing, we can transform
these velocity vector components and their derivatives
from Cartesian coordinates into other coordinate systems.
And specifically we showed here how that works for the case of cylindrical coordinates.
So the the idea is basically geometry and comparing triangles. You end up with some
complicated looking equations. But I just want you to remember the basic idea is just geometry.
The details make it look complicated, but the basic ideas are straightforward.
And given enough time I think you could sit down and figure it out.