OK, in this series of videos we're going to talk about a new topic called coordinate transformations. So we're going to find many
times in fluid mechanics that it's useful if we can express the governing equations in terms of
different coordinate systems. And we talked about already conservation mass.
And the conservation of momentum we'll talk about soon. These are all examples of governing equations
that we may want to express in different coordinate systems because it may make the equations easier to obtain a solution.
So you know as a starting point, you're all probably familiar with Cartesian coordinates.
So here a point in space can be located in terms of three coordinates: x, y and z.
And so to do that we just take its projection onto the x, y, and z axes. And then these 3
coordinates express the position of this point in three dimensional space.
Another useful coordiante system is cylindrical coordinates. And so here again
I'm showing with respect to x, y, and z. But you can locate
a point in space also in terms of three variables. But here you can draw a projection down to the xy axes.
And define r as the distance from the origin
to its projection on the xy plane.
And theta as the angle between
the x axis and this distance r.
And then the height above this plane is expressed by the variable z.
So in cylindrical coordinates, you can express the same point in space in terms of three components. But instead of x, y, z,
we're using r and theta, which describes its location on the xy plane, and z which describes its height above this plane.
OK, so we can use these coordinate systems to write not just position, but also velocity. We know that that's something
we're going to be interested in for fluid mechanics. And so remember
in Cartesian coordinates, we can express the velocity vector in terms of three components u, v, and w.
u times the unit vector in the x direction, plus v times the unit vector in the y direction,
plus w times the unit vector in the z direction.
We can do the same thing in cylindrical coordinates. Except our three components are r, theta, and z.
So we can express the same velocity vector in terms of v,r times the unit vector in the r-direction (er),
plus v,theta times the unit vector in the theta-direction plus v,z times the unit vector in the z-direction (ez).
OK, so we defined two coordinate systems. And the next question becomes how do we transform between them.
So again, l'm going to just look at a two dimensional case here because you can see probably
just by looking at these pictures that both of them have the z component.
It represents the same thing in both of these coordinate systems. So just looking at the xy plane for now
and considering this point.
We can express the position of this point in terms of the Cartesian coordinates x and y, or the cylindrical components r and theta.
And just from trigonometry, then, we can see what the relationship is between these components.
So the r component, the distance from the origin to this point.
You might notice that this is the hypotenuse of a right triangle.
So we can just use Pythagorean theorem to get that distance r. It's just the square root of x^2 + y^2.
Similarly we notice that this angle theta is part of a right triangle.
So we can express that as the inverse tangent of y / x.
And remember the z is the same because it represents the same position
in both coordinate systems.
So this gives us a way to map between these two coordinate systems.
So now, we don't want to only map the variables themselves, but we also
want to map derivatives between them. The reason is that, as you remember from the conservation of mass, we are
going to end up with differential equations, or partial differential equations.
And so we want to know how we can map those derivatives from one coordinate system to another.
So let's look at this one, partial derivative of r with respect to x.
So this is the partial derivative with respect to x of r. So if we substitute r in terms of x and y
[Reading Equation] So we can take that derivative with respect to x.
[Reading Equation]
Notice that the square root of x^2 + y^2 is equal to r.
So I can say that the partial derivative of r with respect to x is equal to x/r.
And also by looking at this right triangle, I can see that for this right triangle.
x/r, that's the definition of cosine theta. Adjacent over hypotenuse.
[Reading equation]
Okay, so let me go to the next slide and continue.
So let me just redraw the geometry that we have here in two dimensions. Expressing the point in terms of position x, y and r, theta.
Let me write my cheat sheet of the transformations between r, theta, z and x, y, z.
This is the result we just got for the partial derivative of r with respect to x.
So now what's the partial derivative of theta with respect to x? So again we take
the partial with respect to x of theta which is [Reading Equation]. So what's the derivative? We have to go back to
your calculus course to remember these identities for derivatives. [Reading Equation]
So for our case, u is y/x. The term we're taking the partial derivative of.
Substitue that in. [Reading Equation]
So now I can combine these terms.
[Reading Equations]
And we notice here that x^2 + y^2 is equal to r^2 becaue we have a right triangle.
Which is equal to -y/r^2. And then I can factor out -1/r.
The reason I'm going to do that is, notice also that y/r is equal to sin theta. So the partial of theta with respect to x is equal to
minus sin(theta)/r. Then for the z component, z represents the same thing in both systems.
In the orthogonal system, z has no projection on the x and y axes. so the partial derivatives with respect to x and y are equal to zero.
We can follow the same argument with these right triangles and take the the partial derivatives of r and theta with respect to y.
[Reading Equations]
Okay, so now with these definitions we have a starting point to transform between cylindrical and Cartesian coordinates.
In terms of these partial derivatives.
But we still have a ways to go, and we'll continue on the next video.