# Question Cyl. Coordinates For Curl

1. Sep 15, 2007

### Jim L

Greetings- out of college for 50 yrs and studying H.M. Schey's book. Cannot understand his derivation of the z component of the curl of a vector function F for a part of a sector of a circle in a plane parallel to xy axis. Cylindrical components. Let me describe equation as three parts for ease in reading.

Eq. is a(b-c).

a= minus delta r/(r (delta r delta theta))
b= F sub r of [r, (theta plus (delta theta/2 ),z](delta r)
c= same as b and change plus to minus.

Author takes the limit of the entire equation as delta r and delta theta approach zero, and has an answer of minus (1/r)times the partial of F subr with respect to theta.

I understand everything except how to take that limit. Can some one help? The example is on page 83 of Schey. Thanks, Jim.

Last edited: Sep 15, 2007
2. Sep 15, 2007

### arildno

Well, I don't have Schey's book, but I can help you along to find the curl of vector function F in cylindrical coordinates.

I assume you are familiar with the del operator, which, written in Cartesian coordinates is given by:
$$\nabla=\vec{i}\frac{\partial}{\partial{x}}+\vec{j}\frac{\partial}{\partial{y}}+\vec{k}\frac{\partial}{\partial{z}}$$

Note that any particular term is composed of a unit vector, whereas the operational "denominator" is the typical infinitesemal length in that unit vector's direction, say $\partial{y}$ is an infinitesemal length in the $\vec{j}$-direction.

This structure holds for the del operator, irrespective of the coordinate system used.

Hence, the expression of the del operator in cylindrical coordinates is:
$$\nabla=\vec{i}_{r}\frac{\partial}{\partial{r}}+\vec{i}_{\theta}\frac{\partial}{r\partial\theta}}+\vec{k}\frac{\partial}{\partial{z}}$$

The curl is given by the cross product between the del operator and the vector function $\vec{F}$:
$$curl(\vec{F})=\nabla\times\vec{F}$$

This is calculated as follows:
$$\nabla\times\vec{F}=\vec{i}_{r}\times\frac{\partial\vec{F}}{\partial{r}}+\vec{i}_{\theta}\times\frac{\partial\vec{F}}{r\partial\theta}+\vec{k}\times\frac{\partial\vec{F}}{\partial{z}}$$
Do remember that:
$$\frac{\partial\vec{i}_{r}}{\partial\theta}=\vec{i}_{\theta}, \frac{\partial\vec{i}_{\theta}}{\partial\theta}=-\vec{i}_{r}$$
and that the other partial derivatives of unit vectors are all $\vec{0}$

You can extract the z-component of the curl from this.

3. Sep 15, 2007

### arildno

To finish this off:

From the first cross product term, note that diff. wrt. to the radius r regards the unit vectors as constant; hence, the only relevant term for the z-component of the curl is:
$$\vec{i}_{r}\times\vec{i}_{\theta}\frac{\partial{F}_{\theta}}{\partial{r}}$$
The third cross product term yields no terms relevant for the z-component of the curl, whereas the second cross-product term yields rise to two:
$$\vec{i}_{\theta}\times\vec{i}_{r}\frac{\partial{F}_{r}}{r\partial\theta}+\vec{i}_{\theta}\times\frac{\partial\vec{i}_{\theta}}{\partial\theta}\frac{F_{\theta}}{r}$$
These three contributions make up the z-component of the curl.

4. Sep 15, 2007

### Jim L

Thanks, but I still do not understand how to take the limit as delta r and delta theta approach zero. Jim

5. Sep 16, 2007

### arildno

Using right-hand convention, we have that the z-component of the curl becomes:
$$(\frac{\partial{F}_{\theta}}{\partial{r}}+\frac{F_{\theta}}{r}-\frac{1}{r}\frac{\partial{F}_{r}}{\partial\theta})\vec{k}$$
Which may be rewritten as:
$$\frac{1}{r}(\frac{\partial}{\partial{r}}(rF_{\theta})-\frac{\partial{F}_{r}}{\partial\theta})\vec{k}$$
Note that this agrees with your book under the assumption that $$F_{\theta}=0$$

6. Sep 16, 2007

### arildno

As for the limiting procedure, it is essentially to measure the relative amount of circulation in some fluid volume, and then make that volume as tiny as it can get.

While this procedure is necessary to formally prove what you're after, it really isn't that crucial to get every detail about it.
The technique I have presented makes it very easy to calculate what you're after; you'll have to decide the level of mathematical rigour that suits you best.

Last edited: Sep 16, 2007
7. Sep 16, 2007

### arildno

To understand the significance of the first term (regarded as 0 in your book's derivation!), let us make F into a velocity field for a rotating rigid object.

Thereby, we have $F_{\theta}=r\omega$, where $\omega$ is the scalar angular velocity. The velocity field is here $\vec{F}=r\omega\vec{i}_{\theta}$

Inserting this into our formula, we get:
$$\nabla\times\vec{F}=\frac{1}{r}\frac{\partial}{\partial{r}}(r^{2}\omega)\vec{k}=2\omega\vec{k}$$

That is, the curl of a velocity field is twice the (vectorial) angular velocity, as is well known.

8. Sep 17, 2007

### Jim L

Thanks,now if you can help me figure out how to convert Cartesian unit vectors to Cyl. Coordinate unit vectors, I will be even more gratefull. Jim

9. Sep 18, 2007

### arildno

Okay, I'll show this in a slightly unusual way, but that might be more intuitively accessible, yet retaining a sufficient level of rigour to avoid meaningless hand waving.

I'll limit myself to derive the expressions for the polar unit vectors in 2-D, since the z-vector is unchanged in cylindrical coordinates.

1. Transformation laws of Cartesian coordinates:
Now, the standard relationships between the Cartesian variables (x,y) and the polar coordinate represention are:
$$x=r\cos(\theta), y=r\sin(\theta)$$

2. Representation of a point's coordinates with aid of polar variables.

The position vector of a point P can be written as:
$$\vec{R}(x,y)=x\vec{i}+y\vec{j}=r\cos\theta\vec{i}+r\sin\theta\vec{j}=\vec{R}(r,\theta)$$
where the unit vectors given are Cartesian.

3. Deduction of polar coordinate unit vectors:
By a TINY increase of a variable, our position vector to a point gains a vectorial addition in the direction of that variable's proper unit vector.

Thus, for the variable r, we make a tiny addition dr, and we get:
$$\vec{R}(r+dr,\theta)-\vec{R}(r,\theta)=(r+dr)(\cos\theta\vec{i}+\sin\theta\vec{j})-r(\cos\theta\vec{i}+\sin\theta\vec{j})=dr(\cos\theta\vec{i}+\sin\theta\vec{j})$$
Hence, we see the proper unit vector in the radial direction:
$$\vec{i}_{r}=\cos\theta\vec{i}+\sin\theta\vec{j}$$

I'll post further tomorrow..

10. Sep 19, 2007

### arildno

Now, we make a tiny change in the angular variable instead, and we get:
$$\vec{R}(r,theta+d\theta)-\vec{R}(r,\theta)=r((\cos(\theta+d\theta)-\cos\theta)\vec{i}+(\sin(\theta+d\theta)-\sin(\theta))\vec{j})=r(\cos(d\theta)-1)\vec{i}_{r}+r\sin(d\theta)(-\sin\theta\vec{i}+\cos\theta)\vec{j}$$
where I have used the summation formulae for the trig function and rearranged.

Now, we have:
$$\cos(d\theta)\approx{1},\sin\theta\approx{d}\theta, d\theta<<1$$

Therefore, the vectorial addition to a small increase in the angular variable is approximately:
$$rd\theta(-\sin\theta\vec{i}+\cos\theta\vec{j})$$

Thus, we see that our proper unit vector in the angular direction is:
$$\vec{i}_{\theta}=-\sin\theta\vec{i}+\cos\theta\vec{j}$$

Note also the general relationship that the partial derivatives of the position vector are parallell to the proper unit vector.

11. Sep 19, 2007

### arildno

We therefore have the relationships between unit vectors:
$$\vec{i}_{r}=\cos\theta\vec{i}+\sin\theta\vec{j},\vec{i}_{\theta}=-\sin\theta\vec{i}+\cos\theta\vec{j}$$

In order to express the Cartesian vectors in terms of the cylindrical ones, multiply the first with cosine to the angle and the second with the negative sine to the angle.

$$\vec{i}=\cos\theta\vec{i}_{r}-\sin\theta\vec{i}_{\theta}$$