Vector Analysis – Parabolic Coordinates

There are many orthogonal curvilinear coordinate systems. One of those is parabolic coordinates. In this post I will look at how parabolic coordinates are related to Cartesian coordinates in 3 dimensional space. I will determine unit vectors in parabolic coordinates in terms of Cartesian coordinate unit vectors. I will determine the arc length of a moving particle following a trajectory as a function of time. Why? That’s one thing I have not determined. Here we go.

Cartesian & Parabolic Coordinate Relationship

Three dimensional parabolic coordinates are made up of (u,v,Φ) components. Cartesian components are (x,y,z). They are related by the following equations:

$$ x = uvcos(\phi),  y = uvsin(\phi),  z=\frac{1}{2}(u^{2}-v^{2}) $$

how did ‘they’ come up with these? Well, in two dimensions (u and v) coordinates are confocal parabolas. In three, u and v are confocal paraboloids rotated about an axis of symmetry (z axis) creating an azimuthal angle Φ where:

$$ tan(\phi)= \frac{y}{x} $$

In other words, try reading this and see if you can make sense of it on your own. Often, even when surrounded by people, I feel alone.

Determining Parabolic Unit Vectors in Terms of Cartesian Coordinates

To determine the unit vectors in parabolic coordinates I will need to:

  1. Create a position vector from the origin in Cartesian coordinates using the (x,y,z) parabolic equalities above.
  2.  Differentiate that position vector with respect to each parabolic coordinate (u,v,Φ). these functions will serve as the parabolic basis vectors.
  3. Normalize the basis vectors to get unit vectors with magnitude equal to 1. This step is equivalent to finding the normalizing scale factors.
  4. Try to mask my emotional void with an unconvincing smile.

Position in Cartesian coordinates:

$$\bar{r} = x\bar{i}+y\bar{j}+z\bar{k} $$

Using the equations above to get the parabolic equivalent:

$$ \bar{r} = uvcos(\phi)\bar{i}+uvsin(\phi)\bar{j}+\frac{1}{2}(u^{2}-v^{2})\bar{k} $$

Differentiating the vector with respect to u,v, and Φ:

$$ \frac{\partial \bar{r}}{\partial u} = vcos(\phi)\bar{i}+vsin(\phi)\bar{j}+u\bar{k} $$

$$ \frac{\partial \bar{r}}{\partial v} = ucos(\phi)\bar{i}+usin(\phi)\bar{j}-v\bar{k} $$

$$ \frac{\partial \bar{r}}{\partial \phi} = -uvsin(\phi)\bar{i}+uvcos(\phi)\bar{j} $$

The above partial derivatives can be normalized by dividing by their respective magnitudes. Calculating the magnitudes:

$$\left \| \frac{\partial \bar{r}}{\partial v} \right \| = \sqrt{u^2+v^2}$$

$$\left \| \frac{\partial \bar{r}}{\partial u} \right \| = \sqrt{u^2+v^2}$$

$$\left \| \frac{\partial \bar{r}}{\partial \phi} \right \| = uv$$

So with these magnitudes, I can divide each partial by its magnitude to get the unit vectors:

$$ \hat{e}_{u} = \frac{\frac{\partial \bar{r}}{\partial u}}{\left \| \frac{\partial \bar{r}}{\partial u} \right \|} = \frac{vcos(\phi)\bar{i}+vsin(\phi)\bar{j}+u\bar{k}}{\sqrt{u^2+v^2}}$$

$$\hat{e}_{v} = \frac{\frac{\partial \bar{r}}{\partial v}}{\left \| \frac{\partial \bar{r}}{\partial v} \right \|} = \frac{ucos(\phi)\bar{i}+usin(\phi)\bar{j}-v\bar{k}}{\sqrt{u^2+v^2}}$$

$$\hat{e}_{\phi} = \frac{\frac{\partial \bar{r}}{\partial \phi}}{\left \| \frac{\partial \bar{r}}{\partial \phi} \right \|} = \frac{-uvsin(\phi)\bar{i}+uvcos(\phi)\bar{j}}{uv}$$

Now we can use the above unit vectors to calculate arc lengths of functions defined in parabolic coordinates.

Determining Arc Length of a Particle Moving as a Function of Time

The function defined parametrically in parabolic coordinates:

$$ u(t) = t, v(t) = t, \phi = 2t $$

We will need to define the arc length in parabolic coordinates. I understand that arc length can be found by integrating along a differential displacement over time. To do this, the magnitude of the arc length can be equated to the square root of the dot product of the velocity vector.

$$s = \int_{a}^{b}ds = \int_{a}^{b}\frac{ds}{dt}dt = \int_{a}^{b}\sqrt{d\bar{r}\cdot d\bar{r}}$$

The next step is then to find what that integral evaluates to. So what is the differential position vector dr? I can use the partial with respect to each parabolic coordinate multiplied by the scale factor to find it.

$$ d\bar{r} = h_{u}\frac{\partial u}{\partial t}\hat{e}_{u} + h_{v}\frac{\partial v}{\partial t}\hat{e}_{v}+h_{\phi}\frac{\partial \phi}{\partial t}\hat{e}_{\phi} $$

$$ ds = \sqrt{h_{u}^2(\frac{\partial u}{\partial t})^2\hat{e}_{u} + h_{v}^2(\frac{\partial v}{\partial t})^2\hat{e}_{v}+h_{\phi}^2(\frac{\partial \phi}{\partial t})^2\hat{e}_{\phi}} $$

My head is going to explode here if we don’t make some progress. A couple steps into this calculation you get something that looks like this:

$$s = \int_{a}^{b}\sqrt{1^2(v^2+u^2)+1^2(u^2+v^2)+2(u^2v^2)} $$

We can replace the v’s and u’s with t using the parametric equations above. Lets integrate with respect to t from 0 to 1 so I can actually get an arc length value out. Moving along another couple steps:

$$s = \int_{0}^{1}\sqrt{4t^2+4t^4}dt$$

Not a particularly pleasant integral, but also not unworkable. That is the same as:

$$ s = \int_{0}^{1}2t\sqrt{1+t^2}dt $$

This will probably be the thousandth time I have had to do a u substitution.

An aside: I have never actually done a u substitution in my career by hand. I have never earned a penny doing u substitutions. I have never connected with another individual regarding u substitutions. U substitutions have never come in handy in my day to day life.  Yet here we are at 6:15PM on a Monday evening in October, writing up blog posts requiring u substitutions.

I think in this case we will use u equal to 1 plus t squared.

$$u = 1 + t^2, du = 2tdt$$

$$s = \int \sqrt{u}du = \frac{u^{\frac{3}{2}}}{\frac{3}{2}} = \frac{2}{3}(1+t^2)^{\frac{3}{2}})|_{0}^{1} = \frac{2}{3}(2^\frac{3}{2}-1)$$

There we have it. That answer represents the length along the arc of the parametrically defined function in parabolic coordinates from time t equal to 0 to time t equal to 1.



Leave a Reply

Your email address will not be published. Required fields are marked *