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:
- Create a position vector from the origin in Cartesian coordinates using the (x,y,z) parabolic equalities above.
- Differentiate that position vector with respect to each parabolic coordinate (u,v,Φ). these functions will serve as the parabolic basis vectors.
- Normalize the basis vectors to get unit vectors with magnitude equal to 1. This step is equivalent to finding the normalizing scale factors.
- 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 s 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.