# I Elliptic Cylinder Coordinates, Acceleration Derivation...options?

#### mishima

I've been deriving ds, velocity and acceleration for an elliptic cylindrical coordinate system. When it comes to ds and velocity, its quite simple and quick.

The acceleration however is tedious by my current method and I'm wondering if there is some shortcut or superior method I'm not aware of.

My current method is to take the time derivative of velocity, which includes taking the time derivative of the basis vectors. This is the standard approach I've seen in mechanics texts for spherical and cylindrical. Is there any other way to find acceleration? We have looked at the metric tensor, but is there a way to get acceleration from it perhaps?

#### fresh_42

Mentor
2018 Award
No, $a=\dot{v}=\ddot{s}$ per definition. I don't see any other approach than this.

#### mishima

That's disheartening. A later section mentions an approach that might give acceleration components, but Im not sure how much they would match the 'standard' method above. Or more importantly, if the calculation is more elegant.

The idea seems to be to imagine a particle in motion under the influence of F = - grad V in the elliptic cylindrical coordinate system, and write the Lagrange equations of motion. Then dividing by scale factors supposedly gives something that resembles the acceleration components.

I'll give it a shot after more studying.

#### Cryo

Gold Member
The idea seems to be to imagine a particle in motion under the influence of F = - grad V in the elliptic cylindrical coordinate system
That does indeed help in some cases the basic reason is that you can state the problem in terms of scalars, and the work in any coordinate system you want naturally.

For example, if you have a simple Lagrangian

$L=m\dot{r}^2-V\left(\mathbf{r}\right)$ you can show that the equations of motion are

$\boldsymbol{\nabla}V=\frac{d}{dt}\left(m\mathbf{\dot{r}}\right)=m\mathbf{\ddot{r}}$

So the acceleration is $\mathbf{a}=m^{-1}\boldsymbol{\nabla}V$

Now, we can actually write the Lagrangian in terms of elliptical coordinates

$L=mg_{\alpha\beta}\dot{r}^\alpha\dot{r}\beta/2-V$

The metric follows from considering the 'infinitessimal length element':
$g_{\alpha\beta}=\mathscr{a}_0^2\left(\sinh^2\mu+\sin^2\nu\right)diag\left(1,1\right)_{\alpha\beta}$

the inverese metric is:
$g^{\alpha\beta}=\frac{1}{\mathscr{a}_0^2\left(\sinh^2\mu+\sin^2\nu\right)}diag\left(1,1\right)^{\alpha\beta}$

where $\mathscr{a}_0$ is the scale-factor for the elliptical coordinates.

The Lagranigian is then:

$L=m\mathscr{a}_0^2\left(\sinh^2\mu+\sin^2\nu\right)(\dot{\mu}^2+\dot{\nu}^2)/2-V\left(\mu,\nu\right)$

Now you simpy apply get the equations of motion from the above Lagrangian, and find $\partial_{\dots} V$:

$\partial_\mu V = \frac{d}{dt}\left(m\mathscr{a}_0^2\left(\sinh^2\mu+\sin^2\nu\right)\dot{\mu}\right) - \partial_\mu\left(m\mathscr{a}_0^2\left(\sinh^2\mu+\sin^2\nu\right)(\dot{\mu}^2+\dot{\nu}^2)/2\right)$

and same for $\partial_{\nu} V$

Then the acceleration is:

$a^\mu=m^{-1}\left(\boldsymbol{\nabla}V\right)=m^{-1}g^{\alpha\beta}\partial_\beta V$, so

$a^\mu=\frac{m^{-1}\partial_\mu V}{\mathscr{a}_0^2\left(\sinh^2\mu+\sin^2\nu\right)}$

and same for $a^{\nu}$. You will have to be careful about the basis vectors, they will not be normalized.

#### mishima

Thanks, I gave this a shot with cylindrical and spherical to get my feet wet. Definitely saves a lot of algebra...going to try elliptic cylindrical now.

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