# Using Hamilton's principle to derive Newton's equations of motion in parabolic coordinates

I have recieved a very hard (optional) assignment on variational calculus, and I have not got a clue where to start other then stating the Euler-Lagrange equations. Here is the problem:

According to Hamilton's principle a particle moving in a potential $$V(x,y,z)$$ in space (three dimensions) in such a way that the functional

$$\quad\quad S\equiv \int_{t_0}^{t_1}(\frac{m}{2}[\dot{x}(t)^2+\dot{y}(t)^2+\dot{z}(t)^2]-V(x(t),y(t),z(t)))dt$$

reaches a extreme ($$m>0$$ is the mass of the particle, the dots denote time derivatives and $$t_0$$ and $$t_1$$ are arbitrary times). Use Hamilton's principle to derive Newton's equations of motion for the coordinates $$(u,v,w)$$ defined as:

$$\quad\quad x=uv\cos(w),\quad\quad y=uv\sin(w),\quad\quad z=(u^2-v^2)/2$$

Now, I believe those coordinates are called parabolic, and they should be orthogonal to each other (if that helps us). We have three positional coordinates, which gives us three Euler-Lagrange equations:

$$\quad\quad \frac{\partial L}{\partial q_i}-\frac{d}{dt}\frac{\partial L}{\partial \dot{q}_i}=0,\quad\quad i=1,2,3\quad and\quad \textbf{q}=(x,y,z)$$.

The function $$L$$ (Lagrangian) is the integrand of $$S$$ in this case.

I really don't know where to go from here; I just get confused about the last part where I am supposed to write them in parabolic coordinates. When do I transform from Cartesian to parabolic, the Euler-Lagrange equations?

Best regards//

• Are there any tricks? It seems that the derivatives will generate a lot of terms, since they are quadratic. – SimpleProgrammer Feb 20 at 10:26

Hint: \left.\begin{align} \dot x&=(u\dot v+\dot u v)\cos w-uv\dot w \sin w\\ \dot y&=(u\dot v+\dot u v)\sin w+uv\dot w \cos w\\ \dot z&=u\dot u-v\dot v \end{align}\right\} \implies \dot x^2+\dot y^2+\dot z^2=(u^2+v^2)(\dot u^2+\dot v^2)+u^2v^2\dot w^2. Now plug in the Euler-Lagrange equation the Lagrangian: $${\cal L}=\frac{m}{2}\left[(u^2+v^2)(\dot u^2+\dot v^2)+u^2v^2\dot w^2\right]-V(u,v,w).$$

• Just what I needed! – SimpleProgrammer Feb 20 at 10:27
• I have derived the Euler-Lagrange equations so far. Now I am left with a compact equation of the form M*d²/dt²(q)=f, where M is a mass matrix, q = (u,v,w)^T and f is a column vector of three components dependent on q and their first derivatives. The mass matrix is diagonal. There are also partial derivatives of the potential V in f. Is this the final solution? Or do I have to calculate a expression of the potential somehow? (I am on my phone right now, I will write out the full equations later, sorry for typos). – SimpleProgrammer Feb 20 at 14:07

Using the above hint and transforming the Lagrangian to parabolic coordinates, I arrive at these Euler-Lagrange equations:

$$\quad\quad \textbf{I}: muv^{2}\dot{w}^{2}=\frac{\partial V}{\partial u}+m(\ddot{u}(u^{2}+v^{2})+\dot{u}(u\dot{u}+v\dot{v}))$$

$$\quad\quad \textbf{II}: mu^{2}v\dot{w}^{2}=\frac{\partial V}{\partial v}+m(\ddot{v}(u^{2}+v^{2})+\dot{v}(u\dot{u}+v\dot{v}))$$

$$\quad\quad \textbf{III}: \frac{\partial V}{\partial w}+muv(2\dot{u}v\dot{w}+2u\dot{v}\dot{w}+uv\ddot{w})=0$$

As I understand it, it is not sufficent to just wirte this is in a compact matrix form; according to the Wikipedia example I also have to solve for the original Euler-Lagrange equations in cartesian coordinates and compare these results. For Cartesian coordinates the Euler-Lagrange equations are much more easily derived:

$$\quad\quad \textbf{I}: \frac{\partial V}{\partial x}+m\ddot{x}=0$$

$$\quad\quad \textbf{II}: \frac{\partial V}{\partial y}+m\ddot{y}=0$$

$$\quad\quad \textbf{III}: \frac{\partial V}{\partial z}+m\ddot{z}=0$$

These equations are Newton's equations of motion, does that imply that the equations above are Newton's equations of motion for parabolic coordinates?