# Lagrangian of a Particle in Spherical Coordinates (Is this correct?)

## Homework Statement

a.) Set up the Lagrange Equations of motion in spherical coordinates, ρ,θ, $\phi$ for a particle of mass m subject to a force whose spherical components are $F_{\rho},F_{\theta},F_{\phi}$.

This is just the first part of the problem but the other parts do not seem so bad.

## Homework Equations

Lagrangian equations of motion.
Since the problem doesn't state it is in any uniform conservative field from which it would have a potential function, I assume I use the more general form of the Lagrangian.
$$\frac{d}{dt} \frac{∂T}{∂\dot{q}}-\frac{∂T}{∂q}=Q_k$$

## The Attempt at a Solution

So my three general forces would be..
$$Q_{\rho}=F_{\rho}$$
$$Q_{\theta}=\rho F_{\theta}$$
$$Q_{\phi}=F_{\phi}$$

The kinetic energy in Spherical coordinates is..
$$T=\frac{1}{2}m\left( \dot{\rho}^2 +\rho^2 \dot{\theta}^2+\rho^2 \dot{\phi}^2 sin^2\theta \right)$$
Thus the three equations of motion are..
$$\frac{d}{dt} \frac{∂T}{∂ \dot{\rho}}-\frac{∂T}{∂\rho}=Q_k=F_{\rho}$$
$$m \ddot{\rho}-m \rho \dot{\rho} \dot{\theta}^2-2 \rho \dot{\rho} \dot{\phi}^2sin^2 \theta=F{\rho}$$
and..
$$\frac{d}{dt} \frac{∂T}{∂ \dot{\theta}}-\frac{∂T}{∂\theta}=Q_k= \rho F_{\theta}$$
$$\frac{d}{dt}\left( m \rho^2 \dot{\theta}^2 \right)-m \dot{\theta} \rho^2 \dot{\phi}^2sin \theta cos \theta$$
$$2m \rho \dot{\theta} \dot{\rho}+m \rho^2 \ddot{\theta}-m \rho^2 \dot{\phi}^2sin\theta cos\theta$$
and finally..
$$\frac{d}{dt} \frac{∂T}{∂ \dot{\phi}}-\frac{∂T}{∂\phi}=Q_k=F_{\phi}$$
$$\frac{d}{dt}\left( m \rho^2 \dot{\phi}^2 sin^2 \theta \right)$$
$$2m \rho \dot{\rho} \dot{\phi}sin^2 \theta+m \rho^2 \ddot{\phi}sin^2 \theta +2m \rho^2 \dot{\phi}\dot{\theta}sin \theta cos \theta$$

Would these be correct? The rest of the problem says to substitute $\dot{\theta}$ for $\omega$ and then find the corresponding centrifugal and Coriolis forces. I do not recall how I would recognize these terms. I know these are both fictitious forces that are only a product of mass times acceleration. Their form is more obvious is polar form though.. If anyone could help me I would very much appreciate it. :]
Thanks!

## Answers and Replies

man i hate not being able to edit some posts....

i said that because it does not look like that you differentiated in spherical coordinates

What do you mean? I started out in Spherical coordinates from the Kinetic energy didn't I? From there on out it was just the chain rule a bunch of times.