# Lagrangian of a spherical pendulum

I am working on a problem that's asking me to express the lagrangian of a mass $$m$$ suspended from a rigid massless rod of length $$L$$, but free to rotate otherwise (a spherical pendulum). The problem text says that one can take generalized coordinates $$\theta\ \text{and}\ \phi$$ as shown in the figure below (please excuse my terrible drawing I'm on my laptop and all I have is the trackpad.) I'm having some difficulty proceeding with this question so I will put the setup image and my work below and then enumerate a few questions I have in the final section. Any help rendered to me will be greatly appreciated.

My Attempt

First I want to determine my generalized coordinates so given this is a spherical setup I believed it to be appropriate to use a form of spherical coordinates such that: $$x = L\sin\theta\cos\phi \rightarrow \dot{x} = L\cos\theta\cos\phi\dot{\theta} - L\sin\theta\sin\phi\dot{\phi}\\ y = L\sin\theta\sin\phi \rightarrow \dot{y} = L\cos\theta\sin\phi\dot{\theta} + L\sin\theta\cos\phi\dot{\phi}\\ z = -L\cos\theta \rightarrow \dot{z} = L\sin\theta\dot{\theta}$$The lagrangian is given by $$\mathcal{L} = T-V$$ where $$T,V$$ are kinetic and potential energies respectively. The potential energy is $$V = mgz = -mgL\cos\theta$$ The kinetic energy is $$T = \frac{1}{2}m[\dot{x}^2 + \dot{y}^2 + \dot{z}^2]\\ = \frac{1}{2}m[(L^2\cos^2\theta\cos^2\phi\dot{\theta}^2 - L^2\sin^2\theta\sin^2\phi\dot{\phi}^2 - 2L\cos\theta\cos\phi\sin\theta\sin\phi\dot{\theta}\dot{\phi}) + (L^2\cos^2\theta\sin^2\phi\dot{\theta}^2 + L^2\sin^2\theta\cos^2\phi\dot{\phi}^2 + 2L\cos\theta\sin\theta\cos\phi\sin\phi\dot{\theta}\dot{\phi}) + (L^2\sin^2\theta\dot{\theta}^2)]\\ = \frac{1}{2}mL^2[\dot{\theta}^2(\cos^2\theta\cos^2\phi + \cos^2\theta\sin^2\phi + \sin^2\theta) + \dot{\phi}^2(\sin^2\theta\sin^2\phi + \sin^2\theta\cos^2\phi)]$$

Questions

1. Was my choice to use (modified) spherical coordinates appropriate? The problem statement certainly seemed to be suggestive of that and the setup is identical to that of the standard spherical polar coordinates barring the fact that the angle with which the polar coordinate $$\theta$$ makes with the $$z$$-axis is inverted. I.e., when $$\theta = 0$$ the vector points directly down the $$z$$-axis rather than up it in the standard convention. I believe for $$z$$ the trig factor is correct due to limit considerations however I am not totally confident with the $$x$$ and $$y$$ coordinates as they depend on products of trig functions of angles and my trig intuition isn't good enough to build those up from first principles. However, given the azimuthal angle $$\phi$$ is defined the same (sweeping out from the $$x$$ axis) as the standard convention and the only change is the "inversion" of the polar angle as noted above I felt confident going forward with these coordinates.

2. How can I simplify the trig expressions in my final expression for the kinetic energy? I've search around for lists of trig identities (I was hoping to find a multidimensional analogue of the commonly used $$\sin^2\theta + \cos^2\theta = 1$$) but to no avail. I didn't even find any identities which involved products of trig functions acting on different variables, let alone the analogue I was hoping to find. So, how would you proceed?

3. This problem is for a GRE prep book and I happen to know the answer is $$\mathcal{L} = T - V = \frac{1}{2}mL^2(\dot{\theta}^2 + \sin^2\theta\dot{\phi}^2) + mgL\cos\theta$$ How is it that the $$\dot{\phi}^2$$ term ends up with a factor of $$\sin^2\theta$$ out front? My mind immediately went to having the wrong angular dependence for the $$z$$ coordinate but that can't be the case since the (gravitational) potential is the same as I have noted. Could this be a result of whatever trig identity was meant to be employed?

• Just simplify your very last expression! Just use the standard identity. Commented Mar 20, 2023 at 17:00

you completely solved the problem, just need some simplifications: $$L =T-V = \frac{1}{2} mL^2 [ \dot{\theta}^2 (\cos^2\theta\underbrace{(\cos^2\phi+\sin^2\phi)}_{1}+\sin^2\theta) +\dot{\phi}^2 \sin^2\theta \underbrace{(\sin^2\phi+\cos^2\phi)}_{1}] -(-mgL\cos\theta) \\ = \frac{1}{2} mL^2 [ \dot{\theta}^2 \underbrace{(\cos^2\theta+\sin^2\theta)}_1 + \dot{\phi}^2 \sin^2\theta] + mgL\cos\theta = \frac{1}{2} mL^2[\dot{\theta}^2+\dot{\phi}^2 \sin^2\theta] + mgL\cos\theta$$