# Can we detect a cyclic coordinate by just inspecting the Lagrangian?

I'm reading through Susskind-Hrabovsky's Theoretical Minimum. On page 126, where they are talking about cyclic coordinates, an example is given:

Suppose two particles moving on a line with a potential energy that depends on the distance between them...

Lagrangian is derived as: $$L = \frac{m}{2}(\dot{x}_1^2 + \dot{x}_2^2) - V(x_1 - x_2).\tag{16}$$

It is suggested that if the Lagrangian doesn't depend on coordinate $$q_i$$, then that coordinate is cyclic and its conjugate momentum is conserved. Then, a coordinate transform is utilized and the Lagrangian in the new coordinate is derived:
$$x_+ = \frac{x_1+x_2}{2}, \qquad x_{-} = \frac{x_1-x_2}{2},$$
$$L = m(\dot{x}_+^2 + \dot{x}_{-}^2) - V(x_{-})$$
Then it was discussed that there is actually a hidden cyclic coordinate and its conjugate momentum is conserved (which is total momentum): $$p_{+} = 2m\dot{x}_{+} = m\dot{x}_1 + m\dot{x}_2$$

1. If there may exist a transformation that reveals a hidden cyclic coordinate (hence a preserved conjugate momentum), then doesn't that make the original statement about we being able to detect cyclic coordinate my merely looking at the Lagrangian, invalid?

2. In general, how can we find the transformation which reveals the cyclic coordinate?

Also, there are some doubts on the derived terms:

1. Shouldn't potential energy in the new coordinate be $$\,V(2\times{x_{-}})$$?

2. Shouldn't $$\,p_+ = m\dot{x}_+$$? Where did that $$2$$ come from?

• For your questions 1 and 2, there are already some factor of two problems in your post going to the second form of the Lagrangian. It seems your real question is 3 and 4, so I'd recommend editing this post to focus on that and make it clearer – octonion Oct 12 '19 at 9:33
• @octonion Fixed it. – Zeta.Investigator Oct 12 '19 at 9:47

## 2 Answers

• 1 & 2. A bit oversimplified a strategy to find candidates for cyclic coordinates is to find coordinates that parametrizes equipotential surfaces of the potential $$V$$.

1. In physics we often use the same notation for a function $$V$$ and its value $$V(x)$$ at a point $$x$$. If we transform the argument $$x=f(y)$$, we often don't bother to write $$V\circ f(y)$$ but just write $$V(y)$$ in a common physics misuse of notation. The transformation $$f$$ is implicitly understood.
1. Use the definition of canonical/conjugate momenta $$p_+:=\frac{\partial L}{\partial \dot{x}^+}$$.

How you can find the cyclic coordinate:

for a cyclic coordinate is:

$$\boxed{\frac{d}{dt}\left(\frac{\partial L }{\partial {\dot{q}_i}}\right)=0\quad \Rightarrow\quad \frac{\partial L }{\partial {q}_i}=0}$$ where $$q_i$$ are the generalized coordinate and $$L=T-V$$

your case

$$T=\frac{m}{2}\left(\dot{x}_1^2+\dot{x}_2^2\right)$$

and $$V=V(x_1-x_2)$$

Ansatz:

we are looking for constant transformation matrix Q where:

$$\underbrace{\begin{bmatrix} q_1\\ q_2\\ \end{bmatrix}}_{\vec{q}}= \underbrace{e\begin{bmatrix} a & b \\ c & d \\ \end{bmatrix}}_{Q}\,\underbrace{\begin{bmatrix} x_1\\ x_2\\ \end{bmatrix}}_{\vec{x}}\tag 1$$ where $$a,b,c,d,e\quad$$ are integer numbers .

with $$q_2=x_1-x_2\quad \Rightarrow\quad c=1\,,d=-1 \quad$$ is $$q_1$$ the cyclic coordinate

from equation (1) we get:

$$\vec{\dot{x}}=Q^{-1}\,\vec{\dot{q}}$$

$$\Rightarrow$$

$$T=\frac{m}{2}\vec{\dot{x}}^T\,\vec{\dot{x}}=m\,\left(Q^{-1}\,\vec{\dot{q}}\right)^T\,\left(Q^{-1}\,\vec{\dot{q}}\right)\overset{!}{=}m\,\vec{\dot{q}}^T\,\vec{\dot{q}}\tag 2$$

equation (2) must fulfill with the three constants $$a,b,e$$ , we choose arbitrary for $$a=1$$ and get :

$$T=m\,\left(2\,{\frac {{{\it q1}}^{2}}{{e}^{2} \left( 1+b \right) ^{2}}}+{\frac { \left( 2\,b-2 \right) {\it q2}\,{\it q1}}{{e}^{2} \left( 1+b \right) ^{2}}}+{\frac { \left( 1+{b}^{2} \right) {{\it q2}}^{2}}{{e}^{2} \left( 1+b \right) ^{2}}} \right)$$

$$\Rightarrow\quad b=1\,,e=1\quad, T=\frac{m}{2}\,(\dot{q}_1^2+\dot{q}_2^2)$$

so the transformation Matrix is:

$$Q=\begin{bmatrix} 1 & 1 \\ 1 &-1 \\ \end{bmatrix}$$

$$Q^{-1}=\frac{1}{2}\begin{bmatrix} 1 & 1 \\ 1 &-1 \\ \end{bmatrix}$$