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):
$$p_{+} = 2m\dot{x}_{+} = m\dot{x}_1 + m\dot{x}_2$$


*

*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?   

*In general, how can we find the transformation which reveals the cyclic coordinate?      
Also, there are some doubts on the derived terms:  


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

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

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

*

*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.


*

*Use the definition of canonical/conjugate momenta $p_+:=\frac{\partial L}{\partial \dot{x}^+}$.   


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