Best Coordinate system - Lagrangian problem

Question

In $\Bbb R^3$ consider an heavy point $P$ whose mass $m$ on a circumference $\Gamma$ of radius $R$, centered in the origin. Now consider that $\Gamma$ lives in the plane $$\Pi = \{( x,y,z) \in \Bbb R^3 : x=y \}$$ We have another point $Q$ same mass $m$ attached to the origin with a spring, $Q$ could move only on the line $\{y=z=0\}$ and is also attached with another string to $P$. Both strings has the same elastic constant. Find the Lagrangian of the system.

I'm a little bit confused on what coordinates I should choose. First thing I thought $(R,\theta)$ living in $\Pi$ plane and cartesian coordinates for $Q$ $$P = (R\cos\theta,R\sin\theta, 0)\\ Q = (0, 0, x)$$ I think it's wrong because here is like I'm forgetting that the plane is rotated of $\frac{\pi}4$, so an adjustment could be $$P = (R\cos\theta,R\sin\theta, 0)\\ Q = \bigg(0, \frac{\sqrt 2}{2}x, \frac{\sqrt 2}{2}x\bigg)$$ Where as the last coordinate I choose the normal of $\Pi$.

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Am I on the right track or missing something? Thank you very much for your help.

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Actually could be a little misleading think $R$ as a coordinates because is fixed, in terms of what I've written above nothing changes but as a clarification $R$ is fixed and the "real" coordinate is $\theta$. Without $\Gamma$ I could also think at variable $R$.

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Edit: Still don't know if what I've thought is correct but after a little bit of thinking spherical coordinates are a valuable choice.

Make use of the setting below

knowing that the plane $\Pi$ correspond to $\phi = \pi/4$

$$P = \bigg(\frac{\sqrt 2}{2} R\sin\theta,\frac{\sqrt 2}{2} R\sin\theta, R \cos \theta\bigg)\\ Q=(r,0,0)$$

Still wondering if the method presented above works just fine...

Answer 1

What I would do is start off with the $P$. To parametrise the circumference I would add a constant vector $\mathbf{r}_0=(x_0, y_0, z_0)$ to the circumference, parametrised by the unit vectors within the plane $x=y$.

Let's proceed to choosing this vectors. Since they might be chosen arbitrarily, I would choose $\mathbf{p}=\left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}, 0\right)$, which is a vector collinear to the line $x=y$, and $\mathbf{q}=\left(0, 0, 1\right)$, which goes along $z$-axis. Now, since I want my circumference to lie on the plane spanned by these two vectors, any point on it should satisfy the following equation for some $\theta$: $$ \mathbf{r}=\mathbf{p}R\cos\theta+\mathbf{q}R\sin\theta $$

What's left is to move the circumference away from the origin to its center by adding the $\mathbf{r}_0$. Finally, the position of $P$ at any given time could be parametrised by: $$ \mathbf{r}=\mathbf{p}R\cos\theta+\mathbf{q}R\sin\theta + \mathbf{r}_0 $$ or equivalently: $$ \mathbf{r}=\left( \begin{array}{c} x_0 +\frac{1}{\sqrt{2}}R\cos\theta \\ y_0 +\frac{1}{\sqrt{2}}R\cos\theta \\ z_0 + R\sin\theta \end{array} \right) $$ where $x_0 = y_0$.