Lagrangian for a Central force problem is:

$$\mathcal{L} = \frac{1}{2}\mu(\dot{r} + r^{2}(\dot{\theta}^{2} + sin^{2}\theta\cdot \dot{\varphi}^{2})) - U(r)$$

We know that angular momentum is defined as: $$\overrightarrow{L} = \mu \cdot \overrightarrow{r} \times \dot{\overrightarrow{r}}$$ $\therefore \hspace{0.5cm} \overrightarrow{r}(t)\cdot \overrightarrow{L} = 0, \hspace{0.5cm}$it means motion takes place in single plane.

By the coordinate transformation we can have motion in x-y plane. Therefore, it means that we have angular momentum in $\hat{Z}$ direction.

$$\therefore\hspace{0.5cm} \theta = cos^{-1}\bigg( \frac{\overrightarrow{L}\cdot \hat{Z}}{||\overrightarrow{L}||}\bigg)$$ where, $$\overrightarrow{L} = \mu \cdot r^{2}\bigg(-\big(\dot{\theta}\cdot \sin\varphi\hspace{0.1cm}+ \hspace{0.1cm} \frac{1}{2}\cdot\varphi\cdot \sin2\theta\cdot\cos\varphi)\hat{X}\hspace{0.1cm} + (\dot{\theta}\cdot \cos\varphi\hspace{0.1cm}- \hspace{0.1cm} \frac{1}{2}\cdot\varphi\cdot \sin2\theta\cdot\sin\varphi)\hat{Y} \hspace{0.1cm}+\hspace{0.1cm} \big(\dot\varphi\cdot\sin^{2}\theta\big)\hat{Z}\bigg)$$ and, $$||\overrightarrow{L}|| = \mu\cdot r^{2}\big(\dot\theta^{2}+\dot\varphi^{2}\sin^4\theta\big)^{1/2}$$

$$\implies \theta = cos^{-1}\Bigg( \frac{\dot\varphi\cdot\sin^{2}\theta}{\big(\dot\theta^{2}+\dot\varphi^{2}\sin^4\theta)^{1/2}}\Bigg)$$

I am not able to how to transform coordinate such that new Lagrangian becomes ,

$$\mathcal{L}_{eff} = \frac{1}{2}\mu\big(\dot{r} + r^{2}\dot{\varphi}^{2}\big) - U(r)$$

  • 2
    $\begingroup$ Algle $\theta$ is not a variable, it is a constant. Just use its value. $\endgroup$ Commented Jan 17, 2021 at 9:44
  • $\begingroup$ Lagrangian is written in spherical coordinates where r,$\theta$,$\varphi$ is variable. Please let me know how $\theta$ is not a variable. $\endgroup$ Commented Jan 17, 2021 at 10:17
  • $\begingroup$ @Eli $\overrightarrow{r}$ is vector which is along unit vector $\hat{r}$, where: $\hat{r} = \sin\theta\cdot\cos\varphi\hat{X} + \sin\theta\cdot\sin\varphi\hat{Y} + \cos\theta\hat{Z}$ $\endgroup$ Commented Jan 17, 2021 at 14:03
  • $\begingroup$ yes i see my mistake $\endgroup$
    – Eli
    Commented Jan 17, 2021 at 15:16
  • 1
    $\begingroup$ Where you have $\dot{r}$ in the Lagrangian, it should be $\dot{r}^2$. $\endgroup$
    – J.G.
    Commented May 2, 2023 at 20:34

2 Answers 2


Since the movement is in a plane, you can indeed assume the motion to be in the $xy$-plane i.e. the $\theta=\pi/2$-plane. The Lagrangian for the system constrained to the $\theta=\pi/2$-plane is given by setting $\dot\theta=0$ and $\theta=\pi/2$ in the given Lagrangian.

  • $\begingroup$ @krupa In first I don't want to assume $\theta$ to any value. Motion in x-y plane, it means that angular momentum is pointing in $\hat{z}$ direction. Can you give me any physical region to give $\dot\theta$ to be zero. From Euler - Lagrangian equation I will get relation of $\theta$, that is : $2\cdot\ddot{\theta} = \sin2\theta\cdot \dot{\varphi}^{2}$. $\endgroup$ Commented Jan 17, 2021 at 12:49
  • 1
    $\begingroup$ You cannot get the Lagrangian below by just changing coordinates on the original Lagrangian. It contains one degree of freedom less than the original Lagrangian. Because the Lagrangian is rotationally symmetric, angular momentum is conserved and you can argue (as you did) that the motion is in a plane (but not in which one). But then you can conclude that it is sufficient to look for a solution in a specific plane. This does not give the general solution of the original problem. However, all solutions can be obtained by rotating solutions of the constrained Lagrangian. $\endgroup$
    – Krup'a
    Commented Jan 17, 2021 at 14:56
  • $\begingroup$ @ Aster Kleel. $\theta$ can be any arbitrary value, however, if the motion is in a plane, then it is not a dynamical variable; in this sense it is constant. $\endgroup$ Commented Jan 1 at 21:57

I don't think you can get the effective lagrangian by a naive change of coordinate without substituting in information about the initial state. The plane where the tracjectory is in, depends on the initial state, so the change of coordinate itself also depends on the initial state.

You can choose, without loss of generality, your z direction of the spherical coordinate to be parallel to the initial angular momentum. Then you can prove the trajectory remains in the plane (all derivatives of theta equals zero, using $2\cdot\ddot{\theta} = \sin2\theta\cdot \dot{\varphi}^{2}$ given theta initially is $\pi /2$ and has zero velocity). With this proven result, you can simplify the lagrangian to the effective one, but only for the initial conditions that satisfy your assumption that initial angular momentum is parallel to the z direction

I think you can also find the change of coordinate (that depends on initial angular momentum) this way:

  1. Change from spherical coordinate to cartesian.
  2. Change the old Cartesian to a new cartesian by a rotational matrix that maps the old z direction to the initial angular momentum direction.
  3. Change the new cartesian back to a new spherical.

In this new coordinate, you can prove the initial angular momentum is parallel to the z direction. This is by construction. Then you can follow the further steps from a previous paragraph, the one starting with "You can choose, without loss of generality, your z direction", to get the effective lagrangian in the new coordinate. However, notice this change of coordinate depends on the initial angular momentum in your old coordinate.

In math, there is no reason why all proof can be done using native change of coordinates (or a small set of rules). Sometimes you need other tools outside the small set of rules.

Furthermore, while thinking in pure mathematical abstraction can help in some cases, thinking physically can also help. The recognition that the change of coordinate itself depends on the initial condition is thinking physically, instead of thinking mathematical in finding clever cancellation by brute force. We should learn and try to use all the tools we have when solving a hard problem. Fixating on whether a small set of rules is sufficient in solving "all" problem, is not as important as learn to use all the tools to solve whatever problem.


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