Spring pendulum Lagrangian has any ignorable coordinates?

[lys04]

Are there any ignorable coordinates in this scenario?
I don’t think so right, because the lagrangian has explicit dependency on both x and theta. Ignorable coordinates means there is no explicit dependence of that coordinate right?
If there are no ignorable coordinates that also means there are no constants of motion?
Since the Euler-Lagrange equation has d/dt(partialL/partial q_i dot) = partialL/partial q_i and none of the partial L/partial q_i’s are 0 then there are no constants of motion right

 

[TSny]

Are there any ignorable coordinates in this scenario?
I don’t think so right, because the lagrangian has explicit dependency on both x and theta.

That's correct.

Ignorable coordinates means there is no explicit dependence of that coordinate right?

Right.

If there are no ignorable coordinates that also means there are no constants of motion?

No. There can be a constant of the motion that is not associated with an ignorable coordinate. Can you think of such a constant of motion for this problem?

 

[lys04]

I read online that the total energy of the system and the angular momentum is conserved, but not sure how they arrived at that.

 

[TSny]

I read online that the total energy of the system and the angular momentum is conserved, but not sure how they arrived at that.

The energy $E = T + V$ is conserved in this problem but the angular momentum is not conserved.

Energy is conserved since the only forces that do work on the system are the conservative force of gravity and the conservative force of the spring. (There is a force at the point of support, but it doesn't do any work.)

Angular momentum about the point of support is not conserved because the force of gravity acting on $m$ produces a torque on the system about the point of support. If you removed the force of gravity, the angle $\theta$ would become an ignorable coordinate in the Lagrangian. Then, the Euler-Lagrange equation of motion for $\theta$ would show that the angular momentum is a constant of the motion.

It is not difficult to show that the following conditions on the Lagrangian are sufficient for $E$ to be conserved (see any standard text covering Lagrangian mechanics):

(i) $L(q_i, \dot q_i)$ does not depend explicitly on time.
(ii) $T$ is of the form $\frac 1 2 \sum_{i, k} a_{ik}\dot q_i \dot q_k$ where the coefficients $a_{ik}$ can depend on the $q_i$ but not on the $\dot q_i$.
(iii) $V$ is a function of the $q_i$ only.

You can check that these conditions are met for this problem.

 

[lys04]

Okay that makes sense, so if there is an ignorable coordinate then the corresponding generalised momentum is also conserved.

Can you give an example of where T is not in the form you described and thus E is not conserved?

Thanks a lot

 

[TSny]

Okay that makes sense, so if there is an ignorable coordinate then the corresponding generalised momentum is also conserved.

Yes.

Can you give an example of where T is not in the form you described and thus E is not conserved?

There are systems where the Lagrangian is of the form $L = \tilde T(q_i, \dot q_i) - V(q_i)$ such that the function $\tilde T(q_i, \dot q_i)$ is not of the form $\sum a_{ik} \dot q_i \dot q_k$ .

An example is the Lagrangian for a relativistic particle moving in a potential: $$L = -mc^2\sqrt{1-\dot x^2/c^2 - \dot y^2/c^2 - \dot z^2/c^2} - V(x, y, z)$$ The first term contains the velocity components, but it doesn't represent the kinetic energy of the particle. So, the Lagrangian in this case is not the difference between the kinetic energy and the potential energy. We could denote the first term by $\tilde T$ so $L = \tilde T(q_i, \dot q_i) - V(q_i)$. The equations of motion derived from $L$ would show that the quantity $\tilde E = \tilde T + V$ is not a constant of the motion. This is not of concern since $\tilde E$ is not the total energy $E$ of the particle.

If the Lagrangian of a system does not depend explicitly on time, then it can be shown that the following quantity is a constant of the motion: $$\sum_i \dot q_i \frac {\partial L}{\partial \dot q_i} - L(q_i, \dot q_i) .$$ For the Lagrangian of the relativistic particle, this quantity can be shown to equal the total relativistic energy.

For a Lagrangian of the form $L = T(q_i, \dot q_i) - V(q_i)$, where $T = \sum a_{ik} \dot q_i \dot q_k$ , the constant of motion given above reduces to $T + V$.