# When do time dependent constraints mean energy conservation?

1. May 30, 2015

### Coffee_

Define energy as E=T+U.

For anyone using different terminology, by rheonomic (time dependent constraints) I mean that if a system has N degrees of freedom, the position vectors of each particle of the system are given by $\vec{r}_i(q_1,q_2,...,q_n,t)$. Where $q_i$ are generalized coordinates and at least ONE of all these vectors has explicit time dependence.

Anyway with this setup there are two possible cases:

1) $\frac{\partial L} {\partial t} = 0$

2) $\frac{\partial L} {\partial t} \ne 0$

For 1) all I'm able to do is prove that the Hamiltonian is NOT equal to the energy and that the Hamiltonian is obviously conserved. This does not yet in general say if energy is conserved or not. I have already encountered examples where this is the case but energy is not conserved. Is there a general statement about energy one can make in these two cases for time dependent constraints?

For 2) I'm pretty sure that energy is never conserved but still note sure how to prove that.

2. May 30, 2015

### vanhees71

Have a look at "Noether's Theorem(s)". The important one for your case is the following: For each one-parameter Lie-symmetry group of symmetry transformations on the phase-space variables there is a conserved quantity. A symmetry transformation is defined as one that keeps the variation of the action invariant. Vary often the action or even the Hamiltonian itself is invariant (but that are just special cases of the more general definition). Of course, you can also formulate this theorem for the Lagrangian formulation of Hamilton's principle. It's only mathematically more elegant in the Hamiltonian formulation, because there you have an elegant formulation in terms of the theory of Lie groups and Lie algebras: The Poisson brackets are a Lie product, and the symmetries are representations of the corresponding groups as canonical transformations on phase space.

Your case is pretty simply stated: If the Lagrangian is not explicitly time dependent, then the Hamiltonian is conserved for the trajectories defined as the equations of motion. You can use Noether's theorem in the general form to prove it, but it's very simple to see by simply using the equations of motion, which in the Lagrangian form of Hamilton's principle are the Euler-Lagrange equations
$$\frac{\partial L}{\partial q^k}-\frac{\mathrm{d}}{\mathrm{d} t} \frac{\partial L}{\partial \dot{q}^k}=0.$$
Now we define the Hamiltonian as
$$H=p_k \dot{q}^k-L \quad \text{with} \quad p_k = \frac{\partial L}{\partial \dot{q}^k}.$$
The we find for the trajectory defined by the equation of motion
$$\frac{\mathrm{d} H}{\mathrm{d} t}=\dot{p}_k \dot{q}^k + p_k \ddot{q}^k - \dot{q}^k \frac{\partial L}{\partial q^k} - \ddot{q}^k \frac{\partial L}{\partial \dot{q}^k} - \frac{\partial L}{\partial t}.$$
Now the 1st and the 3rd term cancel because of the Euler-Lagrange equation as well as the 2nd and the forth due to the definition of the canonical momentum $p_k$. So finally you are left with
$$\frac{\mathrm{d} H}{\mathrm{d} t}=- \frac{\partial L}{\partial t}.$$
So the Hamiltonian is conserved along the trajectory defined by the equation of motion if and only if the Lagrangian is not explicitly time dependent.

This is precisely the condition for the invariance of the action under time translations. Thus time-translation invariance of a system implies the conservation of the Hamiltonian, and usually one identifies the Hamiltonian with the energy of the system.

3. May 30, 2015

### Coffee_

Thanks for taking the time to reply. By energy I meant E=T+U though as defined in my first line. In the rheonolic case where the Lagrangian is time independent the hamiltonian is conserved but not always equal to this E anymore. (Is always equal under scleronomic constraints).

4. May 30, 2015

### cpsinkule

I'm not sure if this is what you're asking, but the Lagrangian can explicitly depend on time as long as it is an exact differential of time because it does nothing but add a constant to the total action. And upon varying the action, this constant contributes nothing.

5. May 30, 2015

### Coffee_

Not really applicable to my problem I suspect but still thanks for the reply.

6. May 30, 2015

### cpsinkule

Your problem supposes that the position of the particle's themselves depend explicitly on time. This is a statement that time is not homogeneous or isotropic. What it would mean is that the position, and thereby the momentum, of a particle would change spontaneously without the presence of a potential or external field. Time dependence is, as far as i know, is only explicitly allowed in the potential function or the addition of a total time differential to the Lagrangian for the reason of the homogeneity and isotropy of time. Even then, what would it mean for the qi to explicitly have time dependence? If you knew the time dependence of the coordinates, then you would have no need for Lagrange's equations in the first place. After all, the whole point of an action principle is to find the time dependence of the coordinates. If you know how the coordinates vary with time from the start, then the trajectory of the system is uniquely determined if you know the initial conditions.

7. May 31, 2015

### Coffee_

No my problem supposes that there is an explicit time dependence present in the position vectors as seen in $R^{3}$ in function of the generalized coordinates, which should not be surprising and comes up in a lot of mechanics problems like a bead on a stick that is rotating with a constant angular velcoity.

8. May 31, 2015

### vanhees71

That's a good example. The point is that you can go to a non-inertial frame, where the problem becomes time-independent. Of course, Noether's theorem doesn't care about whether you are in an inertial or accelerated frame, and still the Hamiltonian in this coordinates is a conserved quantity. You can now argue, whether you call this energy or not. I think, this is pretty unimportant. It's just about naming things.

Just to illustrate it, let's do this exercise. Perhaps then it becomes clear:

So let the bead run (with neglegible friction) along a rot that is rotating with constant angular velocity $\omega$ around the origin of the $xy$ plane. The distance of the bead from the origin may be $r$ (this is the one degree of freedom and the generalized coordinate I use). For the coordinates of the bead in the plane then you have
$$(x,y)=r (\cos \omega t,\sin \omega t).$$
The coordinates $(x,y)$ are relative to an inertial frame and $r$ thus in a rotating, i.e., non-inertial frame. The Lagrangian is simply
$L=T=\frac{m}{2} (\dot{x}^2+\dot{y}^2)=\frac{m}{2} (\dot{r}^2+\omega^2 r^2).$
So you have a Lagrangian which is not explicitly time-dependent and thus the Noether-theorem for time-translation invariance tells you that the Hamiltonian is conserved (no matter whether you call it energy or not). The reason of course simply is that in the non-inertial frame there's a centrifugal force that can be described by the potential $V(r)=-\omega^2 r^2$.

To evaluate the Hamiltonian, we need the canonical momentum:
$$p_r=\frac{\partial L}{\partial \dot{r}}=m\dot{r}$$
and the Hamiltonian thus becomes
$$H=\dot{r} p_r-L=\frac{p_r^2}{2m}-m \omega^2 r^2 \neq T.$$
Of course, it's not the energy as defined in the inertial frame but it's the effective one in the non-inertial frame. Whether or not one calls this energy, is just a question of semantics. I'm not sure, how textbooks usually define energy in such cases. So I'd just leave this open and call the Hamiltonian Hamiltonian, and that's it :-).