9
$\begingroup$

Noether's theorem says that if the following transformation is a symmetry of the Lagrangian

$$t \to t + \epsilon T$$

$$q \to q + \epsilon Q.$$

Then the following quantity is conserved

$$\left( \frac{\partial L}{\partial \dot{q}}\dot{q} - L \right) T - \frac{\partial L}{\partial \dot{q}} Q.$$

Suppose our Lagrangian is given by

$$ L = \frac{1}{2}m \dot{q}^2 - \ln t.$$

Then is not the Lagrangian invariant under the transformation given by

$$T = t$$

$$Q = 0~?$$

Making this transformation contributes only an additive constant to the Lagrangian, which will not affect the dynamics, and so we should conclude that such a transformation is indeed a symmetry of the Lagrangian. However, the quantity

$$ \left( \frac{\partial L}{\partial \dot{q}}\dot{q} - L \right) t = \left(\frac{1}{2}m\dot{q}^2 + \ln t \right)t$$

is clearly not conserved. The E-L equations imply that the kinetic energy is constant, and so this function is clearly an increasing function of time.

Where is my error?

$\endgroup$
  • 2
    $\begingroup$ If $\mathcal{L}$ has explicit time-dependence, then energy is not conserved. $\endgroup$ – Kyle Kanos Dec 30 '13 at 14:22
  • $\begingroup$ No, I'm not trying to argue that energy is conserved. It is clear that this Lagrangian is NOT invariant under t going to t + epsilon. However, it is (I believe) invariant under a 'time dilatation' or a 'rescaling' of time. Hence not energy, but some quantity (energy*time) should be conserved, according to my argument (which is faulty somewhere). $\endgroup$ – gj255 Dec 30 '13 at 14:33
  • $\begingroup$ let us continue this discussion in chat $\endgroup$ – Kyle Kanos Dec 30 '13 at 15:49
3
$\begingroup$

The Lagrangian (and the action as a whole)

$$ L = \frac{1}{2}m \dot{q}^2 - \ln t$$

is not invariant under the transformation given by $$T = t, \qquad Q = 0.$$

The rescaling of $t$ by factor $1+\epsilon$ also modifies the time derivatives: $ \delta \dot{q} = -\epsilon \dot{q}$ (and the measure of integration $dt$), so the suggested quantity is not conserved.

Where is my error?

So, the error is in choosing the wrong Lagrangian / transformation.

Noether's theorem works just fine for explicitly time dependent Lagrangians. Here is another example of Lagrangian with explicit time-dependence: $$ L = \frac{m \dot{q}^2}{2} e^{\alpha t}. $$ Such type of Lagrangian could arise for example, if we try to obtain the equations of motions for dissipative system.

The Euler-Lagrange equation for this system after omitting common factor reads as $$ \ddot{q} = -\alpha \dot{q}. $$

This Lagrangian is invariant under the infinitesimal transformation: $$ t\to t' = t + \epsilon, \qquad q \to q' = q -\epsilon\frac{\alpha q}{2}. $$

Substituting these $T$ and $Q$ in the Noether's theorem we obtain the quantity $$ A=\frac{m }{2} e^{\alpha t}\cdot (\dot{q}^2 + \alpha \dot{q} q). $$

Its time derivative is $$ \dot{A} = \frac{m }{2} e^{\alpha t}\cdot (\alpha \dot{q}^2 + \alpha^2 \dot{q} q +2 \ddot{q}\dot{q} + \alpha \ddot{q} q + \alpha \dot{q}^2) , $$ If we use E-L equation to eliminate $\ddot{q}$ we obtain $\dot{A}=0$, so the quantity is conserved as it should be.

$\endgroup$
  • $\begingroup$ I feel as though I am close to understand but am missing something very fundamental. When you make a transformation t going to t+eps, are you saying that this forces the $q$ and $\dot{q}$ to transform as well, since they depend implicitly on t? $\endgroup$ – gj255 Dec 30 '13 at 17:58
  • $\begingroup$ No, the transformation $t\to t + \epsilon$ is just a shift of $t$ by a constant, it does not modify $\dot{q}$. But transformation $t\to t+\epsilon t$ (in your question) is a rescaling of time, so the time derivatives are multiplied by a factor $(1+\epsilon)^{-1}$. $\endgroup$ – user23660 Dec 30 '13 at 18:01
  • $\begingroup$ In my example the shift of $t$ gives the Lagrangian a constant multiplier $e^{\alpha \epsilon}$. To compensate it we rescale the coordinate by a constant factor (this also of course rescales $\dot{q}$ by the same factor). $\endgroup$ – user23660 Dec 30 '13 at 18:07
  • $\begingroup$ Ah yes, I'm starting to see now, thank you. Let me think about this some more, I imagine I will have another question shortly. $\endgroup$ – gj255 Dec 30 '13 at 18:09
  • $\begingroup$ If the Lagrangian is simply L = 1/2 m (dq/dt)^2 --- kinetic energy only --- then is the Lagrangian invariant under Q = q and T = t? I.e., a rescaling of both the position and time co-ordinates by a factor (1+eps)? I would have thought this is so, since dq'/dt' = dq/dt on account that the (1+eps) factor cancels on the top and bottom. If this is so, however, then I find a conserved quantity to be (1/2 m v^2 t - mv q) --- using the formula I quoted in my original post. But this is clearly not conserved! $\endgroup$ – gj255 Dec 30 '13 at 18:29
2
$\begingroup$

Let's see how we can play with Noether's theorem in the "conservation of energy" example.

First of all, we apply a time-independent variation to the Lagrangian, \begin{equation} t \rightarrow t+ \epsilon \end{equation} For a general Lagrangian, it will change \begin{equation} L(x(t), \dot{x}(t), t)\rightarrow L'(x(t), \dot{x}(t), t ) = L(x(t), \dot{x}(t), t-\epsilon ) \end{equation} what we require in this case, is that the Lagrangian is invariant(not just covariant) as a scalar, \begin{equation} L=L' \implies \frac{\partial L}{\partial t} = 0 \end{equation} That's a property true alone any path.

Then we use an induced time-dependent variation, \begin{equation} t \rightarrow t+ \epsilon(t) \quad x(t) \rightarrow x(t) + \epsilon(t) \dot{x(t)} \end{equation} The change of the action in this case is \begin{equation} \delta S = \int_0 ^T \frac{\partial L }{\partial x} \dot{x(t)} \epsilon(t) + \frac{\partial L } {\partial \dot{x} } \frac{d}{dt}(\epsilon(t) \dot{x}(t)) = \int_0^T dt \epsilon(t) ( \frac{\partial L }{\partial x} \dot{x} + \frac{\partial L } {\partial \dot{x} } \ddot{x} ) + \frac{\partial L } {\partial \dot{x} } \dot{x} \dot{\epsilon} \end{equation} Now adding $\frac{\partial L}{\partial t}=0$ to the first two terms in the parentheses to make it a total derivative, since this is true for all paths in the configuration space . For the second term, we integrate by part to get(variation has zero boundary condtions), \begin{equation} \delta S = \int_0^T dt \epsilon(t) \frac{d}{dt}(L - \frac{\partial L } {\partial \dot{x} } \dot{x}) \end{equation} Alone the classical path, any variation will extremize the action; in particular our special variation $\epsilon(t)$ induced by the symmetry action will. We get the energy conservation equation, \begin{equation} \frac{d}{dt}(L - \frac{\partial L } {\partial \dot{x} } \dot{x}) = 0 \quad \text{along classical path} \end{equation}

In other words, you have to choose $T$ and $Q$ to be constant(independent of time) to dig out the symmetries of the Lagrangian. Then you can apply the time-dependent variation to get a special "equation of motion" along the classical path: conservation law.

I need to elaborate the symmetry equation I have derived.

Under a time-independent variation, we say Lagrangian actually changes \begin{equation} L' = L(x(t-\epsilon), \dot{x}(t-\epsilon), t-\epsilon ) \end{equation} This is due to an identity, \begin{equation} L(x(t),\dot{x}(t),t) = L'(t') \quad t'=t+\epsilon \end{equation} That means the value of Lagrangian of a point in configuration space is independent of the time coordinate describing it. It's like you can use different time zones to describe an event. They only differ by a constant.

However, this is not true if we apply a time dependent variation, \begin{equation} \frac{d}{dt}(x(t-\epsilon)) = \frac{d}{dt}( x(t) - \epsilon \dot{x}(t) ) = \dot{x}(t) - \dot{\epsilon}\dot{x(t)} - \epsilon \ddot{x}(t) \ne \dot{x}(t) - \epsilon \ddot{x}(t) = \dot{x}(\tau)|_{\tau = t-\epsilon} \end{equation} the time dependent variation forces the velocity to change in a another way different from what we like. It's like time is not flowing uniformly, such that even if you displace your velocity vector to the original point before the translation, it still changes, because the rate will depend on how time flows.

In the technical aspect, you variation can't lead to \begin{equation} \frac{\partial L}{\partial t} = 0 \end{equation} which is equivalent to the conservation law.

$\endgroup$
  • $\begingroup$ I don't quite follow why this means you must choose T and Q to be constant. And what about the choice Q = n x r, corresponding to an infinitesimal rotation --- Q is an implicit function of time in this case. Perhaps you mean T and Q must not depend explicitly on time? Either way, I'm still not quite sure where in your argument you conclude T and Q must be constant. Thank you. $\endgroup$ – gj255 Dec 30 '13 at 15:24
  • $\begingroup$ What I mean is T and Q don't depend on time. When we are deriving the equation of symmetry, r doesn't depend on time, because we are not talking about any property of trajectories; we concern about the Lagrangian in any point of the phase space which may or may not be along the classical path. $\endgroup$ – anecdote Dec 30 '13 at 15:30
  • $\begingroup$ OK, I see your point. So where exactly do we conclude that T must not be a function of time? $\endgroup$ – gj255 Dec 30 '13 at 15:36
  • $\begingroup$ The velocity will change if that's an time-dependent variation. $\endgroup$ – anecdote Dec 30 '13 at 16:12
  • $\begingroup$ Sorry, which velocity will change? $\endgroup$ – gj255 Dec 30 '13 at 16:16
2
$\begingroup$
  1. Noether's theorem also works for OP's time-dependent Lagrangian $$ L~=~T-V,\qquad T ~=~\frac{1}{2}m \dot{q}^2, \qquad V(t)~=~\ln t. \tag{A} $$

  2. The time-dependent potential $V(t)=\ln t$ can be viewed as changing conventions for the zero-level of the potential energy. Nevertheless, the kinetic energy $T$ is a constant of motion.

  3. To prove kinetic energy conservation, OP essentially makes the same mistake as OP in this Phys.SE question: Perhaps counter-intuitively, the relevant infinitesimal transformation is not a pure time translation. The key infinitesimal transformation is instead $$ \delta q~=~\varepsilon\dot{q} .\tag{B}$$ Time-translation can be included or excluded as explained in my Phys.SE answer here. Let us for simplicity exclude time-translation $$ \delta t ~=~0. \tag{C}$$ Then the transformation $\delta$ can no longer "feel" the potential term $V$. We are therefore back in a standard application of Noether's theorem.

  4. The infinitesimal transformation of the Lagrangian $$ \delta L ~\stackrel{(A)+(B)+(C)}{=}~\ldots~=~\varepsilon \frac{dT}{dt} \tag{D} $$ is a total derivative, i.e. the transformation (B) & (C) is a quasisymmetry of the Lagrangian (A).

  5. The bare Noether charge is momentum times generator. That is: $p\dot{q}$. The full Noether charge $$p\dot{q}-T~=~T\tag{E}$$ is unsurprisingly the kinetic energy.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.