Monsieur Lagrange pulls a string down through a hole in a horizontal table thereby effecting a rotating (point) mass. A daemon sits on his shoulder and takes careful note of the proceedings. There is no potential function. The Lagrangian is

$$\mathcal{L}= \frac12m \dot{r}^2 +\frac12m r^2\dot{\theta}^2. \tag{1}$$

As Monsieur Lagrange holds himself to a time protocol we have a Lagrangian with constraint

$$\mathcal{L'}= \frac12m \dot{r}^2 +\frac12m r^2\dot{\theta}^2+ \lambda (r-g(t)) .\tag{2}$$

The Euler-Lagrange for $r$ is

\begin{align} \frac{d }{dt }\frac{\partial\mathcal{L'}}{\partial \dot{r}}-\frac{\partial \mathcal{L'}}{\partial {r}}&= m(\ddot{r}-r \dot {\theta} ^2) + \frac{d }{dt }\frac{\partial\lambda(r-g(t))}{\partial \dot{r}}-\frac{\partial \lambda (r-g(t))}{\partial {r}} \\ &\approx m(\ddot{r}-r \dot {\theta} ^2)-\lambda \\ &= 0,\tag{3} \end{align}

where $\approx$ indicates here, and three times subsequently, that possibly some derivatives are multiplied by zero eaux-shell.

The Euler-Lagrange for $\theta$ is

\begin{align}\frac{d }{dt }\frac{\partial\mathcal{L'}}{\partial \dot{\theta}}-\frac{\partial \mathcal{L'}}{\partial {\theta}} &= \frac{d}{dt} (m r^2 \dot{{\theta}}) + \frac{d }{dt }\frac{\partial\lambda(r-g(t))}{\partial \dot{\theta}}-\frac{\partial \lambda (r-g(t))}{\partial {\theta}} \\ &\approx \frac{d}{dt} (m r^2 \dot{{\theta}}) \\ &= 0.\tag{4} \end{align}

We identify $mr^2\dot{\theta}$ as the angular momentum, $L$, a constant of the motion. Without marshalling the full Hamiltonian formalism we can calculate the change in kinetic energy.

$$ T= \frac12m(\dot{r}^2+r^2 \dot{\theta}^2)=\frac12\left(m\dot{r}^2+\frac{L^2}{mr^2}\right).\tag{5} $$

\begin{align} \frac{dT}{dt}&=m\dot{r}\ddot{r} -\frac{L^2}{mr^3}\dot{r}\\ &= \left(m\ddot{r}- \frac{(mr^2\dot{\theta})^2}{mr^3}\right)\dot{r}\\ &=m (\ddot{r} - r \dot{\theta}^2)\dot{r}\\ &= \lambda \dot{r}.\tag{6} \end{align}

No surprises thus far: $\theta$ is "ignorable", angular momentum $L$ is conserved. The Lagrange multiplier $\lambda$ is the total force in the "r" direction equal to the negative of the force exerted by Lagrange who does work as he pulls the mass inward.

Now...the demon has been compiling a table of $\theta$ vs. time and uses the implicit function theory to render r as a function of $\theta$. In the second run Monsieur Lagrange will be so constrained.


$$\mathcal{L''}= \frac12m \dot{r}^2 +\frac12m r^2\dot{\theta}^2+ \mu (r-h(\theta)).\tag{7}$$

The Euler-Lagrange for $r$ is

\begin{align} \frac{d }{dt }\frac{\partial\mathcal{L''}}{\partial \dot{r}}-\frac{\partial \mathcal{L''}}{\partial {r}}&= m(\ddot{r}-r \dot {\theta} ^2) + \frac{d }{dt }\frac{\partial\mu (r-h(\theta))}{\partial \dot{r}}-\frac{\partial \mu (r-h(\theta))}{\partial {r}} \\ &\approx m(\ddot{r}-r \dot {\theta} ^2)-\mu \\ &= 0.\tag{8} \end{align}

The Euler-Lagrange for $\theta$ is

\begin{align} \frac{d }{dt }\frac{\partial\mathcal{L''}}{\partial \dot{\theta}}-\frac{\partial \mathcal{L''}}{\partial {\theta}}&= \frac{d}{dt} (m r^2 \dot{{\theta}}) + \frac{d }{dt }\frac{\partial\mu(r-h(\theta))}{\partial \dot{\theta}}-\frac{\partial \mu (r-h(\theta))}{\partial {\theta}} \\ &\approx \frac{d}{dt} (m r^2 \dot{{\theta}}) + \mu \frac{\partial h}{\partial \theta} \\ &= \frac {d L}{d t} + \mu \frac {\dot r}{\dot \theta} \\ &= 0.\tag{9} \end{align}

Now we immediately see that since $\theta$ is no longer "ignorable" angular momentum is no longer conserved. On the other hand the Lagrange multiplier (now $\mu$)is still the total force operating in the $r$ direction. We can again calculate the rate of change of kinetic energy.

\begin{align} \frac{dT}{dt}&=m\dot{r}\ddot{r} -\frac{L^2}{mr^3}\dot{r}+ \frac1{2mr^2} \cdot \frac{d}{dt} L^2\\ &=\left(m\ddot{r}- \frac{(mr^2\dot{\theta})^2}{mr^3}\right)\dot{r}+ \frac1{2mr^2} \cdot 2L \frac{dL}{dt}\\ &= \mu \dot{r}+ \frac1{2mr^2} \cdot 2 mr^2 \dot{\theta}\cdot (-1) \mu \frac {\dot{r}}{\dot{\theta}}\\ &=\mu \dot{r} - \mu \dot{r}\\ &=0.\tag{10} \end{align}

Well I can't see what the mistake is. There don't seem to be any calculational errors. The Demon will only say things like $\theta$ is proceeding from this to that so you must pull in the string from this $r$ to that $r$.

I seem to recall from Electrical Engineering that "a system is 'Observable' if this matrix is of full rank" and "a system is 'Controllable' if this matrix or some other matrix is of full rank or is less than full rank but I don't even see how it applies.

(absorbed into OP's OP from OP's original "answer" in "anticipation" of the answer of @Qmechanic, comment by @ACuriousMind)

Details? The Daemon is in the Details. The construction in the second part is completely general as per Implicit Function Theory. As for one specific example:

We know $L=mr^2\dot{\theta}$ or $\dot{\theta}= \frac{1}{r^2} \frac{L}{m} $. Let us assume the following trial function of time:

$\frac{1}{r^2}=At+B$, (i.e. $r=g(t)=\frac{1}{\sqrt{At+B}}$) ,with A and B positive. Note also that $L$ is (still) a constant of the motion which we take to be positive.

$$\begin{align}\theta =\frac{L}{m}\int\,(At+B)~\mathrm dt &=\frac{L}{m}\left( \frac{1}{2}At^2+Bt\right) \\ \frac{1}{2}At^2+Bt-\frac{m}{L}\theta &=0\\ \therefore ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ t &= \frac{-B+ \sqrt{B^2 + 4 \frac{1}{2} A \frac{m}{L}} \theta}{A}\end{align}$$

Here we have chosen the positive root.

$$r= \frac{1}{\sqrt{At+B}} = \frac{1}{\sqrt{A \frac{-B+ \sqrt{B^2 + 4 \frac{1}{2} A \frac{m}{L}} \theta}{A}+B }} = \frac{1}{\sqrt{\sqrt{B^2+2A \frac{m}{L}\theta}}}$$

(i.e. $r=h(\theta)= \frac{1}{\sqrt{\sqrt{B^2+2A \frac{m}{L}\theta}}}$)

This does not close the issue at hand for there remains the possibility that even as r is being drawn down the Lagrange multiplier, $\lambda$, is null.

$$\begin{align} r &=(At+B)^ {-\frac{1}{2}}\\ \dot {r} &=-~\frac{1}{2}(At+B) ^{-\frac{3}{2}}A\\ \ddot{r} &=+~\frac{3}{4}(At+B)^{-\frac{5}{2}}A^2= \frac{3}{4} A^2 r^5\end{align}$$

But as

$$\lambda=m\left(\ddot{r}-\dot{\theta^2}r\right)=m\left( \ddot{r}-\left(\frac{L}{m}\right)^2 \frac{1}{r^3}\right)=m\left(\frac{3}{4} A^2 r^5-\left(\frac{L}{m}\right)^2\frac{1}{r^3}\right)$$

I'm sure you see that there is but little chance that this Lagrange multiplier will render no work.

I think my original question contained two cons. Neither entailed the "construction" for the second run. The first con was with regard to the constraint itself: $\lambda(r-g(t))$ Now is this "constraint" holonomic?; non-holonomic? As much as you would like it, even if $\lambda$ is allowed to possess explicit time dependence it can not be cast into the form $\lambda$F(all the generalized coordinates, all the generalized velocities).

Alternativelly Qmechanic likes to think in terms of Potential Energy $V := \lambda (g(\theta,t)-r)$. But, by the same token, I remember Transition Probabilities (in Qm 101 (or 102)) and reviling against time-dependandant Potential Energies even just on the right side of Shroedinger's Equation.

On the one hand the more I think of these conundra (the de-Conservation of Angular Momentum and, as Qmechanic emphasized, the re-Conservation of Kinetic Energy) the less botherred I am. On the other hand there is still something that is being missed. The question isn't being asked properly...


2 Answers 2


Let us put the mass $m=1$ for simplicity. Monsieur Lagrange & the daemon are considering a Lagrangian of the form

$$ L~=~ T -V , \qquad T~:=~\frac{1}{2}v^2~=~ \frac{1}{2}(\dot{r}^2+ r^2\dot{\theta}^2), \qquad V~:=~\lambda(g(\theta,t)-r). \tag{A} $$

The Lagrangian momenta are

$$ p_r~=~\frac{\partial L}{\partial \dot{r}}~=~\dot{r}, \qquad p_{\theta}~=~\frac{\partial L}{\partial \dot{\theta}}~=~r^2\dot{\theta}, \qquad p_{\lambda}~=~\frac{\partial L}{\partial \dot{\lambda}}~=~0. \tag{B}$$

The Lagrangian energy function is $$ h~:=~p_r\dot{r}+p_{\theta}\dot{\theta}+p_{\lambda}\dot{\lambda}-L~=~T+V.\tag{C}$$

  1. In the first part of the experiment, the duo let the function $$g(\theta,t)~=~f(t)\tag{D}$$ be independent of $\theta$ but be a given fixed explicit function of time $t$. This implies that the energy (C) is not necessarily conserved in time.

  2. In the second part of the experiment, the duo tune the function $$g(\theta,t)~=~h(\theta; \text{initial data})\tag{E}$$ to a function without explicit time dependence, but such that the trajectories for $r$ and $\theta$ (but not necessarily $\lambda$) are the same as under the first part. Since there is no explicit time dependence, the energy (C) is conserved in time. OP observes that the energy conservation in the second part is in potential conflict with the first part.

For starters, the construction in the second part is not possible for generically given function $f(t)$.

We predict that for the special functions $f(t)$ where the second part is possible, the constraint force $\lambda$ is zero, so that the energy (C) is conserved in time in both parts of the experiment, and thereby resolves the apparent contradiction. We leave it to the reader to work out the details.

  • $\begingroup$ The Daemon is in the Details. j. $\endgroup$
    – jstark
    Commented Dec 1, 2016 at 16:19

Here's the real answer. Consider the first constrained Lagrangian:

$$\mathcal{L'}= (1/2)m \dot{r}^2 +(1/2)m r^2\dot{\theta}^2+ \lambda (r-g(t))$$

We see the Kinetic Energy, the Lagrange multiplier, the constraint all rendered in polar coordinates. But, the Lagrangian is only INCIDENTAL to the physical setup i.e. Monsieur Lagrange and his string. The efficacy of polar coordinates is even underscored when the direction of the "generalized force" is examined:

$$f_r= \lambda \frac{\partial(r-g(t))}{\partial r}=\lambda$$

$$f_\theta=\lambda\frac{\partial(r-g(t))}{\partial \theta}=0$$

Now when the daemon taunts that the second run should be constrained according to $r=h(\theta)$ Signore Lagrange should rightly respond, "But that is what I have just done."

The second constrained Lagrangian is

$$\mathcal{L''}= (1/2)m \dot{r}^2 +(1/2)m r^2\dot{\theta}^2+ \mu (r-h(\theta))$$

The psychological glitch, and it is a powerful one, is that there is a mezmerization originating with the polar coordinates leading to an expectation that the origin of the constraining force must be the original experimental setup: Signore Lagrange and his string. This misapprehension is allayed only when considering the direction of the "generalized force" :

$$f_r= \mu \frac{\partial(r-h(\theta))}{\partial r}=\mu$$

$$f_\theta=\mu\frac{\partial(r-h(\theta))}{\partial \theta}=\mu (-)\frac{\partial h}{\partial \theta}=- \mu \frac{\partial r}{\partial \theta}=- \mu \frac{\dot{r}}{\dot{\theta}}$$

As $\mathcal{L''}$, which is the Kinetic Energy, the Lagrange multiplier and the constraint, "sniffs" its way through 2+1 space, it does so in such a way that the constraint force can NOT originate with the original experimental setup. $(f_\theta \neq 0)$

This last proves that $L$ must be de-conserved. And since the direction of the "generalized force" is trivially seen to be perpendicular to the actual trajectory: Kinetic Energy is re-conserved!

The simplest experimental setup which manifests $\mathcal{L''}$ is perhaps the simplest of all: a trough (with an ice cube).

(Another iso-speed example is "tether ball", NOT well-explained on this sight.)

Could Signore Lagrange and his string constrain his mass along the trajectory of any such trough? The answer is "Yes" but only so long as the word "trajectory" is understood to mean r as a function of $\theta$. One must re-assume the conservation of $L=mr^2 \dot{\theta}$ and integrate. But, if by "trajectory", one means the joint dependance of r and $\theta$ on t, the answer, as we have seen, is "No."

Finally, might there be ramifications in Real Physics? Could there have been formulations in terms of Lagrangians and then re-formulations in terms which could no longer find "support" within the "context" of the "physical setup" (i.e. the Universe). QFT? GR? Sounds like something Monsieur Dirac would have jumped on. But I will have to leave that to him and other Real physicists.


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