# Hamiltonian, Lagrange multipliers and Dirac's Programme

1. Aug 11, 2007

### Super_Leunam

Hi!
I've been studying Dirac's programme for some time and I realized that there's something missing:
Actually this is missing in every standard book on classical mechanics concerning how constraints are implemented in the lagrangian.
They are usually inserted with some unknown variables called "Lagrange's Multipliers". In Dirac's theory of constrained systems these are derived as a consequence of a "theorem" appearing in Henneaux and teitelboim's "Quantization of gauge systems", page 8. Dirac, in his "Lectures on Quantum Mechanics" didn't even say a word as to how it should be derived, he simply said that it's part of the standard methods of the calculus of variations with constraints and I assume he was refering to the well known method of Lagrange's multipliers. This theorem is proved in the appendix for the first chapter of Henneaux's but I kinda feel like there's something missing.

$$\text{If } \lambda_n\delta q^n + \mu^n \delta p_n = 0$$ for arbitrary variations of the coordinates of the phase (q and p) space tangent to the constraints surface then:

$$\lambda_n = u^m\frac{\partial \phi_m}{\partial q^n}$$

$$\mu^n = u^m\frac{\partial \phi_m}{\partial p_n}$$

Where the u's are the multipliers mentioned and $$\phi_m$$ are the constraints derived from the Lagrangian because of it being "singular" as explained below. By the way, the proof of this theorem in Henneaux's book is almost a copy of the proof for the method of Lagrange's multipliers found in any Analysis book I think the problem stems from the definition of the Hamiltonian. I know that the Hamiltonian is defined as a Legendre Transform
$${\cal H}(q, p) = \textit{Ext}_{v} H (q, v, p) = \textit{Ext}_{v} \{p\cdot v - {\cal L}(q, v)\}$$
Where "Ext" denotes an extremum, I still don't worry about its uniqueness ,though. It follows, according to the rules of calculus and under some technical conditions, that we should have for an extremum ( if they exist, that is) that it satisfies
$$p = \frac{\partial {\cal L}}{\partial v}$$
Then we can replace all the velocities back to the function H
The conditions for this to work out is that the lagrangian be not singular, in order to solve the velocities in a unique way, that is,
$$det\left(\frac{{\partial}^2 {\cal L}}{\partial v_i \partial v_j}\right) \neq 0$$

But what if that doesn't happen? (in certain region of the tangent bundle)
This is precisely the case when we have a constrained system and this is where that little theorem I mentioned is to be used.

My Problem is the following: Lagrange multipliers are to be used when we want to find critical points for a function restricted to some conditions or "constraints" $$\phi_j$$. I thought that the function to be optimized was the H(q, v, p) but so far I just don't seem to make it work since Lagrange's method requires that the gradient of H with respect to the velocities arguments be proportional to the gradient of those restrictions with respect to the same velocities.

The constraints that one derives when having a singular Lagrangian are functions of p and q but not of the velocities. Moreover, if we want to reconcile this assumption with that theorem, the problem is that the gradients considered should be with respect to q and p but not with respect to v . If I want to consider that kind of gradient ( involving derivatives with respect to q and p) that would mean that I'm trying to optimize H considering p and q as variables too, which I don't think is correct.

The final result I want to obtain directly using the method of Lagrange's multipliers as stated in any Analysis book is
$$v = \frac{\partial {\cal H}}{\partial p_n} + u^m \frac{\partial \phi_m}{\partial p_n}$$

$$- \left.\frac{\partial {\cal L}}{\partial q^n}\right|_{v}= \left.\frac{\partial {\cal H}}{\partial q^n}\right|_p + u^m \frac{\partial \phi_m}{\partial q^n}$$
with $$v = \dot{q}$$

which are derived from standard considerations about the hamiltonian using that little theorem. I'll just copy the steps Henneaux considered on page 8 :

Consider arbitrary independent variations of the positions and velocities $$\delta q^n, \hspace{5pt}\delta p_n$$ which should be "tangent" to the surface generated by the $$\phi_m$$ in order to preserve these constraints in first order.(This argument of tangency isn't included in the book, I had to think about it). Then the change induced on H is
$$\delta H = \dot{q}_n\delta p_n +\delta\dot{q}_np_n - \delta\dot{q}\frac{\partial L}{\partial \dot{q}_n} - \delta q_n\frac{\partial L}{\partial q_n} = \dot{q}_n\delta p_n - \frac{\partial L}{\partial q_n}\delta q_n$$
But it should be also that
$$\delta H = \frac{\partial H}{\partial q_n}\delta q_n + \frac{\partial H}{\partial p_n}\delta p_n = \dot{q}_n\delta p_n - \frac{\partial L}{\partial q_n}\delta q_n$$
$$\left(\frac{\partial H}{\partial q_n} + \frac{\partial L}{\partial q_n}\right)\delta q_n + \left(\frac{\partial H}{\partial p_n} - \dot{q}_n\right)\delta p_n = 0$$
where $$\delta p_n$$ are not independent variations but are regarded as linear combinations of $$\delta q^n \text{and } \delta \dot{q^n}$$ (Here I'm using the notation from Henneaux's book where $$\dot{q} = v$$).
And, using that little theorem, Henneaux is able to obtain the desired result for the Hamilton's equations of motion for constrained systems. But I still can't obtain this result from the considerations I've made. ( using directly Lagrange's method so I can justify where those multipliers come from)

What do you think is the origin of those multipliers? In other words, what am I optimizing?

Last edited: Aug 12, 2007
2. Aug 12, 2007

### dextercioby

First of all, it's not H which should be optimized, but the hamiltonian action. What $\delta H$ stands for is the variation of the Hamiltonian due to a set of variations compatible with the primary constraints of the theory. And the hamiltonian doesn't depend on the velocities and niether the primary constraints.

3. Aug 12, 2007

### Super_Leunam

I was thinking of the formal definition of a Hamiltonian as a Legendre Transform. A Legendre Transform IS an optimal value of a new function , say $$H$$ which depends on the two previous sets of variables (q and v) and another set of variables ( p) but the real optimized function ( optimized with respect to one of the old variables, say v) depend only on two variables ( q and p) . Check Wikipedia, its article on Legendre Transform is an excellent explanation.

I wasn't using the principle of Stationary action at all. I just don't seem to understand where those Multipliers come from.

We have two sets of equations:
$$- \left.\frac{\partial {\cal L}}{\partial q^n}\right|_{v}= \left.\frac{\partial {\cal H}}{\partial q^n}\right|_p + u^m \frac{\partial \phi_m}{\partial q^n}$$
$$v = \frac{\partial {\cal H}}{\partial p_n} + u^m \frac{\partial \phi_m}{\partial p_n} \hspace{15pt}v = \dot{q}$$
What I see is that the method of Lagrange Multipliers is being used because of the presence of the $$u^m \frac{\partial \phi_m}{\partial q^n} \hspace{8pt}\text{and}\hspace{8pt} u^m \frac{\partial \phi_m}{\partial p_n}$$. What I see is that , according to Lagrange's Method of Lagrange's Multipliers, there's some function being optimized. I guess it's alright to think that the multipliers come from the optimization of the hamiltonian action. I'll check out if it makes sense according to the statement of the lagrange's multipliers method that the u's come from that particular optimization. but what about the issue of optimizing the function $$H(q, v, p)$$(which is NOT the Hamiltonian yet)? the method of finding stationary points by setting to zero the partial derivatives ( that's how the change of variables $$p = \frac{\partial {\cal L}}{\partial v}$$ is justified) shouldn't work anymore because we don't know whether it's an extremum since the Hessian is singular, it could be neither. there should be another way to extremize the function $$H$$and I thought it was there where the multipliers come from, my assumption is probably wrong but how do I do the optimization "right"?. We don't know if it's actually an extremum what we get after the change of variables from v to p and as such it shouldn't be considered as a legendre transfrorm, or maybe the answer is that the Hamiltonian isn't a Legendre transform always? Actually the Legendre Transform is only defined for functions for which the determinant of its Hessian is positive . I just remebered that. so..The Hamiltonian isn't always a Legendre Transform, properly ... I don't like this kind of Hamiltonians