Why Does Hamilton's Function Equal Zero in Relativistic Lagrangian?

[pmb_phy]

It seems logical to assume that:

$$L(\tau,x^i,\dot{x^i})$$

I disagree. The Lagrangian that you posted is not an explicit function of proper time. Only explicit variables are listed in the Lagrangians variable set. Porper time is implicit in this case. Therefore the Lagrangian is only a function of 4-velocity.

Do you agree that the Hamiltonian corresponding to the above Lagrangian is zero?

Because its a covariant representation and as such the identification of the Hamiltonian as energy is lost. However, that the Hamiltonian is zero makes it useless. I suppose that's why Jackson said it was a problem.

Pete

 

[samalkhaiat]

How do we explain the fact that if we want to get Hamilton's equations out of Lagrangian:
$$L=-mc\sqrt{(\dot x^\mu \dot x_\mu)}$$,
where dot is noting derivative in proper time, we get that Hamilton's function equals zero ?

This is a general feafure of the COVARIANT Hamiltonian formulation of a theory. By taking the time PARAMETER as an EXTRA DYMANICAL VARIABLE, we can transform any theory into the covariant form in which the evolution parameter is arbitrary. As I will show you bellow, the introduction of an arbitrary time into Hamiltonian theory implies the existence of one 1st class constraint and a vanishing Hamiltonian. So, if the action (like yours) is "already covariant", i.e., it treats time as coordinate from the very begining, then the Hamiltonian vanishes identically as a result of parametrization-invariance.
To explain the problem of covariance, let us consider the very simple dynamical system of free non-relativistic particle

$$L(t) = \frac{1}{2} ( \frac{dx}{dt})^{2}$$

Notice that t is an evolution parameter, i.e., it is not a dynamical variable! well, peopel like myself do not like this fact, so we introduce a new evolution parameter (new time) $\delta$ as a monotonic function of t, and take $t \equiv x_{0}$ as an extra coordinate, i.e., we consider the new Lagrangian

$$L( \delta ) = \frac{1}{2} \frac{(dx/d \delta)^{2}}{dx_{0} /d \delta} \equiv \frac{1}{2} \frac{\dot{x}^{2}}{\dot{x}_{0}}$$

as a function of the coordinates $(x_{0},x)$ and time $\delta$, so that

$$\int L(t) \ dt = \int L( \delta ) \ d \delta$$

By introducing the corresponding momenta

$$P^{0} = \frac{ \partial L( \delta)}{ \partial \dot{x}_{0}} = - (1/2) ( \dot{x} / \dot{x}_{0})^{2}$$

$$P = \frac{ \partial L( \delta)}{ \partial \dot{x}} = \dot{x} / \dot{x}_{0}$$

we find that the canonical Hamiltonian vanishes identically;

$$H_{c} \equiv \dot{x}_{0} P^{0} + \dot{x} P - L( \delta) = 0$$

This means that there exists at least one 1st class primary constraint. Indeed, we do have it;

$$\phi = P_{0} + \frac{1}{2} P^{2} \approx 0$$

so, the total Hamiltonian is of the form

$$H_{T} = f \phi$$
where f is an arbitrary multiplier, and the equations of motion are

$$\dot{x}_{0} = \{x_{0}, \phi \} f = f, \ \ \dot{P}_{0} = 0$$
$$\dot{x} = \{x, \phi \} f = P f, \ \ \dot{P} = 0$$

Notice that this system of equation is form invariant (covariant) under the arbitrary infinitesimal transformations

$$x_{0} \rightarrow x_{0} + \epsilon , \ \ P^{0} \rightarrow P^{0}$$
$$x \rightarrow x + \epsilon P , \ \ P \rightarrow P$$

To go back to time t, we choose the "gauge" $x_{0} - \delta \approx 0$. In this gauge the equations take the standard form

$$f = 1 , \ \ \dot{x} = P$$

In short; parametrization-invariance ( which shows up because of the way in which we formulate the action) leads to off-shell mathematical identities, i.e., the identities hold wether or not the action is extremized. It is this fact that makes the identity H = 0 different from the conserved quantity $(1/2)P^{2}$ which holds only on-shell.


regards

sam

 

[samalkhaiat]

The Lagrangian is invariant under reparametrizations along the world line which preserve the parameter values at the endpoints.. A simple consequence of this is that the Hamiltonian is identically zero.

The quote above is the most accurate statement in this thread! Well done.

For the sake of completeness, I will prove the following general theorem:

If the action

$$S[q] = \int_{1}^{2} L(q, \dot{q},t) \ dt$$

is invariant under the infinitesimal transformation

$$t \rightarrow t + \epsilon (t)$$

with $\epsilon = 0$ at the endpoints, but arbitrary othewise, then the Hamiltonian vanishes identically.

Proof:

Given a parametrized state $q(t), \dot{q}(t)$ , we can define a new parametrized state

$$\bar{q}(t) = q( t + \epsilon ) \approx q + \dot{q} \epsilon$$

$$\frac{d}{dt} \bar{q}(t) = \dot{q} + \frac{d}{dt}( \dot{q} \epsilon)$$

Thus, to the 1st order in $\epsilon$ , we have

$$S[ \bar{q}] - S[q] = \int_{1}^{2} \left[ \frac{dL}{dt} \epsilon + ( \dot{q} \frac{ \partial{L}}{ \partial \dot{q}}) \frac{d \epsilon}{dt} \right] dt$$

or
$$0 = [ \epsilon L ]_{1}^{2} + \int_{1}^{2} \left[ \frac{\partial L}{\partial \dot{q}} \dot{q} - L \right] \frac{d \epsilon}{dt} \ dt$$

The 1st term vanishes because $\epsilon = 0$ at the endpoints. And , for arbitrary $d \epsilon /dt$, parametrization-invariance implies

$$H \equiv \frac{\partial L}{\partial \dot{q}} \dot{q} - L = 0$$

Notice that we did not assume that the action is extremal or that q(t) satisfies the E-L equation. Therefore, the above identity holds off-shell, i.e., H = 0 holds for any state (trajectory).


regards

sam

 

[pmb_phy]

...that he's using the standard, extremely well-known method of achieving a vanishing Hamiltonian.

"..extremely well-known method"? Hardly. Unless you're referring to calculating H = 0 when the Lagrangian used is covariant. Is that what you mean??

Lanczos is (or at least should be) standard reading. For the nth time: reparametrization invariance, ...

We'd appreciate it if you'd directly respond to Reilly's question to you regarding this "reparameterization" that you've mentioned here. I.e. Reilly asked you Also, what parameters do you consider in your reparametrization?

Pick your favourite monotonically increasing parametrization along the configuration space path; it'll be just as good as mine. If you so wish, adjoin the time to the configuration space also by considering the time as a function of this parameter. Repeat ad nauseum.

What are you claiming that this reparameterization will accomplish??

Pete

 

[pmb_phy]

It's already been pointed out that your webpage disagrees with Goldstein.

Please stop ignoring my question regarding this bogus claim of yours. I asked you to prove that my web page disagrees with Goldstein. You claim it does but you provide no proof. Let me refresh your memory. I posted the URL - http://www.geocities.com/physics_world/sr/relativistic_energy.htm

The (non-covariant) Lagrangian is in Eq. (15) in my web page. It is the exact same Lagrangian found in Classical Mechanics - 2nd Ed., Goldstein, page 322, Eq. (7.141). So with this clarification, on what basis do you claim my web page disagrees with Goldstein? Especially it was from Goldstein and Jackson from which I took this Lagrangian.

I'll tell you what your mistake is - You erroneously thought I was referring to the covariant Lagrangian. Well you were wrong. I later saw that it was the covariant Lagrangian that the OP was thinking about and the covariant equations produce a zero Hamiltonian and for which the Hamiltonian looses its meaning as the energy of the particle. I even later posted my other web page, which I had to fix first, which speaks about both Lagrangians and both the non-covariant and the covariant equations. That page is here

http://www.geocities.com/physics_world/em/relativistic_charged_particle.htm

As you can see the non-covariant Lagrangian is in Eq. (1) and the non-covariant Lagrange equations are in Eq. (3) while the covariant Lagrangian is located in Eq. (33) and the covariant Lagrange equations are in Eq. (34).

So I'll keep asking for your proof that my web page is wrong or until you admit you made a mistake. The honest humble person would admit his mistake and get on with their life. Give it a try and see how good it feels.

Pete

 

[pervect]

Please stop ignoring my question regarding this bogus claim of yours.

I find your personal style of argumentation to be tedious, shrill, overly argumentative, and overly demanding. Especially since I've already quoted at legnth from a standard textbook which disagrees with you.

Sorry if you don't happen to have that particular textbook, I don't happen to have Lancos.

However, I am still interested in a response by reilly, though I will not be so impolite as to demand one.

 

[pmb_phy]

Especially since I've already quoted at legnth from a standard textbook which disagrees with you.

A totallybogus claim and excactly contradictory to the facts which are plain as day. I guess that's all that can be expected from you. You appear to be too arrogant to admit your error. At least Reilly can see clearly.

The problem with this forum is that you can't be blocked.

Oh. One more thing. Nobody cares about your opinion of me. In fact it was expected.

 

[ZapperZ]

This thread ends NOW.

Zz.