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

[fikus]

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 ?
That same Lagrangian of course gives right equations of motion in Lagrange formulation.

Thanks

 

[pervect]

I'm not sure how to summarize the issue in a post, but it's talked about in "Goldstein, classical mechanics" pg 326, section 7-9, "Covariant Lagrangain formulations".

I suppose the shortest summary would be

This is not a phenomenon of relativistic physics per se: it is a mathematical consequence of enlarging configuration space to include t as a dynamical variable and using some other parameter to mark the system-piont's travel through the space.

 

[country boy]

My copy of Goldstein may be old, but in it the section you refer to is 6-6 on page 207. The footnote on page 209 deals with the relativistic Lagrangian. The key is that the square root (equal to c and thus a constant) has a functional dependence on the velocities u that, through the variational calculus, gives the equations of motion.

 

[shoehorn]

The answer is actually very, very simple. 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 same sort of thing pops up all over physics but is, frustratingly, rarely emphasised in courses in analytical mechanics.

 

[pmb_phy]

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 ?

I disagree. I calculated Hamilton's function here

http://www.geocities.com/physics_world/sr/relativistic_energy.htm

Set U = 0 and you'll get the same Hamiltonian function that you do. The Hamiltonian function in this case is merely the kinetic energy of the particle plus its rest energy. It is not zero as you claim it is.

If you believe that there is an error in my web page then please point it out. I always appreciate people finding errors in my website. It is of great use since in that way people are catching those errors that I became blind to in the derivation. They are extremely few though.

Pete

 

[pervect]

I disagree. I calculated Hamilton's function here ... If you believe that there is an error in my web page then please point it out.

Pete

It's already been pointed out that your webpage disagrees with Goldstein. I'm not sure what more you need? Offhand, I'd expect this is a case of working out the answer to a different (though similar) problem than the OP was asking about, as one can certainly get a relativistic energy function (h) and hence a Hamiltonian (H) that is nonzero by parameterizing with coordinate time rather than proper time.

 

[jostpuur]

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 ?
That same Lagrangian of course gives right equations of motion in Lagrange formulation.

Thanks

You mean
$$\dot{x}^{\mu} = \frac{1}{\sqrt{1-|v|^2/c^2}}(c,v)^{\mu}?$$

Then we have
$$L=-mc\sqrt{\dot{x}_{\mu}\dot{x}^{\mu}} = -mc^2$$

This looks quite useless Lagrange's function.

Wouldn't
$$L=-mc^2\sqrt{1-|v|^2/c^2}$$
be a better Lagrange's function?

 

[shoehorn]

Apologies, I forgot to point out in my reply above that the OP's Lagrangian
is (obviously) incorrect. Taking units where $c=1$ one has

$$L=-m(1-\dot{x}_{a}\dot{x}^{a})^{1/2}.$$
If one introduces some parameter $s$ along the world-line, it's
obvious that one has a constraint of the form

$$\tilde{\mathcal{H}}=\pi_{t}+(\pi_{a}\pi^{a}+m^{2})^{1/2}=0,$$

where $\pi_{a}$ and $\pi_{t}$ are the canonical momenta. It's then
trivially easy to show that the constraint $\tilde{\mathcal{H}}=0$
is actually equivalent to the Hamiltonian

$$\mathcal{H}=\pi_{\alpha}\pi^{\alpha}+m^{2}=0.$$
The real reason why the Hamiltonian vanishes identically is because
the canonical action can be expressed in a form which is invariant
under reparametrizations of the path which preserve the values of
the parameter at the endpoints. This is a very, very well known result
in analytical mechanics and is deeply related to, for example, Jacobi's
principle and Dirac's analysis of constrained Hamiltonian systems.
As I said earlier, any decent course in analytical mechanics will
labour this point ad infinitum.

 

[pmb_phy]

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

I have no recollection of such a comment and I see no inconsistency between those calculations and Goldstein/Lanczos. If you can't point to an error yourself and give a page number etc. then please stop making these unfounded erroneous claims. Even so, just because something disagrees with Goldstein doesn't mean that its wrong. Perhaps you're referring to the relationship between U and V. If so then that comes from Lanczos. Its too bad that you still make claims rather than simply ask first.

Note: The energy function h in my link above has the same value as the Hamiltonian. The only difference is that h is a function of generalized position and generalized velocity while the Hamiltonian is expressed as a generalized position and canonical momentum. There is a slight difference between that page and Goldstein in that I liked some of what Lanczos has done and have encorporated that into my page.

Pete

 

[jostpuur]

Have I understood something wrong when I get
$$\dot{x}_{\mu}\dot{x}^{\mu} = c^2?$$

where dot is noting derivative in proper time

I think this is a mistake. The dot is not supposed to be a derivative in proper time, but just in the local time of the chosen arbitrary frame. The we get
$$\dot{x}_{\mu}\dot{x}^{\mu} = c^2 - |v|^2$$... I immediately assumed, that the proper time is the proper time of the particle. Isn't that the common terminology?

 

[pmb_phy]

Have I understood something wrong when I get
$$\dot{x}_{\mu}\dot{x}^{\mu} = c^2?$$

You've got it correct so no worries.

I think this is a mistake. The dot is not supposed to be a derivative in proper time, but just in the local time of the chosen arbitrary frame. The we get
$$\dot{x}_{\mu}\dot{x}^{\mu} = c^2 - |v|^2$$

You got it right the first time. The dot is with respect to proper time, and not coordinate time.

... I immediately assumed, that the proper time is the proper time of the particle. Isn't that the common terminology?

Yes. That is correct.

Pete

 

[jostpuur]

You've got it correct so no worries.

You got it right the first time. The dot is with respect to proper time, and not coordinate time.
Yes. That is correct.

Pete

But isn't it the

$$-mc^2\sqrt{1-|v|^2/c^2}$$

that is the free particle Lagrange's function, and not

$$-mc^2?$$

 

[pmb_phy]

But isn't it the

$$-mc^2\sqrt{1-|v|^2/c^2}$$

that is the free particle Lagrangen's function, and not

$$-mc^2?$$

Hmmmm! You have a valid point there. Let me think about it. In the mean time check out an application of the Lagrangian at

http://www.geocities.com/physics_world/gr/geodesic_equation.htm

for geodesic motion. You can use proper time for the affine parameter $\lambda$.

Pete

 

[jostpuur]

If we start with the Lagrange's function

$$L=-mc^2\sqrt{1-|v|^2/c^2}$$

then the Hamiltonian is

$$H = \frac{\partial L}{\partial v^i} v^i \;-\; L = \frac{m|v|^2}{\sqrt{1-|v|^2/c^2}} \;+\; mc^2\sqrt{1-|v|^2/c^2} = \frac{mc^2}{\sqrt{1-|v|^2/c^2}}$$

I'm very confused about the claim that the Hamiltonian should be zero.

 

[pervect]

If we start with the Lagrange's function

$$L=-mc^2\sqrt{1-|v|^2/c^2}$$

then the Hamiltonian is

$$H = \frac{\partial L}{\partial v^i} v^i \;-\; L = \frac{m|v|^2}{\sqrt{1-|v|^2/c^2}} \;+\; mc^2\sqrt{1-|v|^2/c^2} = \frac{mc^2}{\sqrt{1-|v|^2/c^2}}$$

I'm very confused about the claim that the Hamiltonian should be zero.

The problem is that you (and probably Pete) are working a different problem than the OP is.

I assume you don't have Goldstein. (I'm surprised Pete doesn't, though, or that if he does, he doesn't just look it up).

I can quote some of the relevant sections, but the entire explanation is just too long to type in.

Here's some of the motivation:

The Lagrangian procedure as given above (ed: with L = $-m c^2 \sqrt{1-\beta^2}$ for a free particle) certainly predicts the correct relativistic equations of motion. Yet it is a relativistic formulation only "in a certain sense." No effort has been made to keep the ideal of a covariant four-dimensional form for all the laws of mechanics. Thus the time t has been treated as a parameter entirely distinct from the spatial coordinates, while a covariant formulation would require that the space and time be considered as entirely similar coordinates in world space. Clearly some invariant parameter should be used instead of t, to trace the progress of the system point in configuration space. Further, the examples of Lagrangian functions discussed in the previous section do not have any particular Lorentz transformation properties. Hamilton's principle must itself be manifestly covariant, which can only mean in this case that the action principle must be a world scaar.

Note that the OP's expression for the Lagrangian is Goldstein's eq (7-162) in the second edition (for those people who have this textbook), and was derived by transforming the action intergal over t from one over an arbitrary parameter $\theta$.

Goldstein also goes on to point out that (using slightly different notation)

$$L = \frac{1}{2} m u_i u^i$$

works correctly as a Lagrangian, which differs by not having a minus sign, and being a quadratic expression, and that there are an infinite number of other possibilities.

As I said earlier, any decent course in analytical mechanics will
labour this point ad infinitum.

Yep.

 

[andrewj]

It's already been pointed out that your webpage disagrees with Goldstein. I'm not sure what more you need? Offhand, I'd expect this is a case of working out the answer to a different (though similar) problem than the OP was asking about, as one can certainly get a relativistic energy function (h) and hence a Hamiltonian (H) that is nonzero by parameterizing with coordinate time rather than proper time.

a point not going to work in a holographic universe

 

[jostpuur]

The problem is that you (and probably Pete) are working a different problem than the OP is.

I assume you don't have Goldstein.

I don't have it.

So I'm lost with the notation or something else. One step at the time: Was this right?

$$\dot{x}^{\mu} = \frac{1}{\sqrt{1-|v|^2/c^2}}(c,v)^{\mu}$$

 

[pmb_phy]

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

No. You think Goldstein disagrees with my web page. But you are most certainly wrong. In fact you never really quoted Goldstein. You gave your version of what you thought a summary should look like. However there is nothing in that part of Goldstein which corresponds to your claim This is not a phenomenon of relativistic physics per se: .

Nothing in Chapter 7 disagrees with my web pages of this. In fact I learned a great deal of relativistic Lagrangian mechanics from Goldstein.

I'm not sure what more you need?

I don't need anything. You're making unfounded assertions which you are unable to back up and thus you have deluded yourself to thinking that you made a meaningful statement which wasn't very meaningful or useful at all.

I'll tell you what you need though. You to pay more attention to what you're reading. Since you're referring to the second edition of Goldstein (which you really should have stated which version you're using. I have all three, you seem to have the second edition while Country Boy has the first edition) then, if you had actually checked my website against the relevant pages/equations of Goldstein then you'd see that the Lagrangian Goldstein gives in his 2nd Ed. in Eq. (7-136) is identical to the Lagrangian as I wrote it, i.e. L = K - U, in the first sentance of

http://www.geocities.com/physics_world/sr/relativistic_energy.htm

where after Eq. (1) I wrote where K = $-m_0 c^2\sqrt{1-\beta^2}$ and U is a function of both position and velocity (in all generality that is). The Hamiltonian for this system is not zero.

The answer to the OP's question is that there are two Lagrangians used in relativistic Lagrangian mechanics. One from the non-covariant form of the equations and one from the covariant equations. I created another page somewhere else to explain how this is done from both ways, i.e. by the non-covariant method and by the covariant method. However I couldn't find it this afternoon. I found it a little while ago an it appears that the page has been corrupted. Since I'm rebuilding the entire website I may have deleted some equation GIFs which are missing. The page is at

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

I will attempt to restore it tonight. In the future please don't paraphrase and please post the exact equation numbers in your source and in my pages which you are criticizing.

Pete

 

[pmb_phy]

If we start with the Lagrange's function

$$L=-mc^2\sqrt{1-|v|^2/c^2}$$

then the Hamiltonian is

$$H = \frac{\partial L}{\partial v^i} v^i \;-\; L = \frac{m|v|^2}{\sqrt{1-|v|^2/c^2}} \;+\; mc^2\sqrt{1-|v|^2/c^2} = \frac{mc^2}{\sqrt{1-|v|^2/c^2}}$$

I'm very confused about the claim that the Hamiltonian should be zero.

I'm just being nitpicky here, however that is not the Hamiltonian. You'd have to have that function in terms of position and momentum.

Pete

 

[shoehorn]

The answer to the OP's question is that there are two Lagrangians used in relativistic Lagrangian mechanics. One from the non-covariant form of the equations and one from the covariant equations.

This is most certainly not the answer to the OP's question. As I have pointed out several times in this thread, the answer to his question is the ability one has to cast the action in a particularly special reparametrization-invariant form. Once again: this is a point of central importance in analytical mechanics and in physics more generally since it is closely related to our freedom to construct explicit, rigourous examples of gauge theories.

And by the way, your claim that there are "two Lagrangians" used in relativistic Lagrangian mechanics is absurd. For each two such Lagrangians you give me, I, and everyone else, can give you an infinite number of equivalent Lagrangians.

 

[pmb_phy]

This is most certainly not the answer to the OP's question.

Thanks for pointing that out. I mistook comments by others as comments made by the Op.

As I have pointed out several times in this thread, the answer to his question ...

Which was why do "we get that Hamilton's function equals zero."

...is the ability one has to cast the action in a particularly special reparametrization-invariant form.

Oy vey!

Once again: ...

Again? Sorry but I don't recall you saying it the first time (or I didn't bother reading your posts).

And by the way, your claim that there are "two Lagrangians" used in relativistic Lagrangian mechanics is absurd. For each two such Lagrangians you give me, I, and everyone else, can give you an infinite number of equivalent Lagrangians.

You should ask a person to clarify what he means by a statement before you go ahead and assume its absurd. I see you're not that kind of person. I'm not the type of person to respond to someone claiming that what I posted was absurd before they even asked me what I meant. Its up to you to figure it out. Hint: It has nothing to do with your little guess here.

 

[shoehorn]

You should ask a person to clarify what he means by a statement before you go ahead and assume its absurd. I see you're not that kind of person. I'm not the type of person to respond to someone claiming that what I posted was absurd before they even asked me what I meant. Its up to you to figure it out. Hint: It has nothing to do with your little guess here.

Look dude, you can either try to be precise with what you say and avoid pointless arguments, or you can be deliberately (or worse, unintentionally) obtuse and enjoy pointless arguments. The statement you made was, deliberately or otherwise, both wrong and obtuse. Deal with it.

It might be of little harm were the admin to lock this thread now given that the OP's question has been answered. There's no point in this degenerating into petty squabbling, and even less point in allowing the thread to degenerate into petty squabbling replete with links to bizarre websites.

 

[jostpuur]

I had a short visit to the library, and took a glance at the Goldstein's book. I didn't have time to start going through the details, but perhaps got the idea:

So the Lagrangian

$$L=-mc^2$$

gives the correct equations of motion, when the path of the particle is thought to be a mapping

$$\mathbb{R}\to\mathbb{R}^4\quad\quad \tau\mapsto x^{\mu}(\tau)$$

instead of the usual

$$\mathbb{R}\to\mathbb{R}^3\quad\quad t\mapsto x^{i}(t)?$$

I somehow got a thought that that would have been the idea, but I don't understand how that's supposed to work either. The constant L still looks useless.

 

[reilly]

Hamiltonians

The only case I can think of with H=0 occurs for a very special canonical transformation. That is, one that maps the system's dynamical variables into constants. In this scheme, the Hamiltonian, H' = H + dS/dt where H is the usual Hamiltonian and S is a solution to the Hamilton Jacobi Eq., which requires H'=0. However, in coordinates we normally use, H is not zero -- except in very special circumstances. If you want to understand all of this, then you have a big job: see Lanczos, The Variational Principles of Mechanics, pp161-290. What am I missing?

To me, the constraint you give is far from obvious. Where does it come from, and why?

Also, what parameters do you consider in your reparametrization?
Regards,
Reilly Atkinson

Apologies, I forgot to point out in my reply above that the OP's Lagrangian
is (obviously) incorrect. Taking units where $c=1$ one has

$$L=-m(1-\dot{x}_{a}\dot{x}^{a})^{1/2}.$$
If one introduces some parameter $s$ along the world-line, it's
obvious that one has a constraint of the form

$$\tilde{\mathcal{H}}=\pi_{t}+(\pi_{a}\pi^{a}+m^{2})^{1/2}=0,$$

where $\pi_{a}$ and $\pi_{t}$ are the canonical momenta. It's then
trivially easy to show that the constraint $\tilde{\mathcal{H}}=0$
is actually equivalent to the Hamiltonian

$$\mathcal{H}=\pi_{\alpha}\pi^{\alpha}+m^{2}=0.$$
The real reason why the Hamiltonian vanishes identically is because
the canonical action can be expressed in a form which is invariant
under reparametrizations of the path which preserve the values of
the parameter at the endpoints. This is a very, very well known result
in analytical mechanics and is deeply related to, for example, Jacobi's
principle and Dirac's analysis of constrained Hamiltonian systems.
As I said earlier, any decent course in analytical mechanics will
labour this point ad infinitum.

 

[reilly]

Equivalent Lagrangians

Note that the OP's expression for the Lagrangian is Goldstein's eq (7-162) in the second edition (for those people who have this textbook), and was derived by transforming the action intergal over t from one over an arbitrary parameter $\theta$.

Goldstein also goes on to point out that (using slightly different notation)

$$L = \frac{1}{2} m u_i u^i$$

works correctly as a Lagrangian, which differs by not having a minus sign, and being a quadratic expression, and that there are an infinite number of other possibilities.



Yep.

The fact that Lagrangians are not unique has been known for some time. However, practical problems tend to be formulated in two fashions: covariant and non-covariant, and as simple as possible. However, the two approaches are completely equivalent, and give the same Hamiltonian, as they must.This is discussed in detail in Panofsky and Phillips, Classical Electricity and Magnetism pp351-358 -- Chap 23. Also, Jackson gives a very succinct discussion -- in my old version, EQ. 12.75 directly shows the equality of the covariant and non-covariant Lagrangians. Both approaches are used, for example, in the design of accelerators, which tend to work fairly well..

But, as discussed by Landau and Lif****z, The Classical Theory of Fields,when interacting E&M fields and charged particles are considered, life gets difficult. L&L derive the approximate Hamiltonian to second order in the electric charge when v/c is small -- rather a tedious derivation. Further, as soon as radiation damping and self energy are considered, then life gets big-time difficult: see Panofsky and Phillips, and Jackson
Regards,
Reilly Atkinson

 

[reilly]

My first look at Pete's web page on Lagrangians and Hamiltonians showed me a correct and standard argument. My memory being not so good anymore, I looked at my lecture notes from when I taught this stuff, I looked at Landau and Lifshetz and Panofsky and Phillips, all of whom seem to give results identical to Pete's. If there is a problem here, lets' see chapter and verse.

All the issues in this thread are well explained in the literature; and after a hundred years of work, virtually everyone in the physics world, uses one of two seemingly different Lagrangians, covariant and non-covariant with absolutely no controversy among those who actually use the theory of relativistic particle dynamics. Sort of like the different representations in QM, Schrodinger, Heisenberg, and Interaction pictures.
Regards,
Reilly Atkinson

 

[shoehorn]

The only case I can think of with H=0 occurs for a very special canonical transformation. That is, one that maps the system's dynamical variables into constants. In this scheme, the Hamiltonian, H' = H + dS/dt where H is the usual Hamiltonian and S is a solution to the Hamilton Jacobi Eq., which requires H'=0. However, in coordinates we normally use, H is not zero -- except in very special circumstances.

As far as I can see, the OP asked how and why the Hamiltonian for a free relativistic particle could be zero. Given the paucity of information he provided on precisely how he achieved this, it's reasonable to assume that he's using the standard, extremely well-known method of achieving a vanishing Hamiltonian. There is nothing whatsoever "special" about this treatment.

If you want to understand all of this, then you have a big job: see Lanczos, The Variational Principles of Mechanics, pp161-290. What am I missing?

Lanczos is (or at least should be) standard reading. For the nth time: reparametrization invariance, Jacobi's principle, and the existence of a means to analyse dynamical systems with constraints should be a bare minimum set of prerequisites in any course on analytical mechanics. These were, for example, core topics in the analytical mechanics course I had in my first year at Cambridge; if they're regarded as important enough to be introduced in the first couple of weeks here, surely they're important enough for others to be familiar with them elsewhere?

To me, the constraint you give is far from obvious. Where does it come from, and why?

The constraint I gave is perfectly standard. Think about it: what do/should you know about classifications of constraints which arise from Lagrangians which are homogeneous of degree n in the velocities? What is the significance of the value of n?

Also, what parameters do you consider in your reparametrization?
Regards,
Reilly Atkinson

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.

 

[pmb_phy]

I had a short visit to the library, and took a glance at the Goldstein's book. I didn't have time to start going through the details, but perhaps got the idea:

So the Lagrangian

$$L=-mc^2$$
...

Where did you get that Lagrangian from?

I somehow got a thought that that would have been the idea, but I don't understand how that's supposed to work either. The constant L still looks useless.

The covariant Lagrangian for a free particle is a function of the particle's 4velocity. You didn't post such a Lagrangian above.

If the non-covariant Lagrangian is used to find the Hamiltonian also in a non-covariant manner, then the Hamiltonian will have the value of the energy of the particle. In the case of a free particle the Hamiltonian will equal the kinetic energy. If one then uses a covariant approach then the Hamiltonian is zero. This is what I meant when I told shoehorn that he had the wrong idea of what I meant. I didn't mean different Lagrangians since Lagrangians are not unique for a given system. I was referring to a non-covariant Lagrangian as compared to the covariant Lagrangian such as that the OP posted.

When I looked at Goldstein I saw his result which was a non-zero Hamilton found through the covariant approach. When I looked a Jackson I saw a different result. Jackson says of his result

While the Hamoltonian above is formally satisfactory, it has several problems. The first is that it is by definition a Lorentz scalar, not an energylike quantity. Second, use of (12.23) and (12.33) shows that H = 0. Clearly such a Hamiltonian formulation differs considerably from the nonrelativistic version. The reader can refer to Barut (pp. 68 ff.) for a discussion of this and other alternative Hamiltonians

I will obviously be getting a cooy of that book!

I have some research to do to understand why Goldstein differs from Jackson. However Goldstein does mention that under certain conditions the Hamiltonian will vanish. I'll have to dig into that too.

I see that the OP has not posted anything in this thread. Perhaps he's satisfied with what he's read so far.

Pete

 

[pervect]

Chapter and verse can be found in Goldstein, pg 327 on the second edition, as I already mentioned.

The problem is to find the form of (the Lagrangian) $\Lambda$ such that (the Euler Lagrange eqs) are equivalent to the equations of motion.

One way of seeking $\Lambda$ is to transform the action integral from the usual integral over t to one over $\theta$, and to treat the time t appearing explicitly in the Lagrangian not as a parameter but as an additional generalized coordinate. Since $\theta$ must be a monotonic function of t as measured in some Lorentz frame, we have

dx_i/dt = (dx_i)/d (theta) d(theta)/dt = i c x'_i/x'_4

Hence the action integral is transformed as

-i/c int L(x_u, ic x'_i/x'_4) x'_4 d(theta)

It would seem therefore that a recipe for a suitable $\Lambda$ is given by the relation

$\Lambda$(x_u, x'_u) = -i x'_4/c L(x_u, ic x'_j/x'_d)

The Lagrangian obtained this way is however a strange creature, unlike any Lagrangian we have so far met. Note that no matter what the functional form of I, the new Lagrangian $\Lambda$ is a homogeneous function of the generalized velocities in the first degree.

This is not a phenomenon of relativistic physics per se: it is a mathematical consequence of enlarging configuration space to include t as a dynamical variable, and using some other parameter to mark the system-point's travel through the space. A Lagrangian obeying Eq. (7-160) is often called (somewhat misleadingly) a homogeneous Lagrangian and the corresponding "homogeneous" problem of the calculus of variations requires special treatment.*

The most serious of the resulting difficulties will arise in the Hamiltonian formulation, but we can glimpse some of them by noting that in consequence the energy function h, according to Eq 2-56 is identically zero.

* For a full exposition, see H. Rund, The Hamilton-Jacobi Thoery in the Calculus of Variations (1966) Chapter 3.

There is a lot more that I haven't quoted, even in this section (on the covariant Lagrngian formulation) and I imagine there is even more in the Hamilton section.

For my purpose of answering the OP's question, it is sufficient to note that if the energy function h is zero, the Hamiltonian will also be zero - the only difference between h and H is a change of variables, and h=0 imples that H=0.

The original quote in Goldstein is of course much more legibly typeset, and I wouldn't be surprised if I made some typos, but the essence should be there.

I really doubt that there is any conflict between Goldstein, and Lanadau & Lifschitz, but I don't happen to have "Classical theory of Fields" and I note that you do not give as nearly a specific reference as I did. If you think there is some conflict, you're welcome to do as much work as I did and give the appropriate pages, or even type in the appropriate section, if you really think this issue is worth discussing.

I really don't have all that much interest in this topic, which isn't related to relativity, but it clearly relates to what the OP was asking about, since he gave the exact equation for what Goldstein says is "somewhat misleadingly called a homogeneous Lagrangian", and also mentioned the fact that H was zero. I thought I'd point it out to him. But the OP seems to have vanish, and this thread seems to have gotten a (time-consuming/wasting) life of its own.

 

[reilly]

References Please

OK; I studied Lanczos twice at Harvard, once, briefly at Stanford, and taught Lanczos at Tufts several times. (Many in the physics trade have not only studied Lanczos, butvenerate the book as well.)But all that was a long time ago-like 40 years -, so there are, I'm sure, many subtle points I don't remember.

To wit, your constraint may be standard, but not to me, so I'd very much appreciate a reference. Ditto for your reparametrization; I guess I'm just not smart enough to grasp your hand waving on that matter. Perhaps you might supply a detailed reference.

If there is such a thing as a zero (null?) Hamiltonian, how do you get the right limit as
v --> 0. I believe the non-rel H for (most) systems is not zero.

Also, how does a zero Hamiltonian generate displacements in time?

Does a zero Hamiltonian allow production of arbitrary numbers of particles without breaking energy conservation?
Regards,
Reilly Atkinson

Lanczos is (or at least should be) standard reading. For the nth time: reparametrization invariance, Jacobi's principle, and the existence of a means to analyse dynamical systems with constraints should be a bare minimum set of prerequisites in any course on analytical mechanics. These were, for example, core topics in the analytical mechanics course I had in my first year at Cambridge; if they're regarded as important enough to be introduced in the first couple of weeks here, surely they're important enough for others to be familiar with them elsewhere?
The constraint I gave is perfectly standard. Think about it: what do/should you know about classifications of constraints which arise from Lagrangians which are homogeneous of degree n in the velocities? What is the significance of the value of n?
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.

 

[shoehorn]

Unfortunately, I find that I've lent my copy of Lanczos to someone and don't have another to hand. I could walk to the library to get one but, well, it's bloody freezing here at the moment. As a result, I'll try to quote the general line of thinking from memory and get back to you later with some references.

Let me begin by looking at a simpler case than the one the OP asked about. Suppose that we have a non-relativistic particle of mass $m$ which moves in a potential $V(x^i,t)$. The canonical action for this theory is

$$S[x^i(t),p_i(t)] = \int dt\, \left( p_i\frac{dx^i}{dt} - H(x^i,p_i,t) \right),$$

where $p_i$ is the momentum conjugate to $x^i$. The Hamiltonian is

$$H = \frac{p_ip^i}{2m} + V(x^i,t).$$

Note that this form of the Hamiltonian makes the implicit assumption that we are dealing with Newtonian theory because of the form of the inner product $p_ip^i=\delta^{ij}p_ip_j$.

Now suppose that we introduce a parametrization of the path in phase space. Let me introduce some arbitrary parametrization of this path, and label it by $\tau$. Moreover, suppose that we look at a larger phase space given by

$$x^\alpha = (t,x^i),\,\, x^\alpha = x^\alpha(\tau),\,\, p_\alpha = p_\alpha(\tau).$$

(By the way, my memory of Lanczos is that this is covered within the context of a discussion of Jacobi's principle and is explained in detail there.) We can then re-express the canonical action as

$$S[x^\alpha(\tau),p_\alpha(\tau)] = \int d\tau\, (p_i\dot{x}^i - H\dot{t}),$$

where dots denote differentiation with respect to the parameter $\tau$. If one varies with respect to $x^i(\tau)$ and $p_i(\tau)$, one obtains the standard Hamiltonian equations of motion, while if one varies with respect to $t(\tau)$ one obtains what (I believe) is widely called the "energy balance equation"

$$\dot{H} = \dot{t}\frac{\partial H}{\partial t}.$$

However, we have still to vary with respect to $p_0(\tau)$, the momentum conjugate to $t(\tau)$. It's easy to do this: we find that $p_0 = -H$ and since $H=H(x^\alpha,p_\alpha)$ we can vary $p_0$ independently in the action. Thus, one obtains the constraint equation

$$\mathcal{H} = p_0 + H(x^\alpha,p_\alpha) = 0.$$

So far, so bog standard. If we want to be able to vary all of the $(x^\alpha,p_\alpha)$ freely, we just follow Dirac's method and add the constraint into the action together with a Lagrange multiplier, $N$. We find that

$$S[x^\alpha(\tau),p_\alpha(\tau);N(\tau)] = \int d\tau\, (p_\alpha\dot{x}^\alpha - N\mathcal{H}).$$

If we vary $N$ we obtain the constraint $\mathcal{H}=0$ again. If we vary $p_0$ we get

$$N = \frac{d(\textrm{absolute "Newtonian" time})}{d(\textrm{label time})} = \dot{t},$$

an expression which assigns physical meaning to the Lagrange multiplier. Again, such occurrences are common all over physics (think about the status of the Lagrange multipliers in general relativity for example).

We can also recover the ordinary Hamiltonian equations of motion and an energy balance equation from the remaining variations. The form of the action is also seen to be invariant under arbitrary reparametrizations of the form

$$\tau\to \tau'(t)$$

whenever

$$N \to N' = N\frac{d\tau}{d\tau'}$$

and the $(x^\alpha,p_\alpha)$ are invariant under this reparametrization. Then, and this is the central point, the Hamiltonian for a non-relativistic point particle of mass $m$ can be taken as $N\mathcal{H}$. What's more, on a classical solution one has $\mathcal{H}=0$, the claim I made earlier in the thread.

All of this depends crucially on the theory being invariant under such (essentially arbitrary) reparametrizations. For the fifth or sixth time in this thread let me repeat it: this is, or at least should be, emphasised ad nauseum in a course on analytical mechanics. To me, it's right up there with the Hamilton-Jacobi equation in importance.

So far I've dealt only with the non-relativistic case. To see how the relativistic case works out, let's take the usual Lagrangian:

$$L = -m\sqrt{1-\dot{x}_i\dot{x}^i}.$$

One can go through essentially the same procedure as in the non-relativistic case, recovering the idea that the Hamiltonian vanishes on-shell. The proof is left as an exercise for anyone who's interested.

Finally, in case you're not convinced of the ubiquity of this method, let me note that it pops up in loads of places. It happens, as a well known example, in bosonic string theory, where heavy use is made of world sheet reparametrization invariance in order to write the Hamiltonian as a quantity which vanishes on shell. Similarly, (vacuum) general relativity has an identically vanishing Hamiltonian of the form

$$H = \int d^3x\, (N\mathcal{H} + N^i\mathcal{H}_i) = 0.$$

where $(N,N^i)$ are the lapse function and shift vector, respectively, and $\mathcal{H},\mathcal{H}^i$ are the Hamiltonian and momentum constraints, respectively.

 

[pmb_phy]

My first look at Pete's web page on Lagrangians and Hamiltonians showed me a correct and standard argument. My memory being not so good anymore, I looked at my lecture notes from when I taught this stuff, I looked at Landau and Lifshetz and Panofsky and Phillips, all of whom seem to give results identical to Pete's. If there is a problem here, lets' see chapter and verse.

Thanks Reilly.

That was a page where I was working in a non-covariant form. The page where I did this in a covariant form was damaged. I fixed it yesterday. It is located here

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

This is the covariant approach. The Hamiltonian derived using the covariant approach is zero whereas the non-covariant approach yields a non-zero Hamiltonian whose value is zero for a free particle and equals the energy.

As far as Goldstein's text quoted above - been there, done that. What's the problem?

Pete

 

[pmb_phy]

Thanks Reilly.

That was a page where I was working in a non-covariant form. The page where I did this in a covariant form was damaged. I fixed it yesterday. It is located here

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

This is the covariant approach. The Hamiltonian derived using the covariant approach is zero whereas the non-covariant approach yields a non-zero Hamiltonian whose value is zero for a free particle and equals the energy.

As far as Goldstein's text quoted above - been there, done that. What's the problem?

Pete

I didn't realize that I have that book Jackson cited, i.e. the one by Barut. I scanned the pages relavent to this discussion and placed them in a pdf file. That file is located at

http://www.geocities.com/physics_world/Barut.pdf

It will be removed next week since its a temporary file and I'm cramped for space on my website.

Best regards

Pete

 

[reilly]

shoehorn -- Better get yourself to the library. See Chap. VII, Canonical Transformations, of Lanczos. In my earlier post I quoted Eq. 74.12 and 74.13 -- I left out EQ.72.17; it gives a somewhat more general xform for the Hamiltonian. A zero Hamiltonian is an artifact of certain canonical transformations.

The situation is somewhat similar to the differences between Schrodinger, Interaction, and Heisenberg pictures in QM. And, you did not answer my questions about time displacements, and non-relativistic limits.
Regards,
Reilly Atkinson

My copy of Lanczos was published in 1957, so chapter and equation numbers might be changed

 

[pervect]

Let's see if we can come to agreement on a few basic things, which I think should be obvious, but may have gotten lost in the smoke, fireworks, and drama.

1) The OP claimed that

$$L=-mc\sqrt{(\dot x^\mu \dot x_\mu)}$$

where the dot notes a derivative with respect to proper time, is a Lagrangian.

Goldstein also agrees that this is a Lagrangian (of a free particle, 7-162).

Do you (reilly) agree that this is a relativistic Lagrangian of a free particle? We note in passing that this is not a unique Lagrangian, I suppose we'd better confirm that you agree with that observation, too.

2) The OP claimed that the Hamiltonian corresponding to the above Lagrangian is zero. To work this out formally, we do need to indicate what variables L is a function of, which the OP did not do, but this is a common omission.

It seems logical to assume that:

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

and also that $x_i = g_{ij} x^i$ where $g_{ij}$ is the space-time metric.

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

Assuming you agree with points 1) and 2), what is your answer as to why the Hamiltonian is zero?