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Relativistic Lagrangian

  1. Nov 11, 2007 #1
    How do we explain the fact that if we want to get Hamilton's equations out of Lagrangian:
    [tex] L=-mc\sqrt{(\dot x^\mu \dot x_\mu)}[/tex],
    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
     
  2. jcsd
  3. Nov 12, 2007 #2

    pervect

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    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

     
  4. Nov 12, 2007 #3
    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.
     
  5. Nov 12, 2007 #4
    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.
     
  6. Nov 12, 2007 #5
    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. :smile: 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
     
  7. Nov 13, 2007 #6

    pervect

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    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.
     
    Last edited: Nov 13, 2007
  8. Nov 13, 2007 #7
    You mean
    [tex]
    \dot{x}^{\mu} = \frac{1}{\sqrt{1-|v|^2/c^2}}(c,v)^{\mu}?
    [/tex]

    Then we have
    [tex]
    L=-mc\sqrt{\dot{x}_{\mu}\dot{x}^{\mu}} = -mc^2
    [/tex]

    This looks quite useless Lagrange's function. :confused:

    Wouldn't
    [tex]
    L=-mc^2\sqrt{1-|v|^2/c^2}
    [/tex]
    be a better Lagrange's function?
     
  9. Nov 13, 2007 #8
    Apologies, I forgot to point out in my reply above that the OP's Lagrangian
    is (obviously) incorrect. Taking units where [itex]c=1[/itex] one has

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

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

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

    [tex]
    \mathcal{H}=\pi_{\alpha}\pi^{\alpha}+m^{2}=0.[/tex]
    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.
     
  10. Nov 13, 2007 #9
    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 refering 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
     
    Last edited: Nov 13, 2007
  11. Nov 13, 2007 #10
    Have I understood something wrong when I get
    [tex]
    \dot{x}_{\mu}\dot{x}^{\mu} = c^2?
    [/tex]

    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
    [tex]
    \dot{x}_{\mu}\dot{x}^{\mu} = c^2 - |v|^2
    [/tex]


    ..... I immediately assumed, that the proper time is the proper time of the particle. Isn't that the common terminology?
     
    Last edited: Nov 13, 2007
  12. Nov 13, 2007 #11
    You've got it correct so no worries. :smile:
    You got it right the first time. The dot is with respect to proper time, and not coordinate time.
    Yes. That is correct.

    Pete
     
  13. Nov 13, 2007 #12
    But isn't it the

    [tex]
    -mc^2\sqrt{1-|v|^2/c^2}
    [/tex]

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

    [tex]
    -mc^2?
    [/tex]
     
    Last edited: Nov 13, 2007
  14. Nov 13, 2007 #13
    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 [itex]\lambda[/itex].

    Pete
     
  15. Nov 13, 2007 #14
    If we start with the Lagrange's function

    [tex]
    L=-mc^2\sqrt{1-|v|^2/c^2}
    [/tex]

    then the Hamiltonian is

    [tex]
    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}}
    [/tex]

    I'm very confused about the claim that the Hamiltonian should be zero.
     
  16. Nov 13, 2007 #15

    pervect

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    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:

    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 [itex]\theta[/itex].

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

    [tex]L = \frac{1}{2} m u_i u^i [/tex]

    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.
     
    Last edited: Nov 13, 2007
  17. Nov 13, 2007 #16
    a point not gonna work in a holographic universe
     
  18. Nov 13, 2007 #17
    I don't have it.

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

    [tex]
    \dot{x}^{\mu} = \frac{1}{\sqrt{1-|v|^2/c^2}}(c,v)^{\mu}
    [/tex]
     
  19. Nov 13, 2007 #18
    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 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 web site 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 = [itex]-m_0 c^2\sqrt{1-\beta^2}[/itex] 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
     
    Last edited: Nov 13, 2007
  20. Nov 13, 2007 #19
    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
     
  21. Nov 14, 2007 #20
    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.
     
  22. Nov 14, 2007 #21
    Thanks for pointing that out. I mistook comments by others as comments made by the Op.
    Which was why do "we get that Hamilton's function equals zero."
    Oy vey!
    Again? Sorry but I don't recall you saying it the first time (or I didn't bother reading your posts).
    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.
     
    Last edited: Nov 14, 2007
  23. Nov 14, 2007 #22
    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.
     
  24. Nov 14, 2007 #23
    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

    [tex]
    L=-mc^2
    [/tex]

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

    [tex]
    \mathbb{R}\to\mathbb{R}^4\quad\quad \tau\mapsto x^{\mu}(\tau)
    [/tex]

    instead of the usual

    [tex]
    \mathbb{R}\to\mathbb{R}^3\quad\quad t\mapsto x^{i}(t)?
    [/tex]

    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.
     
  25. Nov 14, 2007 #24

    reilly

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    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



     
  26. Nov 14, 2007 #25

    reilly

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    Equivalent Lagrangians

    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
     
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