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I Relationship between quadratic and square-root lagrangians

  1. Jun 23, 2016 #1

    stevendaryl

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    For relativistic particle dynamics, there are two different approaches to choosing a Lagrangian that give the same equations of motion:

    The quadratic form is:

    [itex]\mathcal{L} = \frac{m}{2} g_{\mu \nu} U^\mu U^\nu[/itex]

    where [itex]U^\mu = \frac{d x^\mu}{d \tau}[/itex]

    This is for the action that involves integration over proper time:

    [itex]\mathcal{A} = \int d\tau \mathcal{L}[/itex]

    But there is also an integral that is directly in terms of the invariant interval:

    [itex]L = \sqrt{g_{\mu \nu} U^\mu U^\nu}[/itex]

    where [itex]U^\mu = \frac{dx^\mu}{ds}[/itex], and [itex]s[/itex] is an arbitrary path parameter, which we can choose to be coordinate time, leading to a Lagrangian:

    [itex]L = \sqrt{g_{00} + 2 g_{0i} v^i + g_{ij} v^i v^j}[/itex]
    or [itex]L = \sqrt{1 - v^2}[/itex] (using inertial coordinates).

    These two Lagrangians give the same equations of motion, which makes me think that they must be related in some way. The one with the square-root is a lot messier to work with, but its interpretation is a lot more direct: you're getting the equations of motion by extremizing the proper time, while the meaning of the quadratic form is obscure to me.

    The quadratic form is a lot nicer when you want to add interactions:

    [itex]\mathcal{L} = \frac{1}{2} m g_{\mu \nu} U^\mu U^\nu + q g_{\mu \nu} U^\mu A^\nu + \lambda \Phi[/itex]

    where [itex]A^\nu[/itex] is the electromagnetic vector potential, and [itex]\Phi[/itex] is a scalar field (not to be confused with [itex]A^0[/itex]). It's not clear to me where to put the vector and scalar potentials into the square-root form of the Lagrangian.
     
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  3. Jun 23, 2016 #2

    strangerep

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    Afaict, they don't really give the same equations of motion unless you ignore the subtleties of the lightlike case. I've recently been pondering something similar...

    First, let's start with this action: $$S_1 ~:= \int_a^b ds \, F(x,u) ~,~~~~~ \mbox{where}~~ u^\mu := \frac{dx^\mu}{ds} \equiv \dot u^\mu ~, $$with ##s## being an arbitrary parameter. (We can't sensibly start with a proper time parameter in an indefinite-metric space if we want to encompass light-like paths.)

    Since ##s## is unphysical, we (ideally) prefer any physically significant quantities to be independent of reparameterizations of ##s##. So we start off wanting invariance of ##S_1## under rescalings of the form ##s \to s/k##, where ##k## is real and ##>0##. For ##S_1## to be invariant under such rescaling, ##F## must be 1-homogeneous wrt ##u##. I.e., ##F(x,ku) = k F(x,u),## by a theorem of Caratheodory. [Sorry, I can't seem to find a convenient link to the particular thm -- Caratheodory has quite a few.]

    The EL equations of motion obtained by extremizing ##S_1## are
    $$0 ~=~ E_\mu[F] ~:=~ \frac{\partial^2 F}{\partial u^\mu \partial u^\nu}\, \dot u^\nu + \frac{\partial^2 F}{\partial u^\mu \partial x^\nu}\, u^\nu - \frac{\partial F}{\partial x^\mu} ~.$$ Now, contracting the RHS with ##u## we get $$ u^\mu E_\mu[F] ~=~ u^\mu \frac{\partial^2 F}{\partial u^\mu \partial u^\nu}\, \dot u^\nu + u^\mu\frac{\partial^2 F}{\partial u^\mu \partial x^\nu}\, u^\nu - u^\mu \frac{\partial F}{\partial x^\mu} ~=~ 0 ~.$$
    Why is it ##0##? The 1st term vanishes by Euler's Theorem for Homogeneous Functions, since ##\partial F/\partial u^\mu## is 0-homogeneous in ##u##. The 2nd & 3rd terms cancel because ##\partial F/\partial x^\mu## is 1-homogeneous, hence $$u^\mu\, \frac{\partial^2 F}{\partial u^\mu \partial x^\nu} ~=~ \frac{\partial F}{\partial x^\mu} ~,~~~~~~ \mbox{(again by the Euler theorem)} ~.$$
    This means the 4 EL equations are not independent. Indeed, since $$u^\mu\, \frac{\partial^2 F}{\partial u^\mu \partial u^\nu} ~=~ 0 ~,$$ the rank of the matrix of 2nd derivatives is non-maximal. I.e., our "Lagrangian" ##F## is singular, or, iow, we have an irregular variational problem. Your sqrt Lagrangian is essentially of this case.

    In contrast, if we use a different action: $$S_2 ~:= \int_a^b ds \, F^2(x,u) ~,$$we don't encounter that difficulty: the variational problem is now regular, though at the cost of having an ##S_2## which is not invariant under rescalings of ##s##. (The endpoint ##a## is regarded as fixed in this framework.) Nevertheless, it turns out that working with such an action can be easier than the irregular case.

    In definite-metric spaces, one can convert from the arbitrary parameter ##s## to an arc-length parameter (and this is what most DG textbooks seem to do). But in the indefinite-metric space of GR, this becomes problematic in the light-like case.

    HTH. [And apologies if there's still any typos in the above.]

    Edit: I just realized my use the symbol ##u^\mu## above is inconsistent with the treatment in Gourgoulhon (see below). One needs to keep ##\dot x^\mu## distinct from the relativistic 4-velocity ##u^\mu## which satisfies ##u^\mu u_\mu = 1##. This is achieved by defining $$u^\lambda ~=~ \frac{\dot x^\lambda}{\sqrt{-g_{\mu\nu} \dot x^\mu \dot x^\nu}} ~,$$which allows one to treat the lightlike case on the same footing as the timelike case, without worrying about null proper time intervals.
     
    Last edited by a moderator: May 8, 2017
  4. Jun 23, 2016 #3

    haushofer

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    Indeed, if you want to be able to treat massless and massive particles, you need an auxiliary field: the einbein. This action is quadratic, which is the whole point of introducting the einbein in the first place. I guess you can find the interactions in Zee's GR-book for both cases, but I have to check.
     
  5. Jun 24, 2016 #4

    ShayanJ

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    Consider two parametrizations ## x^\mu=x^\mu(s) ## and ## x^\mu=\tilde x^\mu(\tilde s) ##, so we have ## \tilde x^\mu(\tilde s)= x^\mu(s)## and:
    ##\tilde u^\mu \equiv {\Large \frac{d \tilde x^\mu}{d \tilde s} =\frac{dx^\mu}{ds}\frac{ds}{d\tilde s}}=\frac{ds}{d\tilde s}u^\mu##.
    The invariance of the action is equivalent to ## F(\tilde x ^\mu(\tilde s),\tilde u^\mu) d \tilde s=F(x^\mu(s),u^\mu) ds ## which gives ## F(x^\mu(s),\frac{ds}{d\tilde s} u^\mu)=\frac{ds}{d\tilde s} F(x^\mu(s),u^\mu) ##. So in order for the action to be reparametrization invariant under any kind of reparametrization, F should be 1-homogeneous w.r.t. to its second argument.
    (From chapter 11 of the book "Special Relativity in General Frames" by Eric Gourgoulhon)
    I don't see how that follows!
     
  6. Jun 24, 2016 #5

    ShayanJ

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    ## L=\left[ m+\lambda \Phi(x) \right] \sqrt{g_{\mu \nu}U^\mu U^\nu}+qg_{\mu \nu}A^\mu(x) U^\nu ##.
    (From the same book as above. I just adjusted the conventions.)

    P.S.
    Just for completeness, if there is a tensor field involved, then the square root changes to ## \sqrt{\left[ g_{\mu \nu}+\frac{Q}{m} h_{\mu\nu}(x) \right] U^\mu U^\nu}##, so the presence of any tensor interaction is indistinguishable from a change in the metric. But there is also ## \sqrt{g_{\mu\nu} U^\mu U^\nu}+\frac 1 2 Q \frac{h_{\mu \nu}U^\mu U^\nu}{\sqrt{g_{\mu\nu} U^\mu U^\nu}}## which is the first order series expansion of the first Lagrangian when ## |h_{\mu \nu}|\ll |g_{\mu \nu}| ##.
     
  7. Jun 24, 2016 #6

    stevendaryl

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    Thanks. For the two Lagrangians I showed, the quadratic Lagrangian is not the square of the square-root Lagrangian, if you add interactions. According to Shyan, the square-root Lagrangian is:

    [itex]L = -m \sqrt{g_{\mu \nu} U^\mu U^\nu} + q g_{\mu \nu} A^\mu U^\nu[/itex]

    Squaring that doesn't give the quadratic Lagrangian.
     
  8. Jun 24, 2016 #7

    strangerep

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    Yes, of course. I was replying to the earlier parts of your post, not the last bit where you mentioned interactions (and it wasn't clear to me if there was a question there).

    Superficially, your Lagrangian with interactions appears not to have definite homogeneity wrt rescalings of ##s##, (but I'd better check Gourgoulhon's book first to make sure I understand the framework being used there).

    Edit: OK, Gourgoulhon's treatment in ch 11 seems quite clear and informative, explaining well why the sqrt version is the right one to use if one wants to remain independent of rescalings of the arbitrary parameter.
     
    Last edited: Jun 24, 2016
  9. Jun 24, 2016 #8

    strangerep

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    Yes, that's the theorem I was referring to, thanks. (I wasn't familiar with Gourgoulhon's book but I've just now looked at it and the treatment in ch11 seems quite good.)

    So,... er,... do you have a question? (I see only an exclamatory statement.)
     
    Last edited: Jun 24, 2016
  10. Jun 24, 2016 #9

    ShayanJ

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    Never mind, I get it now.
     
  11. Jun 25, 2016 #10

    stevendaryl

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    The quadratic lagrangian (as I understand it) only allows an affine parameter, which is part of the point. The equations of motion for the square-root form are messier-looking.
     
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