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Interpretating the Lagrangian

  1. Nov 24, 2006 #1
    In classical mechanics the Lagrangian is constructed as the difference between kinetic and potential energy, [tex]L=T-V,[/tex] and hence the physical interpretation is fairly straightforward when taken together with Hamiltons principle, i.e the system evolves "along the path of least resistance".

    But now that I've started to study a little QFT I see that there the Lagrangian seems to be simply constructed as "the function that after applying Hamiltons principle gives the correct equation of motion". In other words, I don't see any obvious physical interpretation outside of being a lucky guess giving the correct result.

    Have I missed something important here or is the Lagrangian just a mathematical tool and not something with a distinct physical interpretation like the Hamiltonian?
     
  2. jcsd
  3. Nov 24, 2006 #2

    dextercioby

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    To make an "educated guess" about a Lagrangian density in QFT you need to respect some requirements.

    Though not too many books mention them specifically, i'd say that the custom used lagrangian density

    [tex] \mathcal{L}=\bar{\Psi}\left(i\gamma^{\mu}\partial_{\mu}-m\right)\Psi [/tex]

    for the Dirac field, even though leads to correct field eqns for the Dirac spinors, is not an acceptable Lagrangian density.

    I'll let you figure out why.

    Daniel.
     
  4. Nov 26, 2006 #3
    Not an acceptable Lagrangian? The only restrictions I've seen stated for Lagrangians in QFT is Lorentz-invariance. Is this not the case here?
     
  5. Nov 27, 2006 #4

    dextercioby

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    Nope, it also has to be a bosonic variable (the Lagrangian density i wrote is) , not Lorentz but Poincare' scalar and of course real wrt involution on the algebra the fields are taking values in.
    And it's usually desirable to be local, as well.

    Daniel.
     
  6. Nov 27, 2006 #5

    Demystifier

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    The Lagrangian should be real, which the above Lagrangian is not. This can be fixed by ading a total derivative, which does not change the equations of motion.

    Is there a price for a correct answer? :-)
     
    Last edited: Nov 27, 2006
  7. Nov 27, 2006 #6

    Demystifier

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    This is just a tool. This tool is convenient because it is the simplest way to respect all the required symmetries.
     
  8. Nov 27, 2006 #7

    dextercioby

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    Funny, but i guess the question was for the OP. I assume that this thread is meant for him to learn something [by himself].

    Daniel.
     
  9. Nov 27, 2006 #8

    Demystifier

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    OK, I gave the hint, but at least I have not wrote the correct Lagrangian explicitly. So there is still something to learn by him. :smile:
     
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