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I Constant field in the Lagrangian

  1. Sep 16, 2018 at 2:03 PM #1
    Hello! If you have a Lagrangian (say of a scalar field) depending only on the field and its first derivative and you want to calculate the ground state configuration, is it necessary a constant value? I read about Spontaneous symmetry breaking having this Lagrangian $$L= \frac{1}{2}(\partial_\mu \phi)^2-\frac{1}{2}m^2\phi^2-\frac{\lambda}{4!}\phi^4$$ and they say that to get the lowest energy configuration you set ##\phi## to be a constant and work from there. In the case of a particle it makes sense that a motionless particle would have less energy that a moving one at a point. but is this obvious for fields? Can't a complicate configuration bring the energy lower than a constant value? Thank you!
     
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  3. Sep 16, 2018 at 2:50 PM #2

    Orodruin

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    You can probably answer your own question by computing the energy density of your scalar field.
     
  4. Sep 16, 2018 at 2:57 PM #3
    Sorry I did a mistake, the term mass should be + not - (in order to get spontaneous symmetry breaking). So the energy density would be $$H= \frac{1}{2}(\partial_\mu \phi)^2-\frac{1}{2}m^2\phi^2+\frac{\lambda}{4!}\phi^4$$. Is it obvious that the field must be constant?
     
  5. Sep 16, 2018 at 2:58 PM #4

    Orodruin

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    No, that is not the correct energy density. Are you familiar with how the energy-momentum tensor is derived from the Lagrangian?
     
  6. Sep 16, 2018 at 3:06 PM #5
    I am not sure I am. But we were told that the hamiltonian density is ##H=T+V## which corresponds to the energy density (when you integrate it you get the actual hamiltonian which gives the energy for well behaved potentials). Did I miss-understand that?
     
  7. Sep 16, 2018 at 3:08 PM #6

    Orodruin

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    You are misinterpreting what ##T## is in this case so you have the wrong Hamiltonian. I suggest you look at the full expression for the energy-momentum tensor and extract the time-time component, which is the energy density.
     
  8. Sep 16, 2018 at 5:45 PM #7
    Oh, right! So I got this for the energy density: $$\frac{1}{2}\dot{\phi}^2-\frac{1}{2}m^2\phi^2+\frac{\lambda}{4!}\phi^4$$ Is it obvious that ##\phi## has to be a constant?
     
  9. Sep 16, 2018 at 11:34 PM #8

    Orodruin

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    That is also wrong. You are now missing terms.
     
  10. Sep 17, 2018 at 9:54 AM #9
    So the way I did it is to use $$T_{\mu\nu}=\frac{\partial L}{\partial(\partial_\mu\phi)}\partial_\nu\phi-g_{\mu\nu}L$$ which gives $$T_{\mu\nu}=\partial_\mu\phi\partial_\nu\phi-g_{\mu\nu}(\frac{1}{2}(\partial_\mu \phi)^2+\frac{1}{2}m^2\phi^2-\frac{\lambda}{4!}\phi^4)$$ $$T_{\mu\nu}=\frac{1}{2}\partial_\mu \phi\partial_\nu \phi-\frac{1}{2}m^2\phi^2+\frac{\lambda}{4!}\phi^4$$ Now the energy density is given by $$T_{00}=\frac{1}{2}\partial_0 \phi\partial_0 \phi-\frac{1}{2}m^2\phi^2+\frac{\lambda}{4!}\phi^4$$ Where is my mistake? Thank you!
     
  11. Sep 17, 2018 at 11:54 AM #10

    Orodruin

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  12. Sep 18, 2018 at 10:18 AM #11
    Ah, ok. So it should be like this: $$T_{00}=\frac{1}{2}\partial_0\phi\partial_0\phi+\frac{1}{2}( \partial_x\phi\partial_x\phi+\partial_y\phi\partial_y\phi+\partial_z\phi\partial_z\phi)-\frac{1}{2}m^2\phi^2+\frac{\lambda}{4!}\phi^4$$ I hope this is correct now. But still, why is this a minimum for constant ##\phi##? If the mass term would have been added and not subtracted, that would have made sense, but I am not sure I see it as obvious here? Thank you and sorry for taking so long!
     
  13. Sep 18, 2018 at 10:47 AM #12

    Orodruin

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    The shape of the potential is irrelevant. What is relevant are the derivative terms that are always non-negative and zero only if the field is constant.
     
  14. Sep 18, 2018 at 3:30 PM #13
    But if you have something of the form $$g(x)=f'(x)^2+f(x)$$ the minimum of $$\int g(x)$$ is not necessary when ##f=ct##
     
  15. Sep 18, 2018 at 4:45 PM #14

    Orodruin

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    That’s because ##f(x)## does not have a minimum. Your potential does.
     
  16. Sep 18, 2018 at 5:32 PM #15
    Hmm that makes it more clear. However, I am still a bit confused. The potential energy minimum is not necessarily attained for a constant value of ##\phi##, I mean ##\int V(x)##. So can't it happen that for some non-constant value of ##\phi## the kinetic term will indeed get bigger (as it is non-zero anymore) but the potential gets down by a higher amount and hence ##\int T+V## it's smaller overall? Intuitively makes sense that the more you move the more energy you have, but mathematically is not really obvious to me. Can I derive this using calculus of variation, something like ##\frac{\delta E}{\delta \phi}=0## and from here infer that ##\phi## is a constant?
     
  17. Sep 18, 2018 at 9:41 PM #16

    Orodruin

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    If a field value is a minimum of the potential for one point, it is for all points. Hence, a constant does give you the minimal potential at every point.
     
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