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Finding the equation of motion of a given Lagrangian

  1. Jan 13, 2016 #1
    1. The problem statement, all variables and given/known data

    Given the Lagrangian ##\mathcal{L} = \frac{1}{2}(\partial_{\mu}\Phi)^{2}+\frac{1}{2}\Phi^{2}-\frac{1}{2}\Phi^{3}+\frac{\alpha}{8}\Phi^{4}##, where ##\Phi=\Phi(x)##, find the equation of motion of the system. Assume that the field ##\Phi## is spherically symmetric, i.e. ##\Phi = \Phi(r)##.

    2. Relevant equations

    The Euler-Lagrange equation of motion for a scalar field

    3. The attempt at a solution

    ##S = \int d^{4}x \bigg(\frac{1}{2}(\partial_{\mu}\Phi)^{2}+\frac{1}{2}\Phi^{2}-\frac{1}{2}\Phi^{3}+\frac{\alpha}{8}\Phi^{4}\bigg)##

    The spherical volume element in ##4-##dimensional Euclidean space is ##d^{4}x = r^{3}\ \text{sin}^{2}(\phi_{1})\ \text{sin}(\phi_{2})\ dr\ d\phi_{1}\ d\phi_{2}\ d\phi_{3}##.

    Therefore, ##S = \int d^{4}x \bigg(\frac{1}{2}(\partial_{\mu}\Phi)^{2}+\frac{1}{2}\Phi^{2}-\frac{1}{2}\Phi^{3}+\frac{\alpha}{8}\Phi^{4}\bigg)##

    ##=\int_{0}^{2\pi}d\phi_{3}\ \int_{0}^{\pi}d\phi_{2}\ \text{sin}(\phi_{2}) \int_{0}^{\pi}d\phi_{1}\ \text{sin}^{2}(\phi_{1}) \int_{0}^{r}r'^{3}\ dr'\ \bigg(\frac{1}{2}(\partial_{r'}\Phi)^{2}+\frac{1}{2}\Phi^{2}-\frac{1}{2}\Phi^{3}+\frac{\alpha}{8}\Phi^{4}\bigg)##

    ##=(2\pi)(2)\bigg(\frac{\pi}{2}\bigg)\int_{0}^{r}r'^{3}\ dr'\ \bigg(\frac{1}{2}(\partial_{r'}\Phi)^{2}+\frac{1}{2}\Phi^{2}-\frac{1}{2}\Phi^{3}+\frac{\alpha}{8}\Phi^{4}\bigg)##.

    Am I correct so far?
     
  2. jcsd
  3. Jan 13, 2016 #2

    nrqed

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    Why do you bother with the volume element? Just apply the E-L equation to the Lagrangian density an you will directly get the equation of motion.
     
  4. Jan 13, 2016 #3
    If I don't bother with the volume element, won't the crucial factor of ##r'^{2}## in the Lagrangian density be missing?
     
  5. Jan 14, 2016 #4

    nrqed

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    Hi,

    The least action principle leads to the Euler-Lagrange equations which are conditions to what is integrated over all space. So there is no need to do anything with the volume element, ##d^4x##.

    The E-L equations are

    ##\frac{\partial {\cal{L}}}{\partial \phi} - \partial_\mu \frac{\partial {\cal{ L}}}{\partial (\partial_\mu \phi)} = 0 ##
     
  6. Jan 14, 2016 #5
    I understand that.

    But wouldn't the equation of motion take a simpler form in spherical polar coordinates than in Cartesian coordinates, due to the radial symmetry of the potential?
     
  7. Jan 21, 2016 #6
    Yes, the equation of motion will take a simpler form in spherical polar coordinates. But as nrqed said, you should first derive the Euler-Lagrange equation and then impose the spherical symmetry (actually this just means to replace the d'Alembert operator by a double derivative with respect to "r").
     
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