The Lagrangian for a real scalar field is:

$$\mathcal{L}=\frac{1}{2}\eta^{\mu \nu}\partial_{\mu}\phi\partial_{\nu}\phi-\frac{1}{2}m^2\phi^2 $$

How can I derive the dynamics of this field from this Lagrangian?

  • $\begingroup$ The question formulation (v2) seems to confuse the action $S=\int \! dt~ L$, the Lagrangian $L=\int \! d^3x~ {\cal L}$, and the Lagrangian density ${\cal L}$. $\endgroup$ – Qmechanic Mar 19 '13 at 15:58
  • $\begingroup$ Actually we write the Lagrangian always for free scalar field, like that but how? $\endgroup$ – Unlimited Dreamer Mar 19 '13 at 16:00

This is not the action, the action $S$ is the time integral of the lagrangian $L$, i.e.

$$S=\int L dt $$

The equations of motion for the field $\phi$ is given by the Euler-Lagrange equations for fields (summation over $\mu$ is implicit)

$$\partial_{\mu}\left(\frac{\partial \mathcal{L}}{\partial(\partial_{\mu}\phi)}\right)-\frac{\partial \mathcal{L}}{\partial \phi}=0 $$

which in this case gives the Klein-Gordon equation

$$(\partial_{\mu}\partial^{\mu}+m^2)\phi=0 $$

  • $\begingroup$ thnx. check the editing $\endgroup$ – Unlimited Dreamer Mar 19 '13 at 15:59
  • $\begingroup$ I understand the Euler Lagrangian equation but how can I write for free field? $\endgroup$ – Unlimited Dreamer Mar 19 '13 at 16:01
  • $\begingroup$ For a free scalar field, the equation of motion is the Klein-Gordon equation, that you can derive of the lagrangian density $\mathcal{L}$ via the Euler-Lagrange equation $\endgroup$ – Jorge Lavín Mar 19 '13 at 16:06

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