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If you are unfamiliar with this formalism, consult, for example, Goldstein (1977) or work through the example given in Sakurai (1967), page 3. Instead of writing down a relativistic wave equation, we simply choose a Lagrangian $\mathcal{L}$. Provided our choice is a Lorentz scalar, the equation of motion resulting from (14.4) will be covariant. For example, substituting the Lagrangian $$ \mathcal{L}=\frac{1}{2}\left(\partial_\mu \phi\right)\left(\partial^\mu \phi\right)-\frac{1}{2} m^2 \phi^2 $$ into (14.4) gives the Klein-Gordon equation $$ \partial_\mu \partial^\mu \phi+m^2 \phi=\left(\square^2+m^2\right) \phi=0 . $$ There is no mystery here. The choice of $\mathcal{L}$ was specifically designed to reproduce (14.7).

   If you are unfamiliar with this formalism, consult, for example, Goldstein (1977) or work through the example given in Sakurai (1967), page 3.

Instead of writing down a relativistic wave equation, we simply choose a Lagrangian $\mathcal{L}$. Provided our choice is a Lorentz scalar, the equation of motion resulting from (14.4) will be covariant. For example, substituting the Lagrangian
$$
\mathcal{L}=\frac{1}{2}\left(\partial_\mu \phi\right)\left(\partial^\mu \phi\right)-\frac{1}{2} m^2 \phi^2
$$
into (14.4) gives the Klein-Gordon equation
$$
\partial_\mu \partial^\mu \phi+m^2 \phi=\left(\square^2+m^2\right) \phi=0 .
$$

There is no mystery here. The choice of $\mathcal{L}$ was specifically designed to reproduce (14.7).
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Quarks and leptons: introductory course in modern particle physics
Quarks and leptons: introductory course in modern particle physics
Francis Halzen, Alan… 1st Edition
Chapter 14, Problem 1 ↓

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The term $(\partial_\mu \phi)(\partial^\mu \phi)$ is a Lorentz invariant involving the derivatives of the field, ensuring that the Lagrangian is Lorentz invariant.  Show more…

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If you are unfamiliar with this formalism, consult, for example, Goldstein (1977) or work through the example given in Sakurai (1967), page 3. Instead of writing down a relativistic wave equation, we simply choose a Lagrangian $\mathcal{L}$. Provided our choice is a Lorentz scalar, the equation of motion resulting from (14.4) will be covariant. For example, substituting the Lagrangian $$ \mathcal{L}=\frac{1}{2}\left(\partial_\mu \phi\right)\left(\partial^\mu \phi\right)-\frac{1}{2} m^2 \phi^2 $$ into (14.4) gives the Klein-Gordon equation $$ \partial_\mu \partial^\mu \phi+m^2 \phi=\left(\square^2+m^2\right) \phi=0 . $$ There is no mystery here. The choice of $\mathcal{L}$ was specifically designed to reproduce (14.7).
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