# Invariance of a Lagrangian

• A
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I've just read that for a Lagrangian to be Lorentz Invariant the Lagrangian density cannot have second or higher derivatives.
Last day in class, a professor told us that, for a Lagrangian to be Lorentz Invariant, the Lagrangian density cannot have second or higher derivatives. Is this true?
Because one can write the KG lagrangian as $$\mathscr{L}=\phi(\square + m^2)\phi,$$ which have second derivatives.

And, where can I find a proof of this fact?

Thank you

It's simply not true. You can perfectly write down Lorentz-covariant terms with higher order derivatives (e.g. look at the equations of motion, as you mention!)

• bhobba and vanhees71
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Summary:: I've just read that for a Lagrangian to be Lorentz Invariant the Lagrangian density cannot have second or higher derivatives.

Last day in class, a professor told us that, for a Lagrangian to be Lorentz Invariant, the Lagrangian density cannot have second or higher derivatives. Is this true?
Because one can write the KG lagrangian as $$\mathscr{L}=\phi(\square + m^2)\phi,$$ which have second derivatives.

And, where can I find a proof of this fact?

Thank you

The Lagrangian density for the Klein-Gordon equation is
$$\mathscr{L} = \frac 1 2 (\partial_{\mu} \phi)^2 - \frac 1 2 m^2\phi^2$$
Which involves only first-order derivatives.

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The Lagrangian density for the Klein-Gordon equation is
$$\mathscr{L} = \frac 1 2 (\partial_{\mu} \phi)^2 - \frac 1 2 m^2\phi^2$$
Which involves only first-order derivatives.
Which is equivalent to the Lagrangian given by the OP by partial integration (and up to multiplication by a constant).

• PeroK
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