Graduate Lorentz Invariance of Lagrangian: Proof & Explanation

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SUMMARY

The discussion clarifies that a Lagrangian can indeed be Lorentz invariant even if it contains second or higher derivatives. The Klein-Gordon Lagrangian, represented as $$\mathscr{L} = \frac{1}{2} (\partial_{\mu} \phi)^2 - \frac{1}{2} m^2 \phi^2$$, utilizes only first-order derivatives and is Lorentz invariant. The misconception that higher-order derivatives invalidate Lorentz invariance is addressed, emphasizing that Lorentz-covariant terms can include higher derivatives, particularly in the context of equations of motion.

PREREQUISITES
  • Understanding of Lagrangian mechanics
  • Familiarity with Lorentz invariance principles
  • Knowledge of the Klein-Gordon equation
  • Basic concepts of field theory
NEXT STEPS
  • Study the derivation of the Klein-Gordon Lagrangian
  • Explore the implications of higher-order derivatives in field theories
  • Investigate Lorentz-covariant formulations in classical mechanics
  • Examine the role of boundary conditions in variational principles
USEFUL FOR

Physicists, particularly those specializing in theoretical physics, graduate students studying field theory, and anyone interested in the mathematical foundations of Lagrangian mechanics.

Gaussian97
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TL;DR
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
 
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"I've just read..." where?

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!)
 
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Gaussian97 said:
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.
 
PeroK said:
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).
 
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What could help is to play around with the action of a nonrelativistic point particle, what kind of actions are possible, what the order of derivatives is and what kind of boundary conditions you need.

Just write down (via an inner priduct) some invariant combinations of the position, velocity, acceleration etc. and see what you get upon varying.
 
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