Genady Posted March 5, 2024 Posted March 5, 2024 The Lagrangian, \[\mathcal L(x)= \frac 1 2 \partial^{\mu} \phi (x) \partial_{\mu} \phi (x) - \frac 1 2 m^2 \phi (x)^2\] for a scalar field \(\phi (x)\) is said to be "Lorentz invariant and transforms covariantly under translation." What does it mean that it transforms covariantly under translation?
KJW Posted March 5, 2024 Posted March 5, 2024 (edited) Why don't you try an infinitesimal translation to find out: [math]x^\mu \longrightarrow x^\mu + \delta x^\mu[/math] Note that: [math]f \longrightarrow f + \partial_\mu f\ \delta x^\mu[/math] Edited March 5, 2024 by KJW
Genady Posted March 5, 2024 Author Posted March 5, 2024 18 minutes ago, KJW said: Why don't you try an infinitesimal translation to find out: xμ⟶xμ+δxμ Note that: f⟶f+∂μf δxμ I did, and my answer was that \(\mathcal L\) is invariant under translation.
joigus Posted March 5, 2024 Posted March 5, 2024 3 hours ago, Genady said: I did, and my answer was that L is invariant under translation. Right. A scalar provides a particular type of covariance. Rank zero. L'(x')=L(x) That's what one must prove in this case. 1
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