Genady Posted March 5 Posted March 5 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 Posted March 5 (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 by KJW
Genady Posted March 5 Author Posted March 5 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 Posted March 5 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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