Genady Posted March 5 Share 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? Link to comment Share on other sites More sharing options...
KJW Posted March 5 Share 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 Link to comment Share on other sites More sharing options...
Genady Posted March 5 Author Share 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. Link to comment Share on other sites More sharing options...
joigus Posted March 5 Share 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 Link to comment Share on other sites More sharing options...
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