Transforming covarianly under translation, the meaning here?

[Genady]

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]

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]

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]

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.