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?

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, 20241 yr by KJW

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.

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.