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Posted

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?

Posted (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 by KJW
Posted
18 minutes ago, KJW said:

Why don't you try an infinitesimal translation to find out:

xμxμ+δxμ

Note that:

ff+μf δxμ

 

I did, and my answer was that \(\mathcal L\) is invariant under translation.

Posted
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.

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