What does "transforms covariantly" mean here?

[Hill]

TL;DR

The Lagrangian for scalar field under translation

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 to transform covariantly under translation.
What does it mean that it transforms covariantly under translation?

 

[Demystifier]

This means that under translation $x\to x'(x)=x+a$ it transforms as
$$\phi(x)\to\phi'(x')=\phi(x(x'))$$
where $x(x')=x'-a$ is the inverse of $x'(x)$.

 

[Hill]

This means that under translation $x\to x'(x)=x+a$ it transforms as
$$\phi(x)\to\phi'(x')=\phi(x(x'))$$

I understand that this is how $x$ and how $\phi$ transform. But regarding $\mathcal L$, I think, it makes it rather invariant under translation, doesn't it?

 

[Demystifier]

I understand that this is how $x$ and how $\phi$ transform. But regarding $\mathcal L$, I think, it makes it rather invariant under translation, doesn't it?

Yes, but invariant is a special case of covariant. More precisely, covariance of scalars is invariance.

 

[Hill]

Yes, but invariant is a special case of covariant. More precisely, covariance of scalars is invariance.

Thank you. I thought, there is a reason for him separating the two transformations rather than saying that it is "Lorentz and translational invariant" or "Poincare invariant."