I What does "transforms covariantly" mean here?

AI Thread Summary
The Lagrangian for a scalar field is Lorentz invariant and transforms covariantly under translation. Covariant transformation means that the field transforms according to the relationship between the original and translated coordinates. While the scalar field transforms under translation, the Lagrangian itself is considered invariant, which is a specific case of covariance. The distinction between invariance and covariance is important, as it highlights different aspects of how physical quantities behave under transformations. Understanding these nuances is crucial for proper interpretation in the context of field theory.
Hill
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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?
 
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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)##.
 
Demystifier said:
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?
 
Hill said:
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.
 
Demystifier said:
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."
 
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