What does "transforms covariantly" mean here?

  • #1
Hill
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TL;DR Summary
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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  • #2
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)##.
 
  • #3
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?
 
  • #4
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.
 
  • #5
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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