15
$\begingroup$

Like my title, I sometimes see that my books says something is Lorentz invariant or Lorentz covariant. What's the difference between these two transformation properties? Or are they just the same thing? If they are different, is there a term called "Lorentz contravariant" like tensors?

$\endgroup$
1

1 Answer 1

19
$\begingroup$
  • A Lorentz invariant quantity does not change under a Lorentz transformation. For example, charge is invariant, but energy is not.
  • The components of a vector transform contravariantly, i.e. $A^\mu \to \Lambda^\mu_{\ \ \nu} A^\nu.$
  • The components of a covector transform covariantly, i.e. $\omega_\mu \to \Lambda_\mu^{\ \ \nu} \omega_\nu$.
  • An equation is covariant if both sides transform the same way. This implies that the equation remains true after a Lorentz transformation. For example, $A^\mu = B^\mu$ is covariant, while $A^\mu = B_\mu$ is not.

The two meanings of 'covariant' above are totally distinct, but the root 'co' means the same thing, i.e. 'the same'. In the first case, covariant transformation is contrasted with contravariant transformation (which is 'opposite'). In the second, the 'co' refers to how both sides are transforming the same way.

There are also some other less common usages of these words.

  • Geometrical objects are sometimes called invariant. For example, a vector is invariant; its components change under coordinate transformations but not the vector itself.
  • Upper indices are sometimes generally referred to as contravariant indices, and vice versa.
  • We say an equation is in covariant form if it is written so that both sides clearly transform the same way. For example, Maxwell's equations are already covariant, but it's hard to see. Instead, we say $\partial_\mu F^{\mu\nu} = J^\nu$ is their covariant form.
  • Sometimes, we say covariant equations are invariant.
$\endgroup$
1
  • $\begingroup$ In the first case, covariant transformation is contrasted with contravariant transformation (which is 'opposite') Note that the thing you compare your transformation with is the transformation of the basis of the vector space. In case of differential geometry, that basis transformation is induced by a change of coordinates. A simple example: Scale one coordinate by a factor $\lambda$. Its corresponding basis vector gets scaled accordingly, which means the component of the basis expansion of a given vector has to be scaled by $\lambda^{-1}$, ie contravariantly (the vector itself is invariant) $\endgroup$
    – Christoph
    Commented Jul 28, 2016 at 12:54

Not the answer you're looking for? Browse other questions tagged or ask your own question.