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All examples of lorentz invariant quantities that I have come across seem to be scalars: rest mass, proper time, spacetime interval,dot product of two 4 vectors etc. Another thing is that these are all index contractions.

So, is there any lorentz invariant quantity which isn't a Lorentz scalar?

(My guess is that there isn't: if the quantity isn't scalar, it must have indices. Such a thing must be a tensor of non zero rank. But a thing which is a tensor under lorentz transformation will have its components change from frame to frame and therefore can't be an invariant. One loophole in this reasoning is to assume that the indexed quantity is in fact a tensor of some rank. So, is it possible to have indexed quantities constructed from spacetime which aren't tensors? )

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  • $\begingroup$ it depends if you consider parity exchange transformations part of the Lorentz group or not. If you don't include them, then you have pseudoscalars as well, that invert sign under parity transforms (and probably also T transforms, but not sure) $\endgroup$
    – lurscher
    Commented Oct 18, 2019 at 20:01
  • $\begingroup$ Not all vectors are tensors. See, for example, ntrs.nasa.gov/api/citations/20050175884/downloads/…. Additionally, detailed information on Lorentz scalars is also available at en.wikipedia.org/wiki/Lorentz_scalar. $\endgroup$
    – Michael Levy
    Commented May 24 at 15:00

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“Scalar” and “Lorentz invariant” are synonyms in the context of Special and General Relativity.

However, it is possible to have constant tensors whose components don’t actually change when transformed, such as the Minkowski metric tensor $\eta_{\mu\nu}$ in Special Relativity. We don’t call these “invariant”. Some people call these tensors “isotropic”; others reserve this terminology for constant tensors in Riemannian spaces, such as the Kronecker delta $\delta_{ij}$, rather than those in semi-Riemannian spaces.

As for indexed quantities constructed from spacetime which aren't tensors... The coordinates $x^\mu$ have one index but don’t constitute a tensor in curved spacetime. (But $dx^\mu$ is a tensor.) The Lorentz transformations $\Lambda^\mu{}_\nu$ have two indices but aren’t tensors. Christoffel symbols have three indices but aren’t tensors.

“Scalar” means “rank-0 tensor”. “Vector” means “rank-1 tensor”. Tensors are always defined with respect to a particular transformation group, so you can have rotational tensors, Lorentz tensors, etc.

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  • $\begingroup$ isn't $x^\mu$ a 4 vector and therefore a rank 1 tensor? $\endgroup$
    – lineage
    Commented Oct 18, 2019 at 19:06
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    $\begingroup$ No, but $dx^\mu$ is. $\endgroup$
    – G. Smith
    Commented Oct 18, 2019 at 19:09
  • $\begingroup$ can you please elaborate? how isn't every vector a rank 1 tensor? $\endgroup$
    – lineage
    Commented Oct 18, 2019 at 19:12
  • $\begingroup$ $x^\mu$ is not a vector in General Relativity. Vectors are defined by their transformation rules, not by having multiple components. $\endgroup$
    – G. Smith
    Commented Oct 18, 2019 at 19:13
  • $\begingroup$ “Vector” is a synonym for “rank-1 tensor”. $\endgroup$
    – G. Smith
    Commented Oct 18, 2019 at 19:17
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Every Lorentz invariant is a Lorentz scalar. That is just true by definition.

The trick here is to specify what an object is a scalar, vector, tensor, spinor, etc with respect to.

For instance, the electric charge density is a scalar with respect to spatial rotations, but not with respect to boosts.

Likewise, the four-momentum of a neutral particle can be a vector with respect to Lorentz transformations, but a scalar with respect to gauge transformations, whereas for instance a quark transforms not only under Lorentz transformations but also under U(1) and SU(3) gauge transformations.

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  • $\begingroup$ A quark transforms not only under Lorentz transformations but also under U(1) and SU(3) gauge transformations. Don’t quarks also have an $SU(2)$ gauge transformation related to the weak interaction? $\endgroup$
    – G. Smith
    Commented Oct 18, 2019 at 20:17
  • $\begingroup$ The $SU(2)$ symmetry of the weak interaction is a broken symmetry. Saying that quarks enjoy an $SU(2)$ symmetry is a bit like saying that spins on a ferromagnet enjoy an $O(3)$ symmetry. It really depends on your point of view, so I left it out. $\endgroup$
    – Mason
    Commented Oct 18, 2019 at 20:21
  • $\begingroup$ That makes sense. Thanks. $\endgroup$
    – G. Smith
    Commented Oct 18, 2019 at 20:28

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