# Is every Lorentz invariant a Lorentz scalar?

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? )

• 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) Oct 18 '19 at 20:01

“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.

• isn't $x^\mu$ a 4 vector and therefore a rank 1 tensor? Oct 18 '19 at 19:06
• No, but $dx^\mu$ is. Oct 18 '19 at 19:09
• can you please elaborate? how isn't every vector a rank 1 tensor? Oct 18 '19 at 19:12
• $x^\mu$ is not a vector in General Relativity. Vectors are defined by their transformation rules, not by having multiple components. Oct 18 '19 at 19:13
• “Vector” is a synonym for “rank-1 tensor”. Oct 18 '19 at 19:17

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.

• electric charge is a scalar with respect to spatial rotations, but not with respect to boosts The charge of an electron is $-e$ regardless of whether it is at rest or moving. Maybe you had charge density in mind? Oct 18 '19 at 20:02
• I was being a bit sloppy there, but what I mean is that a charge density will transform to a current density under a boost. Oct 18 '19 at 20:03
• I just edited the sentence in question to say charge density. Are there any glaring mistakes remaining? Oct 18 '19 at 20:10
• Looks good to me. I just upvoted it. Oct 18 '19 at 20:14
• 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? Oct 18 '19 at 20:17