# An indexed lorentz invariant that isn't constant

This is in light of the explanation here regarding lorentz invariants and scalars.

Since its possible to have indexed quantities(that depend on space time) which aren't tensors, the aim is construct an indexed lorentz invariant.

Consider the following quantity given any two 4 vectors $$A$$ and $$B$$:
$$f(A,B)=(A.B,A^{2})=(A^{\mu}B_{\mu},A^{\mu}A_{\mu})$$

$$f$$ isn't a 2D 4 vector since it doesn't transform like $$A$$ i.e.

$$f'_{1}=a A.B +b A^{2}\\ \ne A'.B'$$

where
$$A'=\Lambda A$$ and $$\Lambda^{\mu}_{\nu}= \begin{pmatrix} a&b \\ c&d \end{pmatrix}$$ is the lorentz transformation matrix. Therefore $$f$$ isn't a tensor but is indexed.

Also $$f$$ is clearly lorentz invariant as its components are lorentz scalars though not constants i.e. $$f'=f$$

So, it it correct to say that not every lorentz invariant quantity is a lorentz scalar?

• What about $\eta_{\mu\nu}$? – Prahar Oct 18 at 20:33
• You’re basically asking whether two Lorentz invariants can be called “a” Lorentz invariant. As far as I am aware, no one uses that terminology. This is because there is no point in combining multiple Lorentz invariants into a single “quantity”. It provides no additional insight. – G. Smith Oct 18 at 20:33
• @prahar it's constant – lineage Oct 18 at 20:34
• My bad - misread the title. I can't think of any in Minkowski space but on curved spacetimes there are at least two that I can immediately think of - $g_{\mu\nu}$ and $\epsilon_{\mu_1 \cdots \mu_n}$. – Prahar Oct 18 at 20:37
• As I said in this answer: physics.stackexchange.com/a/508952/106061 you have to specify what an object is scalar with respect to. There is nothing in principal wrong or surprising about an object being scalar with respect to one vector space and tensorial with respect to another. – Mason Oct 18 at 20:38