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This is a simplification of a previous question. I have a function of two 4-vectors and two factors. One factor is on-shell, that is $q_0=\sqrt{M^2+q^2}$. The other factor is off-shell, that is the temporal component can be any real number. The two 4-vectors are either the center of mass 4-momentum and the relative 4-momenentum or the 4-momentum of two particles. I need a test of Lorentz invariance that doesn't depend on the construction of the function to prove or disprove the function as a whole is Lorentz invariant. I suspect that the product of two Lorentz invariant scalars is not itself Lorentz invariant if one of the factors is off-shell and the other is on-shell. I found this answer by Prahar where they gives this test: $$(k_\mu\partial_{k^\nu}-k_\nu\partial_{k^\mu})f(k)$$ As best as I can tell, it works for off-shell functions of one 4-vector. I need an extension to two 4-vectors both on-shell and off-shell.

I have a function that is Lorentz invariant by construction: $$f(q_1,q_2)=-\frac{q_1^\mu q_{2\mu}+M^2}{M^2}$$ Off-shell it simplifies to: $$f(q_1,q_2)=-\frac{s/4-k^\mu k_\mu+M^2}{M^2}$$ It passes the test given by Prahar as it is a function of one 4-vector and also Lorentz invariant by construction. If I put the temporal component of $q_1$ and $q_2$ on-shell, then it simplifies to: $$f(q_1,q_2)=-\frac{\sqrt{M^2+(\vec P/2+\vec k)^2}\sqrt{M^2+(\vec P/2-\vec k)^2}-P^2/4+k^2+M^2}{M^2}$$ It passes the spatial components of Prahar's test so long as $P=0$, but it should pass regardless of $P$. It should also pass the temporal component test.

I tried this extension: $$(k_\mu\partial_{k^\nu}-k_\nu\partial_{k^\mu})(P_\sigma\partial_{P^\rho}-P_\rho\partial_{P^\sigma})f(P,k)$$ I found that it didn't work. I've been banging my head off of doing some kind of chain rule to make something that works. But I have been unsuccessful to that end.

I just need to know where exactly Prahar got their test from as the steps they used to go from the definition of Lorentz invariance to this test are rather cryptic. It would be helpful if I could look up the test by a proper name. Or if an extension to the test could be given that would work on $f(P^\mu,k^\mu)$ and $g(P^\mu,k_0=(\omega_+-\omega_-)/2,k_i)$ and then tell if $f(P^\mu,k^\mu)g(P^\mu,k_0=(\omega_+-\omega_-)/2,k_i)$ satisfies the Lorentz invariant condition.

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I got an answer that I'm satisfied with. I'll find out later if my research advisor agrees.

The test that Prahar gives for one 4-vector is $$(k_\mu\partial_{k^\nu}-k_\nu\partial_{k^\mu})f(k)=0$$ The solution that I have is the sum of two conditions. $$(k_\mu\partial_{k^\nu}-k_\nu\partial_{k^\mu}+P_\mu\partial_{P^\nu}-P_\nu\partial_{P^\mu})f(P,k)=0$$ I'm not particularly happy with this version as dealing with the on-shell $k_0$ is not exactly a consistent exercise for reasons not relevant to the question. So my solution is to examine the particle 4-momenta: $$(q_{-\mu}\partial_{q_-^\nu}-q_{-\nu}\partial_{q_-^\mu}+q_{+\mu}\partial_{q_+^\nu}-q_{+\nu}\partial_{q_+^\mu})f(q_+,q_-)=0$$ To do the on-shell version of the test, I use $\frac{d}{dq_0}=\frac{d}{dq_i}\frac{dq_i}{dq_0}=-2\sqrt{M^2+q_i^2}\frac{d}{dq_i}$ for $q_0=\sqrt{M^2+q_i^2}$.

The on-shell version completely passes my on-shell expressions. The off-shell version completely passes my off-shell expressions. Both versions fail my hybrid expressions, as I expected. Both failed the temporal component of the test. When I take the error and put it on-shell, the on-shell test passes. But I put in values that are off-shell, both fail with very small errors.

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