# Argument of a scalar function to be invariant under Lorentz transformations

I'm trying to prove that a Lorentz scalar object $$\rho(k)$$ which is a function of a cuadri-vector $$k^{\mu}$$ can only have a $$k^2$$ dependency in the argument.

I can imagine that this object has to depend of invariant quantities as the length of $$\rho(k)$$, but I would like to get a explicit derivation because I want to do the same for a (0,2) tensor $$\rho_{\mu\nu}$$:

$$\rho_{\mu\nu}=a(k^2)g_{\mu\nu}+b(k^2)k_\mu k_\nu.$$

• I believe this previous post will help, What exactly does it mean for a scalar function to be Lorentz invariant? . And, to fill in the gaps for transforming a vector, those go as $V^\mu(x)\rightarrow \Lambda^\mu_\nu V^\nu(\Lambda^{-1}x)$. Apr 23, 2022 at 14:37
• Yes, that was something I tried, but I cannot see why would impy that it has only a $k^2$ dependency. Apr 23, 2022 at 14:41
• My guess would then be that when you take the trace, you pick up a factor of $\rho = a(k^2)n + b(k^2)k^2$, where $n$ is the dimension of the space you are in. Apr 23, 2022 at 14:55

A scalar Lorentz invariant function satisfies $$f(k) = f(\Lambda k).$$ for all $$\Lambda$$ satisfying $$\Lambda^T \eta \Lambda = \eta$$. Let us look at the infinitesimal version of this equation. Setting $$\Lambda = 1 + \omega + O(\omega^2)$$ into the equation above, we find the equation $$( k_\mu \partial_{k^\nu} - k_\nu \partial_{k^\mu} ) f(k) = 0 .$$ To solve this differential equation, we change variables. Take, $$k^\mu = \left( \sqrt{ x^ix^i - z } , x^i \right) \quad \Leftrightarrow \quad z = k^2 ,~ x^i=k^i.$$ Using this parameterization, we find $$\partial_{k^0} = - 2 k^0 \partial_z ,\qquad \partial_{k^i} = 2 k_i \partial_z + \partial_{x^i}$$ It follows that $$k_0 \partial_{k^i} - k_i \partial_{k^0} = - k^0 [ 2 k_i \partial_z + \partial_{x^i} ] - k_i ( - 2 k^0 \partial_z ) = - k^0 \partial_{x^i}$$ and $$k_i \partial_{k^j} - k_j \partial_{k^i} = x_i \partial_{x^j} - x_j \partial_{x^i} .$$ The differential equations for $$f(k) \equiv f(z,x^i)$$ now takes the form $$- k^0 \partial_{x^i} f(z,x^i) = 0 , \qquad ( x_i \partial_{x^j} - x_j \partial_{x^i} ) f(z,x^i) = 0 .$$ The first equation immediately implies that $$f$$ doesn't depend on $$x^i$$ so $$f \equiv f (z)=f(k^2)$$.
• One q, shoudnt the first equation be $f'(k)=f(\Lambda^{-1} k)$? Apr 23, 2022 at 15:38