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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 $\rho_{\mu\nu}$:

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

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    $\begingroup$ 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 $V^\mu(x)\rightarrow \Lambda^\mu_\nu V^\nu(\Lambda^{-1}x)$. $\endgroup$
    – MathZilla
    Apr 23, 2022 at 14:37
  • $\begingroup$ Yes, that was something I tried, but I cannot see why would impy that it has only a $k^2$ dependency. $\endgroup$
    – nosumable
    Apr 23, 2022 at 14:41
  • $\begingroup$ 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. $\endgroup$
    – MathZilla
    Apr 23, 2022 at 14:55

1 Answer 1

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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)$.

QED.

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    $\begingroup$ One q, shoudnt the first equation be $f'(k)=f(\Lambda^{-1} k)$? $\endgroup$
    – nosumable
    Apr 23, 2022 at 15:38
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    $\begingroup$ That's what a Lorentz covariant function does under Lorentz transformations. Here, we are talking about Lorentz invariant functions. $\endgroup$
    – Prahar
    Apr 23, 2022 at 15:39

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