It it known that the massive spin-2 irreducible representation of the Poincare group is the traceless symmetrical transverse 4-tensor $h_{\mu \nu}$ with rank 2: $$ (\partial^{2} + m^{2})h_{\mu \nu} = 0, \quad \partial_{\mu}h^{\mu \nu} = 0, \quad h = 0. $$ These conditions may be united into one equation: $$ \partial_{\mu }\partial_{\nu}h - g_{\mu \nu}(\partial^{2} + m^{2})h - (\partial_{\mu}\partial^{\alpha }h_{\alpha \nu} + \partial_{\nu}\partial^{\alpha }h_{\alpha \mu}) + \eta_{\mu \nu}\partial^{\alpha} \partial^{\beta}h_{\alpha \beta} + (\partial^{2} + m^{2})h_{\mu \nu} $$ $$= 0, \tag{1} $$ which is called Pauli-Fierz equation.

Also there is the linearized gravity $g_{\mu \nu} = \eta_{\mu \nu} + h_{\mu \nu}$, and Einstein equations for $h_{\mu \nu}$ takes the form $$ \partial^{2}h_{\nu \sigma} - (\partial_{\nu}\partial^{\alpha}h_{\alpha \sigma} + \partial_{\sigma}\partial^{\alpha}h_{\alpha \nu} ) + (\eta_{\nu \sigma}\partial_{\mu}\partial_{\alpha}h^{\mu \alpha} + \partial_{\nu} \partial_{\sigma}h) - \eta_{\nu \sigma}\partial_{\mu}\partial^{\mu}h = - \Lambda T_{\nu \sigma}, $$ which is absolutely equal to $(1)$if $m = 0 $ and $T_{\nu \sigma} = 0$ (the second condition refers to the free field).

So I have the question: can I simply set $m$ in $(1)$ to zero? Does it automatically reduce the number of degrees of freedom (by the number of spin projection values) of massive spin-2 field to the number of degrees of freedom (by the number of helicity projection values) of massless spin-2 field?

  • $\begingroup$ Could you check your equation (1)? As it is, it has two identical terms in second parenthesis. $\endgroup$
    – user23660
    Dec 29, 2013 at 5:14
  • $\begingroup$ @user23660 : yes, you're right, thank you. I rewrote it. $\endgroup$ Dec 29, 2013 at 10:54
  • $\begingroup$ can anybody please share any good review article on this matter, I am not finding anything suitable. $\endgroup$ Nov 7, 2022 at 3:58

1 Answer 1


Yes, you can simply set $m=0$ in the Fierz-Pauli equation (if it is written correctly :) ). The only thing to remember is that at $m=0$ it becomes gauge invariant under $\delta h_{\mu\nu}=\partial_\mu\xi_\nu+\partial_\nu\xi_\mu$. It is this gauge invariance that reduces properly the number of degrees of freedom. From the gravity point of view the gauge invariance is nothing but linearized diffeomorphisms.


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