As it can be shown, the equations for the irrep with zero mass and helicity 2, -2 respectively can be given in a form $$ \tag 1 \partial^{\dot {b}a}C_{abcd} = 0, \quad \partial^{\dot{b}a}C_{\dot{a}\dot{b}\dot{c}\dot{d}} = 0. $$ Here $C_{abcd} = C_{(abcd)}, C_{\dot{a}\dot{b}\dot{c}\dot{d}} = C_{(\dot{a}\dot{b}\dot{c}\dot{d})}$ are symmetrical spinor tensors or the reps $\left( 2, 0 \right), \left( 0, 2\right)$ of the Lorentz group respectively.

From $C_{abcd}, C_{\dot{a}\dot{b}\dot{c}\dot{d}}$ we may construct 4-tensor $$ C_{\mu \nu \alpha \beta} = (\tilde{\sigma}_{\mu})^{\dot{a}a}(\tilde{\sigma}_{\nu})^{\dot{b}b}(\tilde{\sigma}_{\alpha})^{\dot{c}c}(\tilde{\sigma}_{\beta})^{\dot{d}d}(\varepsilon_{ab}\varepsilon_{cd}C_{\dot{a}\dot{b}\dot{c}\dot{d}} + \varepsilon_{\dot{a}\dot{b}}\varepsilon_{\dot{c}\dot{d}}C_{abcd}) $$ with properties $$ \tag 2 C_{\mu \nu \alpha \beta} = -C_{\nu \mu \alpha \beta} =-C_{\mu \nu \beta \alpha} = C_{\nu \mu \beta \alpha}, \quad C^{\alpha}_{\ \beta \alpha \delta} = 0, \quad \varepsilon^{\alpha \beta \gamma \delta}C_{\alpha \beta \gamma \delta} = 0. $$ In the formal language it is a direct sum of representations with helicity 2 and helicity -2.

From $(1)$ we can also get $$ \tag 3 \varepsilon^{\mu \nu \alpha \beta}\partial_{\nu}C_{\varepsilon \delta \alpha \beta} = 0. $$

$(2), (3)$ gives us the Weyl tensor of the linearized GR.

My question: why we have got the Weyl tensor for massless helicity-2 particles, not the curvature tensor? I.e., why the free gravitons are described by the Weyl tensor, not by the curvature?


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