# What does Weinberg–Witten theorem want to express?

Weinberg-Witten theorem states that massless particles (either composite or elementary) with spin $j > 1/2$ cannot carry a Lorentz-covariant current, while massless particles with spin $j > 1$ cannot carry a Lorentz-covariant stress-energy. The theorem is usually interpreted to mean that the graviton ( $j = 2$ ) cannot be a composite particle in a relativistic quantum field theory.

Before I read its proof, I've not been able to understand this result. Because I can directly come up a counterexample, massless spin-2 field have a Lorentz covariant stress-energy tensor. For example the Lagrangian of massless spin-2 is massless Fierz-Pauli action:

$$S=\int d^4 x (-\frac{1}{2}\partial_a h_{bc}\partial^{a}h^{bc}+\partial_a h_{bc}\partial^b h^{ac}-\partial_a h^{ab}\partial_b h+\frac{1}{2}\partial_a h \partial^a h)$$

We can calculate its energy-stress tensor by $T_{ab}=\frac{-2}{\sqrt{-g}} \frac{\delta S}{\delta g^{ab}}$, so we get

$$T_{ab}=-\frac{1}{2}\partial_ah_{cd}\partial_bh^{cd}+\partial_a h_{cd}\partial^ch_b^d-\frac{1}{2}\partial_ah\partial^ch_{bc}-\frac{1}{2}\partial^ch\partial_ah_{bc}+\frac{1}{2}\partial_ah\partial_bh+\eta_{ab}\mathcal{L}$$ which is obviously a non-zero Lorentz covariant stress-energy tensor.

And for U(1) massless spin-1 field, we can also have the energy-stress tensor $$T^{ab}=F^{ac} F^{b}_{\ \ \ c}-\frac{1}{4}\eta^{ab}F^{cd}F_{cd}$$ so we can construct a Lorentz covariant current $J^a=\int d^3x T^{a 0}$ which is a Lorentz covariant current.

Therefore above two examples are seeming counterexamples of this theorem. I believe this theorem must be correct and I want to know why my above argument is wrong.

• I'm not sure I understand the question: The theorem says "X is not possible". You ask "Why is X not possible?", to which the first answer must be - look at the proof given! If you are dissatisfied with it, tell us with which step. If you believe you have a counterexample, write it down. – ACuriousMind Dec 18 '14 at 14:17
• Again, write down the counterexample! If you can write down a massless charged vector field with a current that is both Poincare and gauge invariant, I'm impressed. The usual conversed current in such theories is gauge covariant, and hence the theorem does not apply. Have you read the section on Wikipedia on cases where the theorem doesn't apply? – ACuriousMind Dec 18 '14 at 15:32
• The theorem says that higher spin particle cannot couple to gravity in a consistent way. You can find some notes on the Weinberg-Witten theorem in David Bailin's website. phys.susx.ac.uk/~mpfg9 – Hraish kumar Dec 20 '14 at 3:21
• @ACuriousMind Counterexample is written. – 346699 Apr 19 '16 at 8:19
• @Hraishkumar More examples have been added. Do you have any comment on this question? – 346699 Apr 19 '16 at 11:24

1. The stress tensor for $h_{ab}$ is not Lorentz covariant, despite the fact that it looks like it is. This is because $h_{ab}$ itself is not a Lorentz tensor. Rather under Lorentz transformations $$h_{ab} \to \Lambda_a{}^c \Lambda_b{}^d h_{cd} + \partial_a \zeta_b + \partial_b \zeta_a ~.$$ The extra term is present to make up for the fact that $h_{ab}$ is not a tensor of the Lorentz group. Plug this into the stress tensor and you will find that the stress tensor also transforms with a inhomogeneous piece thereby making it non-covariant.
2. The photon is not charged under the $U(1)$ gauge symmetry. Thus, its $U(1)$ current is zero. The current you have defined is not the $U(1)$ current. Rather it is the current corresponding to translations. Weinberg-Witten theorem has nothing to say about this current.
1. Pauli-Fierz theory does not violate the Weinberg-Witten theorem because the stress-energy tensor you constructed is not gauge invariant under the infinitesimal transformation $h_{\mu\nu}\mapsto h_{\mu\nu}+\partial_\mu\chi_\nu + \partial_\nu\chi_\mu$ (this is the Lie derivative $\mathcal{L}_\xi h$ of the metric along the vector field $\chi$, i.e. the infinitesimal change of $h$ under the diffeomorphism generated by $\chi$) for $\chi$ any vector field/1-form. Your $T$ transforms as $$\delta T_{ab} = 2\partial_a\partial_c\chi_d\partial^c\partial^d\chi_b-\frac{1}{2}\partial_a h \partial^c\partial_c \chi_b - \partial_a\partial_b \chi_c\partial^c h$$ (unless I made a mistake), which is non-zero even after dropping terms second order in $\chi$. Therefore, $T$ is not a gauge invariant stress-energy, and therefore does not violate the Weinberg-Witten theorem.
2. The $\mathrm{U}(1)$ massless gauge field is not charged under the $\mathrm{U}(1)$ since the adjoint representation of $\mathrm{U}(1)$ is trivial, and hence does not violate the Weinberg-Witten theorem since this theorem explicitly states that the massless field of spin 1 that is forbidden by it is assumed to be charged under the symmetry responsible for the conserved current.