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Consider the following Lagrangian for a massive vector field $A_{\mu}$ in Euclidean space time: $$\mathcal L = \frac{1}{4} F^{\alpha \beta}F_{\alpha \beta} + \frac{1}{2}m^2 A^{\alpha}A_{\alpha}$$ where $F_{\alpha \beta} = \partial_{\alpha}A_{\beta} - \partial_{\beta}A_{\alpha}$ which means $$\mathcal L = \frac{1}{4} (\partial^{\alpha}A^{\beta} - \partial^{\beta}A^{\alpha})(\partial_{\alpha}A_{\beta} - \partial_{\beta}A_{\alpha}) + \frac{1}{2}m^2A^{\alpha}_{\alpha} \tag{1}$$ The canonical energy-momentum tensor is supposed to be, using the relation

$$T^{\mu \nu}_c = -\eta^{\mu \nu} \mathcal L + \frac{\partial \mathcal L}{\partial (\partial_{\mu}\Phi)}\partial^{\nu}\Phi,\,\,\tag{2}$$

$$T^{\mu \nu}_c = F^{\mu \alpha}\partial^{\nu}A_{\alpha} - \eta^{\mu \nu}\mathcal L$$

Then from $T^{\mu \nu}_B = T^{\mu \nu}_c + \partial_{\rho}B^{\rho \mu \nu}$, it is found that $$B^{\alpha \mu \nu} = F^{\alpha \mu}A^{\nu}\tag{3}$$ using the formula $$B^{\mu \rho \nu} = \frac{1}{2}i \left\{\frac{\partial \mathcal L}{\partial (\partial_{\mu}A_{\gamma})} S^{\nu \rho}A_{\gamma} + \frac{\partial \mathcal L}{\partial (\partial_{\rho}A_{\gamma})} S^{\mu \nu}A_{\gamma} + \frac{\partial \mathcal L}{\partial (\partial_{\nu}A_{\gamma})} S^{\mu \rho}A_{\gamma}\right\}$$ My question is how is this equation obtained and how did they obtain $(3)$? Did they make use of the explicit form of the spin matrix for a vector field? (My question is from Di Francesco et al 'Conformal Field Theory' P.46-47).

Here is my attempt: $$B^{\alpha \mu \nu} = \frac{i}{2}\left\{F^{\alpha \gamma}S^{\nu \mu}A_{\gamma} + F^{\mu \gamma}S^{\alpha \nu}A_{\gamma} + F^{\nu \gamma}S^{\alpha \mu}A_{\gamma}\right\}$$ from simplifying the above. Concentrate on the first term. Then $$F^{\alpha \gamma}S^{\nu \mu}A_{\gamma} = F^{\alpha \gamma}\eta_{\gamma c}\eta^{\nu a}\eta^{\mu b} S_{ab}A^c$$ Inputting the form of $S$ for a vector field, I get $$F^{\alpha \gamma}\eta_{\gamma c}\eta^{\nu a}\eta^{\mu b}(\delta^c_a \eta_{bd} - \delta^c_b \eta_{da})A^d$$ But simplifying this and writing the other terms does not yield the result. Did I make a mistake upon insertion of the spin matrix?

Consider the following Lagrangian for a massive vector field $A_{\mu}$ in Euclidean space time: $$\mathcal L = \frac{1}{4} F^{\alpha \beta}F_{\alpha \beta} + \frac{1}{2}m^2 A^{\alpha}A_{\alpha}$$ where $F_{\alpha \beta} = \partial_{\alpha}A_{\beta} - \partial_{\beta}A_{\alpha}$ which means $$\mathcal L = \frac{1}{4} (\partial^{\alpha}A^{\beta} - \partial^{\beta}A^{\alpha})(\partial_{\alpha}A_{\beta} - \partial_{\beta}A_{\alpha}) + \frac{1}{2}m^2A^{\alpha}_{\alpha} \tag{1}$$ The canonical energy-momentum tensor is supposed to be, using the relation

$$T^{\mu \nu}_c = -\eta^{\mu \nu} \mathcal L + \frac{\partial \mathcal L}{\partial (\partial_{\mu}\Phi)}\partial^{\nu}\Phi,\,\,\tag{2}$$

$$T^{\mu \nu}_c = F^{\mu \alpha}\partial^{\nu}A_{\alpha} - \eta^{\mu \nu}\mathcal L$$

Then from $T^{\mu \nu}_B = T^{\mu \nu}_c + \partial_{\rho}B^{\rho \mu \nu}$, it is found that $$B^{\alpha \mu \nu} = F^{\alpha \mu}A^{\nu}\tag{3}$$ using the formula $$B^{\mu \rho \nu} = \frac{1}{2}i \left\{\frac{\partial \mathcal L}{\partial (\partial_{\mu}A_{\gamma})} S^{\nu \rho}A_{\gamma} + \frac{\partial \mathcal L}{\partial (\partial_{\rho}A_{\gamma})} S^{\mu \nu}A_{\gamma} + \frac{\partial \mathcal L}{\partial (\partial_{\nu}A_{\gamma})} S^{\mu \rho}A_{\gamma}\right\}$$ My question is how is this equation obtained and how did they obtain $(3)$? Did they make use of the explicit form of the spin matrix for a vector field? (My question is from Di Francesco et al 'Conformal Field Theory' P.46-47).

Here is my attempt: $$B^{\alpha \mu \nu} = \frac{i}{2}\left\{F^{\alpha \gamma}S^{\nu \mu}A_{\gamma} + F^{\mu \gamma}S^{\alpha \nu}A_{\gamma} + F^{\nu \gamma}S^{\alpha \mu}A_{\gamma}\right\}$$ from simplifying the above. Concentrate on the first term. Then $$F^{\alpha \gamma}S^{\nu \mu}A_{\gamma} = F^{\alpha \gamma}\eta_{\gamma c}\eta^{\nu a}\eta^{\mu b} (S_{ab})^c_dA^d$$ Inputting the form of $S$ for a vector field, I get $$F^{\alpha \gamma}\eta_{\gamma c}\eta^{\nu a}\eta^{\mu b}(\delta^c_a \eta_{bd} - \delta^c_b \eta_{da})A^d$$ But simplifying this and writing the other terms does not yield the result. Did I make a mistake upon insertion of the spin matrix?

Consider the following Lagrangian for a massive vector field $A_{\mu}$ in Euclidean space time: $$\mathcal L = \frac{1}{4} F^{\alpha \beta}F_{\alpha \beta} + \frac{1}{2}m^2 A^{\alpha}A_{\alpha}$$ where $F_{\alpha \beta} = \partial_{\alpha}A_{\beta} - \partial_{\beta}A_{\alpha}$ which means $$\mathcal L = \frac{1}{4} (\partial^{\alpha}A^{\beta} - \partial^{\beta}A^{\alpha})(\partial_{\alpha}A_{\beta} - \partial_{\beta}A_{\alpha}) + \frac{1}{2}m^2A^{\alpha}_{\alpha} \tag{1}$$ The canonical energy-momentum tensor is supposed to be, using the relation

$$T^{\mu \nu}_c = -\eta^{\mu \nu} \mathcal L + \frac{\partial \mathcal L}{\partial (\partial_{\mu}\Phi)}\partial^{\nu}\Phi,\,\,\tag{2}$$

$$T^{\mu \nu}_c = F^{\mu \alpha}\partial^{\nu}A_{\alpha} - \eta^{\mu \nu}\mathcal L$$

but I am not sure how this is obtained.

Here is my attempt:

$(2)$ can be rewritten like $$T^{\mu \nu}_c = -\eta^{\mu \nu}\mathcal L + \frac{\partial \mathcal L}{\partial(\partial_{\mu}A_{\gamma})}\partial^{\nu}A_{\gamma}$$ for the given $\Phi$. Then the derivative acts on the terms in $(1)$ so we have for example, $$\frac{\partial}{\partial(\partial_{\mu}A_{\gamma})} \partial^{\alpha}A^{\beta} = \frac{\partial}{\partial(\partial_{\mu}A_{\gamma})} (\eta^{\alpha c}\eta^{\beta d}\partial_{c}A_{d}) = \eta^{\alpha c}\eta^{\beta d}\delta^{\gamma}_{d}\delta^{\mu}_c$$ and $$\frac{\partial}{\partial(\partial_{\mu}A_{\gamma})}A^{\alpha}A_{\alpha} = 0$$ I think this may be where I go wrong since every term is analogous to these two except if the indices are already lowered as in $$\frac{\partial}{\partial(\partial_{\mu}A_{\gamma})} \partial_{\alpha}A_{\beta} = \delta^{\mu}_{\alpha}\delta^{\gamma}_{\beta}$$ Could someone clarify?

Consider the following Lagrangian for a massive vector field $A_{\mu}$ in Euclidean space time: $$\mathcal L = \frac{1}{4} F^{\alpha \beta}F_{\alpha \beta} + \frac{1}{2}m^2 A^{\alpha}A_{\alpha}$$ where $F_{\alpha \beta} = \partial_{\alpha}A_{\beta} - \partial_{\beta}A_{\alpha}$ which means $$\mathcal L = \frac{1}{4} (\partial^{\alpha}A^{\beta} - \partial^{\beta}A^{\alpha})(\partial_{\alpha}A_{\beta} - \partial_{\beta}A_{\alpha}) + \frac{1}{2}m^2A^{\alpha}_{\alpha} \tag{1}$$ The canonical energy-momentum tensor is supposed to be, using the relation

$$T^{\mu \nu}_c = -\eta^{\mu \nu} \mathcal L + \frac{\partial \mathcal L}{\partial (\partial_{\mu}\Phi)}\partial^{\nu}\Phi,\,\,\tag{2}$$

$$T^{\mu \nu}_c = F^{\mu \alpha}\partial^{\nu}A_{\alpha} - \eta^{\mu \nu}\mathcal L$$

Then from $T^{\mu \nu}_B = T^{\mu \nu}_c + \partial_{\rho}B^{\rho \mu \nu}$, it is found that $$B^{\alpha \mu \nu} = F^{\alpha \mu}A^{\nu}\tag{3}$$ using the formula $$B^{\mu \rho \nu} = \frac{1}{2}i \left\{\frac{\partial \mathcal L}{\partial (\partial_{\mu}A_{\gamma})} S^{\nu \rho}A_{\gamma} + \frac{\partial \mathcal L}{\partial (\partial_{\rho}A_{\gamma})} S^{\mu \nu}A_{\gamma} + \frac{\partial \mathcal L}{\partial (\partial_{\nu}A_{\gamma})} S^{\mu \rho}A_{\gamma}\right\}$$ My question is how is this equation obtained and how did they obtain $(3)$? Did they make use of the explicit form of the spin matrix for a vector field? (My question is from Di Francesco et al 'Conformal Field Theory' P.46-47).

Here is my attempt: $$B^{\alpha \mu \nu} = \frac{i}{2}\left\{F^{\alpha \gamma}S^{\nu \mu}A_{\gamma} + F^{\mu \gamma}S^{\alpha \nu}A_{\gamma} + F^{\nu \gamma}S^{\alpha \mu}A_{\gamma}\right\}$$ from simplifying the above. Concentrate on the first term. Then $$F^{\alpha \gamma}S^{\nu \mu}A_{\gamma} = F^{\alpha \gamma}\eta_{\gamma c}\eta^{\nu a}\eta^{\mu b} S_{ab}A^c$$ Inputting the form of $S$ for a vector field, I get $$F^{\alpha \gamma}\eta_{\gamma c}\eta^{\nu a}\eta^{\mu b}(\delta^c_a \eta_{bd} - \delta^c_b \eta_{da})A^d$$ But simplifying this and writing the other terms does not yield the result. Did I make a mistake upon insertion of the spin matrix?