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[Moderator's note: changed thread title to be more descriptive of the actual question.]
Consider Maxwell's action ##S=\int L## over Minkovski space, where the Lagrangian density is ##L = -\frac{1}{4}F_{\mu\nu}F^{\mu\nu}##, and the Electromagnetic tensor is given by ##F^{\mu\nu} = \partial^\mu A^\nu - \partial^\nu A^\mu## for some potential ##A##. How can we see that this action is invariant under dilations and special conformal transformations?
A conformal transformation of space takes the form $$\xi^\mu(x)=a^\mu+\lambda^{\mu\nu}x_\nu+\lambda_Dx^\mu+x^2\lambda^\mu_K-2x^\mu x\cdot \lambda_K$$
The potential ##A## transforms as $$\delta A(x) = \xi^\mu(x)\partial_\mu A(x) - \Lambda^{\mu\nu}_M(x) \eta_{\mu[\nu}\delta^i_{j]} A^j+\Lambda_D(x) A(x)$$
where ##\eta## is the metric, ##\delta_i^j## is Kronekers' symbol, $${\Lambda_M}_{\mu\nu}(x)=\partial_{[\nu}\xi_{\mu]}=\lambda_{\mu\nu}-4x_{[\mu}\lambda_{K\nu]}$$ $$\Lambda_D(x)=\frac{1}{4}\partial_\rho \xi^\rho=\lambda_D-2x\cdot \lambda_K$$
Above, the extra terms are due to intrinsic parts of conformal transformations that act on the representation indices.
We would know that the electromagnetic tensor has conformal weight ##2##, if we could show that $$\delta F_{\mu\nu}=\xi^\rho\partial_\rho F_{\mu\nu}+2\Lambda_{M[\mu}^\rho F_{\nu]\rho}+2\Lambda_D F_{\mu\nu}$$
Who can we show it and then use it to see that the action is invariant? Do we need the value of ##\Lambda_D## in terms of ##\xi^\rho## to this end?
Consider Maxwell's action ##S=\int L## over Minkovski space, where the Lagrangian density is ##L = -\frac{1}{4}F_{\mu\nu}F^{\mu\nu}##, and the Electromagnetic tensor is given by ##F^{\mu\nu} = \partial^\mu A^\nu - \partial^\nu A^\mu## for some potential ##A##. How can we see that this action is invariant under dilations and special conformal transformations?
A conformal transformation of space takes the form $$\xi^\mu(x)=a^\mu+\lambda^{\mu\nu}x_\nu+\lambda_Dx^\mu+x^2\lambda^\mu_K-2x^\mu x\cdot \lambda_K$$
The potential ##A## transforms as $$\delta A(x) = \xi^\mu(x)\partial_\mu A(x) - \Lambda^{\mu\nu}_M(x) \eta_{\mu[\nu}\delta^i_{j]} A^j+\Lambda_D(x) A(x)$$
where ##\eta## is the metric, ##\delta_i^j## is Kronekers' symbol, $${\Lambda_M}_{\mu\nu}(x)=\partial_{[\nu}\xi_{\mu]}=\lambda_{\mu\nu}-4x_{[\mu}\lambda_{K\nu]}$$ $$\Lambda_D(x)=\frac{1}{4}\partial_\rho \xi^\rho=\lambda_D-2x\cdot \lambda_K$$
Above, the extra terms are due to intrinsic parts of conformal transformations that act on the representation indices.
We would know that the electromagnetic tensor has conformal weight ##2##, if we could show that $$\delta F_{\mu\nu}=\xi^\rho\partial_\rho F_{\mu\nu}+2\Lambda_{M[\mu}^\rho F_{\nu]\rho}+2\Lambda_D F_{\mu\nu}$$
Who can we show it and then use it to see that the action is invariant? Do we need the value of ##\Lambda_D## in terms of ##\xi^\rho## to this end?
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