# Noether current scale transform of EM

I'm trying to solve a question about scale tranform of free EM. I got the next trnaform rules

(these two line where EDITed later)

$\delta x = -bx$

$\delta A = bA$

the current I got $D^\mu = -{T^\mu}_\nu x^\nu +F^{\alpha\mu}A_\alpha$ with ${T^\mu}_\nu = F^{\alpha\mu}\partial_\nu A_\alpha - \eta^\mu_\nu L$

My problems starts when I tried to show that this is actually conserved $\partial_\mu D^\mu = 0$

Since $\partial_\mu {T^\mu}_\nu = 0$ and $\partial_\mu F^{\mu\nu} = 0$ I'm left with

$\partial_\mu D^\mu = -{T^\mu}_\nu\partial_\mu x^\nu+F^{\alpha\mu}\partial_\mu A_\alpha = -{T^\mu}_\mu+F^{\alpha\mu}\partial_\mu = -F^{\alpha\mu}\partial_\mu A_\alpha + \eta^\mu_\mu L+F^{\alpha\mu}\partial_\mu A_\alpha = \eta^\mu_\mu L$

this doesnt make any sense $\Sigma\eta^\mu_\mu = -2$ (or $+2$) so it means that $L = 0$? is that so? did I go wrong somewhere else?

thanks a lot

EDIT

I'm looking at free EM $L$ being $L = (-1/4)F^{\mu\nu}F_{\mu\nu} = (-1/4)F^2$.

Scale tranform is defined like this

$x \rightarrow c^{-1}x$

$A \rightarrow c^1A$.

$F_{\mu\nu} = \partial_\mu A_\nu - \partial_\nu A_\mu = \frac{\partial}{\partial x^\mu} A_\nu + (...) \rightarrow c \frac{\partial}{\partial x^\mu} c A_\nu + (...) = c^2 F_{\mu\nu}$

When I look at the action $S = \int d^4x L = \int (-1/4)F^2$ then I get

$S \rightarrow \int d^4x c^{-4} (-1/4) (c^{2})^2 F^2 = \int d^4x (-1/4)F^2 = S$ so I conclude that this is actually symmetry of $L=(-1/4)F^2$.

When I make infinitesimal expantion of the symmetry, $c = e^b$ then I get the infinitesimal rules

$x \rightarrow (1-b)x$ leading to $\delta x = -bx$

$A \rightarrow (1+b)A$ leading to $\delta A = +bA$

• "the current" would be the Noether current? What is $L$? Define your notation, and tell us what you are trying to do! Are you examining whether or not the transformation $x \mapsto -bx$ and $A \mapsto bA$ is a symmetry? Why would it be a symmetry? Why has this a quantum-field-theory tag? – ACuriousMind Feb 9 '15 at 14:29
• thank @ACuriousMind, I hope I wrote it better this time. – tamir Feb 9 '15 at 15:09
• the queston contains algebra errors $\Sigma\eta^\mu_\mu = 4$ and there's a sign mistake at the current formula. the correct current being $Dμ=−T_{\mu\nu}x_\nu-F_{\alpha\mu}A_\alpha$. Please let me know if I should delete the question. – tamir Feb 9 '15 at 22:58
• HI @tamir: So it turned out that $\Sigma\eta^\mu_\mu$ was basically the culprit behind of the whole question? – Qmechanic Feb 10 '15 at 12:36
• yes, and a minus sign :( – tamir Feb 10 '15 at 16:43