I'm trying to understand Noether's theorem, and it's application to gauge symmetry. Below what I've done so far.
First, the global gauge symmetry. I'm starting with the Lagragian
$$L_{1}=\partial^{\mu}\Psi\partial_{\mu}\Psi^{\ast}-m^{2}\left|\Psi\right|^{2}$$
with classical complex fields. This Lagragian is invariant with respect to the global gauge symmetry $\Psi\rightarrow\tilde{\Psi}=e^{\mathbf{i}\theta}\Psi$, ... such that I end up with
$$\delta S=\int dv\left[\dfrac{\delta L_{1}}{\delta\Psi}\delta\Psi+\dfrac{\delta L_{1}}{\delta\Psi^{\ast}}\delta\Psi^{\ast}+\mathbf{i}\left(\Psi\partial^{\mu}\Psi^{\ast}-\Psi^{\ast}\partial^{\mu}\Psi\right)\partial_{\mu}\delta\theta\right]=\int dv\left[\partial_{\mu}j^{\mu}\right]\delta\theta$$
provided the equations of motion ($\delta L / \delta \Psi = 0$, ...) are valid. All along I'm using that
$$\dfrac{\delta L}{\delta\phi}=\dfrac{\partial L}{\partial\phi}-\partial_{\mu}\dfrac{\partial L}{\partial\left[\partial_{\mu}\phi\right]}$$
and that $\int dv=\int d^{3}xdt$ for short. The conserved current is of course
$$j_{1}^{\mu}=\mathbf{i}\left(\Psi^{\ast}\partial^{\mu}\Psi-\Psi\partial^{\mu}\Psi^{\ast}\right)$$
since $\delta S / \delta \theta =0 \Rightarrow\partial_{\mu}j_{1}^{\mu}=0$.
Here is my first question: Is this really the demonstration for conservation of charge ? Up to now, it seems to me that I only demonstrated that the particle number is conserved, there is no charge for the moment...
Then, I switch to the local gauge symmetry. I'm starting with the following Lagrangian
$$L_{2}=\left(\partial^{\mu}+\mathbf{i}qA^{\mu}\right)\Psi\left(\partial_{\mu}-\mathbf{i}qA_{\mu}\right)\Psi^{\ast} -m^{2}\left|\Psi\right|^{2} -\dfrac{F_{\mu\nu}F^{\mu\nu}}{4}$$
with $F^{\mu\nu}=\partial^{\mu}A^{\nu}-\partial^{\nu}A^{\mu}$. This Lagrangian is invariant with respect to the local gauge transformation
$$L_{2}\left[\tilde{\Psi}=e^{\mathbf{i}q\varphi\left(x\right)}\Psi\left(x\right),\tilde{\Psi}^{\ast}=e^{-\mathbf{i}q\varphi\left(x\right)}\Psi^{\ast},\tilde{A}_{\mu}=A_{\mu}-\partial_{\mu}\varphi\right]=L_{2}\left[\Psi,\Psi^{\ast},A_{\mu}\right]$$
Then I have
$$\delta S=\int dv\left[\dfrac{\delta L_{2}}{\delta\Psi}\delta\Psi+\dfrac{\delta L_{2}}{\delta\Psi^{\ast}}\delta\Psi^{\ast}+\dfrac{\delta L_{2}}{\delta A_{\mu}}\delta A_{\mu}\right]$$
with $\delta\Psi=\mathbf{i}q\Psi\delta\varphi$, $\delta A_{\mu}=-\partial_{\mu}\delta\varphi$, ... such that I end up with
$$\dfrac{\delta S}{\delta\varphi}=\int dv\left[\mathbf{i}q\Psi\dfrac{\delta L_{2}}{\delta\Psi}+c.c.+\partial_{\mu}\left[j_{2}^{\mu}-\partial_{\nu}F^{\nu\mu}\right]\right]$$
with $j_{2}^{\mu}=\partial L_{2}/\partial A_{\mu}$ and $F^{\nu\mu}=\partial L_{2}/\partial\left[\partial_{\nu}A_{\mu}\right]$
Then, by application of the equations of motion, I have
$$\partial_{\mu}\left[j_{2}^{\mu}-\partial_{\nu}F^{\nu\mu}\right]=0\Rightarrow\partial_{\mu}j_{2}^{\mu}=0$$
since $\partial_{\mu}\partial_{\nu}F^{\nu\mu}=0$ by construction. Of course the new current is
$$j_{2}^{\mu}=\mathbf{i}q\left(\Psi^{\ast}\left(\partial^{\mu}+\mathbf{i}qA^{\mu}\right)\Psi-\Psi\left(\partial^{\mu}-\mathbf{i}qA^{\mu}\right)\Psi^{\ast}\right)$$
and is explicitly dependent on the charge. So it seems to me this one is a better candidate for the conservation of charge.
NB: As remarked in http://arxiv.org/abs/hep-th/0009058, Eq.(27) one can also suppose the Maxwell's equations to be valid ($j_{2}^{\mu}-\partial_{\nu}F^{\nu\mu} = 0$, since they are also part of the equation of motion after all, I'll come later to this point, which sounds weird to me), and we end up with the same current, once again conserved.
Nevertheless, I still have some troubles. Indeed, if I abruptly calculate the equations of motions from the Lagrangian, I end up with (for the $A_{\mu}$ equation of motion)
$$j_{2}^{\mu}-\partial_{\nu}F^{\nu\mu}\Rightarrow\partial_{\mu}j_{2}^{\mu}=0$$
by definition of the $F^{\mu \nu}$ tensor.
So, my other questions: Is there a better way to show the conservation of EM charge ? Is there something wrong with what I did so far ? Why the Noether theorem does not seem to give me something which are not in the equations of motions ? said differently: Why should I use the Noether machinery for something which is intrinsically implemented in the Lagrangian, and thus in the equations of motion for the independent fields ? (Is it because my Lagrangian is too simple ? Is it due to the multiple boundary terms I cancel ?)
Thanks in advance.
PS: I've the feeling that part of the answer would be in the difference between what high-energy physicists call "on-shell" and "off-shell" structure. So far, I never understood the difference. That's should be my last question today :-)
This post imported from StackExchange Physics at 2016-06-21 10:18 (UTC), posted by SE-user FraSchelle
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