# Derivative of a Noether current from Dirac Equation

1. Nov 25, 2014

1. The problem statement, all variables and given/known data
Hey guys,
Consider the U(1) transformations
$\psi'=e^{i\alpha\gamma^{5}}\psi$ and $\bar{\psi}'=\bar{\psi}e^{i\alpha\gamma^{5}}$ of the Lagrangian $\mathcal{L}=\bar{\psi}(i\partial_{\mu}\gamma^{\mu}-m)\psi$.

I am meant to find the expression for $\partial_{\mu}J^{\mu}$.

2. Relevant equations
Gamma matrices anticommute
Noether current is $\delta J^{\mu}=\frac{\partial\mathcal{L}}{\partial(\partial_{\mu}\psi)}\delta\psi+\delta x \mathcal{L}$
not sure of anything else...

3. The attempt at a solution
So here's what I've done so far. Since its a U(1) transformation, the coordinates arent changing, so the Noether current is given by $\delta J^{\mu}=\frac{\partial\mathcal{L}}{\partial(\partial_{\mu}\psi)}\delta\psi$. I found that $\delta\psi=i\alpha\gamma^{5}\psi$, so that

$\delta J^{\mu}=-\bar{\psi}\alpha\gamma^{\mu}\gamma^{5}\psi$, then I drop the infinitesimal parameter to get
$J^{\mu}=-\bar{\psi}\gamma^{\mu}\gamma^{5}\psi$.

So the next step is to calculate the derivative of this. Doing so, I get

$\partial_{\mu}J^{\mu}=-(\partial_{\mu}\bar{\psi}\gamma^{\mu}\gamma^{5}\psi+\bar{\psi}\gamma^{\mu}\gamma^{5}\partial_{\mu}\psi)$

And at this point im stuck...im not sure if this is right and/or if I can simplify this or do something neat with it? because I think I'm meant to be using the transformed Lagrangian
$\mathcal{L}'=i\bar{\psi}\partial_{\mu}\gamma^{\mu}\psi-m\bar{\psi}e^{2i\alpha\gamma^{5}}\psi$
for something but I dont really know.

Thanks guys..

2. Nov 25, 2014

### Orodruin

Staff Emeritus
What equations of motion do you have?

3. Nov 25, 2014

I think its just the equations of motion from the untransformed Lagrangian, which are:
$(i\partial_{\mu}\gamma^{\mu}-m)\psi=0$
and
$i(\partial_{\mu}\bar{\psi})\gamma^{\mu}+m\bar{\psi}=0$

4. Nov 25, 2014

### Orodruin

Staff Emeritus
So I suggest using those. :)

5. Nov 25, 2014

I thought of that but I'm not sure if I can use these because I'm considering the transformed Lagrangian...and the above transformations arent a symmetry unless m = 0. So how can I use these equations of motion?

6. Nov 25, 2014

### Orodruin

Staff Emeritus
The fields will follow their equations of motion. Had the transformation of the fields been a symmetry of the Lagrangian, the divergence of the current would be zero. Since it is not, you will simply get a non-zero expression if you anyway chose to write down the current that would be conserved if the symmetry breaking parameter was zero.

7. Nov 25, 2014

I see...so I guess Im doing it wrong? I mean I've found the equations of motion from the transformed Lagrangian
$(i\partial_{\mu}\gamma^{\mu}- m e^{2i\alpha\gamma^{5}})\psi=0$
and
$i \partial_{\mu}\bar{\psi}\gamma^{\mu}+m\bar{\psi}e^{2i\alpha\gamma^{5}}=0$

and now I'm trying to use THESE inside my expression for $\partial_{\mu}J^{\mu}$. I guess you're saying to just use the orignals I posted in post #3?

8. Nov 25, 2014

I get a nonzero answer either way but I'm not sure which one is right...:(

9. Nov 25, 2014

### Orodruin

Staff Emeritus
Yes, this willbe the divergence of the current.