# Proving that a Noether charge is constant

1. Oct 22, 2014

1. The problem statement, all variables and given/known data
Hey guys, so I gota prove that the charge

$Q=\int d^{3}xJ^{0}(\vec{x},t)$

is constant in time, that $\dot{Q}=0$
2. Relevant equations

$J^{\mu}=i[\phi^{\dagger}(\partial^{\mu}\phi)-(\partial^{\mu}\phi^{\dagger})\phi]$

3. The attempt at a solution

So first what I did was find $J^{0}=i[\phi^{\dagger}\dot{\phi}-\dot{\phi^{\dagger}}\phi]$

Then plug this into Q and differentiate it w.r.t. time, which gives us:

$\dot{Q}=i\int d^{3}x(\phi^{\dagger}\ddot{\phi}-\ddot{\phi^{\dagger}}\phi)$

And erm, provided I've done it all right (which I probably havent lol!) i dont know how to show that this is 0?

2. Oct 22, 2014

### Orodruin

Staff Emeritus
What is the equation of motion for $\phi$?

3. Oct 22, 2014

Hmm, I get something like $2\Box\phi+m^{2}\phi^{\dagger}=0$...but I dont know how to use this?

EDIT - actually that's wrong lol you cant have phi and phi^dagger mixing! omg I need to find out what Im doing wrong

4. Oct 22, 2014

Okay so im sure the equation of motion for phi is

$(\Box+m^{2})\phi=0$

5. Oct 22, 2014

### Orodruin

Staff Emeritus
So what does this tell you about $\partial_\mu J^\mu$?

6. Oct 22, 2014

i know that $\partial_{\mu}J^{\mu}=0$ but I dont see how this helps...i know im stupid lol -.-

7. Oct 22, 2014

### Orodruin

Staff Emeritus
There are several ways of using it leading to the conservation of charge (assuming that the current vanishes at infinity). Let us start with the straight forward one: What does $\partial_\mu J^\mu =0$ tell you in terms of components? Can you relate this to something that you can find in an expression for $\dot Q$?

8. Oct 22, 2014

right, so $\partial_{\mu}J^{\mu}=\partial_{0}J^{0}+\partial_{i}J^{i}$ where i goes from 1 to 3 for the 3 spatial components, and the 0th component is time. The problem is that Q dot is just a time derivative...so how do I proceed?

9. Oct 22, 2014

wait...if $\partial_{\mu}J^{\mu}=0$ does that mean that each term in the sum expansion is 0? so that $\partial_{0}J^{0}=0$?

10. Oct 22, 2014

### Orodruin

Staff Emeritus
So you have $\partial_0 J^0 = - \partial_i J^i = - \vec\nabla \cdot \vec J$ and your integral is over space. To paraphrase Monty Python: "Hint, hint, nudge, nudge, know what I mean?" ;)

11. Oct 22, 2014

Hmm its gona be something obvious I just cant see it :S

12. Oct 22, 2014

Hey is it true that $\phi\partial_{0}(\partial^{0}\phi^{\dagger})=\phi^{\dagger}\partial_{0}(\partial^{0}\phi)$? If so then I think I got it lol!

13. Oct 22, 2014

### Orodruin

Staff Emeritus
Yes, this is one of those things you will bang your head into the desk when you realise. ;)
I really find it hard to tell you more without giving the answer away ...

But let me drop a hint on the other approach before I go to bed: You can rewrite the charge $Q$ as
$$Q(t) = \int_S d^3x J^0(\vec x, t) = \int_S d^3x\, n_\mu J^\mu,$$
where $S$ is space at time $t$ and $n$ is the time-like normal vector to $S$ (i.e., (1,0,0,0) in your coordinate system). What is then $Q(t) - Q(0)$?

Edit: Just one more thing which is basically just writing down what you know already ... How can you rewrite the following?
$$\int_V (\vec\nabla \cdot \vec A(\vec x)) d^3x$$

14. Oct 22, 2014