Proving that a Noether charge is constant

[Dixanadu]

Homework Statement

Hey guys, so I gota prove that the charge

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

is constant in time, that $\dot{Q}=0$

Homework Equations

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

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 haven't lol!) i don't know how to show that this is 0?

Thanks in advance guys

 

[Orodruin]

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

 

[Dixanadu]

Hmm, I get something like $2\Box\phi+m^{2}\phi^{\dagger}=0$...but I don't know how to use this?

EDIT - actually that's wrong lol you can't have phi and phi^dagger mixing! omg I need to find out what I am doing wrong

 

[Dixanadu]

Okay so I am sure the equation of motion for phi is

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

 

[Orodruin]

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

 

[Dixanadu]

i know that $\partial_{\mu}J^{\mu}=0$ but I don't see how this helps...i know I am stupid lol -.-

 

[Orodruin]

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 $\dot Q$?

 

[Dixanadu]

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?

 

[Dixanadu]

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

 

[Orodruin]

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?" ;)

 

[Dixanadu]

Hmm its gona be something obvious I just can't see it :S

 

[Dixanadu]

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!

 

[Orodruin]

Yes, this is one of those things you will bang your head into the desk when you realize. ;)
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?
$$
\int_V (\vec\nabla \cdot \vec A(\vec x)) d^3x
$$

 

[Dixanadu]

aww I think I like the other approach lol XD You've helped me a bunch man I owe you one! I think I'll get it eventually :P

Thanks again man sleep well!