Is the Noether Current \(M^{\mu;\nu\rho}\) Conserved in Klein-Gordon Theory?

[Orodruin]

Let me quote your earlier statement:

an index appearing up and down is summed over by Einstein's convention. it is therefore a dummy index and can be replaced by any other letter

You are renaming indices, not swapping them - $\alpha$ only changes sign when you swap indices.

Edit: This may be easier to understand if you do the following renaming in the first term $\rho \to \sigma$, $\nu\to \tau$ and also do the renaming $\rho \to \tau$, $\nu\to \sigma$ in the second, which makes it easier to see that it is only a question of renaming and not of swapping.

 

[Dixanadu]

Yea I knew you'd say that, that's cos I was being vague. Here's what I mean:

Transposing alpha: $\alpha_{\rho\nu}x^{\nu}T^{\mu\rho} \rightarrow -\alpha_{\nu\rho}x^{\nu}T^{\mu\rho}$, we agree so far !

Okay but now instead of transposing the indices (instead of "swapping" them), I am gona rename them as they are being summed. I can do that right? This way I'll get:

$\alpha_{\rho\nu}x^{\nu}T^{\mu\rho} \rightarrow \alpha_{\nu\rho}x^{\rho}T^{\mu\nu}$

Is this what you mean...?

 

[Orodruin]

Yes, nothing has changed, only the name of dummy indices that are being summed over.

 

[Dixanadu]

Okay so moving further on from there, (this might sound stupid) but how do I factor out the alpha without changing the sign in the bracket?

 

[Orodruin]

What do you get after renaming the dummy indices in the second term?

 

[Dixanadu]

I think I skipped a beat. I get alpha with the same indices as the first term. So I have

$J^{\mu}=-\frac{1}{2}\alpha_{\rho\nu}(x^{\nu}T^{\mu\rho}-x^{\rho}T^{\mu\nu})$...is this the long awaited answer?

 

[Dixanadu]

Well doctor, I'm pretty sure this is it - if It's not then I got no idea. I'm gona stick with this. I don't know how to thank you for your help and patience. You rock :D thank you!

 

[Orodruin]

Now note that this has to be a Noether current regardless of $\alpha_{\rho\nu}$, which means that what is inside the parentheses must be the general conserved current (the difference from the expression you started with is that now the expression in the parentheses is also asymmetric in $\nu$ and $\rho$ and you can therefore just read off the current - the alternative is to do what I described earlier with picking a particular transformation).