# Homework Help: Conservation of Noether current

1. Nov 12, 2014

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

$M^{\mu;\nu\rho}=x^{\nu}T^{\mu\rho}-x^{\rho}T^{\mu\nu}$

is conserved. That is:

$\partial_{\mu}M^{\mu;\nu\rho}=0.$
2. Relevant equations
Not given in the question but I'm pretty sure that

$T^{\mu\nu}=\frac{\partial \mathcal{L}}{\partial(\partial_{\mu}\phi)}\partial^{\nu}\phi-\mathcal{L}g^{\mu\nu}$

And we're considering a real Klein-Gordon theory, so we have

$\mathcal{L}=\frac{1}{2}(\partial_{\mu}\phi)(\partial^{\mu}\phi)-\frac{m}{2}\phi^{2}$

3. The attempt at a solution

So here's what I've done so far:

$T^{\mu\rho}=(\partial^{\mu}\phi)(\partial^{\rho}\phi)-\mathcal{L}g^{\mu\rho}$
$T^{\mu\rho}=(\partial^{\mu}\phi)(\partial_{\mu}\phi)g^{\mu\rho}-\mathcal{L}g^{\mu\rho}$
$T^{\mu\rho}=\left[ (\partial^{\mu}\phi)(\partial_{\mu}\phi)-\mathcal{L}\right]g^{\mu\rho}$
$T^{\mu\rho}=\left[ \frac{1}{2}(\partial^{\mu}\phi)(\partial_{\mu}\phi)+\frac{m}{2}\phi^{2}\right]g^{\mu\rho}$

Doing the same thing to $T^{\mu\nu}$ gives

$T^{\mu\nu}=\left[ \frac{1}{2}(\partial^{\mu}\phi)(\partial_{\mu}\phi)+\frac{m}{2}\phi^{2}\right]g^{\mu\nu}$

Now putting it together gives

$M^{\mu;\nu\rho}=( \frac{1}{2}(\partial^{\mu}\phi)(\partial_{\mu}\phi)+\frac{m}{2}\phi^{2}))(x^{\nu}g^{\mu\rho}-x^{\rho}g^{\mu\nu})$

Now I have to hit this with $\partial_{\mu}$. So i get:

$\partial_{\mu}M^{\mu;\nu\rho}=( \frac{1}{2}\partial_{\mu}(\partial^{\mu}\phi)(\partial_{\mu}\phi)+m\phi(\partial_{\mu}\phi))(x^{\nu}g^{\mu\rho}-x^{\rho}g^{\mu\nu})$

And I'm stuck on what to do next. Don't know how to deal with $\partial_{\mu}(\partial^{\mu}\phi)(\partial_{\mu}\phi)$

2. Nov 12, 2014

### Orodruin

Staff Emeritus
You have the same problem as in the other thread of using the same summation index for two different sums.

That aside, have you proven that $T^{\mu\nu}$ is a conserved current? If you have the problem is trivial. If not, I suggest starting by that.

The alternative is deriving the Nother current associated with Lorentz transformations (rotations and boosts).

3. Nov 12, 2014

Proving the currents is the previous question which I am also stuck on lol! I'm basically asked to derive that expression for M using Noether's theorem (I cant get to the end for some reason) and then I have to prove that the charge is 0.

I dont see which index I can change in my equation though...can you help me out please?

4. Nov 12, 2014

• Message originally posted in new thread.
Hey guys,

Consider an infinitesimal Lorentz transformation of coordinates given by $\delta x^{\mu}=\alpha^{\mu}_{\nu}x^{\nu}.$ What would be the corresponding transformation to the field $\delta \phi$?

I'm also working with the real Klein-Gordon theory so i know that $\phi(x)=\phi'(x').$ Also I know:

$\delta\phi(x)=\phi'(x')-\phi(x)-(\delta_{\mu}\phi)\delta x^{\mu}$

So we get $\delta\phi=-(\partial_{\mu}\phi)\alpha^{\mu}_{\nu}x^{\nu}$

Is this right? I mean I have another relation $\phi'(x)=\phi(\Lambda^{-1}x)$ but no idea how to use this, or even if I need it to find $\delta\phi$?

5. Nov 12, 2014

### nrqed

If you see an index appearing three times in a term, you can tell there is something wrong.
Whenever you introduce a new index that is summed over, you must use a letter that is not already present. So your second equation above is incorrect.

6. Nov 12, 2014

### Orodruin

Staff Emeritus
Can you post your working for $T^{\mu\nu}$? Are you using Noether's theorem or just starting from the expression for $T^{\mu\nu}$ in terms of the fields and trying to show $\partial_\mu T^{\mu\nu} = 0$?

Regarding the summation indices, just rename one of your summation indices before insertion of the Lagrangian into the expression for T.

Note: I wrote this some time ago butseemingly never sent it. My apologies if I repeat something nrqed already said.

7. Nov 12, 2014

I havent got any working for T. The previous question requires me to prove the expression for M using Noether's theorem and the infinitesimal Lorentz transformation $\delta x^{\mu} = \alpha^{\mu}_{\nu}x^{\nu}$. Here is my working so far, I get stuck towards the end I think:

http://i.imgur.com/aHW3UXA.png

8. Nov 12, 2014

### Orodruin

Staff Emeritus
Remember that the current is normally defined as the derivative with respect to the infinitesimal symmetry parameter (in your case $\alpha$, infinitesimal Lotentz transformations). I suggest you use this and keep in mind that $\alpha_{\mu\nu}$ has some special property ...

9. Nov 12, 2014

Yes I know $\alpha_{\mu\nu}=-\alpha_{\nu\mu}$ but I dont know what you mean by the derivative being defined with respect to alpha. Also is my expression for $\delta\phi$ even correct...?

10. Nov 12, 2014

### Orodruin

Staff Emeritus
Take the current you have found and differentiate wrt $\alpha_{\mu\nu}$ (but pick indices carefully!!). The resulting current is that generated by that particular Lorentz group generator (the expression you have is for a general Lorentz transformation).

11. Nov 12, 2014

I still dont see what you mean by differentiate with respect to alpha. Yes I have a general current, so are you saying I plug in my delta x and then divide through by the infinitesimal parameter? basically the following:
$\delta J^{\mu}=-T^{\mu\rho}\delta x_{\rho}$ is my general current.

Plug in $\delta x_{\rho}=\alpha^{\nu}_{\rho}x_{\nu}$ to get

$\delta J^{\mu}=-T^{\mu\rho}\alpha^{\nu}_{\rho}x_{\nu}$.

Then divide through by alpha? I have no idea what to do with the indices if this is what I'm meant to do :S

EDIT : i think I have to contract T first by splitting it into symmetric + antisymmetric parts after I plug in alpha...then divide through by alpha?

12. Nov 12, 2014

### Orodruin

Staff Emeritus
Yes, you have a current which is proportional to the infinitesimal boost. You can now select which boost to make, essentially letting $\alpha_{\mu\nu} = \delta^\rho_\mu \delta^\sigma_\nu - \delta^\sigma_\mu \delta^\rho_\nu$ where $\sigma$ and $\rho$ are some fixed indices. The second term is required by the anti-symmetry of $\alpha$.

This essentially is equivalent to differentiating your current wrt $\alpha$.

Again, be careful not to introduce the same name for two different indices!

Edit: By the way, the very last step in the image you linked is not valid - so start from $J^\mu = -\alpha_{\rho\nu} x^\nu T^{\mu\rho}$.

Last edited: Nov 12, 2014
13. Nov 12, 2014

Okay so I have two questions - firstly, how did you swap the nu-indices positions in your expression for J? I mean I had

$J^{\mu}=-\alpha_{\rho}^{\nu} x_{\nu}T^{\mu\rho}$ whereas you have $J^{\mu}=-\alpha_{\rho\nu}x^{\nu}T^{\mu \rho}$? How did you do that?

Secondly, is this statement now correct:

$J^{\mu}=-\frac{1}{2}\alpha_{\rho\nu}x^{\nu}(T^{\mu\rho} - T^{\mu\nu})$ ? I Anti-symmetrised T like my lecturer suggested.

However from this point I dont know how to move forward...so im still stuck. I really am confused about your expression for alpha in terms of those deltas. I understand the antisymmetry of alpha and I'd be tempted to use $\alpha_{\nu\rho}=\frac{1}{2}(\alpha_{\nu\rho}-\alpha_{\rho\nu})$ but dont really know what im doing as far as the differentiation you suggested is concerned.

14. Nov 12, 2014

### Orodruin

Staff Emeritus
Raising and lowering of indices is done as usual, with the metric tensor. If you are contracting a covariant index with a contravariant one, you can switch them without thinking twice (although maybe you should).

I cannot tell exactly what your lecturer said, but T in general is not anti-symmetric and you cannot simply remove the symmetric part. Your approach to alpha should also work, but the easiest by far is just differentiating or picking a particular alpha.

15. Nov 12, 2014

So by differentiating wrt alpha, you mean something like this, right?
$\frac{J^{\mu}}{\alpha_{\mu\nu}}$? But you're saying that due to the antisymmetry of alpha, I have two possibilities - the one I just stated and $-\frac{J^{\mu}}{\alpha_{\nu\mu}}$?

Doing this gives me 2 separate equations and I dont even know how to handle the alpha's indices in the denominator. I guess I dont know enough to apply your approach :( or there's something obvious that I'm missing. Even with my approach I am yet to reach a solution...

EDIT: My lecturer said that the symmetric part vanishes under contraction.

16. Nov 12, 2014

okay I've done what I understand from your approach. Please tell me if this is what you meant:

$J^{\mu}=-\alpha_{\rho\nu}x^{\nu}T^{\mu\rho} = + \alpha_{\nu\rho}x^{\rho}T^{\mu\nu}$

Differentiating the first equality gives:

$\delta J^{\mu}=\frac{\partial J^{\mu}}{\partial \alpha_{\rho\nu}}=-x^{\nu}T^{\mu\rho}$

Differentiating the second equality gives:

$\delta J^{\mu}=\frac{\partial J^{\mu}}{\partial \alpha_{\rho\nu}}=\frac{\partial J^{\mu}}{\partial \alpha_{\nu\rho}}=+x^{\rho}T^{\mu\nu}$

So in total I have

$\delta J^{\mu}=\frac{1}{2}(x^{\rho}T^{\mu\nu}-x^{\nu}T^{\mu\rho})$?

This is different from what I'm required to derive just by a minus sign - but I dont know if this is right mathematically at all.

17. Nov 12, 2014

### Orodruin

Staff Emeritus
So the point is that, due to the anti-symetry, you get the anti-symmetry directly when you differentiate since the components of $\alpha$ are related and you get must get something anti-symmetric out (in the indices of $\alpha$). This you can anti-symmetrize (wrt one of the T indices and the x index), which hopefully is what your lecturer was referring to. The tensor $T^{\mu\nu}$ is not anti-symmetric. In fact, for the KG Lagrangian:
$$T^{\mu\nu} = (\partial^\mu \phi) \frac{\partial \mathcal L}{\partial(\partial_\nu \phi)} - g^{\mu\nu}\mathcal L = (\partial^\mu\phi)(\partial^\nu \phi) - g^{\mu\nu}\mathcal L,$$
which is manifestly symmetric and would disappear on anti-symmetrization.

18. Nov 12, 2014

He might've been but that goes way beyond me...are you saying that I did it wrong...?

I think I've got a solution but it depends on this being true: $\alpha_{\mu\nu}x^{\nu}=-\alpha_{\nu\mu}x^{\mu}$. Basically if I take the transpose of alpha and put in the - sign due to anti-symmetry, do i have to swap the index that's being summed with x? I dont think I should, in which case my solution is invalid and I am officially stumped to no end...could you clarify please?

19. Nov 12, 2014

### Orodruin

Staff Emeritus
This equation cannot be true. On one side you have $\mu$ as a free index and on the other $\nu$ ... I think the most straight-forward way is to simply note that you can pick a transformation such as the one I suggested in post #12. The other is noting that:
$$\frac{\partial \alpha_{\mu\nu}}{\partial \alpha_{\rho\sigma}} = \delta^\rho_\mu \delta^\sigma_\nu - \delta^\sigma_\mu \delta^\rho_\nu$$
for fixed indices, which, naturally, gives exactly the same result.

20. Nov 12, 2014

I just dont see how the index of x changes...can you please explain?

21. Nov 12, 2014

Okay Dr.Orodruin. Here is what I'm doing....I've had enough of this!!!! ARGH

Starting from where you suggested:

$J^{\mu}=-\alpha_{\rho\nu}x^{\nu}T^{\mu\rho}$.

Anti-symmetrising as my lecturer suggested:

$J^{\mu}=-\frac{\alpha_{\rho\nu}}{2}x^{\nu}(T^{\mu\rho}-T^{\mu\nu})$

At this point I have the solution - all I have to do is include the x-factor in the brackets (Yea I'm so funnny - x factor) and set the index appropriately so that the only free index is \mu. So:

$J^{\mu}=-\frac{\alpha_{\rho\nu}}{2}(x^{\nu}T^{\mu\rho}-x^{\rho}T^{\mu\nu})$

Is this reasonable??? I have no idea what else I can do. Your approach makes sense in words but I cant put it on paper for the life of me!

22. Nov 12, 2014

### Orodruin

Staff Emeritus
You are not anti-symmetrizing in this step. The expression you have does not even make sense: You have an expression inside the parentheses where two terms have different indices. This can never occur if you do things correctly. Proper anti-symmetrization would essentially give you the answer.

23. Nov 12, 2014

So do you agree that this is correct: $\alpha_{\mu\nu}T^{\mu\nu}=\frac{1}{2}\alpha_{\mu\nu}(T^{\mu\nu}-T^{\nu\mu})$...? I'm being confused because in the "Hints" section of the question I'm being told to anti-symmetrise exactly like that...

24. Nov 12, 2014

### Orodruin

Staff Emeritus
Yes, but this is not what you have. Your $T$ has an index which is not contracted with an index of $\alpha$.

25. Nov 12, 2014