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

[Dixanadu]

Homework Statement

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.

Homework 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}

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)

 

[Orodruin]

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).

 

[Dixanadu]

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 can't get to the end for some reason) and then I have to prove that the charge is 0.

I don't see which index I can change in my equation though...can you help me out please?

 

[Dixanadu]

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?

 

[nrqed]

Homework Statement

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}

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.

 

[Orodruin]

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.

 

[Dixanadu]

I haven't 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

 

[Orodruin]

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 ...

 

[Dixanadu]

Yes I know \alpha_{\mu\nu}=-\alpha_{\nu\mu} but I don't know what you mean by the derivative being defined with respect to alpha. Also is my expression for \delta\phi even correct...?

 

[Orodruin]

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).

 

[Dixanadu]

I still don't 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?

 

[Orodruin]

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

 

[Dixanadu]

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 don't know how to move forward...so I am 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 don't really know what I am doing as far as the differentiation you suggested is concerned.

 

[Orodruin]

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 don't know how to move forward...so I am 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 don't really know what I am doing as far as the differentiation you suggested is concerned.

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.

 

[Dixanadu]

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 don't even know how to handle the alpha's indices in the denominator. I guess I don't 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.

 

[Dixanadu]

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 don't know if this is right mathematically at all.

 

[Orodruin]

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.

 

[Dixanadu]

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 don't think I should, in which case my solution is invalid and I am officially stumped to no end...could you clarify please?

 

[Orodruin]

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 don't think I should, in which case my solution is invalid and I am officially stumped to no end...could you clarify please?

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.

 

[Dixanadu]

I just don't see how the index of x changes...can you please explain?

 

[Dixanadu]

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 can't put it on paper for the life of me!

 

[Orodruin]

Anti-symmetrising as my lecturer suggested:

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

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.

 

[Dixanadu]

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...

 

[Orodruin]

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

 

[Dixanadu]

Mhmm okay. So how will I ever end up in a situation where my alpha has the same indices as T? I'm so, so annoyed jeez!

 

[Orodruin]

You will and should not. I do not believe that your lecturer told you to anti-symmetrize $T$, but $\alpha$, simply because $\alpha$ is anti-symmetric, while $T$ is not.

To be quite honest, if you are not 100 % sure about how indices may be renamed or raised/lowered, how they are summed over, when and how you can symmetries or anti-symmetrize, you really should review this before attempting problems like this one. You cannot do things simply because you believe that your lecturer has told you to do a particular thing without understanding whether what you believe he told you is a valid operation or not.

 

[Dixanadu]

Im fairly okay with basic indices stuff but this problem is probably the first one I've ever done that's so complex in that sense.

I guess I should try anti-symmetrising alpha then. Is there any hint you can give me with that...? please dig deep in your mind and try to find the easiest possible approach :(

EDIT: I guess I should be less vague and tell you what the root of my problem is. I cannot apply the anti-symmetrisation to alpha either because there is no tensor I can multiply it with where both indices are being summed. Same problem I had with T.

 

[Orodruin]

So the very very basic thing about symmetrizing or anti-symmetrizing is the following:

Take a rank-2 tensor $A_{ij}$ (it does not even have to be a tensor, it suffices to be an object with 2 indices - also I am going to use roman indices since I can write them faster). The following operations are then trivially allowed (just by adding and subtracting the same quantity and then using basic addition/subtraction):
$$
A_{ij} = \frac 12(A_{ij}+A_{ij}) + \frac 12 (A_{ji} - A_{ji}) = \frac 12 (A_{ij}+A_{ji}) + \frac 12 (A_{ij} - A_{ji}).
$$
Here, the first term is the symmetric part of $A$ and the second the anti-symmetric part. Now, if $A$ is anti-symmetric, then $A_{ij} = - A_{ji}$ and the first term vanishes. Thus, for an anti-symmetric object:
$$
A_{ij} = \frac 12 (A_{ij} - A_{ji}).
$$
This is a replacement you can legally do if (and only if) $A_{ij}$ is anti-symmetric.

In general, $A$ could also carry other indices that would be carried along in the entire operation, the main part is the anti-symmetrization over the anti-symmetric indices. You can also do a similar argumentation for symmetric objects, in which case you would keep the first term instead of the second.

 

[Dixanadu]

Thank you. I understand the antisymmetrisation a lot better now, I just don't know how to apply it given what you told me in post #22. I'm never gona have both indices of alpha being summed anywhere...so I don't know why I'm being asked to antisymmetrise

 

[Orodruin]

I'm never gona have both indices of alpha being summed anywhere...so I don't know why I'm being asked to antisymmetrise

Well, first of all:

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

Both indices of $\alpha$ in the above expression are summation indices. Second, I did not say anything about a sum over the indices of the anti-symmetrized object. There is no need for the indices to be summation indices in order to make that replacement - the only requirement is that the object itself is anti-symmetric.

In order to get all the way, you will also need to rename some summation indices, so do you understand why $A_i B^i$ and $A_k B^k$ represent exactly the same quantity?

 

[Dixanadu]

Yes. 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 :D yay i know something!

 

[Orodruin]

So what happens when you replace $\alpha$ in the expression quoter in #30 with its anti-symmetrized expression?

 

[Dixanadu]

I end up with J^{\mu}=-\frac{1}{2}(\alpha_{\rho\nu}-\alpha_{\nu\rho})x^{\nu}T^{\mu\rho} I think

 

[Orodruin]

Yes, so given what I just told you about renaming indices ... Do you not feel that there are some indices that just scream for a change of name in one of the terms?

 

[Dixanadu]

Well, putting everything inside the brackets gives me:

J^{\mu}=-\frac{1}{2}(\alpha_{\rho\nu}x^{\nu}T^{\mu\rho}-\alpha_{\nu\rho}x^{\nu}T^{\mu\rho}). So I am guessing it's the second term but there is a pair of summed indices there. The only change I can see that would get me to my answer is if i let rho -> nu and vice versa. But then I'll be picking up a minus sign cos alpha is antisymmetric? Or is it legal for me to make that swap without doing anything to the signs? I hope that's what you mean

 

[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).