Conservation of Noether current

[Dixanadu]

Homework Statement

Hey guys. So I gota prove that the currents given by

[itex]M^{\mu;\nu\rho}=x^{\nu}T^{\mu\rho}-x^{\rho}T^{\mu\nu}[/itex]

is conserved. That is:

[itex]\partial_{\mu}M^{\mu;\nu\rho}=0.[/itex]

Homework Equations

Not given in the question but I'm pretty sure that

[itex]T^{\mu\nu}=\frac{\partial \mathcal{L}}{\partial(\partial_{\mu}\phi)}\partial^{\nu}\phi-\mathcal{L}g^{\mu\nu}[/itex]

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

[itex]\mathcal{L}=\frac{1}{2}(\partial_{\mu}\phi)(\partial^{\mu}\phi)-\frac{m}{2}\phi^{2}[/itex]

The Attempt at a Solution

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

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

Doing the same thing to [itex]T^{\mu\nu}[/itex] gives

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

Now putting it together gives

[itex]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})[/itex]

Now I have to hit this with [itex]\partial_{\mu}[/itex]. So i get:[itex]\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})[/itex]

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

 

[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 [itex]\delta x^{\mu}=\alpha^{\mu}_{\nu}x^{\nu}.[/itex] What would be the corresponding transformation to the field [itex]\delta \phi[/itex]?

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

[itex]\delta\phi(x)=\phi'(x')-\phi(x)-(\delta_{\mu}\phi)\delta x^{\mu}[/itex]

So we get [itex]\delta\phi=-(\partial_{\mu}\phi)\alpha^{\mu}_{\nu}x^{\nu}[/itex]

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

 

[nrqed]

Homework Statement

The Attempt at a Solution

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

[itex]T^{\mu\rho}=(\partial^{\mu}\phi)(\partial^{\rho}\phi)-\mathcal{L}g^{\mu\rho}[/itex]
[itex]T^{\mu\rho}=(\partial^{\mu}\phi)(\partial_{\mu}\phi)g^{\mu\rho}-\mathcal{L}g^{\mu\rho}[/itex]

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 [itex]\delta x^{\mu} = \alpha^{\mu}_{\nu}x^{\nu}[/itex]. 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 [itex]\alpha_{\mu\nu}=-\alpha_{\nu\mu}[/itex] but I don't know what you mean by the derivative being defined with respect to alpha. Also is my expression for [itex]\delta\phi[/itex] 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:
[itex]\delta J^{\mu}=-T^{\mu\rho}\delta x_{\rho}[/itex] is my general current.

Plug in [itex]\delta x_{\rho}=\alpha^{\nu}_{\rho}x_{\nu}[/itex] to get

[itex]\delta J^{\mu}=-T^{\mu\rho}\alpha^{\nu}_{\rho}x_{\nu}[/itex].

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

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

Secondly, is this statement now correct:

[itex] J^{\mu}=-\frac{1}{2}\alpha_{\rho\nu}x^{\nu}(T^{\mu\rho} - T^{\mu\nu})[/itex] ? 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 [itex]\alpha_{\nu\rho}=\frac{1}{2}(\alpha_{\nu\rho}-\alpha_{\rho\nu})[/itex] 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

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

Secondly, is this statement now correct:

[itex] J^{\mu}=-\frac{1}{2}\alpha_{\rho\nu}x^{\nu}(T^{\mu\rho} - T^{\mu\nu})[/itex] ? 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 [itex]\alpha_{\nu\rho}=\frac{1}{2}(\alpha_{\nu\rho}-\alpha_{\rho\nu})[/itex] 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?
[itex]\frac{J^{\mu}}{\alpha_{\mu\nu}}[/itex]? But you're saying that due to the antisymmetry of alpha, I have two possibilities - the one I just stated and [itex]-\frac{J^{\mu}}{\alpha_{\nu\mu}}[/itex]?

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:

[itex]J^{\mu}=-\alpha_{\rho\nu}x^{\nu}T^{\mu\rho} = + \alpha_{\nu\rho}x^{\rho}T^{\mu\nu}[/itex]

Differentiating the first equality gives:

[itex]\delta J^{\mu}=\frac{\partial J^{\mu}}{\partial \alpha_{\rho\nu}}=-x^{\nu}T^{\mu\rho}[/itex]

Differentiating the second equality gives:

[itex]\delta J^{\mu}=\frac{\partial J^{\mu}}{\partial \alpha_{\rho\nu}}=\frac{\partial J^{\mu}}{\partial \alpha_{\nu\rho}}=+x^{\rho}T^{\mu\nu}[/itex]

So in total I have

[itex]\delta J^{\mu}=\frac{1}{2}(x^{\rho}T^{\mu\nu}-x^{\nu}T^{\mu\rho})[/itex]?

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: [itex]\alpha_{\mu\nu}x^{\nu}=-\alpha_{\nu\mu}x^{\mu}[/itex]. 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: [itex]\alpha_{\mu\nu}x^{\nu}=-\alpha_{\nu\mu}x^{\mu}[/itex]. 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:

[itex]J^{\mu}=-\alpha_{\rho\nu}x^{\nu}T^{\mu\rho}[/itex].

Anti-symmetrising as my lecturer suggested:

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

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:

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

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:

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

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: [itex]\alpha_{\mu\nu}T^{\mu\nu}=\frac{1}{2}\alpha_{\mu\nu}(T^{\mu\nu}-T^{\nu\mu})[/itex]...? 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:

[itex]J^{\mu}=-\alpha_{\rho\nu}x^{\nu}T^{\mu\rho}[/itex].

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 [itex]J^{\mu}=-\frac{1}{2}(\alpha_{\rho\nu}-\alpha_{\nu\rho})x^{\nu}T^{\mu\rho}[/itex] 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:

[itex]J^{\mu}=-\frac{1}{2}(\alpha_{\rho\nu}x^{\nu}T^{\mu\rho}-\alpha_{\nu\rho}x^{\nu}T^{\mu\rho})[/itex]. 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