I did it for you

It is very easy to show, from Yang-Mills equations, how to obtain a scalar field equation through the Smilga’s choice. Let us write down Yang-Mills equations

$\partial^\mu\partial_\mu A^a_\nu-\left(1-\frac{1}{\alpha}\right)\partial_\nu(\partial^\mu A^a_\mu)+gf^{abc}A^{b\mu}(\partial_\mu A^c_\nu-\partial_\nu A^c_\mu)+gf^{abc}\partial^\mu(A^b_\mu A^c_\nu)+g^2f^{abc}f^{cde}A^{b\mu}A^d_\mu A^e_\nu = 0$

using the choice $A_1^1=A_2^2=A_3^3=\phi$ . This is really a great simplification. Smilga, in his book, already checked this for us but we give here the full computation. From above eqautions, the only critical term is the following

$f^{abc}A^{b\mu}(\partial_\mu A^c_\nu-\partial_\nu A^c_\mu)$

as this term would produce terms deviating from the known form of the scalar theory. For SU(2) we have $f^{abc}=\epsilon^{abc}$ the fully-antisymmetric Levi-Civita tensor. This means that we will have

$\epsilon^{a1c}A^{11}(\partial_1A_\nu^c-\partial_\nu A_1^c)+$

$\epsilon^{a2c}A^{22}(\partial_2A_\nu^c-\partial_\nu A_2^c)+$

$\epsilon^{a3c}A^{33}(\partial_3A_\nu^c-\partial_\nu A_3^c).$

Where we have used largely Smilga’s choice. Now do the following. Take the following components to evolve $\nu=1$ $a=1$ , $\nu=2$ $a=2$ and $\nu=3$ , $a=3$ . It easy to see that the possible harmful term is zero with the Smilgaì’s choice. Now, for the cubic term you should use the useful relation

$\epsilon^{abc}\epsilon^{cde}=\delta_{ad}\delta_{be}-\delta_{ae}\delta_{bd}$

and you will get back the quartic term.

The gauge fixing term can be easily disposed of through a rescaling of spatial variables while the kinematic term gives the right contribution. You will get three identical equations for the scalar field.

Of course, Smilga in his book already did this and I repeated his computations after the Editor of PLB asked for a revision having the referee already put out this problem. The Editorial work was done very well and two referees read the paper emphasizing errors where they were.

Finally, Tao’s critcism does not apply as I said. This does not mean that what he says is wrong. This means that does not apply to my case.

Update: As the question of the gauge fixing term appears so relevant, let me fix it once and for all. Firstly, I would like to point out that these solutions belong to a class of solutions in the Maximal Abelian Gauge (MAG). But let us forget about this and consider the question of gauge fixing. This term is arbitrarily introduced in the Lagrangian of the field in order to fix the gauge when a quantization procedure is applied. Due to gauge invariance and the fact that becomes an exact differential after partial integration, it useful to have it there for the above aims. The form that it takes is

$\frac{1}{\alpha}(\partial A)^2$

and is put directly into the Lagrangian. How does this term become with the Smilga’s choice? One has

$\frac{1}{\alpha}\sum_{i=1}^3(\partial_iA_i^i)^2$

and the final effect is a pure rescaling into the space variables of the scalar field. In this way the argument is made consistent. One cannot take the other way around for the very nature of this term and claiming the result is wrong.

This particular class of solutions belongs to the subgroup of SU(N) given by the direct product of U(1). This is a property of MAG and all the matter is really consistent and works.

Finally, I invite people commenting this and other posts to limit herself to polite responses and in the realm of scientific discussion. Of course, doing something wrong happens and happened to anyone working in a scientifc endeavour for the simple reason that she is really doing things. People that only do useless criticisms boiling down to personal offenses are kindly invited to refrain from further interventions.

Update 2: I will get a paper published about this matter. Please, check here.

This entry was posted on Wednesday, March 11th, 2009 at 1:11 pm and is filed under mathematics, Physics, QCD. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.

19 Responses to I did it for you

Luboš Motl says:

11/03/2009 at 1:20 pm

The gauge fixing term cannot be “disposed of” in any way (certainly not by the nonsensical words that follow your verb “dispose”). It is exactly the failure of your configuration to solve the Yang-Mills equations.

The term is proportional to things like partial_nu partial_a phi, so it is not delta(nu,a) times the scalar quartic equation, which kills your “solution” and closes this debate for me.

You can’t be throwing away gauge-fixing terms exactly because your Ansatz doesn’t satisfy the Lorentz gauge. The correct “gauge” is alpha=infinity, the gauge-fixing term is as finite and comparable to the other “correct” terms as you can get, but it has a wrong tensor structure and kills all the remaining hopes that you have found new Yang-Mills solutions.

I just started my stopwatch to measure how much time you need to understand this trivial point. An average undergrad exposed to physics would need 2 hours.

Best wishes
Lubos

Reply
mfrasca says:

11/03/2009 at 1:26 pm

Let us write the gauge fixing term as

$\partial_\nu\partial^\mu A_\mu^a$

I will do the computation only for $\nu=1$ and $a=1$ being always the same. This will give you, remembering that you only have $A^1_1=A_2^2=A_3^3\ne 0$ ,

$\partial_1\partial^1A_1^1$

that should be added to the kinematic term. You have just to rescale spatial variables to dispose of it.

Sorry, Lubos. It works and works very well.

Marco

Reply
mfrasca says:

11/03/2009 at 1:59 pm

Here is the Smilga’s equations:

http://arxiv.org/abs/hep-ph/9901412v2

Look at page 8. These lectures were improved and become a book:

http://www.amazon.com/Lectures-Quantum-Chromodynamics-V-Smilga/dp/9810243316/ref=pd_bbs_sr_1?ie=UTF8&s=books&qid=1236772757&sr=8-1

Marco

Reply
unit says:

11/03/2009 at 2:16 pm

As far as I understand at page 8 the author let all the components vanish except one, look at the first line after eqn 1.29. Also, I don’t see how your “rescaling” (for the 2 factor?) can work with mixed partial derivatives, see for example a=1, nu=2. Can you do the full computation for this component?

Reply
Luboš Motl says:

11/03/2009 at 2:21 pm

Dear Marco,

all values of “nu, a” are not the same thing as “1, 1”. For example, for “nu, a” equal to “1, 2” :-), the second choice in an alphabetical sorting that you may have thought about if you worked just a little bit harder, all other terms in the equation are zero, being proportional to delta(nu,a)

However, the gauge-fixing term is proportional do “d_1 d_2 phi” which is not equal to zero and the equation fails. Not all 3×3 tensors are proportional to the identity matrix, believe me or not.

By the way, you wouldn’t solve the equation even for a=nu=1 because the required “rescaling of space” needed to make various terms work would differ for different terms but the “a not equal to nu” terms make the story much clearer.

Incidentally, only if the Lorentz gauge is satisfied, you could use the alpha=1 (Feynman) gauge and drop the gauge fixing term. In that case, the system would be indeed similar to a system of scalars. That shouldn’t be a news for you because for the Feynman gauge, the electromagnetic action is d_m A_n d^m A^n and resembles a system of course scalar fields, a feature that the interactions kind of keep.

However, if you impose both Lorentz gauge and your Ansatz, A_m^a = delta_m^a phi, then the Lorentz gauge tells you that the derivatives of phi vanish, meaning that phi is constant, and the only solution to Yang-Mills or scalar equations in this class is phi=0, as mentioned previously: it’s the only Yang-Mills solution that can be obtained by your method.

One hour is gone from your time. You’re already slower than 40% of undergraduate physics students and you already needed additional help. If you’re gonna need more than 3 hours, I will delist you from my blogroll and include among average crackpots next to Woits and similar crap.

Believe me that an overall respectability of basic physics skills and integrity is more important than the attempt to defend a particular indefensible solution.

All the best
Lubos

Reply
mfrasca says:

11/03/2009 at 2:25 pm

Dear unit,

Are you the same that writes in Italian newsgroups fighting against crackpots that live there?

After eq.(1.29) you will find the following ansatz

$A_i^a=\delta_{ia}A$

that is the one I did.

The component a=1 and nu=2 does not exists because the equations of motion can only have the values I put. In order to have in the equations of motion a=1 nu=2 for the gauge fixing term such a component should be different from zero.

Marco

Reply
Luboš Motl says:

11/03/2009 at 2:35 pm

Dear Marco,

if the component of the current j_1^2 doesn’t “exist”, then it is not a source of a gauge field. And even the fact that it is not a gauge theory doesn’t change the fact that the equation is violated. You know, in physics, you can’t obey equations with just a subset of terms for a subset of equations you choose, do you know?

I am delisting you already because I don’t have patience to ever stupider discussions.

Best
Lubos

Reply
Luboš Motl says:

11/03/2009 at 2:48 pm

“In order to have in the equations of motion a=1 nu=2 for the gauge fixing term such a component should be different from zero.”

This statement is fully equivalent to the statement that an apple can be sitting in the middle of the room (the origin of my coordinates), without falling down by the force of gravity, because coordinates x=y=z=0 that are equal to zero don’t have to obey the equations of motion.

Are you really THAT stupid, Marco?

Reply
unit says:

11/03/2009 at 4:08 pm

Dear Marco,

yes, probably I am the one you’re referring to, or at least I was when I was younger 🙂 . Anyway I’m just trying to understand, I can be wrong here.

I do not fully understand your “rescaling”, but I didn’t the calculation and I can’t see as easily as lubos if there are terms that behave differently. But the “off-diagonal affair” is more serious for me, honestly I can’t see how can you avoid to consider that set of equations. In Smilga’s book the trick works because the “field” doesn’t depend on x, so the equations are all the same, I understand correctly?
You’re right about the ansatz and I was wrong. But your solution \phi depend also on x.

thanks for your patience.

Bye

Reply
mfrasca says:

11/03/2009 at 4:38 pm

Dear unit,

I remember some exchange between us. I have written on Italian newsgroups for some years then I used avatars.

About Lubos, for him is always hunting time. He looks for crackpots here and there. It does not matter if this people are active researchers with far larger list of publications than him. He left academia and is still convinced that people at MIT or Harvard look at his blog to decide a career. Not so smart, indeed.

Of course, you are right about this point as is also Terry Tao. I have made an update to this post explaining the point. But the question is that if you take just three components of the field you cannot pretend that your arbitrary gauge fixing term should contain all others that you considered to be zero. So, this term should be properly defined.

The reason is that this class of solutions belong to a subgroup of SU(N) and enters into the Maximal Abelian Gauge. You should take your gauge fixing term properly to be consistent.

Marco

Reply
unit says:

11/03/2009 at 5:45 pm

Hi Marco,

Unfortunately I’m not anough skilled to follow your explanation. If it’s all about the quantization step, and somehow the theorem still works at that stage, I just believe you. Otherwise if your explanation it’s at the classical level honestly I’m still unconvinced. Whatever name you put on that term and that component, is still there and you have to satisfy that equation for the solution to be valid. For example you can cook up a toy (wrong) solution using similar argument in electrodynamics, obtaining static waves which obviously aren’t solution to the (full) set of Maxwell equations. It’s not exactly the same thing, because of course the U(1) representation is 1-dimensional and so you don’t have a “gauge-index”, but in my opinion the point remains. Maybe I’ll try to reproduce your line of thought with a different ansatz, trying to obtain a manifestly wrong solution for the SU(2) case.

Clearly, there is still the possibility that I’m wrong.

bye and thanks for the useful discussion.

Reply
mfrasca says:

11/03/2009 at 6:03 pm

Hi unit,

It is clear that there is a consistency problem here with the introduction of the gauge fixing term. This term must be consistently introduced. If you work with a Lagrangian you are done as you get the right gauge fixing term and this was the main motivation to use functionals to work out the mapping.

You are welcome.

Cheers,

Marco

Reply
carlbrannen says:

11/03/2009 at 11:45 pm

I am having great difficulty understanding why there is any argument over this. The question at hand is “does a classical [Yang-Mills] wave equation possess a solution or not?” At least this is what Marco is claiming.

One could ask a similar question about Maxwell’s equations. When one writes down something that is supposed to be a solution of these equations one considers only the terms which are non zero in the solution. It’s just not a very difficult problem; it does not require the subtlety that Lubos and Terry are throwing at it. This are considered as classical equations of motion only. This is very explicit in Smilga’s article. (And thanks Marco, for linking the arXiv article, it saves me the money for the book.)

Reply
Michael Dinsdale says:

12/03/2009 at 1:02 am

hi Marco

Thanks for taking the time to go through this. But I’m afraid I’m still not convinced that these solutions exist 🙂

I think my problem is more or less the same one other commenters have picked up on – you seem to only consider the equation of motion with a=nu, but it really needs to hold for all a,nu. Both the gauge-fixing term and the term with one derivative seem to me to give “off-diagonal” contributions which impose a whole heap of other restrictions on the allowed values of phi (like the a=1,v=2 example Terry Tao brought up in the Wikipedia discussion).

I’d guess that the only way to satisfy all these equations simultaneously would be to take phi independent of \vec{x} (as Smilga does) – though I could be wrong about this. In any case, though, merely solving the cubic EOM for phi seems not to be enough to guarantee that the corresponding A solves the YM EOM. (Unfortunately!)

Do you think that’s right, or am I missing something?

michael

Reply
mfrasca says:

12/03/2009 at 10:04 am

Hi Michael,

Thank you for taking time to look into my work.

The term with derivative does not give contribution when you take all equal components. Indeed, if you fix the gauge to the Lorenz gauge, it is easily seen that that equations of motions give consistently zero. On the other side, if you work with the functional and take only $A_i^i$ with i=1,2,3, the gauge fixing term takes a consistent form $\sum_i (\partial_iA_i^i)^2$ and the equations of motions you obtain in this way are consistent too. But in this latter case you incur in the problem of the extremum pointed out by Terry.

There is a curious point here. For the argument given in my paper http://arxiv.org/abs/0709.2042 all is needed is exactly Smilga solutions without spatial dependence as all I did in the following on the scalar field is a gradient expansion. The kind of generality I have pretended with the mapping theorem is not needed for my derivation of the spectrum of the Yang-Mills equations. But of course, this is all the point as I have not showed this fully and this is something I must do as soon as I will have some spare time.

Marco

Reply
unit says:

12/03/2009 at 12:04 pm

Dear Marco
I’ve spent some time reading your paper and I’m more and more convinced that there is a serious problem in it. I’ve already espressed my concerns about theorem 1 and as long as I undestand correctly you fully need it (and probably more, actually) to reach your results. For example you say “all is needed is exactly Smilga solutions without spatial dependence as all I did in the following on the scalar field is a gradient expansion”. Since the solution in Smilga’s book is constant throughout space, how can you use it for a gradient expansion ?

Reply
unit says:

12/03/2009 at 4:07 pm

Hi Marco,

How can u do a gradient expansion on a function (your phi) which is constant in space, i.e. doesn’t depend on x?

Reply
anonymous says:

12/03/2009 at 8:23 pm

@carlbrannen

You can only neglect the equations for the zero components if it is clear that zero is a consistent solution of their equations of motion. In some cases this is obvious, e.g. if the equations decouple, or if you have some symmetry (e.g. a gauge symmetry) that allows you to set these components to zero at all times.

Apparently this is not so obvious in the case discussed here. M. Frasca claims that with a suitable change of variables the equations for the zero components decouple. Since I don’t have the time to check his arguments thoroughly, I don’t know whether this is true.

Cheers.

Reply
mfrasca says:

12/03/2009 at 8:33 pm

Hi unit,

I am currently writing a paper to fix this question as Terry asked to me. I think this is the only serious procedure to correct the problem. I hope to submit it on arxiv and under review in a few days. I will give a notice in the blog with a proper link.

Thank you all, folks!

Marco

Reply