# Terms in the Yang-Mills Lagrangian

1. Feb 10, 2013

### dm4b

I've been doing some self-study in Peskin and Schroeder and been struggling a bit in Part III.

Right now, I am stuck on the last two terms in 16.6 (Lagrangian for Yang-Mills).

Presumably these come from (-1/4) (F$^{a}_{\mu\nu}$)^2, but I am getting stuck on getting the indices to work out.

Can anybody fill in the details?

Also, I find his reasoning a bit confusing in the paragraph below 16.7 where he justifies the terms in Figure 16.1 for some of the Feynman rules of a Non-abelian gauge theory. Any help here would also be appreciated. Perhaps working an example for the three gauge boson vertex. I am sure I can get the four-gauge vertex once I see the three-gauge clearly.

Thanks for any and all help.

2. Feb 11, 2013

### Einj

The last term in (16.6) is just the product of the two pieces like the last one in (16.2). You just have to relabel the indices a little.

As far as the Feynman rules are concerned, is not easy to understand them in an intuitive way. However, just think that, anytime you get a derivative it correspond to an impulse. You can then saturate any of your "legs" with the A fields in every possibile combination. Every time you saturate a leg that has, for example, an index μ with a field with has index κ this leads to a gμκ and so on. If you try every combination you can get your rule.

However, if you want the whole explicit calculation of the Feynman rules for a non-abelian theory you can take a look to the Muta - "Foundations of Quantum Chromodynamics". It is a really good book and compute all the Feynman rules in the path integral formalism.

3. Feb 11, 2013

### dm4b

I got that much. As I mentioned in the OP, I was unable to get the indices to work out and agree with what they have in the text.

I'd latex up what I had but I threw it out in frustration ;-)

Sorry, I wasn't clear here on what I found confusing. I pretty much understand the various contractions. But, the terms in the figure were once again confusing me. I'll latex up the ones in question and try to clarify more later 2nite when I get home.

4. Feb 12, 2013

### Einj

As far as your first question is concerned, the last term in the Yang-Mills tensor is $gf^{abc}A^b_\mu A^c_\nu$. In the lagrangian you have (apart for some multiplicative factors) $F_{\mu\nu}^a F^{\mu\nu}_a$. Now, if you consider the product of the two last terms like I wrote above, when you do this you obtain:

$$g^2 f^{abc}A^b_\mu A^c_\nu f^{alm}A_l^\mu A_m^\nu$$

Now, since all the indices are contracted, you can relabel them any way you want. In particular, if you make the substitution: a -> e , b -> a, c -> b, l -> c, m ->d, μ ->κ and v->λ, then you get exactly the same expression reported by Peskin.
You can do a completely analogous thing for the three-vertex term.

I hope this was what you were asking.

5. Feb 12, 2013

### dm4b

Hi Einj,

I appreciate you attempt to help, but it's still not coming together for me. There's just a lot more details going on when you try and work out the square of the Field Strength term.

However, I did notice PnS say they are only writing out the non-linear terms.

Also, on page 230 of Zee I seem to be matching up with him a little better as he includes more terms.

Still, I'm not matching up completely with either of them .... ugh.

I'd love to latex up what I have, but I'd be here for hours using the physicsforums way. Guess I need to get an app for that.

6. Feb 13, 2013

### andrien

after multiplication for the third term you get
(∂kAλa-∂λAka)(gfabcAbkAcλ)+(gfabcAbkAcλ)(∂kAλa-∂λAka)
after multiplying relabel the indices in λ and k in second and fourth term and then interchange b and c in the same and note fabc=-facb,all terms will sum up to give 4 times one term and that cancels out with the 1/4 of outside.you get the last second.the last term is not any trick you should be able to do it.

7. Feb 13, 2013

### dm4b

Andrien, I think this might do the trick for me. Yes, I've got the last term good already. I think I have something close to the term you mention above (with is in Zee's book). Perhaps a little relabeling will get a match and cpl more steps you mention and I'm there. I'll try later when I get home again. Thanks.

8. Feb 13, 2013

### dm4b

Yes, this is all working out for me. Well, one thing was different. In your second term, I have the lambda and kappa indices up above on the As in parentheses. Other than the raising and lowering of operators on the As, everything else fell into place.

Thanks for the help.

9. Feb 13, 2013

### dm4b

So now I'm working on the terms for the three gauge boson vertex. I see a clear pattern for writing those terms out, but I don't clearly see the connection to the contractions he talks about below, which leads him to 16.8. Actually, if I understand him correctly, which apparently I'm not, I would have thought 16.8 had a typo.

I think this single page has given me more problems than the prior 15 chapters combined.

10. Feb 14, 2013

### andrien

that does not matter for flat spacetime,neither one needs a metric tensor for raising and lowering indices in flat spacetime.so if you even work with all indices down or up,it does not make difference.For example,lagrangian is written as (Fμv)2 rather than (Fμv)(Fμv) in book,so you can see it does not make difference in flat spacetime.

11. Feb 14, 2013

### andrien

I hope you know how to calculate vertex factor.For three gauge boson vertex,there are three A's in the second last term.You have to substitute plane wave form for those A's.You can see there is no speciality with one line as compared to other in the diagram for three gauge boson vertex.you of course have 3! ways of permuting it i.e. any A can take any values.You just have to count for it.it gives those six terms.(for getting vertex factor you have to multiply lagrangian by i,put the plane wave form,take out normalization constant and factors already taken care by external lines,the rest is vertex factor)

12. Feb 14, 2013

### dm4b

I know it doesn't make a difference in flat spacetime, but if you take a look at eq 16.6 they have mixed indices on the A terms, as if they did have raised indices on one of the field strength terms. In fact if you raise them on one just like you did in your post, it produces the last term in the Lagrangian perfectly. Anyhow, if they did, I just like to understand all the details especially since stuff learned in this chapter is foundational for all the rest that follow it.

Last edited: Feb 14, 2013
13. Feb 14, 2013

### dm4b

Thanks, I understand all that though, that's basically what they said in the book. Once again, I think I am just having problems matching up indices with what they have.

Here's the first problem I am having. They say they contract k with the first A factor, i.e. the derivative. Why do we not have (-ik$_{\mu}$) then? Why do they choose $\nu$?

Last edited: Feb 14, 2013
14. Feb 15, 2013

### andrien

OK,they do differentiate in upper and lower indices while I have not done that.So you can go with the book because when they write (Fμv)2 they mean FμvFμv and not (Fμv)(Fμv) which I have supposedly mean.So you can go with that but it is not a big deal.

15. Feb 15, 2013

### andrien

well,that not easy to do.I will tell you to look in greiner 'gauge theory of weak interaction' page 132 for it.

16. Feb 15, 2013

### dm4b

It's funny you mention that, was thinking about maybe ordering that book. I used his QM book on the side a ways back. Figured maybe his other ones might be good references too. Thanks for the suggestion ;-)

17. Feb 16, 2013

### andrien

the book has calculated three gauge boson factor in a slight different way,the way of peskin and schroeder goes like this
after substituting plane wave form you will get
-igfabc(-ikk)AaλAkbAλc
=-igfabc(-ikv)AaρAvbAρc(just changed some dummy indices)
=-igfabc(-ikv)gμρ AAvbAρc
(a should always be with μ,b with v and c with ρ also I got it in subscripts as compared to superscripts of book but you can do just the same thing to get superscripts ,again it is immaterial then you will get subscripts in A's)
if you contract the first with μ but second and third with ρ and v then you will get
-igfabc(-ikρ)AavAρbAvc
=-igfabc(-ikρ)gμvAAρbAvc
=-igfacb(-ikρ)gμvAAvbAρc
=+igfabc(-ikρ)gμvAAvbAρc
remember a is with μ, b with v, c with ρ.
similarly you get other 4 terms also after contraction with ρ and v in first A .you should work out the four vertex by your own or take the help of greiner book.

Last edited: Feb 16, 2013
18. Feb 19, 2013