# Anyone familiar with Srednicki ch72?

1. Sep 9, 2011

### LAHLH

Hey,

Just wondering if anyone is familiar with this chapter or the subject matter. I'm trying to understand why there are an additional 5 permutations of the the three gluon vertex making 6 terms in all (equation 72.5).

I know you have to label external propagators with all different momenta in all topologically inequiv ways, so I guess permuting the 3 4-momentum r,p,q into the positions gives 3!=6 combos in total...but I'm not sure really what's going on here, what happens to the other indexes and how does one get down to the final line of (72.5)

thanks for any input

2. Sep 9, 2011

### Polyrhythmic

You're right, you sum over permutations of the three sets of indices. You can see why they indices in each set are associated with eachother in Fig. 72.1. You get additional minus signs through resetting the structure constant's indices to abc.

3. Sep 10, 2011

### LAHLH

OK, I understand now how to generate these terms from the original (my mistake was thinking only 4-momenta q,p,r was being swapped not the whole unit of (a,mu,p)<-->(b,nu,q) say). So for example I can generate his terms of (72.5) from performing this en bloc swapping:

The original term is $$(-gf^{abc})(r_{\mu}g_{\nu\rho})$$. A permutation would be $$(a,\mu,p) \leftrightarrow (b,\nu,q)$$ and from this one gets $$(-gf^{bac})(r_{\nu}g_{\mu\rho})$$ which is the same as $$(+gf^{abc})(r_{\nu}g_{\rho\mu})$$....the third term of his (72.5). OK fine.

But I still have a couple of questions.

1) Why aren't some of these permutations topologically equivalent. For example calling our 3 sets of indices 1,2,3. Then the permutations are 123,213,321,132,312,231. But {123,312,231} are all equivalent if you draw the diagrams by the 3-way cyclic symmetry? as are {213,321,132} with each other.

Compare this to say ch17.

2) Secondly, I'm not sure I understand the motivation for why we're permuting the entire set of indices here as oppose to just the momenta. I see the justification for the momenta permutations as set out by Srednicki at the beginning of ch10, because you need to hit with functional derivatives to get correlation functions essentially, and these func derivatives act in all possible ways.

Any help massively appreciated. thanks again

4. Sep 12, 2011

### LAHLH

Anyone know?

5. Sep 12, 2011

### Avodyne

They're not algebraically equivalent, so they all have to be included explicitly.

Let's take a simpler example. Suppose we have 3 real scalar fields $\varphi_i$, $i=1,2,3$, and the interaction term is $g\varphi_1\varphi_2\varphi_3$. We might naively say that the vertex factor for three lines carrying indices $i,j,k$ is $-ig\delta_{i1}\delta_{j2}\delta_{k3}$. But this is wrong: it's actually the sum of six terms like this, will all possible permutations of $i,j,k$.

The functional derivatives carry the same indices as the field, so the same argument applies, now to momentum+index instead of just momentum.

6. Sep 20, 2011

### LAHLH

Thanks for the reply Avodyne. Sorry that I couldn't respond sooner as I've been away.

hmm so in ch10 where we have the three groups of 8 tree diagrams (p75, fig 10.1), taking one of these three there are 7 others "topologically equivalent" but crucially also "algebraically equivalent": for example the first diagram which I will denote as 12..1'2' is both topologically equivalent to 21..1'2' (because you simply are exchanging incoming particles which is doing nothing) and algebraically equiv because the term you obtain in both cases would be $$\Delta(x_1-y)\Delta(x_2-y)\Delta(x^{'}_{1}-z)\Delta(x^{'}_{2}-z)$$

whereas in ch72 swapping any of the sets of indices/momenta leads to a different algebraic term so diagrams that look topologically equivalent (can be easily rotated into each other, e.g. diagrams 123 and 312 by 120 degree rotation) contribute differently algebraically.

I mean yes, they do contribute differently algebraically, but this seems to me to be only because we said so via this permutation argument? I can't quite see why we want to do this still, it seems in contrast to what was done in ch17...

Why though? I don't understand why in ch16 this doesnt happen....you start with the diagram that looks like Fig 16.1 but with barbells on the end for sources, then by the argument at start of ch10 to get the vertex function you need to apply functional derivatives in all possible ways, so naively you should start with 123,213,321,132,312,231 diagrams, but then I'd expect to seperate these into two groups of three based on topological equivalence (members of each group rotate into each other)

Why doesnt he do this there? why does he do it ch17 for the 4-valent vertex?

7. Sep 20, 2011

### LAHLH

A related question which may help shed some light is: why in (72.6) for the four-gluon vertex does he only consider 6 permutations, why not the full 4!=24? I had assumed earlier cyclic symmetry (i.e. you get 6 groups each containing 4 topologically equivalent members e.g. 1234, 4123,3412, 2341 is such a group) but by the logic of the previous posts such groups would have members that are not algebraically equivalent so one couldn't lump them together like this.

If this is not the argument then what is? and if cyclic symmetry is the argument in 72.6 why doesnt it also apply in 72.5?

8. Sep 21, 2011

### Avodyne

Take the 1st line of 72.6 (for the 4-gluon vertex factor), and write out all 24 permutations. You will find that, after using the symmetry properties of $f^{abc}$ and $g_{\mu\nu}$, you have 4 copies of each of the 6 terms on the final right-hand-side of 72.6. The 4 copies of each term provide a factor of 4 that cancels the factor of 1/4 in the lagrangian (last term of 72.3).

On the other hand, take the 1st line of 72.5 (for the 3-gluon vertex factor), and write out all 6 permutations. You will find that no two are equivalent, and so all must be written down explicitly. There is no extra numerical factor, but also no factor in the lagrangian (next-to-last term in 72.3).

9. Sep 22, 2011

### LAHLH

OK, yes I see what you mean about my second question, and that makes sense, but I'm still wondering about my first question, which I guess I should make a little clearer really.

Taking your example with interaction term $$g\phi_1\phi_2\phi_3$$ and going back to basics. So in ch9 one learns that the path integral is going to look like:

$$Z(J_1,J_2,J_3) \propto exp\left[ig\int\,\mathrm{d}^4x\, \left(\frac{1}{i}\frac{\delta}{\delta J_1(x)}\right)\left(\frac{1}{i}\frac{\delta}{\delta J_2(x)}\right) \left(\frac{1}{i}\frac{\delta}{\delta J_3(x)}\right) \right] Z_0(J_1,J_2,J_3)$$

where now we have three fields distinct so need three sources.

OK what is the vertex factor here?

I'm struggling to see exactly why it would involve the delta functions you mentioned, why not simply $$ig\int\,\mathrm{d}^4x\$$ ? As long as the three different types of propagator corresponding to our three fields each end on 'x' like $$\Delta_1 (x-x') ,\Delta_2 (x-x''), \Delta_3 (x-x''')$$ then what is the issue?

Anyway you could spell this out for me as for some reason it's not sinking in...

10. Sep 26, 2011

### Avodyne

These prescriptions give the same result. That is, requiring 3 different lines to meet at a vertex with vertex factor $ig$ is the same as taking the indices of the 3 lines to be $i,j,k$ with vertex factor $ig(\delta_{i1}\delta_{j2}\delta_{k3}+{}$5 permutations$)$.

The advantage of the latter prescription is that it generalizes more easily. Consider an interaction term $\frac{1}{6}g d_{ijk}\varphi_i\varphi_j\varphi_k$, where the indices are summed, and $d_{ijk}$ is a completely symmetric tensor. The vertex factor is then $ig d_{ijk}$. Note that $d_{ijk}=\delta_{i1}\delta_{j2}\delta_{k3}+{}$5 permutations is the previous example.

Last edited: Sep 26, 2011