# Mandl & Shaw page 361

1. Aug 3, 2013

### Vic Sandler

In the second edition of Mandl and Shaw QFT, on page 361, it's fairly obvious that there is a sign discrepancy between eqn (15.125) and eqn (15.127). In eqn (15.125), there are 5 imaginary i's and 4 explicit minus signs for a final product of +i. However, the first equal sign in eqn (15.129) gives eqn (15.127) an explicit minus sign, i.e. a final product of -i. Eqn (15.125) is created from fig (15.17b) and seems to be correct, but I am not completely sure. Can someone confirm that eqn (15.125) is an accurate application of the Feynman rules to Fig (15.17b)?

2. Aug 5, 2013

### andrien

There is a little mistake in 15.125,the structure constant finm should read fimn.Rest just follow(verify it)

3. Aug 5, 2013

### Vic Sandler

I don't understand your suggestion. In figure 15.17(b) on page 351, I think there is a typo. The leftmost gluon is labeled $p - k, n, \nu$ but in eqn (15.125), there is a factor $T^m_{ad}\gamma_{\nu}$. This suggests to me that the label on the leftmost gluon should be $p - k, m, \nu$. Am I correct? If not, where does the m in eqn (15.125) come from?

In figure 14.4(b) on page 315, the subscripts on f are in the same order as the superscripts on V. That is, lmn follows the figure in clockwise order, and so does $\sigma\tau\nu$. If I make the change from n to m in figure 15.17(b) as I suggested above, and change $f_{inm}$ to $f_{imn}$ as you suggested in your post, then the indices on f will follow the figure in counterclockwise order while the indices on V follow it in clockwise order. I don't think this can be.

Can you explain why you think the order of inm is wrong by means of the application of the Feynman rules to figure 15.17(b)?

PS: The application of rule 4 on page 319 suggests to me that the loop factors should be in clockwise order, but I'm not sure about it.

Last edited: Aug 5, 2013
4. Aug 6, 2013

### andrien

yes,you are correct on this.
Take a look carefully at 14.68 and you will see it.Also when he writes 15.129 ,he says that he is using 15.66.Why do you think it to be the case.

5. Aug 6, 2013

### Vic Sandler

Note that there appears to be a typo in eqn (14.67) on page 317. The final factor of $D_F^{\tau\gamma}$ should be $D_F^{\nu\gamma}$

You make a good point. However, in the case of eqn (14.68) on page 317, comparing it to fig 14.5 on the same page, you will see that the subscripts on f, namely lmn are in counterclockwise order. The subscripts on V namely $\sigma\tau\nu$ which are tied in eqn (14.67) to $\alpha\beta\gamma$ are also in counterclockwise order. So in both figure 14.4(b) on page 315, and in figure 14.5 on page 317 we find that the subscripts on f should follow the same order as the subscripts on V. In figure 14.4(b), the order is clockwise, and in figure 14.5 it is counterclockwise, but my point is that in each case, the order of f matches the order of V. Now I have two questions.

1. If the order of subscripts of f in eqn (15.125) should be changed from clockwise to counterclockwise, then shouldn't the order of subscripts of V also be similarly changed?

2. In any given Feynman diagram, how do you determine whether the order should be clockwise or counterclockwise?

6. Aug 7, 2013

### andrien

why do you care about this anticlockwise and clockwise.The order in which three gauge boson vetrex is written does not matter.You don't even need to use such unnecessary thing.If in the diagram all arrows are pointing inward(K1+K2+K3=0) then take the factor flmn or any such and go in the sense you counted it(no need to write Vστν type thing).If l is with μ(k1),m with v(k2),n with σ(k3) then you can write simply gflmn(gμv(k1-k2)σ+g(k2-k3)μ+gσμ(k3-k1)v.You can see clearly the way it is written.It is correct.Follow this and you will never do wrong.(Beware M&S has defined covariant derivative a little different,so be cautious if you read other books such as Peskin and Schroeder).The problem you are trying to handle is treated in chapter 16 of P&S.

Last edited: Aug 7, 2013
7. Aug 7, 2013

### Vic Sandler

In one sense, I don't care about it. Since both f and V are totally anti-symmetric in their indices, it doesn't matter whether those indices are arranged clockwise or counterclockwise with respect to the Feynman diagram. The Feynman rule for Fig 14.4(b) has both clockwise and the rule in eqn (14.68) has both counterclockwise and I do not consider this to be a problem.

However, there is a sense in which I do care about it. Namely if you use the counterclockwise orientation for f and clockwise for V. That is what you are suggesting in order to fix eqn (15.125). Of course this fixes the minus sign problem, but why is it alright to do it? When must the the sets of indices both be oriented in the same order and when are they allowed to be oriented in opposite order?

For the 3-gluon vertex in figure 15.17b, not all the arrows are pointing inward. Is this an issue?

8. Aug 7, 2013

### andrien

I will say again that use the vertex factor as I have said earlier(no use of Vσμv type notation),you are getting too much confused now.
you should convert them into all pointing inward.

9. Aug 7, 2013

### Vic Sandler

If I applied your method to the Feynman diagram in fig 14.4(b), I would get $f_{lnm}$. But the author gets $f_{lmn}$. In fig 14.4(b) he has the indices in the order lmn which is clockwise in the diagram. In figure 14.5, he has the indices in the order lmn which is counterclockwise in the diagram. If, in applying the Feynman rules to a particular diagram, I use the the wrong orientation I will introduce a wrong minus sign because f is totally antisymmetric. I don't know when to apply the fig 14.4b clockwise orientation and when to apply the figure 14.5 counterclockwise orientation.

10. Aug 8, 2013

### andrien

See this.if you write flnm then k1-k2 will appear as k2-k1,k2-k3 as k3-k2 and so.So an extra minus sign will appear which combined with flnm give flmn and you get the original formula.It does not matter how you go.(These things you should figure out yourself,take your time or consult other books such as those of greiner's which will provide you step by step methods)

11. Aug 8, 2013

### Vic Sandler

This is not correct. You will simply reverse two of the k's, The third will remain unchanged. In other words for instance, swapping $k_2$ with $k_3$,

$$(k_1 - k_2)_{\sigma} \rightarrow (k_1 - k_3)_{\sigma}$$
$$(k_2 - k_3)_{\tau} \rightarrow (k_3 - k_2)_{\tau}$$
$$(k_3 - k_1)_{\nu} \rightarrow (k_2 - k_1)_{\nu}$$

So you won't get the minus sign you need unless you also swap the indices on V.

Here is what I think about the 3 gluon Feynman diagram and rules.

In the 3 gluon diagram, since all arrows point inward, there are no asymmetries related to the direction of momentum. For this reason, if you turn the diagram 120 degrees, the equation you get from applying the Feynman rules should not change it's value, nor if you turn it 240 degrees. Also, if you hold the diagram up to a strong light and read it from the back of the page, again, you should get the same equation. This means that not only will the indices on f suffer a transposition, but so will the momenta and so will the indices on V. However, if the equation is to be totally symmetric and f is to be totally antisymmetric, the only possibility is that V also be totally antisymmetric. What's more, if you permute the indices of V and calculate the result directly using eqn (15.68) and remembering to impose the momentum changes, you also find that V is totally antisymmetric. That is why figures 14.4(b) and 14.5 are not contradictory. It doesn't matter whether you traverse the gluon lines in clockwise or counterclockwise direction, but it does matter that you do the same for f and for the momenta, and for V. All three things must be done together. When this principle is applied to diagram 15.17(b), you get eqn (17.125) exactly as it is written, with no need to alter the indices on f.

Here where I think the problem lies. If eqn (15.128) is changed to
$$N^{\mu}(p', p, k) = V^{\mu\tau\nu}(p - p', p' - k, k - p)\gamma_{\tau}(\not{k} + m)\gamma_{\nu}$$
then the whole problem may be solved. I will not know for sure until I finish deriving the rest of the proof down to eqn (15.90b) on page 362. One thing is for sure, eqn (15.128) cannot stand as it is in the book.

This is not an entirely satisfactory solution. Why not simply leave V unchanged from eqn (17.125) when formulating eqn (15.128) and drop the minus sign in (15.129)? And why mention eqn (15.66) if it isn't being invoked?

None the less, for the time being, I consider this a tentative solution to the problem and I thank you for your input.

12. Aug 8, 2013

### andrien

No,you are doing it wrong again.Those indices will be changed.When you write flnm,then
If l is with μ(k1),m with v(k2),n with σ(k3) then again the factors will go as
[gμσ(k1-k3)v+gσv(k3-k2)μ+g(k3-k1)σ],then owing to symmetric property of metric tensor gμσ=gσμ and so you get the vertex factor as I have written before(getting a minus sign,so flnm becomes flmn).GOT IT.

13. Aug 8, 2013

### Vic Sandler

The formula you have written here simply cannot be right since it has the term $k_3 - k_1$ twice. What's more, it uses the subscript $\mu$ which is nowhere to be found in any of the figures 14.4(b), 14.5, or 15.17(b). This makes it difficult for me to follow what you mean. Would you be so kind as to resubmit it using the subscripts from figure 14.5 and with care not to include typos. Since you claim that all you are doing is moving the subscripts on k, can you please tell me which of the following transformations you are utilizing:

$(k_1, k_2, k_3) \rightarrow (k_1, k_2, k_3)$ (of course not, this is no transformation at all.)
$(k_1, k_2, k_3) \rightarrow (k_1, k_3, k_2)$
$(k_1, k_2, k_3) \rightarrow (k_2, k_1, k_3)$
$(k_1, k_2, k_3) \rightarrow (k_2, k_3, k_1)$
$(k_1, k_2, k_3) \rightarrow (k_3, k_1, k_2)$
$(k_1, k_2, k_3) \rightarrow (k_3, k_2, k_1)$

It would be most helpful if you would show what V looks like before the transformation (presumably this would be eqn 14.68), and what V looks like after the transformation. Please include all of the g's, k's, and subscripts so I can make a careful comparison and understand what you mean.

14. Aug 8, 2013

### Vic Sandler

I've decided, while I'm waiting for your reply, to try one of them myself. Here is what happens when I apply the transformation
$(k_1, k_2, k_3) \rightarrow (k_2, k_3, k_1)$
to eqn (14.68)

$$g_{\nu\tau}(k_3 - k_2)_{\sigma} + g_{\tau\sigma}(k_2 - k_1)_{\nu} + g_{\sigma\nu}(k_1 - k_3)_{\tau}$$
$$\rightarrow$$
$$g_{\nu\tau}(k_1 - k_3)_{\sigma} + g_{\tau\sigma}(k_3 - k_2)_{\nu} + g_{\sigma\nu}(k_2 - k_1)_{\tau}$$

The result is neither V nor -V because for instance, if we compare the $\tau$ components $g_{\sigma\nu}(k_1 - k_3)_{\tau} \ne \pm g_{\sigma\nu}(k_2 - k_1)_{\tau}$ so it seems that $(k_1, k_2, k_3) \rightarrow (k_2, k_3, k_1)$ is not what you had in mind. It must be one of the other ones, but which one?

15. Aug 9, 2013

### andrien

I made a typo there,the term will look
[gμσ(k1-k3)v+gσv(k3-k2)μ+g(k2-k1)σ].Now it is OK.I thought you see the patern of writting it.But all right,(time for spoon feeding).Assuming the same If l is with μ(k1),m with v(k2),n with σ(k3)(μ can be τ or anything,it is upto the index you write).
now when one writes flmn,then it means that write glm(kl(1)-km(2))n+(mn term)+(nl term)[cyclic order just in the sense you write).Since l is with μ,m with v.Replace l by μ,m by v,n by σ and you get the term.Now if it is flnm,then again gln[kl(1)-kn(3)]m+(nm term)+(ml term)(cyclic order).Replace l by μ,m by v,n by σ and you get the term.Use symmetric property of metric tensor.

16. Aug 9, 2013

### Vic Sandler

You are using lmn for subscripts on V instead of $\sigma, \tau, \nu$ so when you permute lmn, you also permute $\sigma, \tau, \nu$. I've been saying since post #3 that if you permute lmn, you also have to permute $\sigma, \tau, \nu$.

I asked you to pick a Feymnan diagram, either 14.4(b), 14.5, or 15.17(b). I would prefer 14.5, but you are free to choose. You did not do so. I asked you to change your symbols to the symbols of the diagram, You did not do so. And I am beginning to suspect that you don't do these things because you can't. But I can.

Start with Feynman diagram 14.4(b), apply its associated Feynman rule to figure 17.17(b) without any deviation of any kind and get eqn (17.125) exactly as it is written.

As for eqn (14.68), you cannot change $V_{\sigma\tau\nu} \rightarrow -V_{\sigma\tau\nu}$ by any change you make to lmn because eqn (14.68) doesn't contain l, m, or n. It only contains $\sigma, \tau, \nu$ and $k_1, k_2, k_3$. You have said that you can make the change $V{\sigma\tau\nu} \rightarrow -V{\sigma\tau\nu}$ without making any change to $\sigma, \tau, \nu$, only permuting $k_1, k_2, k_3$. I don't think you can do so in spite of your arcane notation, and here is why.

There are only 6 ways to permute $k_1, k_2, k_3$, namely
$(k_1, k_2, k_3) \rightarrow (k_1, k_2, k_3)$
$(k_1, k_2, k_3) \rightarrow (k_1, k_3, k_2)$
$(k_1, k_2, k_3) \rightarrow (k_2, k_1, k_3)$
$(k_1, k_2, k_3) \rightarrow (k_2, k_3, k_1)$
$(k_1, k_2, k_3) \rightarrow (k_3, k_1, k_2)$
$(k_1, k_2, k_3) \rightarrow (k_3, k_2, k_1)$

I will start with eqn (14.68) and take care to see that I make no changes to it other than those permutations.
Here is (14.68)
$g_{\nu\tau}(k_3 - k_2)_{\sigma} + g_{\tau\sigma}(k_2 - k_1)_{\nu} + g_{\sigma\nu}(k_1 - k_3)_{\tau}$

Here is what it looks like after each one of the permutations above.
$g_{\nu\tau}(k_3 - k_2)_{\sigma} + g_{\tau\sigma}(k_2 - k_1)_{\nu} + g_{\sigma\nu}(k_1 - k_3)_{\tau}$
$g_{\nu\tau}(k_2 - k_3)_{\sigma} + g_{\tau\sigma}(k_3 - k_1)_{\nu} + g_{\sigma\nu}(k_1 - k_2)_{\tau}$
$g_{\nu\tau}(k_3 - k_1)_{\sigma} + g_{\tau\sigma}(k_1 - k_2)_{\nu} + g_{\sigma\nu}(k_2 - k_3)_{\tau}$
$g_{\nu\tau}(k_1 - k_3)_{\sigma} + g_{\tau\sigma}(k_3 - k_2)_{\nu} + g_{\sigma\nu}(k_2 - k_1)_{\tau}$
$g_{\nu\tau}(k_2 - k_1)_{\sigma} + g_{\tau\sigma}(k_1 - k_3)_{\nu} + g_{\sigma\nu}(k_3 - k_2)_{\tau}$
$g_{\nu\tau}(k_1 - k_2)_{\sigma} + g_{\tau\sigma}(k_2 - k_3)_{\nu} + g_{\sigma\nu}(k_3 - k_1)_{\tau}$

I never get $-V_{\sigma\tau\nu}$ unless I also permute $\sigma, \tau, \nu$. If I do permute $\sigma, \tau, \nu$ taking care to do so properly, I find that $V_{\sigma\tau\nu}$ is totally antisymmetric in its indices. The above constitutes a proof that V is totally antisymmetric. You can't just simply say that the proof is wrong, you have to point out which step is wrong.

Last edited: Aug 9, 2013
17. Aug 9, 2013

### andrien

I am just giving the general idea.No book uses the notation such as vστv or such to determine vertex factor.So I said earlier,use the formula I have given.Indices will change accordingly as order of lmn changes,which is clear from the vertex factor.You are not permuting any k1,k2 or k3.Only ordering of l,m,n will decide which order of σ,τ or v will occur and accordingly k1,k2,k3.I can tell one other thing and that is to check out other books,such as peskin and schroeder(eqn. 16.82).This discussion is over for now.

18. Aug 9, 2013

### Vic Sandler

No matter, you will never convince me and I will never convince you. Even so, I know where the problem is because you have helped me to find the answer. In the OP, I asked "Can someone confirm that eqn (15.125) is an accurate application of the Feynman rules to Fig (15.17b)?". The fact is, it is not. And I know that because you got me thinking about the effects of permuting the labels on the diagram. When I wrote that OP, I had not considered those effects and so I calculated incorrectly. It is possible that the author made the same mistake I did, I don't know. However, if you carefully apply the rules you will get an extra minus sign.

I appreciate the fact that you stayed with this for as long as you did and I am not surprised that you finally have had enough. I am convinced that I am correct about the permutation of the labels. You must permute labels like $(l, \sigma, k_1)$ as a unit. That is keep all 3 together as you permute. I know you don't agree, but the reason I got confused in the first place was because I didn't realize that it was true. By your help, I now believe it and understand it and even if you don't, it is the key to solving the problem.

Tomorrow I will show you how to apply this rule correctly to the diagram 15.17(b) and get the extra minus sign that belongs there. It is late and I am going to bed.

19. Aug 10, 2013

### Vic Sandler

Yes, I can help you. eqn (15.125) has a flaw in it. Compare figure 14.4(b) to figure 14.5. In the eqn (14.67) use the g factors in the Ds to convert $\alpha$ to $\sigma$, $\beta$ to $\tau$ and $\gamma$ to $\nu$ in figure 14.5. You will see that the labels $(\tau, m, k_2)$ and $(\nu, n, k_3)$ are reversed between the two figures. When I checked to see if eqn(15.125) was correct, I was not thinking in terms of such transpositions but with the help of andrien, I did. Armed with that information I now realize that when I checked eqn (15.125) I blindly used eqn (14.68) just as it is without change. This is not correct. Eqn (14.68) is not compatible with figure 14.4(b) because the transpositions have not been accounted for. What is required is a new eqn (14.68') as follows

$$V_{\sigma \tau \nu} = g_{\tau \nu}(k_2 - k_3)_{\sigma} + g_{\nu \sigma}(k_3 - k_1)_{\tau} + g_{\sigma \tau}(k_3 - k_1)_{\nu}$$

When this corrected version of (14.68) is used, eqn (15.125) remains as it is, but eqn (15.126) gains a minus sign. Of course, the failure to include eqn (14.68'), the reference to figure 15.15(b), the factor of $(k + m_0)$ in eqn (15.125), the n label in figure 15.17(b), the incorrect middle term on the rhs of eqn (15.126), the bogus minus sign in the first parameter on the rhs of eqn (15.128), and the invocation of eqn (15.66) which isn't being used are also all typos for a total of 8 in the derivation of a single equation. Not good.

Last edited: Aug 10, 2013
20. Aug 10, 2013

### Vic Sandler

One more thing has caught my attention. The author mentions on page 316 that he interchanges the sets of variables in eqn (14.61) and these sets are linked to the sets in eqn (14.65). So if these variables are not permuted as complete sets, you may not be able to rely on his proof of the vertex factor (14.68).