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Another question from Srednicki's QFT book 
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#1
Jul1614, 09:39 AM

P: 3,243

Sorry for my questions, (it does seem like QFT triggers quite a lot of questions :D).
Anyway, on page 103 (it has a preview in google books), I am not sure how did he get equation (14.40), obviously it should follow from (14.39), but I don't understand where did ln(m^2) disappear ? Shouldn't we have an expression like (14.40) but the term (linear in k^2 and m^2) be instead: (linear in k^2,m^2 and ln(m^2)). Cause as far as I can tell from (14.39) [tex]\Pi(k^2)[/tex] depends also on ln(m^2), and thus the term "linear in..." should be replaced with "linear also in ln(m^2)", cause as far as I can tell ln(m^2) isn't linear function of m^2, right? Hope someone can enlighten me. 


#2
Jul1814, 04:19 AM

P: 906

No that's not possibly the case, because even if you have the [itex] \ln (m^{2}) [/itex] it's multiplied with a [itex]D[/itex].. After integration of [itex]D dx [/itex] you will get [itex](k^{2}+m^{2})ln(m^{2})/6[/itex]
so I think the [itex]ln(m^{2})[/itex] is absorbed within the [itex]κ_{A,B}[/itex] But I hope someone can be more helpful 


#3
Jul1814, 01:42 PM

P: 3,243

Actually the integral on Ddx yields 1/6 k^2 +m^2, it's written previously.



#4
Jul1814, 01:45 PM

P: 906

Another question from Srednicki's QFT book
and that's what I've written? I just didn't take in account the minus from the ln....Ah yes, you are write, just put a 6 in front of m^2 then



#5
Jul1814, 02:59 PM

P: 3,243

I think the task of solving the problem of mass gap from clay institute which relates to QFT looks a lot more intimidating as I keep reading. :D



#6
Jul2014, 04:43 AM

Sci Advisor
P: 303

The mass gap problem is essentially that you have to prove YangMills exists and after you have done that, you must prove it describes massive particles (rather than massless ones as you would naively think from the Lagrangian). So, yes it is very difficult!



#7
Jul2114, 01:36 AM

P: 3,243

Ok, I plan to ask all of my questions from QFT books here (if the moderators want to include my other posts into this thread it's ok by me, but make it chronological orderd (or as we say use the timeordering operator).



#8
Jul2114, 01:49 AM

P: 3,243

In this case we don't have a book preview of pages 166167 .
So I'll write the equations: [tex](27.23) ln \mathcal{T}^2_{obs} = C_1+2ln \alpha +3\alpha (ln \mu +C_2)+O(\alpha^2)[/tex] Now he says that "Differentiating wrt ln \mu then gives": [tex](27.24)0=\frac{d}{dln \mu} ln \mathcal{T}^2_{obs} = \frac{2}{\alpha} \frac{d\alpha}{dln \mu} +3\alpha +O(\alpha^2) [/tex] Now as far as I can tell when you differentiate: [tex]\frac{d}{dln \mu} (3\alpha(ln \mu +C_2))=3\alpha + 3 \frac{d\alpha}{dln \mu} (ln \mu +C_2)[/tex] so where did [tex]3 \frac{d\alpha}{dln \mu} ln \mu[/tex] disapper from eq. (27.24)? Don't see why he didn't include this term in eq. (27.24). Anyone? 


#9
Jul3014, 02:49 PM

P: 3,243

Hi, so I hope there are still some folks who look at this thread of mine.
So now I am looking at Srednicki's solution to question 48.4b, here: http://www.scribd.com/doc/87916496/S...SolutionsRev I don't understand how in eq (48.53) he pulled from the trace the term (1+s1s2), shouldn't it be (1s1s2)? I mean if you simplify the term inside the trace in the first line you should get: Tr((1s1s2)(...)) = (1s1s2)Tr(...) So how did he get the plus sign there? Perhaps x^2=1 and not 1 as he wrote there? Puzzled. 


#10
Jul3014, 03:16 PM

P: 906

I think there is a mistake in the signs...
[itex] \bar{x} ( \bar{p}_{1} m) \bar{x}[/itex] [itex]\bar{x} \bar{p}_{1} \bar{x}  m \bar{x} \bar{x} [/itex] [itex]+\bar{x}\bar{x} \bar{p}_{1} 2 xp_{1} + m [/itex] [itex] \bar{p}_{1}+m [/itex] So I think the minus in front of s1s2 in the first line of 48.53 is wrong? Obviously by bared quantities I mean slash, it's just \slash{} didn't work... 


#11
Aug2114, 01:56 PM

P: 3,243

On page 316, I am not sure how did he get equation (51.10) from eq. (51.9)?
Here's what I have done : eqaution (51.10) is the same as the next equation: (I am writing it down without a slash, since I am not sure how to use it here) [tex] \frac{(p^{\mu}\gamma_{\mu}+m)}{(p^{\mu}\gamma_{\mu}+mi\epsilon)(p^{\mu}\gamma_{\mu}+m)}+\int \frac{....}{p^2+s\epsilon^22i\epsilon\sqrt{s}}[/tex] So as you can see it's a bit different than (51.9), the denominator in the integral, the epsilon doesn't have as a factor, the coefficient, i . And the first term after rearranging it we should have: [tex] \frac{ p^{\mu}\gamma_{\mu} + m}{p^2+m^2i\epsilon \cdot (p^{\mu}\gamma_{\mu} +m)}[/tex] It's not the same term as in eq (51.9), perhaps he stated previously that it's meaningless factor multiplies the epsilon, I might have forgotten it. What do you think? P.S There's a preview in google books: http://books.google.co.il/books?id=5...page&q&f=false 


#12
Aug2114, 06:34 PM

P: 906

the page is unfortunately not previewed.



#13
Aug2114, 08:51 PM

P: 3,243

I can see the preview in google books of pages 316315 does exist, I think you have a problem in your computer.
Cheers! 


#14
Aug2214, 08:59 AM

P: 3,243

Now I am not sure as for the solution of question 51.2) in Srednicki.
It's in here: http://www.scribd.com/doc/87916496/S...onsRev#logout If I am not mistaken he uses the next first order approximation: [tex] V_Y(p,p') = V_Y(0,0) + V_Y'(0,0) p\cdot p' [/tex] But I don't understand what did he plug instead of [tex]V_Y'(0,0)[/tex] Anyone, or if you have a more lengthy detailed calculation as to how did he get eq. (51.58) 


#15
Aug2314, 08:32 PM

Sci Advisor
HW Helper
P: 2,948

Hope this helps. 


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