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Another question from Srednicki's QFT book

  1. Jul 16, 2014 #1
    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. jcsd
  3. Jul 18, 2014 #2


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    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
  4. Jul 18, 2014 #3
    Actually the integral on Ddx yields 1/6 k^2 +m^2, it's written previously.
  5. Jul 18, 2014 #4


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    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 :smile:
    Last edited: Jul 18, 2014
  6. Jul 18, 2014 #5
    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
  7. Jul 20, 2014 #6


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    The mass gap problem is essentially that you have to prove Yang-Mills 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!
  8. Jul 21, 2014 #7
    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 time-ordering operator).
  9. Jul 21, 2014 #8
    eqs. (27.23)-(27.24) on page 166 of Srednicki's QFT.

    In this case we don't have a book preview of pages 166-167 .

    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).

  10. Jul 30, 2014 #9
    Another question from Srednicki.

    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:

    I don't understand how in eq (48.53) he pulled from the trace the term (1+s1s2), shouldn't it be (1-s1s2)?

    I mean if you simplify the term inside the trace in the first line you should get:
    Tr((1-s1s2)(...)) = (1-s1s2)Tr(...)

    So how did he get the plus sign there?

    Perhaps x^2=1 and not -1 as he wrote there?

  11. Jul 30, 2014 #10


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    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...
  12. Aug 21, 2014 #11
    Another question from Srednicki's text.

    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}+m-i\epsilon)(-p^{\mu}\gamma_{\mu}+m)}+\int \frac{....}{p^2+s-\epsilon^2-2i\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^2-i\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?

    There's a preview in google books:
    Last edited: Aug 21, 2014
  13. Aug 21, 2014 #12


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    the page is unfortunately not previewed.
  14. Aug 21, 2014 #13
    I can see the preview in google books of pages 316-315 does exist, I think you have a problem in your computer.

  15. Aug 22, 2014 #14
    Now I am not sure as for the solution of question 51.2) in Srednicki.

    It's in here:

    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)
  16. Aug 23, 2014 #15


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    The key point is that if you multiply any positive quantity by epsilon, you may still write the result as epsilon since it is taken much smaller than any other quantity. [itex] \sqrt{s} [/itex] is positive so we may write [itex] \sqrt{s} \epsilon \simeq \sqrt{s} [/itex]. Likewise for the other term, multiplying or dividing epsilon by [itex] p^2 + m^2 [/itex] does not matter.

    Hope this helps.
  17. Oct 1, 2014 #16
    Well, if you can write [tex]\sqrt{s} \epsilon \approx \sqrt{s} [/tex] then presumably you can divide by $\sqrt{s}$ both sides (besides when s=0 which we have equality), and thus you'll get: [tex] \epsilon \approx 1[/tex] but we've taken it to be approaching zero from above, so it should be [tex]\approx 0[/tex].

    Best regards.
  18. Oct 1, 2014 #17


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    Sorry, I had a typo in the equation. As I wrote in words, I meant that [tex]\sqrt{s} \epsilon \approx \epsilon [/tex]
  19. Oct 1, 2014 #18
    ok, now it makes sense.
  20. Oct 2, 2014 #19
    Another question from Srednicki's.
  21. Oct 2, 2014 #20
    I have got another question from Srdnicki's book.
    from chapter 54, pages 338-337.

    He arrives at the next form of the lagrangian:

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