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Constraints on a fourth rank tensor

  1. Jul 27, 2014 #1

    CAF123

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    1. The problem statement, all variables and given/known data
    Consider a theory which is translation and rotation invariant. This implies the stress energy tensor arising from the symmetry is conserved and may be made symmetric. Define the (Schwinger) function by ##S_{\mu \nu \rho \sigma}(x) = \langle T_{\mu \nu}(x)T_{\rho \sigma}(0)\rangle##. By the above restrictions on the theory, we have that $$S_{\mu \nu \rho \sigma} = S_{\nu \mu \rho \sigma} = S_{\mu \nu \sigma \rho} = S_{\nu \mu \sigma \rho}\,\,\,(1),$$ $$S_{\mu \nu \rho \sigma}(x) = S_{\rho \sigma \mu \nu}(-x)\,\,\,\,(2).$$ If we now impose parity invariance and scale invariance with ##T_{\mu \nu}## of dimension 2, we have further $$S_{\mu \nu \rho \sigma}(x) = S_{\rho \sigma \mu \nu}(x)\,\,\,(3)$$ and $$S_{\mu \nu \rho \sigma}(\lambda x) = \lambda^{-4}S_{\mu \nu \rho \sigma}(x)\,\,\,(4)$$

    All these contraints restrict the form that ##S_{\mu \nu \rho \sigma}## can take. What I want to do is to write out the most generic form first and then build in the constraints given by (1),(2),(3) and (4) above.

    2. Relevant equations
    Problem from Di Francesco et al, 'Conformal Field Theory' P.108

    3. The attempt at a solution
    We can build the 4th rank tensor from only x and delta's (or the metric g). We cannot use the levi-civita tensor because that is antisymmetric in any two indices so fails (1) immediately. The most general 4th rank tensor I was getting is $$S_{\mu \nu \rho \sigma}(x) = a_1x_{\mu}x_{\nu}x_{\rho}x_{\sigma}f_1(x^2) + a_2g_{\mu \nu}g_{\rho \sigma}f_{2,1}(x^2) + a_3g_{\mu \rho}g_{\nu \sigma}f_{2,2}(x^2) + a_4g_{\mu \sigma}g_{\rho \nu}f_{2,3}(x^2) + a_5g_{\mu \nu}x_{\rho}x_{\sigma}f_{3,1}(x^2) $$$$+ a_6g_{\rho \sigma}x_{\mu}x_{\nu}f_{3,2}(x^2) + a_7g_{\mu \sigma}x_{\rho}x_{\nu}f_{3,3}(x^2) + a_8g_{\mu \rho}x_{\sigma}x_{\nu}f_{3,4}(x^2) + a_9 g_{\nu \rho}x_{\sigma}x_{\mu}f_{3,5}(x^2) + a_{10}g_{\nu \sigma}x_{\rho}x_{\mu}f_{3,6}(x^2)$$ where ##a_i## are constants and ##f_i(x^2)## are Lorentz invariant functions.

    Consider the first term above. By (4), this becomes ##S_{\mu \nu \rho \sigma}(\lambda x) = \lambda^4x_{\mu}x_{\nu}x_{\rho}x_{\sigma}f_1(\lambda^2 x^2) = \lambda^{-4}S_{\mu \nu \rho \sigma}(x) \Rightarrow f_1(x^2) = \lambda^8f_1(\lambda^2x^2)##. The answer in the book given above has ##f_1(x^2) = A_4/(x^2)^4##. How did they obtain this? Once I have the general idea about how to correctly impose the constraints, I should be able to continue.

    Thanks!
     
  2. jcsd
  3. Jul 27, 2014 #2

    strangerep

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    OK, you have:
    $$ \lambda^8 f_1(\lambda^2x^2 ~=~ f_1(x^2) ~,$$and you must find the form of ##f##. So try an analytic series expression for ##f##, i.e., like a Taylor series, or Laurent series -- involving powers of its argument. E.g., consider the ##n##'th term in that series and substitute into your constraint above. You'll get something like $$ \lambda^8 \, a ( \lambda^2 x^2)^n ~=~ a (x^2)^n ~,$$where ##a## is some constant (which cancels out here). Can you determine the allowed value(s) of ##n## from this?
     
  4. Jul 28, 2014 #3

    CAF123

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    Hi strangerep,
    Nice! So what you did was rewite ##f## as a power series in its argument so like $$\lambda^8 \sum_{n=-\infty}^{\infty}a_n (\lambda^2 x^2)^n = \sum_{n=-\infty}^{\infty} a_n (x^2)^n$$ For the LHS to be equal to the RHS, the coefficients in the sequence of terms have to be equal, so we compare the ##n^{\text{th}}## term in each series. For this particular example, I found that only when n=-4 are the two sides equal (which is the result of the constraint) which means the function ##f_1## is completely specified by the (n=-4)th coefficient in the series namely that ##f_1 = a_4/(x^2)^4##. Would that be right?

    Using this method, I was able to get all the terms that are present in the solution on P.108. Although I am still not sure why they disregarded the following term: $$a_7g_{\mu \sigma}x_{\rho}x_{\nu}f_{3,3}(x^2) + a_8g_{\mu \rho}x_{\sigma}x_{\nu}f_{3,4}(x^2) + a_9 g_{\nu \rho}x_{\sigma}x_{\mu}f_{3,5}(x^2) + a_{10}g_{\nu \sigma}x_{\rho}x_{\mu}f_{3,6}(x^2)$$ As far as I can see it satisfies all constraints.

    Is it correct to say that ##a_{11}(\epsilon_{\mu \sigma}\epsilon_{\nu \rho} + \epsilon_{\mu \rho}\epsilon_{\nu \sigma})(x^2)^{-2}## is not a further possibility since under any permutation of the indices, we can see that this is equivalent to ##g_{\mu \nu}g_{\rho \sigma} + g_{\mu \rho}g_{\nu \sigma} + g_{\mu \sigma}g_{\nu \rho}## or at least some linear combination of those quantities. (in answer to Ex.4.9 P.110)

    Finally, this section deals with tracelessness of ##T_{\mu \nu}## in two dimensions. Where have I used that the conformal invariance is in two dimensions?

    Thanks, strangerep.
     
    Last edited: Jul 28, 2014
  5. Jul 28, 2014 #4

    strangerep

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    Yes.

    I haven't worked it through in detail, but are you sure you've used all the available index symmetries? E.g., (4.70) seems to imply that your 1st and 3rd terms can be coalesced into 1 term.

    I think that's what Ex 4.9a is asking you to show explicitly. Since it's only in 2D I suppose it's reasonable to write out all the possibilities explicitly (but I haven't actually done it, so caveat emptor.)

    Isn't that what Ex 4.9b is showing you? I.e., in higher dimensions there's another term in the most general Schwinger function, which prevents the conclusion in eq(4.78). Note that they implicitly used the results of Ex 4.9 to arrive at eq(4.74).

    Their logical sequence could probably have been made more transparent. :frown:
    The sentence preceding eq(4.74) should be something like:
    "...restrict the most general form that ##S_{\mu\nu\rho\sigma}## can take, and, by Ex 4.9, that form in 2D is: [...]
     
  6. Jul 29, 2014 #5

    CAF123

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    I am not entirely sure. But here was my thinking: We only have x and g along with some Lorentz/translational invariant functions at our disposal to form the tensor. Then I formed all possible combinations of these quantities into the long equation I posted in the OP. (i.e 4 x's, 2 x's and 1 g, 2 g's). These are all the possible possibilities to yield the fourth rank tensor in this case. Is that right?

    I would see that this satisfies ##S_{\mu \nu \rho \sigma} = S_{\nu \mu \sigma \rho}## but not that e.g ##S_{\mu \nu \rho \sigma} = S_{\nu \mu \rho \sigma}## so would it not fail (1) given this? That was my reasoning for including four terms above, so I would cover all the possible permutations in (1).

    hmm, would you be able to start me off? I am not really sure what to consider first.

    I see, thanks.
    Thanks.
     
    Last edited: Jul 29, 2014
  7. Jul 29, 2014 #6

    strangerep

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    There's also the ##\epsilon## terms, illustrated in ex 4.9 for the 2D and 3D case explicitly. So one must pick the dimension and solve for that case.

    I don't have much spare time right now, so you'll have to work it out explicitly and show me. Use all 3 equations that make up (4.70), to impose constraints on your a's.

    Do it systematically. First confirm that the expression in ex 4.9a is compatible with the index symmetries. Then just start writing out all terms explicitly. E.g., ##\mu## takes the values 1 and 2. So that expands the expression to 4 terms:
    $$
    \epsilon_{1\sigma} \epsilon_{\nu \rho} ~+~ \epsilon_{1\rho} \epsilon_{\nu \sigma}
    ~+~ \epsilon_{2\sigma} \epsilon_{\nu \rho} ~+~ \epsilon_{2\rho} \epsilon_{\nu \sigma}
    $$Then, ##\nu## takes the values 1 and 2, which expands the above into 8 terms. Then do the same with ##\sigma##, but now some of the ##\epsilon##'s will vanish. Then proceed to ##\rho## similarly. Then substitute (e.g.,) ##\epsilon_{12} = 1##, and ##\epsilon_{12} = -1##, and so on.
    Finally, show that every remaining term is equivalent to one involving ##g##'s in (4.77).
     
  8. Jul 30, 2014 #7

    CAF123

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    If ##S_{\mu \nu \rho \sigma} = a_7 g_{\mu \sigma}x_{\rho}x_{\nu} f_{3,3}(x^2) + a_8 g_{\mu \rho}x_{\sigma}x_{\nu}f_{3,4}(x^2) + a_{9}g_{\nu \rho}x_{\sigma}x_{\mu} f_{3,5}(x^2) + a_{10}g_{\nu \sigma}x_{\rho}x_{\mu}f_{3,6}(x^2)##, then $$S_{\nu \mu \rho \sigma } = a_7 g_{\nu \sigma}x_{\rho}x_{\mu} f_{3,3}(x^2) + a_8 g_{\nu \rho}x_{\sigma}x_{\mu}f_{3,4}(x^2) + a_{9}g_{\mu \rho}x_{\sigma}x_{\nu} f_{3,5}(x^2) + a_{10}g_{\mu \sigma}x_{\rho}x_{\nu}f_{3,6}(x^2),$$ which means ##a_7 =a_{10}, a_8 = a_9## and ##f_{3,3} = f_{3,6}, f_{3,4} = f_{3,5}##. Similarly, $$S_{\mu \nu \sigma \rho} = a_7 g_{\mu \rho}x_{\sigma}x_{\nu} f_{3,3}(x^2) + a_8 g_{\mu \sigma}x_{\rho}x_{\nu}f_{3,4}(x^2) + a_{9}g_{\nu \sigma}x_{\rho}x_{\mu} f_{3,5}(x^2) + a_{10}g_{\nu \rho}x_{\sigma}x_{\mu}f_{3,6}(x^2),$$ which together with the above constraints imply ##a_7 = a_8 = a_9 = a_{10}## and ##f_{3,i} = f_{3,j}## for all i,j.
    Do you mean to say '[..] into 8 more terms'. E.g when ##\mu = 1##, then ##\nu = 1## or ##2## so this generates another four terms. When ##\mu = 2,## then ##\nu = 1## or ##2##, so altogether we have an additional eight terms, which along with the four you had makes 12. Or did I misinterpret your method?
     
  9. Jul 31, 2014 #8

    strangerep

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    And... ?

    There's one more index symmetry at the right hand end of (4.70) that you seem to have ignored. I.e., swap ##\mu, \nu## again and see what happens.

    Well, you definitely misinterpreted (or misread) something. Look again at the expression I wrote, in which ##\mu## is already expanded into the 2 cases ##\mu=1## and ##\mu=2##. I.e., 4 terms in total. Then, when you expand this for the values of ##\nu##, the number of terms doubles, i.e., 8 in total.
     
  10. Jul 31, 2014 #9

    CAF123

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    This gives $$S_{\nu \mu \sigma \rho} = a_7g_{\nu \rho}x_{\sigma}x_{\mu}f_{3,3}(x^2) + a_8g_{\nu \sigma}x_{\rho}x_{\mu}f_{3,4}(x^2)+ a_9 g_{\mu \sigma}x_{\rho}x_{\nu}f_{3,5}(x^2) + a_{10}g_{\mu \rho}x_{\sigma}x_{\nu}f_{3,6}(x^2),$$ which, as far as I can see, gives no new information regarding the constraints on the ##a_i## and the ##f's##. That is why I didn't write it down explicitly.

    Could I not expand the terms in various ways though? E,g given ##\epsilon_{1 \sigma}\epsilon_{\nu \rho} + \epsilon_{1 \rho}\epsilon_{\nu \sigma} + \epsilon_{2 \sigma}\epsilon_{\nu \rho} + \epsilon_{2 \rho}\epsilon_{\nu \sigma}##, this can be expanded differently by what choice of ##\nu## I put with ##\mu##. I could expand like ##\epsilon_{1 \sigma}\epsilon_{1 \rho} + \epsilon_{1 \rho}\epsilon_{1 \sigma} + \epsilon_{2 \sigma}\epsilon_{1\rho} + \epsilon_{2 \rho}\epsilon_{1 \sigma}## or ## \epsilon_{1 \sigma}\epsilon_{2 \rho} + \epsilon_{1 \rho}\epsilon_{2 \sigma} + \epsilon_{2 \sigma}\epsilon_{2 \rho} + \epsilon_{2 \rho}\epsilon_{2 \sigma}##. But then I notice that two terms are common above, so would the general expansion not look something like $$\epsilon_{1 \sigma}\epsilon_{1 \rho} + \epsilon_{1 \rho}\epsilon_{1 \sigma} + \epsilon_{2 \sigma}\epsilon_{1\rho} + \epsilon_{2 \rho}\epsilon_{1 \sigma} + \epsilon_{2 \sigma}\epsilon_{2 \rho} + \epsilon_{2 \rho}\epsilon_{2 \sigma}?$$ i.e one involving six terms?


    Unless of course you asked me to write down ##\epsilon_{\mu \sigma}\epsilon_{1\rho} + \epsilon_{\mu \rho}\epsilon_{1 \sigma} + \epsilon_{\mu \sigma}\epsilon_{2 \rho} + \epsilon_{\mu \rho}\epsilon_{2\sigma}##, then that is an extra four terms, but if I do the same with ##\sigma## then I won't get the cancellation of ##\epsilon##'s that you mentioned.

    Thanks!
     
    Last edited: Jul 31, 2014
  11. Jul 31, 2014 #10

    strangerep

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    You have an explicit expression for ##S_{\mu \nu \sigma \rho}## from your post #7, and now you have an expression ##S_{\nu \mu \sigma \rho}##. Equate them. What can you deduce?

    OK, I realize now that my explanation in post #6 was too sketchy and generic. So let's come at this from a different direction...

    Pick some specific values for ##\nu \mu \sigma \rho##. Write out the 2-term ##\epsilon## expression, and replace the ##\epsilon##'s with 1, 0, or -1, as appropriate. Compare the result against the terms in ##S_{\nu \mu \sigma \rho}##, and see whether terms of that form are already accounted for in the ##S_{\nu \mu \sigma \rho}## expression.

    Then pick other specific values for ##\nu \mu \sigma \rho## and repeat the exercise.
     
  12. Aug 1, 2014 #11

    CAF123

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    This tells me that $$a_7 f_{3,3} = a_8 f_{3,4}, a_8 f_{3,4} = a_7 f_{3,3}, a_{10}f_{3,6} = a_9f_{3,5}, a_9 f_{3,5} = a_{10}f_{3,6}$$ so ##a_7 = a_8, a_{10} = a_9## and ##f_{3,3} = f_{3,4}, f_{3,6} = f_{3,5}##.

    From ##S_{\nu \mu \rho \sigma} = S_{\mu \nu \sigma \rho}##, I have that ##a_7 = a_9, a_8 = a_{10}## and ##f_{3,5} = f_{3,3}##, ##f_{3,6} = f_{3,4}##.

    I.e all constants are the same and all L-invariant f's are the same. I am not sure this is what you wanted me to write since this was my conclusion already, but I don't see what else I can deduce by equating them.
    Why does ##\epsilon## take three values? But I understand this method and I see that for permutations of ##\nu \mu \sigma \rho##, I obtain the correct results on both sides of the equation for a suitable choice of constants (assuming ##\mu, \nu = 1,2##) in two dimensions.

    If we can now move to the bottom of P. 108, where they say 'It is then straightforward ...'. Up to two extra terms, I have managed to show (4.75) from which (4.76) and (4.77) follow. Writing $$S^{\mu \sigma}_{\mu \sigma}(x) = \frac{A}{(x^2)^4} \left\{(3g^{\mu}_{\,\,\,\mu}g^{\sigma}_{\,\,\,\sigma} - g^{\mu \sigma}g_{\mu \sigma} - g^{\mu}_{\,\,\,\sigma}g_{\mu}^{\,\,\,\sigma})(x^2)^2 - 4x^2(g^{\mu}_{\,\,\,\mu}x^2 + g^{\sigma}_{\,\,\,\sigma}x^2) + 8(x^2)^2\right\}$$ how did FMS then conclude that this vanishes?
    (Apologies for the incorrect formatting of the LHS, I can't obtain the correct LaTeX to get it to come out correctly)
    Thanks.
     
    Last edited: Aug 1, 2014
  13. Aug 1, 2014 #12

    CAF123

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    [..] continued from above (I was too late to edit)
    For showing that the term with epsilons reduces to one in the metric, what should I take to be the metric in these calculations? I managed to find relations among the constants for the metric describing a Euclidean space, I.e the identity (only diagonal elements non zero, equal to 1).

    (Continuing from the bottom of the last post)
    It doesn't seem to be the case that ##S^{11}_{11} + S^{12}_{12} + S^{21}_{21} + S^{22}_{22} = 0##
     
    Last edited: Aug 1, 2014
  14. Aug 1, 2014 #13

    strangerep

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    Hmm. Maybe my quick off-the-cuff guesses are just wasting your time. (Remember that I cannot spend much time working this out fully for myself.)

    Have you used the parity invariance condition (4.72) yet?

    Write out the 2x2 antisymmetric matrix.

    Insert tildes (blankspace) at appropriate places: ##S^{\mu~\sigma}_{~\mu ~\sigma}##. There's also \; and \, for halfspace and quarterspace.

    In N dimensions, ##g^{\mu}_{~\mu} = \delta^{\mu}_{~\mu} = N##. So in 2 dimensions, ##g^{\mu}_{~\mu} = 2##.

    Also, ##g^{\mu \sigma}g_{\mu \sigma} = g^{\sigma}_{~\sigma} = 2##.

    Does that make it work out?

    Well, what is the usual metric in special relativity? (Actually, you should probably check Di Francesco's conventions for that -- they're probably mentioned somewhere.) Then lose 2 spatial dimensions...
     
  15. Aug 2, 2014 #14

    CAF123

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    Yes, that is just another permutation of the indices which, along with the constraints arising from (4.70), gives nothing new. Then by (4.73) I find explicit forms for the f's in the method you initially proposed. They are ##f_{3,3} = f_{3,4} = f_{3,5} = f_{3,6} = A(x^2)^{-3}## and this is not the function in the first two terms of (4.74) so I cannot see how the form I have been dealing with is not present in the result (4.74). (i.e it doesn't seem to be able to be recast into a linear combination of terms already present in (4.74))

    Yes, sorry, I thought you meant the indices on ##\epsilon## took values 0,1,-1.
    Thanks for the tip!

    It does!
    The Minkowski metric is the usual metric in special relativity. But I think what FMS is doing is valid for an arbritary number of dimensions and valid for both Minkowski and Euclidean metrics modulo some subtleties for d=2. On page 45 (not part of the current section), they say they are using the Euclidean formulism. I am not sure if this is being carried throughout. But when you write ##g^{\mu}_{\,\,\,\mu} = \delta^{\mu}_{\mu}##, that is only valid for the Euclidean metric right? (since the Minkowski metric includes negative timelike components)

    Just before (4.75), FMS write '..the conservation law ##\partial^{\mu}T_{\mu \nu} = 0## obviously extends to the Schwinger function'. How so?

    And after (4.78), we know that ##S^{\mu ~\,\,\, \sigma}_{\,\,\,\mu \,\,\,\sigma} = 0## everywhere because by using the explicit expression in (4.77) (as I did above) it vanished independent of the value I took for ##x##?
     
    Last edited: Aug 2, 2014
  16. Aug 3, 2014 #15

    strangerep

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    I'm running out of guesses.

    BTW, have you tried keeping the extra terms, and trying to apply the steps that lead from (4.74) to the eventual conclusion that (4.78) vanishes everywhere?

    No. ##(g^{\mu\nu})## is always defined to be the matrix which is inverse to ##(g_{\mu\nu}) ##. I.e., ##g^{\mu\lambda} g_{\lambda\nu} = \delta^\mu_\nu## .

    Just differentiate it. You can pass the ##\partial_{x^\mu}## inside the ##<\cdots>##.

    Yes.
     
  17. Aug 3, 2014 #16

    CAF123

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    I just did right now, and indeed (if I did it right) then since all the constants and fi's used in that expression were the same then when I apply the derivative onto the term in step (4.75), I get something like##A[\dots] = 0##. As I said before, I have recovered all the terms in (4.75), and since the coefficients must vanish, this would imply A=0. So yes, it would vanish as far as I can tell.

    That is what I thought initially and would give the right result. However, if we apply this logic to the generic ward identity expression (2.157, P.44) or (4.63, P.106) ,for example, the LHS is something like ##\partial_{\mu}\langle j^{\mu} X \rangle## where despite the notation, the derivative acts only on ##j## and not also on ##X##, the product of the fields. If we pass the derivative through ##\langle \dots \rangle## then the LHS would be zero identically by imposing the conservation law. While I agree the derivative can be passed through, (although I believe when we write it, we keep the derivative outside the brackets) I don't think the derivative would vanish since the vanishing of the derivative is only valid in a classical treatment and the ward identities are classical quantities mirrored in a QFT.

    See bottom of post #10 here https://www.physicsforums.com/showthread.php?t=762515. I don't think Sam got rid of the ##\partial_{\mu}\langle T^{\mu \nu} X \rangle## term which is meant to be the term where the derivative acts on the current.

    What do you think?
     
    Last edited: Aug 3, 2014
  18. Aug 3, 2014 #17

    strangerep

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    I think Sam understands all this stuff better than I do. I doubt that I can help you much more in the limited time available to me. :frown:
     
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