Sign up ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

This is an approach to anomalies which seems unfamiliar to me..

  • Firstly what is this function $W$ which seems to satisfy the equation, $\frac{\partial W }{\partial g^{\mu \nu} } = \langle T_{\mu \nu} \rangle $. Is there any standard such $W$?

  • Now one says that this $W$ must satisfy the following R.G equation,

$(\mu \partial_\mu + 2\int d^dx g^{\mu\nu}\frac{ \delta }{\delta g^{\mu \nu} } )W = 0$

Where does this come from?

  • So substituting the first equation into the second and defining $A_{anomaly} = \langle T^\mu_\mu \rangle$ we get, $\mu \partial_\mu W = -2\int d^dx A_{anomaly}$. And one one wants to take two derivatives of the L.H.S w.r.t the metric to get the LHS to look like, $\mu \partial_\mu \langle T_{ab}(x) T_{cd}(0) \rangle $.

But I don't know what is this theorem which is now being used to say that this derivative of the 2-point function of the stress-tensor will necessarily have the following form,

$\mu \partial_\mu \langle T_{ab}(x) T_{cd}(0) \rangle = \frac { C_T} { 4(d-2)^2 (d-1)}\Delta^T_{abcd} \mu \partial_\mu (1/x^{2d-4} ) $

where $C_T$ is some number and $\Delta^T_{abcd} = \frac{1}{2}[S_{ac}S_{bd} + S_{ad}S_{bc}] - \frac{S_{ab}S_{cd} }{d-1} $ , where $S_{ab} = \partial_a \partial_b - \delta_{ab} \partial^2 $

Can someone kindly help as to what is the motivation/proof of the above structure? Is something similar known for derivatives of higher point correlations of the stress-tensor?

share|cite|improve this question
What is the source you are using? That would help us... $W$ must be the standard generating functional of the connected Green's functions $W[g]=\ln Z[g]$. – Adam Jun 13 '14 at 16:08
@Adam But how does that fit in? With W being as you are saying it is how does one show that it satisfies, $\frac{\partial W }{\partial g^{\mu \nu} } = T_{\mu \nu} $ ? [...the closest thing I know is the equation, $T_{\mu \nu} = \frac{1}{\sqrt{-g}}\frac{\delta S }{\delta g^{\mu \nu} }$...but this is not the same as what is being claimed..] – user6818 Jun 13 '14 at 16:38
@user68618: we don't know what's claimed, because you don't give us the source. I think that here $T_{\mu\nu}$ should be understood as $\langle T_{\mu\nu}\rangle$. – Adam Jun 13 '14 at 16:50
The initial part of Appendix D of 1405.7862 is what I am paraphrasing here. – user6818 Jun 13 '14 at 17:03
@Adam Yes it is indeed the expectation value - I have corrected the question - but how does that help? – user6818 Jun 13 '14 at 17:04

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.