I am studying CFT, where I encounter Stokes' theorem in complex coordinates: $$ \int_R (\partial_zv^z + \partial_{\bar{z}}v^{\bar{z}})dzd\bar{z} = i \int_{\partial R}(v^{z}d\bar{z} - v^{\bar{z}}dz). $$ I am trying to prove this by starting from the form of Stokes'/Greens theorem: $$ \int_R(\partial_xF^y - \partial_yF^x)dxdy = \int_{\partial R}(F^xdx + F^ydy $$ and transforming to complex coordinates. The reason I ask this here and not on Math exchange is that in CFT we have the distinction between indices up and indices down: $$ v^z = v^{\tau}+iv^{\sigma} \\ v^{\bar{z}} = v^{\tau} - iv^{\sigma} \\ v_z = v^{\tau}-iv^{\sigma} \\ v_{\bar{z}} = v^{\tau} + iv^{\sigma}$$, with $z = \tau +i\sigma$ and $\bar{z} = \tau - i\sigma$. The substitution is kind of straight forward, but I get: $$ \int_R(\partial_zv^{z} - \partial_{\bar{z}}v^{\bar{z}})dzd\bar{z} = i\int_{\partial R}v^zd\bar{z}+v^{\bar{z}}dz, $$ so I get two relative minus signs wrong. As the calculation is not so difficult, I have the feeling I am missing something crucial here. Does anyone have an idea?

  • 2
    $\begingroup$ Your expression of standard Stokes theorem is false : index are not coherent and not correct. The correct expression is : $\int_R(\partial_xF_y - \partial_yF_x)dxdy = \int_{\partial R}(F_xdx + F_ydy)$ $\endgroup$
    – Trimok
    Jun 29, 2013 at 11:22
  • 3
    $\begingroup$ I would say that the complex form of the theorem is simpler so you are carrying coal to Newscastle if you're converting it to the higher-dimensional real case. $\endgroup$ Jun 29, 2013 at 11:34
  • 3
    $\begingroup$ And, for your final formula, if you make the transformation $w^z = v^z$, $w^{\bar z} = -v^{\bar z}$, it seems that you get the correct formula. $\endgroup$
    – Trimok
    Jun 29, 2013 at 11:46
  • $\begingroup$ Hi @Erik, which references are you using? $\endgroup$
    – Qmechanic
    Jun 29, 2013 at 13:22
  • 1
    $\begingroup$ @Qmechanic David Tong's chapter 4, damtp.cam.ac.uk/user/tong/string.html and Polchinski's String Theory (Volume 1) $\endgroup$
    – Funzies
    Jun 29, 2013 at 14:28

1 Answer 1


I think I can help. Pick up the expression you started with. Convert to coordinates $\tau$ and $\sigma$ as you did. Use the Green's theorem in these coordinates and then convert back to the complex coordinates again. I am not good with word by this transliterates in the following expression

$$2∫∫dσ dτ (∂_{σ}v^{σ} + ∂_{τ}v^{τ})$$ = (now use Green’s theorem and the next integral is a contour counterclockwise bounding the area defined by the previous double integral $$= 2∫ [v^{σ} dτ - v^{τ} dσ] .$$ Now convert back to $v^{z}$ and $v^{\bar z}$ and to the differentials ($dz$, $d\bar z$) and you get to the correct result.

  • 1
    $\begingroup$ Hello, and welcome to Physics Stack Exchange. This site supports mathjax for writing mathematical symbols. Please use it instead of unicode. $\endgroup$
    – DanielSank
    Nov 25, 2015 at 18:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.