In an article by Penrose in Hughston and Ward "Advances in Twistor Theory", it is claimed that the twistor function

$$ f(Z^\alpha) = \log{\frac{Z^1Z^2 - Z^0Z^3}{Z^2Z^3}}$$

produces an anti-self-dual Coulomb field. To my knowledge this is precisely a Maxwell field with $\mathbf{B} = -i\mathbf{E}$ and $\mathbf{E} \propto \mathbf{r}/r^3$. I am trying to verify this using the contour integral formula but am getting the wrong result. Could someone point out my mistake?

Consider an anti-self-dual (ASD) Coulomb field $F$. Then we may write $$F_{ab} = F_{AA'BB'} = \phi_{AB}\epsilon_{A'B'}$$ by a standard argument. In particular then $$E_x = F_{01} = \frac{1}{2}(F_{00'00'} - F_{00'11'} + F_{11'00'} - F_{11'11'}) = - \phi_{01}$$

Using the usual integral formula for the Penrose transform we have

$$\phi_{01}(t,x,y,z) = \frac{1}{2\pi i}\oint \rho_x \frac{\partial}{\partial \omega^0}\frac{\partial}{\partial \omega^1} f(Z^\alpha) \pi_{E'}d\pi^{E'}$$


$$\phi_{01}(t,x,y,z) = \frac{1}{2\pi i} \oint \frac{\pi_{0'}\pi_{1'}}{(x^{10'}\pi_{0'}^2+(x^{11'}-x^{00'})\pi_{0'}\pi_{1'} - x^{01'}\pi_{1'}^2)^2} \pi_{E'}d\pi^{E'}$$

Now choosing local coordinates $\pi_{E'}=(1,\xi)$ and recalling that

$$\left(\begin{array}{cc} x^{00'} & x^{01'} \\ x^{10'} & x^{11'} \end{array}\right) =\frac{1}{\sqrt{2}} \left(\begin{array}{cc}t+x & y+iz \\ y-iz & t-x \end{array}\right) $$

we get

$$\phi_{01} = \frac{1}{2\pi i}\oint d\xi \frac{\xi}{(1/\sqrt{2}(y-iz)+\sqrt{2}x\xi - 1/\sqrt{2}(y+iz)\xi^2)^2}$$

which has double poles at

$$\xi = \frac{-\sqrt{2}x \pm \sqrt{2x^2 + 2y^2 + 2z^2}}{-\sqrt{2}(y+iz)} = \frac{x \mp r}{y + iz}$$

Denote these $\xi_1$ and $\xi_2$. Then we have (mistake here, very obviously, in hindsight!)

$$\phi_{01}(t,x,y,z) = \frac{1}{2\pi i}\oint d\xi \frac{\xi}{(\xi -\xi_1)^2(\xi - \xi_2)^2}$$

The residue at $\xi_1$ is \begin{align*} r_1 &= \rho_{\xi_1}\frac{d}{d\xi}\frac{\xi}{(\xi - \xi_2)^2} \\ &= \frac{1}{(\xi_1 - \xi_2)^2} -2 \frac{\xi_1}{(\xi_1-\xi_2)^3} \\ &= \frac{\xi_1 - \xi_2 - 2\xi_1}{(\xi_1-\xi_2)^3} \\ &= -\frac{x(y+iz)^2}{2r^3} \end{align*}

It is now manifest that no contour integral will reproduce the required Coulomb field. What have I done wrong? A hint would be greatly appreciated. Many thanks in advance!

(P.S. apologies for the length of this question! Also apologies for cross-posting from MathOveflow, I just get the feeling I'm more likely to find a response here!)


Your first mistake is $$ E_x = F_{01} = \phi_{01}$$ You have apparently confused spinor and vector indices here. The identity $E_x=F_{01}$ holds if $0,1$ are interpreted as the vector indices with four possible values corresponding to $0123=txyz$. But then you can't write that it's equal to $\phi_{01}$ because the latter has the spinor indices with two possible values. To transform them into each other, you need the spintensor $\sigma^\mu_{A\bar B}$ or whatever is their symbol for it. It's not true that only one component of the $\sigma$ spintensor is non-vanishing for each $\mu$.

You could have seen this mistake by noticing that your "translation" of $\phi_{01}$ only "turned on" the electric field components. But as you say yourself, the formula for $F_{\mu\nu}$ in terms of $\phi_{AB}$ creates an anti-self-dual field for any values of $\phi_{AB}$. It follows that in terms of $E_i,B_i$, all components of $\phi_{AB}$ are combinations of the components of the vector $E+iB$ (I hope that my relative sign agrees with your conventions). Similarly, the components of $E-iB$ would be encoded in the extra term $\phi_{\bar A\bar B}$ if you added it.

In particular, your conventions for the rotations are such that components of the spintensors are eigenstates of $J_x$ that rotates the $yz$-plane (see the matrix after "recalling that" in your question). The component $\phi_{01}$ that you calculated contains the "average" value of $J_x$ which is $J_x=0$; the other "extreme" components $\phi_{00}$ and $\phi_{01}$ carry $J_x=\pm 1$.

The component of $E+iB$ with $J_x=0$ is clearly a multiple of $E_x+iB_x$. So this is the combination you were calculating by your contour integral if it were right. (The components $\phi_{00}$ and $\phi_{11}$ are proportional to the $y\pm iz$ components of $E+iB$; there are four terms in each.) However, even dimensional analysis is enough to see that your evaluation of the integral wasn't right. The variable $\xi$ is dimensionless (ratio of two components of a spinor) and so is the measure $d\xi$. But if you look at the integral formula for $\phi_{01}$ and trace the dimensionful things such as $x,y,z$, you see that $\phi_{01}$ goes like $1/{\rm length}^2$, just like what is expected for the $\vec r / r^3$ result. Instead, your claimed residues are dimensionless – this result is off by two powers of the length.

Whatever way you got the "residues" had to be incorrect. I suspect that you did some mistake – not identifiable because you didn't show your calculation of the residues – related to residues of higher-order singularities. Probably some sign error in the power of the added $z^k$ in the contour integral $\oint dz$, or differentiation instead of integration, I don't know. A valid integration procedure couldn't have yielded a result that violates dimensional analysis. See


for the right way to calculate residues of higher singularities.

One may see by general arguments, not just specific calculation, that the integral has to produce the result proportional to what you expect.

  • $\begingroup$ Thank you so much for your answer! I disagree with you about the indices - I don't think I have mixed them up. I was implicitly using the spintensor. I have expanded my calculation in the question above, do you agree with me now? Thank you for pointing out the dimensional analysis - I should have spotted that. I'll take another look at how I calculated the residues, and add it to the question if I still can't see the error! $\endgroup$ – Edward Hughes Apr 25 '13 at 10:22
  • $\begingroup$ I've added my calculation for the residues and even though I now know it is wrong I cannot see the mistake! Can you see an obvious error? I'm sure it must be something trivial that my brain is refusing to see! Many thanks again. $\endgroup$ – Edward Hughes Apr 25 '13 at 10:32
  • $\begingroup$ Dear @EdwardHughes, you did mix up the two types of indices - otherwise the translation of components of $\phi_{AB}$ into combinations of $E_i$ and $B_i$ would only give you the anti-self dual combinations i.e. components of $E+iB$. ... Your "calculation of the residues" is totally wrong - you haven't used the function whose residues you're calculating at all! ;-) What you should have $\xi$-differentiated is not just the dull and universal (function-independent) expression $\xi / (\xi-\xi_2)^2$ but $(\xi-\xi)^2 f(z)$ where $f(z)$ is the function you're integrating. $\endgroup$ – Luboš Motl Apr 25 '13 at 16:37
  • $\begingroup$ Why don't you just use the right formula from en.wikipedia.org/wiki/… instead of the totally invalid, naive formula you used? $\endgroup$ – Luboš Motl Apr 25 '13 at 16:38
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    $\begingroup$ Excellent to hear on good news, @EdwardHughes $\endgroup$ – Luboš Motl Apr 27 '13 at 5:56

I simply misfactorised the quadratic - I knew it was a stupid mistake. I'm amazed that I didn't see it, but even more amazed that nobody else did! Here is the correct solution.

$$\phi_{01}(t,x,y,z) = \frac{1}{2\pi i}\oint d\xi \frac{\xi}{(x^{01'})^2(\xi -\xi_1)^2(\xi - \xi_2)^2}$$

The residue at $\xi_1$ is \begin{align*} r_1 &= \rho_{\xi_1}\frac{d}{d\xi}\frac{\xi}{(x^{01'})^2(\xi - \xi_2)^2} \\ &= \frac{1}{(\xi_1 - \xi_2)^2(x^{01'})^2} -2 \frac{\xi_1}{(x^{01'})^2(\xi_1-\xi_2)^3} \\ &= \frac{\xi_1 - \xi_2 - 2\xi_1}{(x^{01'})^2(\xi_1-\xi_2)^3} \\ &= -\frac{x(y+iz)^2}{2r^3(y+iz)^2} \end{align*}

This is now the correct answer :).


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