Straightforward twistor theory/spinor calculus question

[Bballer152]

Not sure where to post this, but this question has been frustrating me for hours now because I'm pretty sure it has a very straightforward answer. I'll reproduce the problem in a moment, but for those with the Huggett and Tod's Intro to Twistor Theory, I don't understand how we get the expression immediately before eq. 8.5 from the given info.

We're given the twistor function $$f(Z^\alpha)=\frac{1}{(A_\alpha Z^\alpha)(B_\beta Z^\beta)}$$ and we restrict the function to the line $$L_x$$ so that

$$A_\alpha Z^\alpha = (iA_A x^{AA'}+A^{A'})\pi_{A'}\equiv \alpha^{A'} \pi_{A'}$$ and
$$B_\alpha Z^\alpha = (iB_A x^{AA'}+B^{A'})\pi_{A'}\equiv \beta^{A'} \pi_{A'}$$.

Now comes the part I don't understand. Apparently, by supposing that the dual twistors $$A_\alpha$$ and $$B_\alpha$$ meet in the line $$L_y,$$ then we should get $$\alpha^{A'}\beta_{A'}=\frac{1}{2}A_AB^A(x-y)^2.$$ When I do this I get a similar answer but my contractions don't work out this way. In particular, my spinors indices contract in a different order and I have no idea how to "split them up", if that makes sense. Any help would be GREATLY appreciated!

 

[AndyW]

Thank you for your question regarding twistor theory. I understand that you are struggling with understanding how to obtain the expression before equation 8.5 in Huggett and Tod's Intro to Twistor Theory. I will try my best to explain this to you.

Firstly, let's review the given information. We have the twistor function f(Z^\alpha)=\frac{1}{(A_\alpha Z^\alpha)(B_\beta Z^\beta)}, which we are restricting to the line L_x. This means that A_\alpha Z^\alpha and B_\alpha Z^\alpha are both on the line L_x. We can rewrite these expressions as A_\alpha Z^\alpha = (iA_A x^{AA'}+A^{A'})\pi_{A'} and B_\alpha Z^\alpha = (iB_A x^{AA'}+B^{A'})\pi_{A'}.

Next, we are supposing that the dual twistors A_\alpha and B_\alpha meet on the line L_y. This means that we can set \alpha^{A'}=\beta_{A'} and obtain the expression \alpha^{A'}\beta_{A'}=\frac{1}{2}A_AB^A(x-y)^2.

Now, let's look at how we can get this expression. We can start by expanding the expression for A_\alpha Z^\alpha and B_\alpha Z^\alpha on the line L_x. This gives us A_\alpha Z^\alpha = (iA_A x^{AA'}+A^{A'})\pi_{A'}=(iA_A x^{AA'}+A^{A'})\pi_{A'} and B_\alpha Z^\alpha = (iB_A x^{AA'}+B^{A'})\pi_{A'}=(iB_A x^{AA'}+B^{A'})\pi_{A'}.

Next, we can substitute in the expression \alpha^{A'}=\beta_{A'} and obtain A_\alpha Z^\alpha = (iA_A x^{AA'}+A^{A'})\pi_{A'}=(iA_A x^{AA'}+A^{A'})\pi_{A'} and B_\alpha Z^\alpha = (iB_A x^{AA'}+B^{A'})\pi_{A'}=(iB_A x^{AA'}+B^{A'})\pi_{A'}