I am going through the calculations in arXiv:1312.7856. These involve a conformal map between the Minkowski Rindler Wedge ($\mathcal{R}$), given by

$X^1 \geq 0,X^+\geq 0,X^-\geq 0 \quad$ (with $X^{\pm}=X^1 \pm X^0$),

and the causal diamond ($\mathcal{D}$) given by:

$|x|+|t|\leq R$.

See the figure below: Mapping from the Rindler Wedge to the causal diamond.

The mapping is (supposedly) given by (equation 2.11 in arXiv:1102.0440):

$x^\mu = \frac{X^\mu - X \cdot X C^\mu} {1-2(X \cdot C) + (X \cdot X) (C \cdot C)} + 2 R^2 C^\mu \qquad$ with $C^\mu = (0,1/2R)$.

This is a special conformal transformation followed by a translation.

We can check where certain points are mapped to. For instance, we instantly see that $X^\mu = (X^0,X^1)=(0,0)$ maps to $(0,R)$.

However $(0,X^1)$ seems to map to $\left(0, \frac{X^1 - \tfrac{1}{2R}(X^1)^2}{1-\tfrac{X^1}{R} +\tfrac{X^2}{4R^2}} + R \right)$.

We know that it should end up withing $\mathcal{D}$ because our original point was in $\mathcal{R}$. So the first term needs to be negative (or zero) for all $R$. Taking $R=1$, we find that the first term is positive: $(0,1)$ goes to $(0,2+1),2 > R = 1)$, so this lies outside $\mathcal{D}$.

But it should be inside. Can anyone please tell me what is going wrong and how I can see that this map (or another conformal map) does have the desired effect of mapping $R$ to $D$?

P.S.: I have here presented the case for $D=2$ to simplify things, although the mapping should apply for an arbitrary number of dimensions $(1, 1, d-2)$ with $d-2$ 'trivial' dimensions.


There's a mistake in the paper. The correct transformation is $$x^\mu = 2R \frac{X^\mu - b^\mu X^2}{1-2b \cdot X + b^2 X^2} + Rb^\mu,$$ with $b^\mu = (0,-1,\vec{0})$.

I picked the minus sign in the definition of b to make the fact that it's made of a special conformal transformation manifest.

I found it in Rudolf Haag's book Local Quantum Physics in section V.4.2.

EDIT: Also notice that the equation makes no dimensional sense; the coordinates on the right should actually be $\frac{X^\mu}{2R}$.
Absorbing this sign into the parametrising vector $b^\mu$ gives exactly Casini-Huerta-Myers' transformation except for a sign.
[Chronology note: this edit came after OP's comment.]

  • $\begingroup$ Thanks for the answer, sorry for the delay in accepting it. I stopped using this account and forgot I had asked the question when I had found the answer myself. I should have given an update. Indeed there was a sign error in the paper. If I remember correctly the proposed transformation does map the left Rindler wedge to the causal diamond. (Although the authors stated that it would map the right wedge to the causal diamond. This difference results in a much needed minus sign later on in the calculation if you continue on with the "wrong" transformation. $\endgroup$ – Marten Jun 12 '17 at 8:52
  • $\begingroup$ Thanks for pointing out that the one in the paper is just off by a sign! The one I wrote here and the one in the paper with the corrected sign are related by a rescaling; I hadn't noticed this fact, and therefore thought the one in the paper was just completely wrong. Calling my transformation $f_S$ and the corrected one from Casini-Huerta-Myers $f_{CHM}$, the relation is $f_{CHM} (X^\mu) = f_S \left( \frac{X^\mu}{2R} \right)$. $\endgroup$ – Ronak M Soni Oct 2 '17 at 19:56

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