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I had some questions while reading the Chandrasekhar textbook "The Mathematical Theory of Black Holes", in particular about the scalars introduced to reformulate the Maxwell equations ($g^{ik} F_{ij;k}=0 \text{ and } F_{[ij;k]}=0$), given a complex null tetrad $(l^\mu,n^\mu,m^\mu,\bar{m}^\mu)$: $$ \phi_0 = F_{13} = F_{ij} l^i m^j \quad ; \quad \phi_1=\frac{1}{2}(F_{12}+F_{43})=\frac{1}{2} F_{ij}(l^i n^j + \bar{m}^i m^j) \quad ; \quad \phi_2=F_{42}=F_{ij}\bar{m}^i n^j$$

leading to these four equations (in Chandrasekhar's notation for intrinsic derivatives): $$\phi_{1|1}-\phi_{0|4}=0 \quad ; \quad \phi_{2|1}-\phi_{1|4}=0 \quad ; \quad \phi_{1|3}-\phi_{0|2}=0 \quad ; \quad \phi_{2|3}-\phi_{1|2}=0 \quad $$

How come the scalars can be written in terms of a subset of the $F_{ij}$ tensor elements, assuming this is without knowing which of the tetrad components vanish (e.g. for $\phi_0$ it seems it is taken that only $l^1$ and $m^3$ are non-vanishing) ?

Also, is it possible that these equations are equivalent to contracting $g^{ik} F_{ij;k}=0 $ with each of the tetrad vectors $(l^j,n^j,m^j,\bar{m}^j)$ ?

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The short answer to your first question is that the Newman-Penrose Maxwell equations are the different components of the general relativistic Maxwell equations in spinor form.

To see this, note that your second equation ($F_{[ab;c]} = 0$) can be expressed as ${}^*F^{ab}{}_{;b} = 0$, similar to the first one $F^{ab}{}_{;b} = 0$ (these are of course the source free equations). Then, using the symmetries of the field tensor ($F_{ab} = F_{[ab]}$) we can expand it in spinor form as (see below for a derivation if you are unfamiliar with spinor algebra): $$ F_{ab} = \phi_{AB}\epsilon_{A'B'} + \overline{\phi}_{A'B'}\epsilon_{AB}, $$ where $\phi_{AB} = \phi_{(AB)}$, and $\epsilon_{AB}$ is the spinor metric (Levi-Civita symbol). And since the dual in spinor form becomes (again see further below for a derivation) $$ {}^*F_{ab} = i(\phi_{AB}\epsilon_{A'B'} - \overline{\phi}_{A'B'}\epsilon_{AB}), $$ the two Maxwell equations in spinor form reduce to $$\tag{1} \phi^{AB}{}_{;A'B} = 0. $$

Now, since the field spinor $\phi_{AB}$ is symmetric its independent components are given by $$ \begin{aligned} \phi_0 &\equiv \phi_{AB}o^Ao^B, & \phi_1 &\equiv \phi_{AB}o^A\iota^B, & \phi_2 &\equiv \phi_{AB}\iota^A\iota^B, \\ &= F_{ab}l^am^b, & &= \frac{1}{2}F_{ab}(l^an^b + \overline{m}^am^b), & &= F_{ab}\overline{m}^an^b, \\ &= F_{13}, & &= \frac{1}{2}(F_{12} + F_{43}), & &= F_{42}, \end{aligned} $$ where $(o^A,\iota^A)$ is our spinor dyad. Here the second row follows by the identification of null tetrad with spinor pairs $(l^a,n^a,m^a,\overline{m}^a) = (o^Ao^{A'},\iota^A\iota^{A'},o^A\iota^{A'},\iota^Ao^{A'})$, and the third line is obtained by simply noting that e.g. $l^a$ is the first null frame basis vector and $m^a$ is the third so $F_{ab}l^am^b = F_{13}$ by definition.

The components of the spinor Maxwell equation $(1)$ will include directional derivatives of the components along with products of the components and different spin-coefficients, according to the standard definition of the covariant derivative (which Chandrasekhar calls intrinsic derivatives). The scalar versions are given on the next page in his book, so I will not write them out here.

To clarify then: the four spinor equations you give form a system of equations equivalent to the source-free Maxwell equations. So you are on the right track in your last question, but the equations are in fact formed from both $F^{ab}{}_{;b} = 0$ and $F_{[ab;c]} = 0$.


Expanding the field tensor in spinor form: Using the fact that the spinor metric $\epsilon_{AB}$ is skew-symmetric (it is usually taken to be normalized as the Levi-Civita symbol, but this is not necessary) we immediately get $$ \epsilon_{AB}\epsilon_{CD} + \epsilon_{AD}\epsilon_{BC} + \epsilon_{AC}\epsilon_{DB} = 0, $$ or more suggestively, upon raising a few indices $$ \delta_A^C\delta_B^D - \delta_B^C\delta_A^D = \epsilon_{AB}\epsilon^{CD}. $$ So for some spinor $Q_{\cdots AB \cdots}$ we get by spliting the index pair $AB$ into the symmetric and skew-symmetric part $$\tag{2} Q_{\cdots AB \cdots} = Q_{\cdots (AB) \cdots} + \frac{1}{2}\epsilon_{AB}Q_{\cdots C}{}^C{}_{\cdots}. $$ Doing some explicit algebra on the spinor form of an skew-symmetric tensor $T_{ab} = T_{[ab]}$ we get \begin{align} T_{ab} &= \frac{1}{2}(T_{ab} - T_{ba}) \\ &= \frac{1}{2}(T_{AA'BB'} - T_{BB'AA'}) \\ &= \frac{1}{2}\left((T_{ABA'B'} - T_{ABB'A'}) + (T_{ABB'A'} - T_{BAB'A'})\right) \tag{3}\\ &= T_{(AB)[A'B]} + T_{[AB](A'B')} \\ &= \frac{1}{2}(T_{(AB)C'}{}^{C'}\epsilon_{A'B'} + T_{(A'B')C}{}^C\epsilon_{AB}), \end{align} where the symmetrization of the remaining index pair follows since the trace of an anti-symmetric tensor is zero. If $T_{ab}$ is real, the two terms must be complex conjugates. So for $F_{ab}$ we define $\phi_{AB} = \frac{1}{2}F_{(AB)C'}{}^C$. To get the desired form.


The spinor form of the dual: To do this we must note that the spinor equivalent of a tensor is given by maps $\varsigma_p : T_pM \to (\mathcal{S} \otimes \overline{\mathcal{S}})$, where $\mathcal{S}$ is our spin space. Dropping, the $p$ subscript and adopting index notation for this map, we demand that $\varsigma$ satisfies $$\tag{4} g_{ab} = \varsigma_a{}^{AA'}\varsigma_b{}^{BB'}\epsilon_{AB}\epsilon_{A'B'}. $$ The dual ${}^*T_{ab}$ of a skew-symmetric tensor $T_{ab} = T_{[ab]}$ is defined as \begin{align*} {}^*T_{ab} = \frac{1}{2}\sqrt{-g}\epsilon_{abcd}T^{cd}, \end{align*} where $\epsilon_{abcd}$ is the Levi-Civita symbol and $g$ is the determinant of the metric. To write this in spinor form note that the product $\sqrt{-g}\epsilon_{abcd}$ forms the Levi-Civita (pseudo-)tensor, whence from the definition \begin{align}\tag{5} \sqrt{-g}\epsilon_{abcd} &= \sqrt{-g}\epsilon_{abcd}\varsigma^a{}_{AA'}\varsigma^b{}_{BB'}\varsigma^c{}_{CC'}\varsigma^d{}_{DD'} \\ &= \sqrt{-g}\varsigma\epsilon_{AA'BB'CC'DD'}, \end{align} where $\varsigma$ is the determinant of $\varsigma^a{}_{AA'}$ with $AA'$ treated as a single vector index, and $\epsilon_{AA'BB'CC'DD'}$ correspondingly is the Levi-Civita symbol with the indices treated the same way. Now, from the skew-symmetry of $\epsilon_{AB}$ we have $\epsilon_{AB} = \lambda\widetilde{\epsilon}_{AB}$ for some $\lambda$, where $\widetilde{\epsilon}_{AB}$ is the Levi-Civita symbol. Therefore, one can easily verify that \begin{align}\tag{6} \pm|\lambda|^4\epsilon_{AA'BB'CC'DD'} &= \epsilon_{AD}\epsilon_{BC}\epsilon_{A'C'}\epsilon_{B'D'} - \epsilon_{AC}\epsilon_{BD}\epsilon_{A'D'}\epsilon_{B'C'}, \end{align} with the sign dependent on how the orientation of $TM$ is related to the spinor index pairs by $\varsigma_a{}^{AA'}$. We can always choose to have e.g. a plus sign (which corresponds to the construction $(l^a,n^a,m^a,\overline{m}^a) = (o^Ao^{A'},\iota^A\iota^{A'},o^A\iota^{A'},\iota^Ao^{A'})$ or its equivalents), but in either case $(4)$ and $(6)$ yield $g = \varsigma^{-2}|\lambda|^8$. Plugging this result into $(5)$, using $(6)$ again, and raising $c$ and $d$ finally yields \begin{align}\tag{7} \sqrt{-g}\epsilon_{abef}g^{ec}g^{df} = \pm i\left(\delta^D_A\delta^C_B\delta^{C'}_{A'}\delta^{D'}_{B'} - \delta^C_A\delta^D_B\delta^{D'}_{A'}\delta^{C'}_{B'}\right), \end{align} whence we can write \begin{align*} {}^*T_{ab} &= \pm\frac{i}{2}\left(T_{(AB)C'}{}^{C'}\epsilon_{A'B'} - {T}_{(A'B')C}{}^C\epsilon_{AB}\right), \end{align*} by applying $(7)$ to $(3)$, in effect adding $\pm i$ as a factor to $\epsilon_{A'B'}$ and $\mp i$ as a factor to $\epsilon_{AB}$. Using the previously derived definition of $\phi_{AB}$, and selecting a "$+$-orientation," we get the desired expression.

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  • $\begingroup$ I appreciate your detailed answer in the form of the spinor formalism. I had trouble understanding that initially in the textbook and was more comfortable with using tetrads. I think the point of my first question was not about how come the scalars are written in terms of $F_{ij}$ but rather by "subset" I meant why is the only "surviving" term of $F_{ij}l^i m^j$ for instance the 13 term, and not say 14 or 21 $\endgroup$ – N.E. Jul 16 '18 at 13:46
  • $\begingroup$ @N.E. I tried adding more information, but I must confess I'm not entirely sure what you mean. Perhaps you have mistakenly identified the index numbering with the corresponding orthonormal frame (tetrad)? For example $F_{ab}l^am^b = F_{13}$ by definition of the different components. $\endgroup$ – Erik Jörgenfelt Jul 16 '18 at 14:04
  • $\begingroup$ Based on $F_{ab} l^a m^b = F_{13}$ for instance, it seems as if only $l^a=(l^1,0,0,0)$, $m^b=(0,0,m^3,0)$ and $l^1 m^3=1$, which if I'm not mistaken are unwarranted assumptions about which tetrad components are non-vanishing $\endgroup$ – N.E. Jul 16 '18 at 15:03
  • $\begingroup$ @N.E. note that the index numbers in e.g. $F_{13}$ are in the null frame. So that the only non-vanishing component of $l^a$ is $l^1 = 1$ by definition. Similarly $n^a = (0,1,0,0)$, $m^a = (0,0,1,0)$, and $\overline{m}^a = (0,0,0,1)$, all by definition. $\endgroup$ – Erik Jörgenfelt Jul 16 '18 at 15:11

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