Making sense of Dirac's rotation operator in "General Theory of Relativity"

[Kostik]

TL;DR Summary

Dirac defines a rotation operator $R$ which appears to be a simple rotation in the $x$-$y$ plane, while analyzing plane waves traveling in the $z$ direction. However, when he "applies" $R$ to the quantity $u_{\mu\nu}$ his results make no sense.

In Dirac's "General Theory of Relativity", Chap. 34 on the polarization of gravitational waves, he introduces a rotation operator $R$, which appears to be a simple $\pi/2$ rotation, since
$$R
\begin{pmatrix}
A_0 \\
A_1 \\
A_2 \\
A_3
\end{pmatrix}
=
\begin{pmatrix}
1 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & -1 & 0 & 1 \\
0 & 0 & 0 & 1
\end{pmatrix}
\begin{pmatrix}
A_0 \\
A_1 \\
A_2 \\
A_3
\end{pmatrix}
=
\begin{pmatrix}
A_0 \\
A_2 \\
-A_1 \\
A_3
\end{pmatrix}$$ He then "applies" this operator to $$u_{\mu\nu} = \frac{dg_{\mu\nu}(\xi)}{d\xi} \quad\quad \xi=l_\sigma x^\sigma.$$ We are working with plane waves, so $g_{\mu\nu}$ is a function of the single variable $\xi=l_\sigma x^\sigma$, where $l_\sigma$ is the wave vector. Here, Dirac is looking at waves in the $x^3$ direction, so one may take $\xi = x^0-x^3$.

The obvious thing to do (and this is exactly how Weinberg proceeds in his text, see pp. 256-257) is to write
$${R_\mu}^\nu =
\begin{pmatrix}
1 & 0 & 0 & 0 \\
0 & \cos\theta & \sin\theta & 0 \\
0 & -\sin\theta & \cos\theta & 1 \\
0 & 0 & 0 & 1
\end{pmatrix}
$$ and, corresponding to the vector equation $A'_\mu = {R_\mu}^\rho A_\rho$ we can define
$$u'_{\mu\nu} = {R_\mu}^\rho {R_\nu}^\sigma u_{\rho\sigma}.$$ We have at hand various relations among the $u_{\mu\nu}$ arising from using harmonic coordinates, namely $u_{22}=-u_{11}$, with the result being that we can calculate $$u'_{11} = \cos{2\theta} \, u_{11} + \sin{2\theta} \, u_{12}$$ $$u'_{12} = -\sin{2\theta} \, u_{11} + \cos{2\theta} \, u_{12} .$$ Dirac, however, "applies" his rotation operator to $u_{11}$ and $u_{12}$, and he states (without proof or explanation) $$Ru_{11}=u'_{11}=u_{12}+u_{21}=2u_{12}$$ $$Ru_{12}=u'_{12} = u_{22}-u_{11}=-2u_{11}.$$ There is no rotation angle that will produce these relations stated by Dirac. Therefore, I am at a loss to understand how Dirac "applies" his rotation $R$ to $u_{\mu\nu}$. Does anyone who has read his text have a clue?

 

[martinbn]

It seems that $R$ is applied to the first index, then to the second, and the results are added.

For example $R u_{12}$ appliednto the first index gives $u_{22}$ and to the second $-u_{11}$. So $R u_{12} = u_{22} - u_{11}$.

 

[Kostik]

It seems that $R$ is applied to the first index, then to the second, and the results are added.

For example $R u_{12}$ appliednto the first index gives $u_{22}$ and to the second $-u_{11}$. So $R u_{12} = u_{22} - u_{11}$.

I noticed this also, but why? What is the point of it?

 

[martinbn]

I thought your question is how the operator works. Are you asking why this is the rule or why Dirac needs this opwrator?

 

[Kostik]

I thought your question is how the operator works. Are you asking why this is the rule or why Dirac needs this opwrator?

Both. What does this operator accomplish. It is not a rotation, and not any kind of sensible generalization of its action on a vector.

 

[martinbn]

Both. What does this operator accomplish. It is not a rotation, and not any kind of sensible generalization of its action on a vector.

Of course it is a rotation! On a vector $A$ it acts as $R A_1=A_2$ and $RA_2=A_1$. This is a rotation.

 

[Kostik]

Obviously. I'm talking about the "prescription" for acting on $u_{\mu\nu}$. The relations Dirac gives, and the formula that it apparent "It seems that $R$ is applied to the first index, then to the second, and the results are added" is NOT a rotation. The only sensible extension to $u_{\mu\nu}$ is the one I gave above, and that does not produce Dirac's "prescription" for any $\theta$.

 

[martinbn]

Obviously. I'm talking about the "prescription" for acting on $u_{\mu\nu}$. The relations Dirac gives, and the formula that it apparent "It seems that $R$ is applied to the first index, then to the second, and the results are added" is NOT a rotation. The only sensible extension to $u_{\mu\nu}$ is the one I gave above, and that does not produce Dirac's "prescription" for any $\theta$.

NO. It is an infinitesimal rotation. He says so in the book! When you differentiate what you thought it shoud have been you get what it has to be.

 

[Kostik]

Can you be a little more explicit? Differentiate what with respect to what to get what? Differentiate ${R_\rho}^\mu (\theta) {R_\sigma}^\nu (\theta) u_{\mu\nu} = u’_{\rho\sigma}$?

 

[martinbn]

Can you be a little more explicit? Differentiate what with respect to what to get what? Differentiate ${R_\rho}^\mu (\theta) {R_\sigma}^\nu (\theta) u_{\mu\nu} = u’_{\rho\sigma}$?

This is how it acts on $g_{\mu\nu}$, and the $u$ is the derivative of $g$. That's why $R$ operates on it that way.

 

[Kostik]

This is how it acts on $g_{\mu\nu}$, and the $u$ is the derivative of $g$. That's why $R$ operates on it that way.

That doesn’t make sense. $g_{\mu\nu}$ is a function of $\xi = l_\sigma x^\sigma$, but not the rotation operators ${R_\mu}^\rho$. So the derivative will cruise right through those.

 

[Kostik]

Is there any discussion of the $Ru_{\mu\nu}$ case?

 

[anuttarasammyak]

Tensor $u_{\mu\nu}$ is regarded as vector for each index $\mu$ and $\nu$, or like a product $a_\mu b_\nu$.
$$R a_\mu b_\nu = (R a_\mu) b_\nu + a_\mu (R b_\nu )$$
With the rule explained " From 1 to 2 : +, From 2 to 1 : - ", Dirac's result is ovbious.

 

[martinbn]

That doesn’t make sense. $g_{\mu\nu}$ is a function of $\xi = l_\sigma x^\sigma$, but not the rotation operators ${R_\mu}^\rho$. So the derivative will cruise right through those.

The $g$ lives in a tensor bundle, the $u$ in its tangent bundle. $R$ is the derivative (or if you preffer the dfferential) of the operator that you think it should be.

 

[Kostik]

Tensor $u_{\mu\nu}$ is regarded as vector for each index $\mu$ and $\nu$, or like a product $a_\mu b_\nu$.
$$R a_\mu b_\nu = (R a_\mu) b_\nu + a_\mu (R b_\nu )$$
With the rule explained " From 1 to 2 : +, From 2 to 1 : - ", Dirac's result is ovbious.

Yes the rule is obvious. But how is it a rotation in this sense?

 

[Vanadium 50]

OP, you have a lot of questions. Perhaps Dirac's book is confusing you, and another one would be better. Many people like Schutz, and there's always MTW, which adopts a unique approach that some really like.

 

[Kostik]

I’ve made it to the end of Ch 34 with just 2 pages to go, dotting every i and crossing every t, so I think I’ve done pretty well. It would be a shame not to get all the way through it.

 

[renormalize]

Yes the rule is obvious. But how is it a rotation in this sense?

@Kostik you have good reason to be confused! Dirac is employing here what is, at least in my experience, a rare formalism for small rotations that I will label "Dirac Infinitesimal" rotations $R_{DI}$. He introduces these in his classic Principles of Quantum Mechanics 4th Ed. on pg. 141:

Starting from the usual 2D passive rotation matrix:$$
R\left(\phi\right)=\left(\begin{array}{cc}
\cos\phi & \sin\phi\\
-\sin\phi & \cos\phi
\end{array}\right)
$$he defines his infinitesimal rotation to be the constant matrix:$$
R_{DI}\equiv\lim_{\delta\phi\rightarrow0}\left(\frac{R\left(\delta\phi\right)-I}{\delta\phi}\right)=\left(\begin{array}{cc}
0 & 1\\
-1 & 0
\end{array}\right)
$$So for the vector $A$ and symmetric tensor $u$:$$
A\equiv\left(\begin{array}{c}
A_{1}\\
A_{2}
\end{array}\right),\quad u\equiv\left(\begin{array}{cc}
u_{11} & u_{12}\\
u_{12} & u_{22}
\end{array}\right)
$$a Dirac infinitesimal rotation gives:$$
R_{DI}\left(A\right)\equiv R_{DI}A=\left(\begin{array}{cc}
0 & 1\\
-1 & 0
\end{array}\right)\left(\begin{array}{c}
A_{1}\\
A_{2}
\end{array}\right)=\left(\begin{array}{c}
A_{2}\\
-A_{1}
\end{array}\right)
$$and$$
R_{DI}\left(u\right)\equiv R_{DI}u+uR_{DI}^{T}=\left(\begin{array}{cc}
0 & 1\\
-1 & 0
\end{array}\right)\left(\begin{array}{cc}
u_{11} & u_{12}\\
u_{12} & u_{22}
\end{array}\right)+\left(\begin{array}{cc}
u_{11} & u_{12}\\
u_{12} & u_{22}
\end{array}\right)\left(\begin{array}{cc}
0 & -1\\
1 & 0
\end{array}\right)=\left(\begin{array}{cc}
2u_{12} & u_{22}-u_{11}\\
u_{22}-u_{11} & -2u_{12}
\end{array}\right)
$$which agrees with the results on pg. 67 of his General Theory of Relativity.

 

[Kostik]

Thank you for this. What is the idea or motivation behind the strange definition of $R_{DI}(u)$? Weinberg’s generalization is so much more intuitive (and leads to the identification of wave La with helicity $\pm 2, \pm 1, 0$).

 

[martinbn]

Thank you for this. What is the idea or motivation behind the strange definition of $R_{DI}(u)$?

It is called the derivative!!!

 

[renormalize]

What is the idea or motivation behind the strange definition of $R_{DI}(u)$? Weinberg’s generalization is so much more intuitive (and leads to the identification of wave La with helicity $\pm 2, \pm 1, 0$).

In §34 Dirac has already used coordinate (gauge) freedom to eliminate the unphysical helicity-0 and -1 components, which is why he is working with a 2x2 tensor. To compare their approaches, consider Weinberg's eq.(10.2.11) for the rotation of the polarization tensor $e_{\rho\sigma}$ as restricted to the same 2x2 piece that Dirac treats, but take the rotation to be infinitesimal:$$\delta e\equiv e^{\prime}-e=R\left(\delta\theta\right)eR^{T}\left(\delta\theta\right)-e=\left(I+R_{DI}\delta\theta\right)e\left(I+R_{DI}^{T}\delta\theta\right)-e=\left(R_{DI}e+eR_{DI}^{T}\right)\delta\theta=R_{DI}\left(e\right)\delta\theta\tag{10.2.11'}$$This illustrates why $R_{DI}$ isn't really "strange", it simply captures the essential "core" of an infinitesimal rotation by ignoring the usual identity-matrix part. In components we have:$$
\left(\begin{array}{cc}
\delta e_{11} & \delta e_{12}\\
\delta e_{12} & \delta e_{22}
\end{array}\right)=\left(\begin{array}{cc}
2e_{12} & e_{22}-e_{11}\\
e_{22}-e_{11} & -2e_{12}
\end{array}\right)\delta\theta\tag{1}
$$Following Weinberg eq.(10.2.15), now define the complex polarization parameter $\mathbf{e}_{\pm}\equiv e_{11}\mp ie_{12}=-e_{22}\mp ie_{12}$. In view of (1), under an infinitesimal rotation $\mathbf{e}_{\pm}$ changes by:$$\delta\mathbf{e}_{\pm}=\pm2i\mathbf{e}_{\pm}\delta\theta$$or:$$\mathbf{e}_{\pm}^{\prime}=\exp\left(\pm2i\delta\theta\right)\mathbf{e}_{\pm}$$ which is exactly the infinitesimal form of Weinberg's eq.(10.2.12) and verifies that the 2x2 piece of the polarization tensor carries helicity-2. This of course is the same result that Dirac gets. So to me, the choice of using the Dirac rotation $R_{DI}$ versus Taylor-expanding the usual rotation matrix $R$ is really just a matter of convenience and/or familiarity.

 

[DrGreg]

Dirac's terminology is unusual. The more usual terminology for $r_z$ is "infinitesimal generator of the rotation group $R(\phi)$".

 

[Kostik]

@Kostik you have good reason to be confused! Dirac is employing here what is, at least in my experience, a rare formalism for small rotations that I will label "Dirac Infinitesimal" rotations $R_{DI}$. He introduces these in his classic Principles of Quantum Mechanics 4th Ed. on pg. 141:
View attachment 349572
Starting from the usual 2D passive rotation matrix:$$
R\left(\phi\right)=\left(\begin{array}{cc}
\cos\phi & \sin\phi\\
-\sin\phi & \cos\phi
\end{array}\right)
$$he defines his infinitesimal rotation to be the constant matrix:$$
R_{DI}\equiv\lim_{\delta\phi\rightarrow0}\left(\frac{R\left(\delta\phi\right)-I}{\delta\phi}\right)=\left(\begin{array}{cc}
0 & 1\\
-1 & 0
\end{array}\right)
$$So for the vector $A$ and symmetric tensor $u$:$$
A\equiv\left(\begin{array}{c}
A_{1}\\
A_{2}
\end{array}\right),\quad u\equiv\left(\begin{array}{cc}
u_{11} & u_{12}\\
u_{12} & u_{22}
\end{array}\right)
$$a Dirac infinitesimal rotation gives:$$
R_{DI}\left(A\right)\equiv R_{DI}A=\left(\begin{array}{cc}
0 & 1\\
-1 & 0
\end{array}\right)\left(\begin{array}{c}
A_{1}\\
A_{2}
\end{array}\right)=\left(\begin{array}{c}
A_{2}\\
-A_{1}
\end{array}\right)
$$and$$
R_{DI}\left(u\right)\equiv R_{DI}u+uR_{DI}^{T}=\left(\begin{array}{cc}
0 & 1\\
-1 & 0
\end{array}\right)\left(\begin{array}{cc}
u_{11} & u_{12}\\
u_{12} & u_{22}
\end{array}\right)+\left(\begin{array}{cc}
u_{11} & u_{12}\\
u_{12} & u_{22}
\end{array}\right)\left(\begin{array}{cc}
0 & -1\\
1 & 0
\end{array}\right)=\left(\begin{array}{cc}
2u_{12} & u_{22}-u_{11}\\
u_{22}-u_{11} & -2u_{12}
\end{array}\right)
$$which agrees with the results on pg. 67 of his General Theory of Relativity.

Many thanks @renormalize, your two posts above are extremely helpful!!

Everything makes sense, although one thing remains perplexing. From the definition of $R_{DI}$ as a limit, it is clear that, as you wrote above:
$$R_{DI}\equiv\lim_{\delta\phi\rightarrow0}\left(\frac{R\left(\delta\phi\right)-I}{\delta\phi}\right)=\left(\begin{array}{cc}
0 & 1\\
-1 & 0
\end{array}\right)$$ Yet the "straightforward" matrix multiplication $R_{DI}u$ where
$$u\equiv\left(\begin{array}{cc}
u_{11} & u_{12}\\
u_{12} & u_{22}
\end{array}\right)$$ does not give the correct answer. Instead, one must use the form
$$R_{DI}u \equiv R_{DI}u+uR_{DI}^{T}$$ which you rigorously derived by matrix multiplication. This is puzzling.

 

[renormalize]

Yet the "straightforward" matrix multiplication $R_{DI}u$ where
$$u\equiv\left(\begin{array}{cc}
u_{11} & u_{12}\\
u_{12} & u_{22}
\end{array}\right)$$ does not give the correct answer. Instead, one must use the form
$$R_{DI}u \equiv R_{DI}u+uR_{DI}^{T}$$ which you rigorously derived by matrix multiplication. This is puzzling.

Don't forget that $u$ is a rank-2 tensor, which means that you have to apply the rank-1 (vector) rule $R_{DI}A$ twice, once to each index of $u$, and then add the results to find the total change $R_{DI}u+uR_{DI}^{T}$ under an infinitesimal rotation by $\delta\phi$. (As an aside, it doesn't make sense to ponder products like $R_{DI}uR_{DI}^T$, since those are of quadratic-order $\delta\phi^2$, not linear).
And notably, since $R_{DI}$ is antisymmetric, $R_{DI}^T=-R_{DI}$, the change in $u$ under an infinitesimal rotation $\delta u=R_{DI}\left(u\right)\delta\phi=\left(R_{DI}u+uR_{DI}^{T}\right)\delta\phi$ can be written in the suggestive "matrix mechanics" form:$$\frac{du}{d\phi}=\left[R_{DI},u\right]$$In words, this says that the "evolution" of the operator $u$ under rotation is given by minus the commutator of $u$ with the operator $R_{DI}$. So it's quite natural for Dirac to consider this operator in his treatise on quantum mechanics.

 

[Kostik]

@Kostik you have good reason to be confused! Dirac is employing here what is, at least in my experience, a rare formalism for small rotations that I will label "Dirac Infinitesimal" rotations $R_{DI}$. He introduces these in his classic Principles of Quantum Mechanics 4th Ed. on pg. 141:
View attachment 349572
Starting from the usual 2D passive rotation matrix:$$
R\left(\phi\right)=\left(\begin{array}{cc}
\cos\phi & \sin\phi\\
-\sin\phi & \cos\phi
\end{array}\right)
$$he defines his infinitesimal rotation to be the constant matrix:$$
R_{DI}\equiv\lim_{\delta\phi\rightarrow0}\left(\frac{R\left(\delta\phi\right)-I}{\delta\phi}\right)=\left(\begin{array}{cc}
0 & 1\\
-1 & 0
\end{array}\right)
$$So for the vector $A$ and symmetric tensor $u$:$$
A\equiv\left(\begin{array}{c}
A_{1}\\
A_{2}
\end{array}\right),\quad u\equiv\left(\begin{array}{cc}
u_{11} & u_{12}\\
u_{12} & u_{22}
\end{array}\right)
$$a Dirac infinitesimal rotation gives:$$
R_{DI}\left(A\right)\equiv R_{DI}A=\left(\begin{array}{cc}
0 & 1\\
-1 & 0
\end{array}\right)\left(\begin{array}{c}
A_{1}\\
A_{2}
\end{array}\right)=\left(\begin{array}{c}
A_{2}\\
-A_{1}
\end{array}\right)
$$and$$
R_{DI}\left(u\right)\equiv R_{DI}u+uR_{DI}^{T}=\left(\begin{array}{cc}
0 & 1\\
-1 & 0
\end{array}\right)\left(\begin{array}{cc}
u_{11} & u_{12}\\
u_{12} & u_{22}
\end{array}\right)+\left(\begin{array}{cc}
u_{11} & u_{12}\\
u_{12} & u_{22}
\end{array}\right)\left(\begin{array}{cc}
0 & -1\\
1 & 0
\end{array}\right)=\left(\begin{array}{cc}
2u_{12} & u_{22}-u_{11}\\
u_{22}-u_{11} & -2u_{12}
\end{array}\right)
$$which agrees with the results on pg. 67 of his General Theory of Relativity.

Many thanks @renormalize, your two posts above are extremely helpful!!

Everything makes sense, although one thing remains perplexing. From the definition of $R_{DI}$ as a limit, it is clear that

Don't forget that $u$ is a rank-2 tensor, which means that you have to apply the rank-1 (vector) rule $R_{DI}A$ twice, once to each index of $u$, and then add the results to find the total change $R_{DI}u+uR_{DI}^{T}$ under an infinitesimal rotation by $\delta\phi$. (As an aside, it doesn't make sense to ponder products like $R_{DI}uR_{DI}^T$, since those are of quadratic-order $\delta\phi^2$, not linear).
And notably, since $R_{DI}$ is antisymmetric, $R_{DI}^T=-R_{DI}$, the change in $u$ under an infinitesimal rotation $\delta u=R_{DI}\left(u\right)\delta\phi=\left(R_{DI}u+uR_{DI}^{T}\right)\delta\phi$ can be written in the suggestive "matrix mechanics" form:$$\frac{du}{d\phi}=\left[R_{DI},u\right]$$In words, this says that the "evolution" of the operator $u$ under rotation is given by minus the the commutator of $u$ with the operator $R_{DI}$. So it's quite natural for Dirac to consider this operator in his treatise on quantum mechanics.

Yes, got it. It's a notational issue. $R_{DI}$ is a rank-2 operator on vectors and a rank-4 operator on rank-2 tensors. In the expression $R_{DI}u$, $R_{DI}$ is not the 4x4 matrix shown above - it is a rank-4 quantity.

One last question, if I may. How exactly are Weinberg's helicities connected to Dirac's eigenvalues? Clearly they are connected. But in Weinberg's treatment, one needs to construct specific functions $u_{11}-iu_{12}$, etc., which don't show up in Dirac's treatment. His eigenvalues apply directly to various portions of $u_{\mu\nu}$.

 

[renormalize]

One last question, if I may. How exactly are Weinberg's helicities connected to Dirac's eigenvalues? Clearly they are connected. But in Weinberg's treatment, one needs to construct specific functions $u_{11}-iu_{12}$, etc., which don't show up in Dirac's treatment. His eigenvalues apply directly to various portions of $u_{\mu\nu}$.

Actually, both the Weinberg and Dirac helicities are eigenvalues, just those of two distinct (but related) eigensystems.
In terms of Dirac's $R_{DI}$, Weinberg's helicities are the eigenvalues $\omega_i$ of the eigensystem:$$iR_{DI}\left(w_{i}\right)\equiv i\left(R_{DI}w_{i}+w_{i}R_{DI}^{T}\right)=\omega_{i}w_{i}\tag{W}$$where the $w_i$ are the eigentensors of $iR_{DI}$. After some linear algebra, the 4 solutions of eigensystem $\left(\text{W}\right)$ are:$$
w_{+}=\left(\begin{array}{cc}
-1 & i\\
i & 1
\end{array}\right),\:w_{-}=\left(\begin{array}{cc}
-1 & -i\\
-i & 1
\end{array}\right),\:w_{a}=\left(\begin{array}{cc}
1 & 0\\
0 & 1
\end{array}\right),\:w_{b}=\left(\begin{array}{cc}
0 & -1\\
1 & 0
\end{array}\right)
$$along with their associated eigenvalues:$$\omega_{\pm}=\pm2,\:\omega_{a}=\omega_{b}=0$$Using this Weinberg eigentensor basis, we can decompose a 2D general (not necessarily symmetric) real rank-2 tensor $u$ as:$$
u\equiv\left(\begin{array}{cc}
u_{11} & u_{12}\\
u_{21} & u_{22}
\end{array}\right)=u_{2}^{+}w_{+}+u_{2}^{-}w_{-}+u_{0}^{a}w_{a}+u_{0}^{b}w_{b}=\left(\begin{array}{cc}
u_{0}^{a}-u_{2}^{+}-u_{2}^{-} & -u_{0}^{b}+i\left(u_{2}^{+}-u_{2}^{-}\right)\\
u_{0}^{b}+i\left(u_{2}^{+}-u_{2}^{-}\right) & u_{0}^{a}+u_{2}^{+}+u_{2}^{-}
\end{array}\right)
$$Solving for the mode coefficients of the eigentensors in terms of the matrix entries gives:$$u_{2}^{\pm}=\frac{1}{4}\left(u_{22}-u_{11}\mp i\left(u_{12}+u_{21}\right)\right),\:u_{0}^{a}=\frac{1}{2}\left(u_{11}+u_{22}\right),\:u_{0}^{b}=\frac{1}{2}\left(u_{21}-u_{12}\right)$$So we have the following table of eigenmode coefficients and associated helicity eigenvalues:
$$
\begin{array}{|c|c|}
\hline \text{Mode} & \text{Weinberg helicity } h=\omega\\
\hline u_{22}-u_{11}\mp i\left(u_{12}+u_{21}\right) & \pm2\\
\hline u_{11}+u_{22} & 0\\
\hline u_{21}-u_{12} & 0\\
\hline
\end{array}
$$As you note, Weinberg's helicity-2 modes are necessarily complex.
Dirac, on the other hand, considers the eigensystem constructed from the square of $iR_{DI}$, solves to find that system's eigenvalues $\delta_i$ and then defines $\pm\sqrt{\delta}$ to be the helicity of a mode. Specifically, the Dirac eigensystem is:$$iR_{DI}\left(iR_{DI}\left(d_{i}\right)\right)=\delta_{i}d_{i}\tag{D}$$with the 4 solutions:$$
d_{1} =\left(\begin{array}{cc}
-1 & 0\\
0 & 1
\end{array}\right),\:d_{2}=\left(\begin{array}{cc}
0 & 1\\
1 & 0
\end{array}\right),\:d_{3}=\left(\begin{array}{cc}
1 & 0\\
0 & 1
\end{array}\right),\:d_{4}=\left(\begin{array}{cc}
0 & -1\\
1 & 0
\end{array}\right)
$$$$\delta_{1}=4,\:\delta_{2}=4,\:\delta_{3}=\delta_{4}=0$$Decomposing $u$ in this Dirac basis yields:$$
u\equiv\left(\begin{array}{cc}
u_{11} & u_{12}\\
u_{21} & u_{22}
\end{array}\right)=u_{1}d_{1}+u_{2}d_{2}+u_{3}d_{3}+u_{4}d_{4}=\left(\begin{array}{cc}
-u_{1}+u_{3} & u_{2}-u_{4}\\
u_{2}+u_{4} & u_{1}+u_{3}
\end{array}\right)
$$so that:$$u_{1}=\frac{1}{2}\left(u_{22}-u_{11}\right),\:u_{2}=\frac{1}{2}\left(u_{12}+u_{21}\right),\:u_{3}=\frac{1}{2}\left(u_{11}+u_{22}\right),\:u_{4}=\frac{1}{2}\left(u_{21}-u_{12}\right)$$The helicity table for Dirac eigenmodes is therefore:
$$
\begin{array}{|c|c|}
\hline\text{Mode} & \text{Dirac helicity } h=\pm\sqrt{\delta}\\
\hline u_{22}-u_{11} & \pm2\\
\hline u_{12}+u_{21} & \pm2\\
\hline u_{11}+u_{22} & 0\\
\hline u_{21}-u_{12} & 0\\
\hline
\end{array}
$$All of Dirac's modes are real.
Weinberg's approach is natural for describing complex gravitational plane waves while Dirac's definition has the virtue of yielding strictly real eigenmodes for a given helicity. In my opinion, which is the "right" version of eigenmodes and helicities comes down to a matter of convenience and/or taste.

 

[Kostik]

Thank you, this is hugely helpful. Presumably if one repeated this analysis with the full 4x4 matrix $R_{DI}$ instead of just the central 2x2 sub-matrix, one would pick up the helicities/eigenvalues $\pm1$?

 

[Kostik]

Thank you, this is hugely helpful. Presumably if one repeated this analysis with the full 4x4 matrix $R_{DI}$ instead of just the central 2x2 sub-matrix, one would pick up the helicities/eigenvalues $\pm1$?

Yep. The helicity/eigenvalue $\pm 1$ case is actually more work, since there are 8 eigenmatrices $w$ (and another 4 for helicity/eigenvalue $0$, making 16 in all, of course). But it all works out. When the gravity wave $u_{\mu\nu}$ is expanded in the 16 eigenmatrix basis $u=a^1 w_1 + \cdots + a^{16} w_{16}$, the coefficients of the basis matrices have helicity $h=-\omega$ where $\omega$ is the eigenvalue of the matrix.

 

[Kostik]

\begin{array}{|c|c|}
\hline\text{Mode} & \text{Dirac helicity } h=\pm\sqrt{\delta}\\
\hline u_{22}-u_{11} & \pm2\\
\hline u_{12}+u_{21} & \pm2\\
\hline u_{11}+u_{22} & 0\\
\hline u_{21}-u_{12} & 0\\
\hline
\end{array}

Strictly speaking, Dirac's real modes do not have any "helicity" using Weinberg's definition, which is based on transformation under a rotation. The operator $iR^2$ has eigenvalues $4$ and $0$ as you showed, which implies the operator $iR$ has eigenvalues $\pm 2, 0$, which we saw before.

By the way, nice notation using $w, \omega$ and $d, \delta$.

 

[renormalize]

Strictly speaking, Dirac's real modes do not have any "helicity" using Weinberg's definition, which is based on transformation under a rotation. The operator $iR^2$ has eigenvalues $4$ and $0$ as you showed, which implies the operator $iR$ has eigenvalues $\pm 2, 0$, which we saw before.

At the bottom of pg. 67 of GTR Dirac says:
"Thus $u_{11}+u_{22}$ is invariant, while $iR$ has the eigenvalues $\pm2$ when applied to $u_{11}-u_{22}$ or $u_{12}$. The components of $u_{\alpha\beta}$ that contribute to the energy (34.2) thus correspond to spin 2."
Since gravitational plane waves propagate at $c$, Dirac's statement implies that such waves correspond to helicity 2.

 

[Kostik]

Actually, both the Weinberg and Dirac helicities are eigenvalues, just those of two distinct (but related) eigensystems.
In terms of Dirac's $R_{DI}$, Weinberg's helicities are the eigenvalues $\omega_i$ of the eigensystem:$$iR_{DI}\left(w_{i}\right)\equiv i\left(R_{DI}w_{i}+w_{i}R_{DI}^{T}\right)=\omega_{i}w_{i}\tag{W}$$where the $w_i$ are the eigentensors of $iR_{DI}$. After some linear algebra, the 4 solutions of eigensystem $\left(\text{W}\right)$ are:$$
w_{+}=\left(\begin{array}{cc}
-1 & i\\
i & 1
\end{array}\right),\:w_{-}=\left(\begin{array}{cc}
-1 & -i\\
-i & 1
\end{array}\right),\:w_{a}=\left(\begin{array}{cc}
1 & 0\\
0 & 1
\end{array}\right),\:w_{b}=\left(\begin{array}{cc}
0 & -1\\
1 & 0
\end{array}\right)
$$along with their associated eigenvalues:$$\omega_{\pm}=\pm2,\:\omega_{a}=\omega_{b}=0$$Using this Weinberg eigentensor basis, we can decompose a 2D general (not necessarily symmetric) real rank-2 tensor $u$ as:$$
u\equiv\left(\begin{array}{cc}
u_{11} & u_{12}\\
u_{21} & u_{22}
\end{array}\right)=u_{2}^{+}w_{+}+u_{2}^{-}w_{-}+u_{0}^{a}w_{a}+u_{0}^{b}w_{b}=\left(\begin{array}{cc}
u_{0}^{a}-u_{2}^{+}-u_{2}^{-} & -u_{0}^{b}+i\left(u_{2}^{+}-u_{2}^{-}\right)\\
u_{0}^{b}+i\left(u_{2}^{+}-u_{2}^{-}\right) & u_{0}^{a}+u_{2}^{+}+u_{2}^{-}
\end{array}\right)
$$Solving for the mode coefficients of the eigentensors in terms of the matrix entries gives:$$u_{2}^{\pm}=\frac{1}{4}\left(u_{22}-u_{11}\mp i\left(u_{12}+u_{21}\right)\right),\:u_{0}^{a}=\frac{1}{2}\left(u_{11}+u_{22}\right),\:u_{0}^{b}=\frac{1}{2}\left(u_{21}-u_{12}\right)$$So we have the following table of eigenmode coefficients and associated helicity eigenvalues:
$$
\begin{array}{|c|c|}
\hline \text{Mode} & \text{Weinberg helicity } h=\omega\\
\hline u_{22}-u_{11}\mp i\left(u_{12}+u_{21}\right) & \pm2\\
\hline u_{11}+u_{22} & 0\\
\hline u_{21}-u_{12} & 0\\
\hline
\end{array}
$$As you note, Weinberg's helicity-2 modes are necessarily complex.
Dirac, on the other hand, considers the eigensystem constructed from the square of $iR_{DI}$, solves to find that system's eigenvalues $\delta_i$ and then defines $\pm\sqrt{\delta}$ to be the helicity of a mode. Specifically, the Dirac eigensystem is:$$iR_{DI}\left(iR_{DI}\left(d_{i}\right)\right)=\delta_{i}d_{i}\tag{D}$$with the 4 solutions:$$
d_{1} =\left(\begin{array}{cc}
-1 & 0\\
0 & 1
\end{array}\right),\:d_{2}=\left(\begin{array}{cc}
0 & 1\\
1 & 0
\end{array}\right),\:d_{3}=\left(\begin{array}{cc}
1 & 0\\
0 & 1
\end{array}\right),\:d_{4}=\left(\begin{array}{cc}
0 & -1\\
1 & 0
\end{array}\right)
$$$$\delta_{1}=4,\:\delta_{2}=4,\:\delta_{3}=\delta_{4}=0$$Decomposing $u$ in this Dirac basis yields:$$
u\equiv\left(\begin{array}{cc}
u_{11} & u_{12}\\
u_{21} & u_{22}
\end{array}\right)=u_{1}d_{1}+u_{2}d_{2}+u_{3}d_{3}+u_{4}d_{4}=\left(\begin{array}{cc}
-u_{1}+u_{3} & u_{2}-u_{4}\\
u_{2}+u_{4} & u_{1}+u_{3}
\end{array}\right)
$$so that:$$u_{1}=\frac{1}{2}\left(u_{22}-u_{11}\right),\:u_{2}=\frac{1}{2}\left(u_{12}+u_{21}\right),\:u_{3}=\frac{1}{2}\left(u_{11}+u_{22}\right),\:u_{4}=\frac{1}{2}\left(u_{21}-u_{12}\right)$$The helicity table for Dirac eigenmodes is therefore:
$$
\begin{array}{|c|c|}
\hline\text{Mode} & \text{Dirac helicity } h=\pm\sqrt{\delta}\\
\hline u_{22}-u_{11} & \pm2\\
\hline u_{12}+u_{21} & \pm2\\
\hline u_{11}+u_{22} & 0\\
\hline u_{21}-u_{12} & 0\\
\hline
\end{array}
$$All of Dirac's modes are real.
Weinberg's approach is natural for describing complex gravitational plane waves while Dirac's definition has the virtue of yielding strictly real eigenmodes for a given helicity. In my opinion, which is the "right" version of eigenmodes and helicities comes down to a matter of convenience and/or taste.

Just a last note, while you compute the four eigenbasis matrices $d_i \, (i=1,2,3,4)$ of the operator $(iR)^2$ which have eigenvalues $4, 0$, you do not immediately see the eigenbasis matrices $w_i$ of $iR$ which have eigenvalues $\pm 2, 0$. These were, of course, obtained previously and directly by solving the eigenvalue equation with the operator $iR$.