Orthogonality of Gravitational Wave Polarizations

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SUMMARY

The discussion centers on the orthogonality of gravitational wave polarizations, specifically examining two plane gravitational waves with transverse-traceless (TT) amplitudes, ##A^{\mu\nu}## and ##B^{\mu\nu}##. It is established that these waves are orthogonal if the condition ##(A^{\mu\nu})^*B_{\mu\nu}=0## holds true. The conversation explores the implications of a 45-degree rotation of ##B^{\mu\nu}##, leading to the conclusion that this rotation makes ##B^{\mu\nu}## proportional to ##A^{\mu\nu}##. The participants discuss the transformation of rank two tensors under rotations and the use of complex rotation matrices.

PREREQUISITES
  • Understanding of transverse-traceless (TT) gauge in gravitational wave physics
  • Familiarity with complex conjugates and their application in tensor analysis
  • Knowledge of matrix rotation and its implications for rank two tensors
  • Basic grasp of gravitational wave polarization concepts
NEXT STEPS
  • Study the mathematical properties of transverse-traceless (TT) tensors in gravitational wave theory
  • Learn about the transformation of rank two tensors under various rotation matrices
  • Explore complex rotation matrices and their applications in physics
  • Investigate the implications of gravitational wave polarization on astrophysical observations
USEFUL FOR

Physicists, astrophysicists, and students studying gravitational waves, particularly those interested in the mathematical foundations of wave polarization and tensor transformations.

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Homework Statement


Two plane gravitational waves with TT (transverse-traceless) amplitudes, ##A^{\mu\nu}## and ##B^{\mu\nu}##, are said to have orthogonal polarizations if ##(A^{\mu\nu})^*B_{\mu\nu}=0##, where ##(A^{\mu\nu})^*## is the complex conjugate of ##A^{\mu\nu}##. Show that a 45 degree rotation of ##B^{\mu\nu}## makes it proportional to ##A^{\mu\nu}##.

Homework Equations


For waves propagating in the z direction under the TT gauge, ##A^{xx}##, ##A^{xy}##, and ##A^{yy}=-A^{xx}## are the only non-zero components.

The Attempt at a Solution


$$(A^{\mu\nu})^*B_{\mu\nu}=2(A^{xx})^*B_{xx}+2(A^{xy})^*B_{xy}=0$$
$$B_{xy} = -B_{xx}\frac{(A^{xx})^*}{(A^{xy})^*}$$
$$ (B_{\mu\nu})=B_{xx}\begin{pmatrix}
1 & -(A^{xx})^*/(A^{xy})^*\\
-(A^{xx})^*/(A^{xy})^* & -1\\
\end{pmatrix} = \frac{B_{xx}}{(A^{xy})^*}\begin{pmatrix}
(A^{xy})^* & -(A^{xx})^*\\
-(A^{xx})^* & -(A^{xy})^*\\
\end{pmatrix}$$

At this point, I'm not really sure what to do. I don't know what it means to "rotate" a matrix so I assume the correct thing to do is complex rotate each component, but I'm not entirely sure what this entails either. I tried multiplying each component by ##e^{i\pi/4}=(1+i)/\sqrt{2}## but I'm not sure how this helps. Any advice?
 
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How do rank two tensors transform under rotations?
 
Orodruin said:
How do rank two tensors transform under rotations?

Ahh of course, use the rotation matrix twice. This gets me B proportional to A* if I use the normal (real valued) rotation matrix. Is there a complex version that will get me B proportional A?
 

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