Nullity of wave 4 - vector for grav. plane wave

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SUMMARY

The discussion centers on the nullity of the wave 4-vector in the context of the plane wave solution \(\bar{h^{\mu \nu }} = A^{\mu \nu }e^{ik_{\alpha }x^{\alpha }}\) derived from the equation \(\square ^{2}\bar{h^{\mu \nu }} = 0\). Participants confirm that the dispersion relation \(\omega ^{2} = |k|^{2}\) indicates that the wave travels at the speed of light (c), leading to the conclusion that \(k^{\mu}k_{\mu} = 0\). The relationship between the wave equation and the nullity of the wave vector is established, affirming that the wave vector is indeed null as a direct consequence of the dispersion relation.

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  • Understanding of wave equations in physics
  • Familiarity with the concept of 4-vectors in relativity
  • Knowledge of dispersion relations in wave mechanics
  • Basic grasp of the speed of light and its implications in physics
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  • Study the derivation of the wave equation \(\square^{2}\bar{h^{\mu \nu }} = 0\)
  • Explore the properties of 4-vectors in special relativity
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WannabeNewton
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How can one tell from [tex]\square ^{2}\bar{h^{\mu \nu }} = 0[/tex] that in the plane wave solution [tex]\bar{h^{\mu \nu }} = A^{\mu \nu }e^{ik_{\alpha }x^{\alpha }}[/tex] the wave 4 - vector is null. If you plug in the solution you just end up with the dispersion relation [tex]\omega ^{2} = \left | k \right |^{2}[/tex]. Is it implied from this that it is null or am I just missing something obvious or is it not possible to deduce its nullity from the wave equation itself? Intuitively it would have to be null because the dispersion relation implies the waves travel at c but is this sufficient to conclude that [tex]k^{\mu }k_{\mu } = 0[/tex]?
 
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You've got it all right there, kμkμ = |k|2 - ω2/c2 = 0
 

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