Weak Gravitational Field & Wave Eq. - Analyzing Effects on Massless Scalar Field

[Haorong Wu]

TL;DR Summary

I derive the wave equations in a curved spacetime. I do not understand why it is strongly affected by the gravitational field.

A massless scalar field in a curved spacetime propagates as $$(-g)^{-1/2}\partial_\mu(-g)^{1/2}g^{\mu\nu}\partial_\nu \psi=0 .$$

Suppose the gravitational field is weak, and ##g_{\mu\nu}=\eta_{\mu\nu}+\epsilon \gamma_{\mu\nu}## where ##\epsilon## is the perturbation parameter. And let the field be ##\psi=A e^{ik(x0-x3)}##.

Then the wave equation can be solved up to the first order of ##\epsilon##, giving $$2ik\partial_3 A-\nabla ^2 A+\epsilon [-\frac k 2 (2k(\gamma_{00}+\gamma_{33})+i(\partial_3 \gamma_{00}-\partial_3 \gamma_{11}-\partial_3 \gamma_{22}+\partial_3 \gamma_{33}+2\partial_2 \gamma_{23}+2\partial_1 \gamma_{13}))A+\cdots] =0$$ where terms with the derivatives of ##A## is omitted.

I note that there are ##k^2## inside the perturbation terms for ##A##. For a light with wavelength of ##1000 ~\rm{nm}##, ##k## will be about ##6 \times 10^6##. The metric for the Earth will let ##\gamma_{00}+\gamma_{33}## be about ## \frac {mz^2}{(x^2+y^2+z^2)^{3/2}} \approx 1.4\times 10^{-9}## for ##x=y=0, z=6.37\times10^6## is the radius of the earth, ##m=8.87\times 10^{-3}## is the Schwarzschild radius of earth.

Hence it appears that the perturbation for ##A## is quite strong even in a weak gravitational field. But this should be wrong. But where did I make a mistake?

 

[haushofer]

Maybe it's me, but I don't get it what you're doing here. What exactly do you mean by "the quite strong perturbation for A"? Also, some text appears to be missing.

Maybe if you rephrase your question more carefully, you'll gain more insight yourself also

 

[Haorong Wu]

Hi, @haushofer . Thank you for your advice. I apologize for not stating clearly.

I am studying the propagation of light in curved spacetime, especially the propagation of a Gaussian beam defined by ##\psi=A(\mathbf x) e^{ik(x0-x3)}## where ##A(\mathbf x)## is the transverse profile which does not depend on time ##x^0##.

In a flat spacetime, the wave equation can be solved to be $$2ik\partial_3 A-\nabla ^2 A=0 .$$

Meanwhile, if the spacetime is curved, the metric in the wave equation $$(g^{\mu\nu}\partial_\mu\partial \nu-\xi R) \psi=0$$ with ## R=0## for Schwarzschild metrics will induce extra terms, i.e., \begin{align}&2ik\partial_3 A-\nabla ^2 A+\epsilon [-\frac k 2 (2k(\gamma_{00}+\gamma_{33})\nonumber \\&~~~~+i(\partial_3 \gamma_{00}-\partial_3 \gamma_{11}-\partial_3 \gamma_{22}+\partial_3 \gamma_{33}+2\partial_2 \gamma_{23}+2\partial_1 \gamma_{13}))A+\cdots] =0. \nonumber \end{align}

I am troubled by the ##k^2## factor in the term $$-k^2( \gamma_{00}+\gamma_{33} )A,$$ since it will give a quite large number. Then when the beam is sent from the surface of earth, its transverse profile will be strongly affected by gravitational field. But that should not happen.

I must make some mistakes somewhere, but I still could not figure it out.

 

[hutchphd]

Won't it give a number of size similar to ##\nabla^2 A##? Then there is an ##\epsilon## in front.
If I am totally wrong, this is not my field !

 

[Haorong Wu]

Thanks, @hutchphd . There are terms for ##\nabla^2 A## from the gravitational field, but they do not contain any ##k##. Hence their factors will be quite small and I do not include them.

 

[haushofer]

Without having done the calculation: dimensionally this k^2 comes from the second spatial derivative on psi, right? So it should be there.

Btw, your partial derivative with mu in your second post should be a covariant one.

 

[Haorong Wu]

@haushofer

Yes, it comes from the second spatial derivatives with respect to ##x^0## and ##x^3##. Well, I am trying another approach. Hope it will somehow work out.

And, yes, that should be ##\nabla_\mu \nabla_\nu##. It will add some Christoffel symbols and the terms with ##k^2## will be the same. It seems in some early literature, people would use ##\partial## to represent covariant derivatives. I will be more careful.

 

[haushofer]

How large is epsilon compared to k?

 

[Haorong Wu]

How large is epsilon compared to k?

Hi, @haushofer . I have complete the calculation.

##\epsilon## is the perturbation parameter which will be set to identiy at the end. It turns out that somehow the effect of ##k^2## is very very small. The decoherence induced by gravitational fields will be negligible.

I was misled by a paper which gives a wrong equation. With that equation, I get the result that the decoherence is surprisingly large. Finally, I realize that the equation has a wrong dimension. After I use a correct one, I get the expected results.

Cheers.

 

[haushofer]

Ok. Maybe it's nice for future reference for others to show some details,but don't feel obliged ;)