I am reading following research paper - Feynman propagator in curved spacetime: A momentum-space representation. In page 2, Eq.2.8 and later they solve following differential equation for propagator (the equation is in Riemann-normal coordinates) -

$$\eta^{\mu \nu}\partial_{\mu}\partial_{\nu}G -[m^2 + (\xi - \frac{1}{6})R]G - \frac{1}{3}R_{\alpha}^{\hspace{5pt}\nu}y^{\alpha}\partial_{\nu}G + \frac{1}{3}R^{\mu\hspace{5pt}\nu}_{\hspace{5pt}\alpha \hspace{5pt} \beta}y^{\alpha}y^{\beta}\partial_{\mu}\partial_{\nu}G + \mathcal{O}(\partial^{3}g^{\mu\nu}) = -\delta^{n}(y) $$

They use iterative procedure by first converting to momentum space and writing -

$$G(k) = G_0(k) + G_1(k) + G_2(k) + ...$$

Where, $G_i(k)$ represents $G$ with geometrical coefficients involving $i$-th derivative of the metric $g^{\mu\nu}$.

Can someone explain what is the iterative procedure? And what do they mean by Adiabatic order expansion for this method? (I got this terminology from the book on same topic by the same author, explaining the same procedure)

I'll outline what they have done, below -

In the lowest order, the solution is - $$G_0(k) = (k^2 + m^2)^{-1}$$

This is intuitive to me, because zeroth order corresponds to usual scalar field solution.

Because we are using normal coordinates the first derivative of the metric is zero and hence there is no such co-efficient and thus - $G_1(k) = 0$

However, for the second order, they write -

$$\eta^{\mu \nu}\partial_{\mu}\partial_{\nu}G_2 -m^2 G_2 + (\xi - \frac{1}{6})R G_0 - \frac{1}{3}R_{\alpha}^{\hspace{5pt}\nu}y^{\alpha}\partial_{\nu}G_0 + \frac{1}{3}R^{\mu\hspace{5pt}\nu}_{\hspace{5pt}\alpha \hspace{5pt} \beta}y^{\alpha}y^{\beta}\partial_{\mu}\partial_{\nu}G_0 = 0$$

(Here $G_2$ is just the Fourier transform of $G_2(k)$)

How have they written this? Why do we have $G_2$ only in first two terms and not in others? I think that $\delta^n(y)$ is absent as it has been cancelled from first two two terms involving G_0, which is just be a Fourier transform of it.

Please explain the step and, if possible, the general principle of such methods of solving differential equations.


1 Answer 1


What do they mean by adiabatic order expansion?

The physical idea behind this adiabatic expansion of $G$ is that when the metric varies slowly in space and time, you can keep just a few terms. The adiabatic order is the number of derivatives of the metric. Terms with more derivatives are smaller when the metric varies slowly. For example, if we think of periodic time variation with frequency $\omega$, terms of adiabatic order $n$ contain factors of $\omega^n$, which will get smaller and smaller as $n$ increases if $\omega$ is small.

Why do we have $G_2$ only in first two terms and not in others?

The first two terms have no curvature factors. These terms are of adiabatic order 2 because that is the order of $G_2$.

But the last three terms contain curvature factors in the form of the Ricci scalar, the Ricci tensor, and the Riemann tensor. These factors all have adiabatic order 2 because they involve two derivatives of the metric. So, for the last three terms to have adiabatic order 2, they only need to contain $G_0$. If they contained $G_2$, the terms would have adiabatic order 4, and the equation you asked about is supposed to be the order-2 equation.

Can someone explain what is the iterative procedure?

You first find $G_0$. Next you use $G_0$ to find $G_2$. Then you use $G_2$ to find $G_4$. Etc.

  • $\begingroup$ @Indigo1729 Thanks for accepting my answer. The 50-point bounty you offered disappeared into cyberspace because the bounty expired before you accepted. $\endgroup$
    – G. Smith
    May 10, 2020 at 19:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.