Weak-field approximation

Now we assume the gravitational field to be very weak. So we put

$$g_{\mu \nu} = \eta_{\mu \nu} + h_{\mu \nu}$$ (49)

where $h_{\mu \nu}$ is a small perturbation of the flat spacetime metric $\eta_{\mu \nu}$, and we systematically discard the terms of order h²,h³,... in the following calculations. Thereafter, we suppose that any quantity T (scalar or tensor) may be written as

T = T⁽⁰⁾ + T⁽¹⁾ + O(h²)

where T⁽⁰⁾ is the unperturbed quantity in flat spacetime and T⁽¹⁾ denotes the perturbation of first-order with respect to $h_{\mu \nu}$. Henceforth, indices will be lowered with $\eta_{\mu \nu}$ and raised with $\eta^{\mu \nu}=\eta_{\mu \nu}$.

We shall put for the sake of simplicity

$$K_{\mu} = k^{(0)}_{\mu} = S^{(0)}_{\verb*+ +,\mu}$$ (51)

Neglecting the first order terms in h, Eq.(12) gives $K^{\alpha}K_{\alpha} = 0$, whereas Eq.(17) reduces to the equation of a null geodesic in flat spacetime related to Cartesian coordinates

$$K^{\alpha} K_{\beta, \alpha} = 0$$ (52)

In agreement with the assumptions made in Sect.3 to obtain Eqs.(29) and (31), we consider that at the zeroth order in $h_{\mu \nu}$, the light emitted by the source is described by a plane monochromatic wave in a flat spacetime. So we suppose that the quantities $K_{\mu}$, $a^{(0) \mu}$ and consequently a⁽⁰⁾ are constants throughout the domain of propagation.

Moreover, we regard as negligible all the perturbations of gravitational origin in the vicinity of the emitter (this hypothesis is natural for a source at spatial infinity) and the quantity a₀ in Eqs.(28) and (29) is given consequently by

a₀ = a⁽⁰⁾ = const.

Furthermore, it results from $K_{\mu}=const.$ that $k_{\alpha;\beta}~=~O(h)$. Therefore, terms like $k^{\alpha}_{;\mu}k_{\alpha;\nu}$ or $R_{\rho \sigma}k^{\rho} v^{\mu} k^{\sigma}_{;\mu}$ are of second order and can be systematically disregarded.

According to our general assumption in this section, the unit 4-velocity of the observer may be expanded as

$$u^{\alpha}_{obs} = U^{\alpha} + u^{(1)\alpha}_{obs} + O(h^2)$$ (54)

at any point of ${\cal{C}}_{obs}$, with the definition

$$U^{\alpha} = u^{(0)\alpha}_{obs}$$ (55)

It follows from (48) and from $g_{\alpha \beta} u^{\alpha} u^{\beta} = 1$ that

$$U^{\alpha} = const.$$ (56)

and

$$\eta_{\alpha \beta} U^{\alpha} U^{\beta} = 1$$ (57)

From these last equations, we recover the fact that the unperturbed motion of a freely falling observer is a time-like straight line in Minkowski space-time.

Now we have to know the quantities $v^{\mu}$ occurring in Eqs.(47) and (52) at the lowest order. An elementary calculation shows that, in Eqs.(46), the covariant differentiation may be replaced by the ordinary differentiation. So we have to solve the system

$$\frac{dv^{\alpha}}{dv} = v^{\mu} k^{\alpha}_{,\mu}$$ (58)

together with the boundary conditions (34).

Assuming the expansion

$$v^{\mu} = v^{(0)\mu} + v^{(1)\mu} + O(h^2)$$ (59)

it is easily seen that the unique solution of (62) and (34) is such that at any point of the light ray $\gamma$, the components $v^{(0)\mu}$ are constants given by

$$v^{(0)\mu} = U^{\mu}$$ (60)

Neglecting all the second order terms in (47) and (52), we finally obtain

and

$$(1+z)\frac{d}{ds}\left( \frac{1}{1+z} \right)_{obs} = -\fra... ...)}_{\mu \rho \nu \sigma} U^{\mu} U^{\nu} K^{\rho} K^{\sigma} dv$$ (61)

all the integrations being performed along the unperturbed path of light.

In Eq.(66) $R^{(1)}_{\mu \rho \nu \sigma}$ denotes the linearized curvature tensor of the metric $g_{\mu \nu} = \eta_{\mu \nu} + h_{\mu \nu}$, i.e.

$$R^{(1)}_{\mu \rho \nu \sigma} = - \frac{1}{2}(h_{\mu \nu,\rh... ...u \nu} - h_{\mu \sigma , \nu \rho} - h_{\nu \rho , \mu \sigma})$$ (62)

and $R^{(1)}_{\mu \nu}$ is the corresponding linearized Ricci tensor

$$R^{(1)}_{\mu \nu} = \eta^{\alpha \beta} R^{(1)}_{\alpha \mu \beta \nu}$$ (63)

It is worthy to note that the components $R^{(1)}_{\mu \rho \nu \sigma}$and $R^{(1)}_{\mu \nu}$ are gauge-invariant quantities. Indeed, under an arbitrary infinitesimal coordinate transformation $x^{\alpha} \rightarrow x'^{\alpha} = x^{\alpha}+ \xi^{\alpha}(x)$, $h_{\mu \nu}(x)$ transforms into $h'_{\mu \nu}(x) = h_{\mu \nu}(x) -\xi_{\mu, \nu} - \xi_{\nu, \mu}$, and it is easily checked from (67) and (68) that

$$R^{(1)}_{\mu \rho \nu \sigma}(h'_{\alpha \beta}) = R^{(1)}_{\mu \rho \nu \sigma}(h_{\alpha \beta})$$ (64)

$$R^{(1)}_{\mu \nu}(h'_{\alpha \beta}) = R^{(1)}_{\mu \nu}(h_{\alpha \beta})$$ (65)

This feature ensures that the right-hand sides of Eqs. (65) and (66) are gauge-invariant quantities.

Equation (65) reveals that the first order geometrical scintillation effect depends upon the gravitational field through the Ricci tensor only. On the other side, it follows from (66) that the part of the scintillation due to the spectral shift depends upon the curvature tensor.

These properties have remarkable consequences in general relativity. Suppose that the light ray $\gamma$ travels in regions entirely free of matter. Since the linearized Einstein equations are in a vacuum

$$R^{(1)}_{\mu \nu} = 0$$ (66)

it follows from Eq.(65) that

$$2\frac{\dot{a}}{a} = 0 + O(h^2)$$ (67)

As a consequence, $\dot{{\cal{N}}}/{\cal{N}}$ reduces to the contribution of the change in the spectral shift

$$\frac{\dot{{\cal{N}}}}{{\cal{N}}} = -\frac{1}{U^{\lambda}K_... ...}_{\mu \rho \nu \sigma} U^{\mu} U^{\nu} K^{\rho} K^{\sigma} dv$$ (68)

From (72), we recover the conclusion previously drawn by Zipoy (1966) and Zipoy & Bertotti (1968): within general relativity, gravitational waves produce no first order geometrical scintillation.