General theory of the gravitational scintillation

In a region of spacetime free of electric charge, the propagation equations for the electromagnetic vector potential $A_{\mu}$ are (e.g., Misner et al. 1973)

$$A^{\mu ;\alpha}_{\verb*+ +;\alpha} - R_{\verb*+ +\alpha}^{\mu} A^{\alpha} = 0$$ (3)

when $A^{\mu}$ is chosen to obey the Lorentz gauge condition

$$A^{\mu}_{\verb*+ +;\mu} = 0$$ (4)

It is convenient here to treat $A_{\mu}$ as a complex vector. Hence the electromagnetic field tensor $F_{\mu \nu}$ is given by

$$F_{\mu \nu} = {\cal{R}}e(A_{\nu;\mu}- A_{\mu;\nu})$$ (5)

The corresponding electromagnetic energy-momentum tensor is defined by

$$T^{\mu \nu} = \frac{1}{4 \pi} \left[-F^{\mu \rho} F^{\nu}_{.... ...rac{1}{4} F^{\alpha \beta} F_{\alpha \beta} g^{\mu \nu} \right]$$ (6)

where $F_{. \rho}^{\nu} = g^{\nu \lambda}~F_{\lambda \rho}$. The components of this tensor satisfy the conservation equations $T^{\mu \nu}_{\verb*+ +;\nu} = 0$ as a consequence of Eqs.(3).

For an observer located at the spacetime point x and moving with the unit 4-velocity $u^{\alpha}$, the density of electromagnetic energy flux is given by the Poynting vector

$${\cal{P}}^{\mu}(x,u) = c T^{\mu \nu}(x) u_{\nu}(x)$$ (7)

and the density of electromagnetic energy as measured by the observer is

$$\mu_{el}(x,u) = T^{\mu \nu}(x) u_{\mu} u_{\nu}$$ (8)

In this paper, we use the geometrical optics approximation. So we assume that there exist wave solutions to Eqs.(3) which admit a development of the form

$$A^{\mu}(x,\xi) = [a^{\mu}(x) + O(\xi)] \exp(\frac{i}{\xi} \hat{S}(x))$$ (9)

where $a^{\mu}(x)$ is a slowly varying complex vector amplitude, $\hat{S}(x)$ is a real function and $\xi$ a dimensionless parameter which tends to zero as the typical wavelength of the wave becomes shorter and shorter. A solution like (9) represents a quasi plane, locally monochromatic wave of high frequency (Misner et al. 1973).

Let us define the phase S and the vector field $k_{\alpha}$ by the relations

$$S(x,\xi) = \frac{1}{\xi} \hat{S}(x)$$ (10)

and

$$k_{\alpha} = S_{, \alpha}$$ (11)

Inserting (9) into Eqs.(3) and (4), then retaining only the leading terms of order $\xi^{-2}$ and $\xi^{-1}$, yield the fundamental equations of geometrical optics

$$k^{\alpha} k_{\alpha} = 0$$ (12)

$$k^{\alpha}a^{\mu}_{; \alpha} = -\frac{1}{2} a^{\mu} k^{\alpha}_{; \alpha}$$ (13)

with the gauge condition

$$k_{\alpha} a^{\alpha} = 0$$ (14)

Light rays are defined to be the curves whose tangent vector field is $k^{\alpha}$. So the parametric equations $x^{\alpha} = x^{\alpha}(v)$ of the light rays are solutions to the differential equations

$$\frac{dx^{\alpha}}{dv} = k^{\alpha}(x^{\lambda}(v))$$ (15)

where v is an affine parameter. Differentiating Eq.(12) and noting that

$$k_{\alpha;\beta} =k_{\beta ;\alpha}$$ (16)

follows from (11), it is easily seen that $k^{\alpha}$ satisfies the propagation equations

$$k^{\alpha}k_{\beta;\alpha}=0$$ (17)

These equations, together with (12), show that the light rays are null geodesics.

Inserting (9) into (5) and (6) gives the approximate expression for $F_{\mu \nu}$

$$F_{\mu \nu} = {\cal{R}}e[i(k_{\mu} a_{\nu} - k_{\nu} a_{\mu}) e^{iS}]$$ (18)

and for $T^{\mu \nu}$ averaged over a period

$$T^{\mu \nu} = \frac{1}{8 \pi} a^2 k^{\mu} k^{\nu}$$ (19)

where a is the scalar amplitude defined by

$$a = (-a^{\mu}\overline{a}_{\mu})^{1/2}$$ (20)

From (7) and (19), it is easily seen that the Poynting vector is proportional to the null tangent vector $k^{\mu}$.This means that the energy of the wave is transported along each ray with the speed of light. Let us denote by ${\cal{F}}(x,u)$ the energy flux received by an observer located at x and moving with the 4-velocity $u^{\alpha}$: by definition, ${\cal{F}}(x,u)$ is the amount of radiating energy flowing per unit proper time across a unit surface orthogonal to the direction of propagation. It follows from (8) and (19) that

$${\cal{F}}(x,u) = c \mu_{el} (x,u) = \frac{c}{8 \pi} a^2(x) (u^{\mu}k_{\mu})^2_{obs}$$ (21)

This formula enables us to determine the photon flux ${\cal{N}}(x,u)$ received by the observer located at x and moving with the 4-velocity $u^{\alpha}$. Since the 4-momentum of a photon is $p^{\mu} = \hbar k^{\mu}$, the energy of the photon as measured by the observer is $cp^{\mu} u_{\mu} = c\hbar (u^{\mu} k_{\mu})$. We have therefore

$${\cal{N}}(x,u)=\frac{1}{8 \pi \hbar} a^2(x) (u^{\mu} k_{\mu})_{obs}$$ (22)

The spectral shift z of a light source (emitter) as measured by an observer is given by (e.g. G.F.R. Ellis, 1971)

$$1+z = \frac{(u^{\mu} k_{\mu})_{em}}{(u^{\nu} k_{\nu})_{obs}}$$ (23)

Consequently, the photon flux ${\cal{N}}(x,u)$ may be written as

$${\cal{N}}(x,u)=\frac{1}{8 \pi \hbar} a^2(x) \frac{( u^{\mu} k_{\mu})_{em}}{1+z}$$ (24)

The scalar amplitude a can be written in the form of an integral along the light ray $\gamma$ joining the source to the observer located at x. Multiplying Eq.(13) by $\overline{a}_{\mu}$ yields the propagation equation for a

$$k^{\alpha} a_{; \alpha} \equiv \frac{da}{dv} = -\frac{1}{2} a k^{\alpha}_{; \alpha}$$ (25)

where d/dv denotes the total differentiation of a scalar function along $\gamma$. Then, integrating (25) gives

$$a_{\vert obs} = a_{\vert x_0} \exp \left( -\frac{1}{2} \int ... ..._{v_{x_0}}^{v_{obs}} k^{\alpha}_{; \alpha} \verb*+ +dv \right)$$ (26)

where x₀ is an arbitrary point on the light ray $\gamma$.

In the following, we consider that the light source is at spatial infinity. We suppose the existence of coordinate systems $x^{\alpha}$ such that on any hypersurface x⁰ = const., $\vert g_{\mu \nu}~-~\eta_{\mu \nu}\vert~=~O(1/r)$ when $r~=~[\sum_{i=1}^3~(x^i)^2]^{1/2}~\rightarrow~\infty$, with $\eta_{\mu \nu}~=~diag(1,-1,-1,-1)$. We require that in such coordinate systems the quantities $k_{\alpha ; \beta}$, $k_{\alpha ; \beta ; \gamma}$ and $a_{;\alpha}$ respectively fulfill the asymptotic conditions

$$\left\lbrace \begin{array} {ccc} k_{\alpha; \beta} (x_0) & ... ...a} (x_0) & = & O(1/\vert v_{x_0}\vert^{1+p}) \end{array}\right.$$ (27)

when $v_{x_0} \rightarrow - \infty$, with p > 0. Moreover, we assume that the scalar amplitude a_|x0 in Eq.(26) remains bounded when $v_{x_0} \rightarrow - \infty$ and we put

$$\lim_{v_{x_0} \rightarrow - \infty} a_{x_0} = a_0$$ (28)

It results from these assumptions that a_|obs may be written as

$$a_{\vert obs} = a_{0} \exp \left( -\frac{1}{2} \int \limits_{-\infty}^{v_{obs}} k^{\alpha}_{; \alpha} \verb*+ +dv \right)$$ (29)

Now, let us differentiate $k^{\alpha}_{;\alpha}$ with respect to v along $\gamma$. Applying (1) and (2), then taking (16) and (17) into account, we obtain the relation (Sachs 1961)

$$\frac{d}{dv}(k^{\alpha}_{;\alpha}) = - k^{\alpha;\beta} k_{\alpha;\beta} - R_{\alpha \beta} k^{\alpha} k^{\beta}$$ (30)

As a consequence, we can write

$$\int \limits_{-\infty}^{v_{obs}} k^{\alpha}_{; \alpha} \verb... ... k^{\alpha} k^{\beta} + k^{\alpha;\beta} k_{\alpha;\beta}] dv'$$ (31)

The convergence of the integrals is ensured by conditions (27).

Equations (29) and (31) allow to determine the factor a²(x) in ${\cal{N}}(x,u)$ from the energy content of the regions crossed by the light rays and from the geometry of the rays themselves.

It is well known that 1/(1+z) (or (1+z)) can also be obtained in the form of an integral along the light ray $\gamma$ (see e.g. Ellis 1971 or Schneider et al. 1992). However, the corresponding formula will not be useful for our discussion and we will not develop it here.

In fact, the scintillation phenomenon consists in a variation of ${\cal{N}}$ with respect to time. For this reason, it is more convenient to calculate the total derivative of ${\cal{N}}$ along the world-line ${\cal{C}}_{obs}$ of a given observer, moving at the point x with the 4-velocity $u^{\alpha}$.

Given a scalar or tensorial quantity F, we denote by $\dot{F}$ the total covariant differentiation along ${\cal{C}}_{obs}$ defined by

$$\dot{F} \equiv u^{\lambda} F_{;\lambda} = \frac{\nabla F}{ds}$$ (32)

where $ds = (g_{\mu \nu} dx^{\mu} dx^{\nu})^{1/2}$ is the line element between two events $x^{\mu}$ and $x^{\mu}+dx^{\mu}$ on ${\cal{C}}_{obs}$.

In Eq.(24), the quantity $c\hbar(u^{\mu}k_{\mu})_{em}$ is the energy of a photon emitted by an atom of the light source as measured by an observer comoving with this atom. So $(u^{\mu}k_{\mu})_{em}$ is a constant which depends only on the nature of the atom (this constant characterizes the emitted spectral line). Consequently, the change in the photon flux with respect to time is simply due to the change in the scalar amplitude a and to the change in the spectral shift z. From (24), we obtain at each point x of ${\cal{C}}_{obs}$

$$\frac{\dot{{\cal{N}}}}{{\cal{N}}} = 2 \frac{\dot{a}}{a} + (1+z) \frac{d}{ds} \left( \frac{1}{1+z} \right)$$ (33)

Henceforth, we shall call the contribution $2\dot{a}/a$ in Eq.(33) the geometrical scintillation because the variations in a are related to the focusing properties of light rays by gravitational fields (see G.F.R.Ellis 1971 and references therein; see also Misner et al. 1973).

Let us now try to find expressions for $\dot{a}/a$ and $\frac{d}{ds} (1+z)^{-1}$ in the form of integrals along $\gamma$. In what follows, we assume that the ray $\gamma$ hits at each of its points x(v) a vector field $v^{\mu}$ which satisfies the boundary condition

$$v^{\mu}(x_{obs}) = u^{\mu}_{obs}$$ (34)

Let us emphasize that $v^{\mu}$ can be chosen arbitrarily at any point x which does not belong to the world line ${\cal{C}}_{obs}$ (for example, $v^{\mu}(x)$ could be the unit 4-velocity of an observer at x, an assumption which is currently made in cosmology; however we shall make a more convenient choice for $v^{\mu}$ in what follows).

It results from the boundary conditions (27) and (34) that $\dot{a}/a$ may be written as

$$\left.\frac{\dot{a}}{a}\right\vert _{obs} = \int \limits_{-\infty}^{v_{obs}} \frac{d}{dv} [v^{\mu}(\ln a)_{;\mu}] dv$$ (35)

Thus we have to transform the expression

$$\frac{d}{dv} [v^{\mu}(\ln a)_{;\mu}] = k^{\alpha}( v^{\mu} (\ln a)_{;\mu})_{;\alpha}$$ (36)

taken along $\gamma$. Of course, we must take into account the propagation equation (25) which could be rewritten as

$$k^{\alpha}(\ln a)_{;\alpha} = - \frac{1}{2} k^{\alpha}_{;\alpha}$$ (37)

Noting that

$$k^{\alpha}( v^{\mu} (\ln a)_{;\mu})_{;\alpha} = k^{\alpha} v... ...a)_{;\mu ;\alpha} + k^{\alpha} v^{\mu}_{;\alpha} (\ln a)_{;\mu}$$ (38)

then using the relation

$$F_{;\alpha;\beta} = F_{;\beta;\alpha}$$ (39)

which holds for any scalar F, we find

$$\frac{d}{dv} [v^{\mu}(\ln a)_{;\mu}] = v^{\mu} [k^{\alpha} (\ln a)_{;\alpha}]_{;\mu} + [k,v]^{\mu} (\ln a)_{;\mu}$$ (40)

where the bracket [k,v] of $k^{\alpha}$ and $v^{\beta}$ is the vector defined by

$$[k,v] ^{\mu} \equiv k^{\alpha} v^{\mu}_{;\alpha} - v^{\alpha} k^{\mu}_{;\alpha}$$ (41)

Taking (37) into account, it is easily seen that

$$\frac{d}{dv} [v^{\mu}(\ln a)_{;\mu}] = -\frac{1}{2} v^{\mu} k^{\alpha}_{;\alpha ;\mu} + [k,v]^{\mu} (\ln a)_{;\mu}$$ (42)

Now, using the identity (1) and the definition (2) yields

$$\frac{d}{dv} [v^{\mu}(\ln a)_{;\mu}] = \frac{1}{2} R_{\mu \n... ... v^{\mu} k^{\alpha}_{;\mu ;\alpha} +[k,v]^{\mu} (\ln a)_{;\mu}$$ (43)

Let us try to write the term $- \frac{1}{2} v^{\mu} k^{\alpha}_{;\mu ;\alpha}$ in the form of an integral along $\gamma$. In agreement with (27), we have at any point x(v) of $\gamma$:

$$v^{\mu} k^{\alpha}_{;\mu ;\alpha} = \int \limits_{-\infty}^{... ...v} k^{\lambda} (v^{\mu}k^{\alpha}_{;\mu ;\alpha})_{;\lambda}dv$$ (44)

A tedious but straightforward calculation using (1), (2) and (17) leads to the following result

In the above formulae $v^{\mu}$ is an arbitrary vector. So we can choose $v^{\mu}$ so that the transport equations

$$[k,v] ^{\mu} = 0$$ (45)

are satisfied along the ray $\gamma$. Since (46) is a system of first order partial differential equations in $v^{\mu}$, there exists one and only one solution satisfying the boundary conditions (34). With this choice, $2\dot{a}/a$ is given by the integral formula:

Now we look for an integral form for the total derivative $\frac{d}{ds} (1+z)^{-1}$ along ${\cal{C}}_{obs}$. Henceforth, we suppose for the sake of simplicity that the observer is freely falling, i.e. that ${\cal{C}}_{obs}$ is a timelike geodesic. So we have

$$\dot{u}^{\alpha} = u^{\lambda} u^{\alpha}_{;\lambda} = 0$$ (46)

Since $(u^{\mu}k_{\mu})_{em}$ is a constant characterizing the observed spectral line (see above), it follows from (23) and (48) that

$$\frac{d}{ds}\left( \frac{1}{1+z} \right)_{obs} = \frac{1} {(u^{\alpha}k_{\alpha})_{em}} (u^{\mu} u^{\nu} k_{\mu;\nu})_{obs}$$ (47)

Given an arbitrary vector field $v^{\mu}$ fulfilling the boundary condition (34), Eq.(49) may be written as

$$\frac{d}{ds}\left( \frac{1}{1+z} \right)_{obs} = \frac{1}{(... ...v_{obs}} k^{\lambda}(v^{\mu} v^{\nu} k_{\mu;\nu})_{;\lambda} dv$$ (48)

Using (1), (17) and (41), a straightforward calculation gives the general formula

which holds for any freely falling observer.

Now let us choose for $v^{\mu}$ the vector field defined by (46) and (34). We obtain