Effect of solar radiation on interplanetary dust grains

An interplanetary dust grain orbiting the Sun absorbs solar radiation, and immediately re-radiates the absorbed energy isotropically in its instantaneous rest frame. (Here, we are assuming that the size of the grain is much larger than the wavelength of the radiation, and, also, that the grain is rapidly rotating, so the radiative heating of its surface is even.) Because the angular distributions of the absorbed and emitted radiation patterns are different, and because electromagnetic radiation possesses momentum (Fitzpatrick 2008), a net radiation pressure force is exerted on the grain. Although this force is very small, when integrated over a sufficiently long period of time, it can significantly modify the dust grain orbit, causing the grain to spiral in toward the Sun. Let us investigate this effect.

Let $r$, $\theta $, $z$ be cylindrical polar coordinates in a frame of reference, centered on the Sun, that is aligned with the orbital plane of the dust grain, as described in Section I.1. Let ${\bf r}$ and ${\bf v}=\dot{\bf r}$ be the (relative) position and velocity of the grain, respectively. Consider a photon emitted by the Sun. Let $E$ and ${\bf p}$ be the photon's energy and momentum, respectively, in the heliocentric frame. Let $E'$ and ${\bf p}'$ be the corresponding quantities in the dust grain's instantaneous rest frame. According to standard relativistic theory (Rindler 1977),

$\displaystyle E'$ $\displaystyle =\gamma\,(E-{\bf v}\cdot{\bf p}),$ (10.160)
$\displaystyle {\bf p}'$ $\displaystyle = {\bf p} +\left[(\gamma-1)\,\frac{{\bf v}\cdot{\bf p}}{v^{\,2}}-\frac{\gamma\,E}{c^{\,2}}\right]{\bf v},$ (10.161)

where $\gamma=(1-v^{\,2}/c^{\,2})^{-1/2}$. Assuming that the grain is moving non-relativistically, so that $v/c\ll 1$, the previous expressions reduce to

$\displaystyle E'$ $\displaystyle \simeq E - {\bf v}\cdot{\bf p},$ (10.162)
$\displaystyle {\bf p}'$ $\displaystyle \simeq {\bf p} - \frac{E}{c^{\,2}}\,{\bf v},$ (10.163)

respectively, to first order in $v/c$. The corresponding inverse transforms are

$\displaystyle E$ $\displaystyle \simeq E'+ {\bf v}\cdot{\bf p}',$ (10.164)
$\displaystyle {\bf p}$ $\displaystyle \simeq {\bf p}' + \frac{E'}{c^{\,2}}\,{\bf v}.$ (10.165)

To first order in $v/c$, the previous four expressions also apply to energy and momentum measured per unit time.

Let ${\cal E}_{\rm abs}'$ and ${\bf P}_{\rm abs}'$ be the electromagnetic energy and momentum, respectively, absorbed by the grain per unit time in its instantaneous rest frame. Let ${\cal E}_{\rm em}'$ and ${\bf P}_{\rm em}'$ be the corresponding quantities emitted by the grain. Now, we are assuming that

$\displaystyle {\cal E}_{\rm em}' = {\cal E}_{\rm abs}'.$ (10.166)

In other words, all energy absorbed by the grain is immediately re-emitted. We also assuming that

$\displaystyle {\bf P}_{\rm em}' = {\bf0}.$ (10.167)

In other words, the emitted radiation pattern is isotropic in the grain's instantaneous rest frame, and consequently carries off zero net momentum. It follows that the grain's net rate of energy gain in its instantaneous rest frame is

$\displaystyle {\mit\Delta}{\cal E}' = {\cal E}_{\rm abs}'-{\cal E}_{\rm em}'= 0,$ (10.168)

whereas the corresponding rate of momentum gain is

$\displaystyle {\mit\Delta}{\bf P}' = {\bf P}_{\rm abs}'-{\bf P}_{\rm em}'= {\bf P}_{\rm abs}'.$ (10.169)

Transforming to the heliocentric frame, we obtain

$\displaystyle {\mit\Delta} {\bf P} \simeq {\mit\Delta} {\bf P}' + \frac{{\mit\Delta}{\cal E}'}{c^{\,2}}\,{\bf v} = {\mit\Delta} {\bf P}' = {\bf P}_{\rm abs}'.$ (10.170)

where use has been made of Equations (10.165), (10.168), and (10.169). However, according to Equation (10.163),

$\displaystyle {\bf P}_{\rm abs}' \simeq {\bf P}_{\rm abs} - \frac{{\cal E}_{\rm abs}}{c^{\,2}}\,{\bf v}.$ (10.171)

Furthermore, to first order in $v/c$,

$\displaystyle {\cal E}_{\rm abs}$ $\displaystyle \simeq E\,N\,A\,(c-v_r),$ (10.172)
$\displaystyle {\bf P}_{\rm abs}$ $\displaystyle \simeq \frac{E}{c}\,N\,A\,(c-v_r)\,{\bf e}_r,$ (10.173)

where $E$ is the mean energy of a photon emitted by the Sun, $(E/c)\,{\bf e}_r$ the corresponding mean momentum (Rindler 1977), $N$ the local photon number density, and $A$ the cross-sectional area of the grain (normal to ${\bf e}_r$). It follows that, to first order in $v/c$,

$\displaystyle {\mit\Delta} {\bf P} \simeq E\,N\,A\left(1-\frac{v_r}{c}\right) {\bf e}_r - E\,N\,A\,\frac{{\bf v}}{c}.$ (10.174)

Of course, $E\,N\,c=L/4\pi\,r^{\,2}$ is the local solar electromagnetic energy flux, where $L= 3.846\times 10^{26}\,{\rm W}$ is the solar luminosity (Yoder 1995).

In the heliocentric frame, the net force per unit mass acting on the grain is

$\displaystyle {\bf F} = \frac{{\mit\Delta}{\bf P}}{m},$ (10.175)

where $m$ is the grain's mass. It follows from the previous analysis that

$\displaystyle {\bf F} = {\bf F}_{\rm rad} + {\bf F}_{\rm PR},$ (10.176)


$\displaystyle {\bf F}_{\rm rad} = \frac{L\,A}{4\pi\,r^{\,2}\,m\,c}\left(1-\frac{v_r}{c}\right){\bf e}_r,$ (10.177)


$\displaystyle {\bf F}_{\rm PR} = - \frac{L\,A}{4\pi\,r^{\,2}\,m\,c^{\,2}}\,{\bf v}.$ (10.178)

[A more rigorous derivation of the previous two equations is given in Robertson 1937 and Klačka 1993.] Here, ${\bf F}_{\rm rad}$ is the radiation pressure force per se, and is everywhere directed radially outward from the Sun. On the other hand, ${\bf F}_{\rm PR}$ is the so-called Poynting-Robertson drag, and is always oppositely directed to the grain's instantaneous velocity (Poynting 1904; Robertson 1937). The origin of the latter force, which is much smaller in magnitude than the former, is the slightly non-isotropic angular distribution of the radiation re-emitted by the grain, as seen in the Sun's rest frame. Finally, if we write ${\bf F} = F_r\,{\bf e}_r+F_\theta\,{\bf e}_\theta+F_z\,{\bf e}_z$ then the previous three equations imply that

$\displaystyle F_r$ $\displaystyle = \frac{L\,A}{4\pi\,r^{\,2}\,m\,c}\left(1-\frac{2\,\skew{5}\dot{r...
...rac{a}{r}\right)^2\left[1-\frac{2\,h\,e\,\sin\theta}{c\,a\,(1-e^{\,2})}\right],$ (10.179)
$\displaystyle F_\theta$ $\displaystyle =-\frac{L\,A}{4\pi\,r^{\,2}\,m\,c}\,\frac{r\,\skew{5}\dot{\theta}}{c} =-\frac{L\,A}{4\pi\,a^{\,2}\,m\,c}\left(\frac{a}{r}\right)^3\frac{h}{c\,a},$ (10.180)
$\displaystyle F_z$ $\displaystyle = 0,$ (10.181)

where use has been made of Equations (I.3), (I.7), and (I.10).

Assuming that the total radiation pressure force, (10.176), is small compared to the force of gravitational attraction between the Sun and the dust grain—and can, thus, be treated as a perturbation—the grain's orbit can be modeled as Keplerian ellipse whose six elements evolve slowly in time under the influence of the pressure. The six elements in question are chosen to be the major radius, $a$, the mean anomaly at epoch, ${\cal M}_0$, the eccentricity, $e$, the argument of the perigee, $\omega$, the inclination (to the ecliptic plane), $I$, and the longitude of the ascending node (measured with respect to the vernal equinox), ${\mit \Omega }$. (See Section 4.12.) The evolution of these elements is governed by the Gauss planetary equations, (I.53)–(I.58). Now, in the perturbative limit, the evolution of the dust grain's orbital elements takes place on a timescale that is much longer than its orbital period. We can concentrate on this evolution, and filter out any relatively short-term oscillations in the elements, by averaging the Gauss planetary equations over an orbital period. A suitable orbit-average operator is

$\displaystyle \langle\cdots\rangle \equiv \frac{1}{T}\int_0^T (\cdots)\,dt = (1-e^{\,2})^{-1/2}\oint (\cdots)\left(\frac{r}{a}\right)^2\,\frac{d\theta}{2\pi}.$ (10.182)

Here, we have made use of the fact that $h\equiv r^{\,2}\,\skew{5}\dot{\theta}=(1-e^{\,2})^{1/2}\,n\,a^{\,2}$ is a constant of the unperturbed motion. (See Chapter 4.) Note that $n=(\mu/a^{\,3})$ is the unperturbed mean orbital angular velocity. Moreover, $\mu=G\,M$, where $M$ is the solar mass. (Of course, we are ignoring the mass of the grain with respect to that of the Sun.) Making use of Equations (10.179)–(10.181), as well as some of the formulae appearing in Appendix I, the orbit-averaged Gauss planetary equations become

$\displaystyle \left\langle\frac{\skew{3}\dot{a}}{a}\right\rangle$ $\displaystyle = - \frac{L\,A}{4\pi\,a^{\,2}\,m\,c^{\,2}}\,\frac{2+3\,e^{\,2}}{(1-e^{\,2})^{3/2}},$ (10.183)
$\displaystyle \langle\skew{5}\dot{\cal M}_0 \rangle$ $\displaystyle = -\frac{L\,A}{4\pi\,a^{\,2}\,m\,c^{\,2}}\,\frac{2\,c}{n\,a},$ (10.184)
$\displaystyle \langle \skew{3}\dot{e}\rangle$ $\displaystyle =-\frac{L\,A}{4\pi\,a^{\,2}\,m\,c^{\,2}}\,\frac{5\,e}{2\,(1-e^{\,2})^{1/2}},$ (10.185)
$\displaystyle \langle \skew{3}\dot{\omega} \rangle$ $\displaystyle = 0,$ (10.186)
$\displaystyle \langle\dot{I}\rangle$ $\displaystyle = 0,$ (10.187)
$\displaystyle \langle\skew{5}\dot{\mit\Omega}\rangle$ $\displaystyle = 0.$ (10.188)

It can be seen that the radiation pressure force causes the grain's orbital major radius and eccentricity to both decrease monotonically in time [because the left-hand sides of Equations (10.183) and (10.185) are both negative.] On the other hand, radiation pressure has no effect on the either the location of the grain's perihelion point or the orientation of its orbital plane (because $\langle\skew{3}\dot{\omega}\rangle$, $\langle\dot{I}\rangle$ and $\langle\skew{5}\dot{\mit\Omega}\rangle$ are all zero). Finally, given that the right-hand side of Equation (10.184) is negative, the radiation pressure force produces a slight reduction in the grain's mean orbital angular velocity: the modified angular velocity being $d{\cal M}/dt = n+ \langle\skew{5}\dot{\cal M}_0\rangle$. Note that, to lowest order in $v/c$, the radial radiation pressure force, ${\bf F}_{\rm rad}$, averages out of the expressions for $\langle\skew{3}\dot{a}/a\rangle$ and $\langle\skew{3}\dot{e}/e\rangle$. Hence, the reduction in the grain's orbital major radius and eccentricity is due to the Poynting-Robertson drag, ${\bf F}_{\rm PR}$, combined with the first-order component of ${\bf F}_{\rm rad}$. ( ${\bf F}_{\rm PR}$ is similar in magnitude to the first-order component of ${\bf F}_{\rm rad}$.) On the other hand, the lowest-order component of ${\bf F}_{\rm rad}$ makes the dominant contribution to the expression for $\langle\skew{5}\dot{\cal M}_0\rangle$ (in fact, we have neglected any first-order contributions to this expression).

In order to be in the perturbative limit, the relative changes in the dust grain's orbital elements induced by the radiation pressure force in an orbital period must all be small. Because $\langle\skew{5}\dot{\cal M}_0\rangle\gg \langle\skew{3}\dot{a}/a\rangle,
\langle\skew{3}\dot{e}\rangle$ [given that $c/(n\,a)\gg 1$, which follows because the grain is moving non-relativistically], this requirement yields $\langle\skew{5}\dot{\cal M}_0\rangle/n\ll 1$, or

$\displaystyle \frac{L\,A}{2\pi\,m\,c\,\mu}\ll 1,$ (10.189)

which is equivalent to the requirement that the radiation pressure force be much smaller than the force of gravitational attraction between the grain and the Sun. Let us model the grain as a sphere of radius $b$ and mass density $\rho $. It follows that $A=\pi\,b^{\,2}$ and $m= (4\pi/3)\,\rho\,b^{\,3}$. Thus, the previous expression yields

$\displaystyle b\gg \frac{3}{8\pi}\,\frac{L}{G\,M\,\rho\,c}.$ (10.190)

Given that $L= 3.85\times 10^{26}\,{\rm W}$, $G= 6.67\times 10^{-11}\,{\rm m}^{\,3}\,{\rm kg}^{-1}\,{\rm s}^{-2}$, $M= 1.99\times 10^{30}\,{\rm kg}$, $c=3.00\times 10^8\,{\rm m}\,{\rm s}^{-1}$ (Yoder 1995), and assuming that $\rho\simeq 1\times 10^3\,{\rm kg}\,{\rm m}^{-3}$, we obtain

$\displaystyle b \gg 10^{-6}\,{\rm m}.$ (10.191)

We conclude that the radiation pressure force is perturbative for “large” dust grains whose radii are much greater than a micron. On the other hand, the radiation pressure force—whose dominant component is directed radially outward from the Sun—exceeds the force of gravitational attraction for “small” dust grains whose radii are less than, or of order, a micron. We would, therefore, expect radiation pressure to eventually expel small dust grains from the solar system.

For the case of a large dust grain in a circular orbit around the Sun, Equation (10.183) gives

$\displaystyle \frac{d a^{\,2}}{dt} = -\frac{L\,A}{\pi\,m\,c^{\,2}}.$ (10.192)

Hence, the time required for a grain whose orbit has an initial radius $a$ to spiral into the Sun is

$\displaystyle \tau = \frac{\pi\,m\,c^{\,2}\,a^{\,2}}{L\,A}.$ (10.193)

Again modeling the grain as a sphere of radius $b$ and mass density $\rho\simeq 1\times 10^3\,{\rm kg}\,{\rm m}^{-3}$, we obtain

$\displaystyle \tau = \frac{4\pi}{3}\,\frac{c^{\,2}\,\rho\,a^{\,2}\,b}{L}\simeq 7\times 10^8\,a^{\,2}\,b\,{\rm years}.$ (10.194)

Here, $a$ is measured in astronomical units, and $b$ in meters. Thus, a dust grain of radius $10^{-4}\,{\rm m}$, and initial orbital radius 1 AU, takes about $70,000$ years to spiral into the Sun.