#  Quantum Self-Propulsion of an Inhomogeneous Object out of Thermal
Equilibrium

Kimball A. Milton<EMAIL_ADDRESS>H. L. Dodge Department of Physics and
Astronomy, University of Oklahoma, Norman, OK 73019, USA  Nima Pourtolami
<EMAIL_ADDRESS>National Bank of Canada, Montreal, Quebec H3B 4S9,
Canada  Gerard Kennedy<EMAIL_ADDRESS>School of Mathematical Sciences,
University of Southampton, Southampton, SO17 1BJ, UK

###### Abstract

In an earlier paper, we explored how quantum vacuum torque can arise: a body
or nanoparticle that is out of thermal equilibrium with its environment
experiences a spontaneous torque. But this requires that the body be composed
of nonreciprocal material, which seems to necessitate the presence of an
external influence, such as a magnetic field. Then the electric polarizability
of the particle has a real part that is nonsymmetric. This effect occurs to
first order in the polarizability. To that order, no self-propulsive force can
arise. Here, we consider second-order effects, and show that spontaneous
forces can arise in vacuum, without requiring exotic electromagnetic
properties. Thermal nonequilibrium is still necessary, but the electric
susceptibility of the body need only be inhomogeneous. We investigate four
examples of such a body: a needle composed of distinct halves; a sphere and a
ball, each hemisphere being made of a different substance; and a thin slab,
each face of which is different. The results found are consistent with
previous numerical investigations. Here, we take into account the skin depth
of metal surfaces. We also consider the frictional forces that would cause the
body to acquire a terminal velocity, which might be observable. More likely to
be important is relaxation to thermal equilibrium, which can still lead to a
readily observable terminal velocity. A general treatment of such forces on a
moving body, expressed in momentum space, is provided, which incorporates both
propulsive and frictional forces. The source of the propulsive force is the
nonsymmetric pattern of radiation from different parts of the body, the higher
reflectivity of the metal portion playing a crucial role.

##  I Introduction

The subject of quantum vacuum forces on neutral particles or bodies has a long
history. The classic analysis of Einstein and Hopf [1] shows that a body
moving nonrelativistically through vacuum experiences a frictional force
opposing its motion that depends on the temperature of the environment.
Dissipation is required, which may arise either from intrinsic processes
within the body, producing an imaginary part of the electric susceptibility of
the constituent material, or from the radiation field itself. For a short
representative list of papers on quantum vacuum friction, see Refs. [2, 3, 4,
5, 6, 7].

Much more work has been done on quantum electrodynamic friction near
dissipative surfaces; for example, see Refs. [8, 9, 10, 11, 12, 13, 14, 15,
16, 17, 19, 21, 20, 22, 23, 18, 24, 25, 26]. In particular, attention has been
directed to the force on a body or particle moving parallel to a dielectric or
conducting surface. This configuration will typically give rise to a much
larger frictional force, although even such friction has yet to be observed in
the laboratory. These are nonequilibrium processes, but usually a non-
equilibrium steady state (NESS) configuration has been considered, where the
moving object neither gains nor loses energy, but absorbs momentum from the
radiation field. This automatically happens if the body is nondissipative,
with the radiation field being the only source of dissipation. If the body is
dissipative, satisfaction of the NESS condition means that the temperature of
the body bears a definite ratio to that of the environment, that ratio
depending of the velocity of the body. For low velocities, where NESS reduces
to thermal equilibrium, the ratio of temperatures tends to unity.

We recently revisited the subject of quantum vacuum friction, considering the
relativistic regime and nonequilibrium phenomena [27, 28]. We proposed that
the NESS temperature ratio might be an accessible signature for experimentally
observing such effects. We considered both dipole fluctuations and field
fluctuations as sources for the dissipative effects.

But, even more remarkably, fluctuation phenomena may lead to spontaneous
torques and forces on stationary bodies out of equilibrium with their
environment [29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40]. These effects,
breaking time-reversal symmetry, typically require that the bodies be
constituted of nonreciprocal material, that is, that the electric
susceptibility be non-Hermitian.111For much more on nonreciprocal materials,
see Ref. [41]. This may be achieved through the application of an external
magnetic field; see, for example, Ref. [36]. We explored such anomalous
torques and forces in a recent paper [42]; see also Ref. [43]. The effects
considered there were first-order in the susceptibility or the polarizability.
Although vacuum torques were found, no self-propulsive forces could appear in
first order, unless the body were adjacent to another body, a dielectric or
metal slab, for example.

To obtain a vacuum force for a body at rest requires not only nonequilibrium,
but also nonuniformity of the electric suseptibility, although not necessarily
nonreciprocity [44, 45]. This requires going to second order in the
susceptibility. Even if the body is spatially nonsymmetric but possesses
uniform susceptibilty, no spontaneous force can arise, although it is claimed
that a torque for such a body can appear if axial symmetry is broken, as for a
pinwheel [44]. (We will explore the quantum vacuum torque in a subsequent
paper.) In this paper, we demonstrate that a net force emerges if the
susceptibility is nonuniform across the object. We analyze the general
situation for an arbitrary susceptibility (Sec. II), and apply the analysis to
four particular configurations: an inhomogeneous needle (Sec. III); an
inhomogeneous spherical shell (Sec. IV); a ball with two halves made of
different materials (a “Janus ball”) [44] (Sec. V); and a thin plate
consisting of metal on one side and blackbody material222By blackbody material
we mean a substance whose mean electric polarizability per unit surface area
is such as to yield the Planck spectrum and Stefan’s law. See Sec. VI. on the
other (Sec. VI). Of course, once a body is set in motion, a frictional force
will arise, resulting, eventually, in the body reaching a terminal velocity if
the friction is large enough; this is explored in Sec. VII. However, it is
likely that thermal relaxation effects will be much more important, and we
show in Sec. VIII that these might still result in an readily observable
terminal velocity. An outlook is provided in the Conclusions. Appendix A
provides a momentum-space treatment of the spontaneous force, and shows how
second-order friction emerges naturally. The thermal radiation from the body
is analyzed more fully in Appendix B; the nonsymmetric thermal radiation
pattern is responsible for the propulsive force. An explicit example of a
dilute dielectric body attached to a perfectly reflective mirror is given in
Appendix C.

##  II Self-propulsive force

We follow the procedure detailed in Ref. [42, 43], where we studied only
nonreciprocal bodies and worked only to first order in susceptibility, and
found spontaneous torques, but no vacuum self-propulsive force.

We started with the formula for the Lorentz force on a dielectric body [Eq.
(7.1) of Ref. [42]]:333 The notation used in the second term here indicates,
for its $i$th component, $P_{j}\nabla_{i}E_{j}$.

$\mathbf{F}=\int(d\mathbf{r})\int\frac{d\omega}{2\pi}\frac{d\nu}{2\pi}e^{-i(\omega+\nu)t}\left\\{-\left(1+\frac{\omega}{\nu}\right)\left[\bm{\nabla}\cdot\mathbf{P}(\mathbf{r};\omega)\right]\mathbf{E}(\mathbf{r};\nu)-\frac{\omega}{\nu}\mathbf{P}(\mathbf{r};\omega)\cdot(\bm{\nabla})\cdot\mathbf{E}(\mathbf{r};\nu)\right\\}.$
(1)

Here, the first term will not contribute to the corresponding quantum force,
because the expectation value will make the sum of the frequencies vanish,
$\omega+\nu=0$. [See Eq. (4) below.] Then, as in the application of the
principle of virtual work employed in Refs. [27, 28], we see that the force
density may be thought of as the negative gradient of the interaction (free)
energy density,

$\mathfrak{f}=-\mathbf{P(r},t)\cdot\mathbf{E(r},t),$ (2)

where the gradient acts only on the second, $\mathbf{E}$, factor. In Ref. [42]
we expanded, to first order, the electric field in terms of the electric
polarization,

$\mathbf{E}^{(1)}(\mathbf{r};\nu)=\int(d\mathbf{r^{\prime}})\,\bm{\Gamma}(\mathbf{r,r^{\prime}};\nu)\cdot\mathbf{P}(\mathbf{r}^{\prime};\nu),$
(3a) where $\bm{\Gamma}$ is the retarded electromagnetic Green’s dyadic, or we
expanded the polarization in terms of the electric field,
$\mathbf{P}^{(1)}(\mathbf{r};\omega)=\bm{\chi}(\mathbf{r};\omega)\cdot\mathbf{E}(\mathbf{r};\omega),$
(3b)

where $\bm{\chi}=\bm{\varepsilon}-\bm{1}$ is the (local) electric
susceptibility of the body.

In Ref. [42], we found a torque on a nonreciprocal body. We did so by applying
the fluctuation-dissipation theorem (FDT) to evaluate field products:

$\displaystyle\langle\mathscr{S}E_{i}(\mathbf{r};\omega)E_{j}(\mathbf{r}^{\prime};\nu)\rangle$
$\displaystyle=$ $\displaystyle
2\pi\delta(\omega+\nu)(\Im\bm{\Gamma})_{ij}(\mathbf{r,r^{\prime}};\omega)\coth\frac{\beta\omega}{2},\quad\beta=\frac{1}{T},$
(4a)
$\displaystyle\langle\mathscr{S}P_{i}(\mathbf{r};\omega)P_{j}(\mathbf{r}^{\prime};\nu)\rangle$
$\displaystyle=$ $\displaystyle
2\pi\delta(\omega+\nu)\delta(\mathbf{r-r^{\prime}})(\Im\bm{\chi})_{ij}(\mathbf{r};\omega)\coth\frac{\beta^{\prime}\omega}{2},\quad\beta^{\prime}=\frac{1}{T^{\prime}},$
(4b)

where $\mathscr{S}$ denotes a symmetrized product, and $\Im$ signifies the
anti-Hermitian part, to obtain first-order effects. Here, $T$ is the
temperature of the blackbody background, while $T^{\prime}$ is that of the
body. But there is, in vacuum, no first-order force, because the gradient of
the Green’s dyadic is then evaluated at coincident points.

In the current paper, we consider second-order effects. Iterating Eqs. (3a)
and (3b) results in

$\displaystyle\mathbf{E}^{(2)}(\mathbf{r};\nu)$ $\displaystyle=$
$\displaystyle\int(d\mathbf{r^{\prime}})\,\bm{\Gamma}(\mathbf{r,r^{\prime}};\nu)\cdot\bm{\chi}(\mathbf{r}^{\prime};\nu)\cdot\mathbf{E}(\mathbf{r^{\prime}};\nu),$
(5a) $\displaystyle\mathbf{E}^{(3)}(\mathbf{r};\nu)$ $\displaystyle=$
$\displaystyle\int(d\mathbf{r^{\prime}})(d\mathbf{r^{\prime\prime}})\,\bm{\Gamma}(\mathbf{r,r^{\prime}};\nu)\cdot\bm{\chi}(\mathbf{r}^{\prime};\nu)\cdot\bm{\Gamma}(\mathbf{r^{\prime},r^{\prime\prime}};\nu)\cdot\mathbf{P(r^{\prime\prime};\nu)},$
(5b) $\displaystyle\mathbf{P}^{(2)}(\mathbf{r};\omega)$ $\displaystyle=$
$\displaystyle\int(d\mathbf{r^{\prime}})\,\bm{\chi}(\mathbf{r};\omega)\cdot\bm{\Gamma}(\mathbf{r,r^{\prime}};\omega)\cdot\mathbf{P}(\mathbf{r^{\prime}};\omega),$
(5c) $\displaystyle\mathbf{P}^{(3)}(\mathbf{r};\omega)$ $\displaystyle=$
$\displaystyle\int(d\mathbf{r^{\prime}})\,\bm{\chi}(\mathbf{r};\omega)\cdot\bm{\Gamma}(\mathbf{r,r^{\prime}};\omega)\cdot\bm{\chi}(\mathbf{r^{\prime}};\omega)\cdot\mathbf{E}(\mathbf{r^{\prime}};\omega),$
(5d)

the order referring to the number of susceptibility factors, either
$\bm{\chi}$ or $\bm{\Gamma}$. The second-order terms in the interaction energy
(fourth order in susceptibilities) are obtained by using (1)
$\mathbf{P}^{(0)}$ and $\mathbf{E}^{(3)}$, (2) $\mathbf{P}^{(2)}$ and
$\mathbf{E}^{(1)}$, both of which give quadratic structures in $\mathbf{P}$,
while quadratic structures in $\mathbf{E}$ emerge by using (3)
$\mathbf{P}^{(1)}$ and $\mathbf{E}^{(2)}$, and (4) $\mathbf{P}^{(3)}$ and
$\mathbf{E}^{(0)}$. The FDT (4b) is applied for the former, while the FDT (4a)
is applied for the latter. The resulting terms in the force will be denoted
$\mathbf{F}^{(0,3)}$, $\mathbf{F}^{(2,1)}$, $\mathbf{F}^{(1,2)}$, and
$\mathbf{F}^{(3,0)}$, respectively, in an obvious notation.

The EE fluctuations give rise to a force contribution in the $z$ direction of

$\displaystyle F_{z}^{\rm EE}$ $\displaystyle=$
$\displaystyle\int(d\mathbf{r})(d\mathbf{r^{\prime}})\frac{d\omega}{2\pi}\chi_{ij}^{\vphantom{q}}(\mathbf{r};\omega)\bigg{[}\nabla_{z}\Gamma_{ik}(\mathbf{r,r^{\prime}};-\omega)\chi^{\vphantom{q}}_{kl}(\mathbf{r}^{\prime};-\omega)(\Im\Gamma)_{jl}(\mathbf{r,r^{\prime}};\omega)$
(6a)
$\displaystyle\quad\mbox{}+\Gamma_{jk}(\mathbf{r,r^{\prime}};\omega)\chi_{kl}^{\vphantom{q}}(\mathbf{r^{\prime}};\omega)\nabla_{z}(\Im\Gamma)_{li}(\mathbf{r^{\prime},r};\omega)\bigg{]}\coth\frac{\beta\omega}{2},$
while the PP fluctuations give $\displaystyle F_{z}^{\rm PP}$ $\displaystyle=$
$\displaystyle\int(d\mathbf{r})(d\mathbf{r^{\prime}})\frac{d\omega}{2\pi}\nabla_{z}\Gamma_{jk}(\mathbf{r,r^{\prime}};-\omega)\bigg{[}\chi_{kl}^{\vphantom{q}}(\mathbf{r^{\prime}};-\omega)\Gamma_{lm}(\mathbf{r^{\prime},r};-\omega)(\Im\chi)_{jm}(\mathbf{r};\omega)$
(6b)
$\displaystyle\mbox{}+(\Im\chi)_{mk}(\mathbf{r^{\prime}};\omega)\Gamma_{lm}(\mathbf{r,r^{\prime}};\omega)\chi_{jl}(\mathbf{r};\omega)\bigg{]}\coth\frac{\beta^{\prime}\omega}{2}.$

Equation (6a) displays the (1,2) and (3,0) contributions, respectively, while
Eq. (6b) displays the (0,3) amd (2,1) contributions, respectively.

Henceforth, we restrict attention to reciprocal bodies. The general result is
particularly simple if the susceptibility is isotropic,

$\chi_{ij}^{\vphantom{q}}(\mathbf{r};\omega)=\delta_{ij}\chi(\mathbf{r};\omega),$
(7)

and the background is not only reciprocal,

$\Gamma_{ij}(\mathbf{r,r^{\prime}};\omega)=\Gamma_{ji}(\mathbf{r^{\prime},r};\omega),$
(8)

but also homogeneous, in which case the Green’s dyadic is a function only of
$\mathbf{R=r-r^{\prime}}$. We then use symmetry under
$\mathbf{r}\leftrightarrow\mathbf{r^{\prime}}$ and also $\omega\to-\omega$,
noting that the imaginary part of the susceptibility is odd in $\omega$, to
derive the quantum vacuum force in the $z$-direction, which is, including the
contributions from EE and PP fluctuations,

$F_{z}=\int(d\mathbf{r})(d\mathbf{r^{\prime}})\int_{-\infty}^{\infty}\frac{d\omega}{2\pi}X(\mathbf{r,r^{\prime}};\omega)\operatorname{Im}\Gamma_{ji}(\mathbf{r^{\prime}-r};\omega)\nabla_{z}\operatorname{Im}\Gamma_{ij}(\mathbf{r-r^{\prime}};\omega)\left[\coth\frac{\beta\omega}{2}-\coth\frac{\beta^{\prime}\omega}{2}\right],$
(9)

where the second-order susceptibility product is

$X(\mathbf{r,r^{\prime}};\omega)=\operatorname{Im}\left[\chi(\mathbf{r};\omega)\chi(\mathbf{r^{\prime}};\omega)^{*}\right]=\operatorname{Im}\chi(\mathbf{r};\omega)\operatorname{Re}\chi(\mathbf{r^{\prime}};\omega)-\operatorname{Re}\chi(\mathbf{r};\omega)\operatorname{Im}\chi(\mathbf{r^{\prime}};\omega).$
(10)

(Note that only the first term of Eq (6a), $F^{(1,2)}$, contributes in the
isotropic case, because the integrand of the second term is antisymmetric
under $\mathbf{r}\leftrightarrow\mathbf{r}^{\prime}$.) It is clear that if the
susceptibility of the body is homogeneous, so $\chi$ is independent of
$\mathbf{r}$ within the body, the force is zero, since $X=0$ then. So, we need
an inhomogeneous body in order to generate a spontaneous quantum vacuum force.

The next simplest possibility is to imagine the object as consisting of two
separate parts, labeled $A$ and $B$, in each of which $\chi$ is spatially
constant, as illustrated in Fig. 1.

Figure 1: Cartoon of the inhomogeneous bodies considered in this paper. The
upper part, $A$, will typically be a (nondispersive) insulator, while the
lower part, $B$, will be a Drude-type metal. The figure is intended to
illustrate a solid body, rotationally invariant about the $z$-axis. Because of
this symmetry, the force must be either parallel or antiparallel to the
$z$-axis.

Then the formula for the force reduces to

$F_{z}=8\int_{0}^{\infty}\frac{d\omega}{2\pi}X_{AB}^{\vphantom{q}}(\omega)\left[\frac{1}{e^{\beta\omega}-1}-\frac{1}{e^{\beta^{\prime}\omega}-1}\right]I_{AB}(\omega),$
(11a) where
$I_{AB}(\omega)=\int_{A}(d\mathbf{r})\int_{B}(d\mathbf{r^{\prime}})\,\frac{1}{2}\nabla_{z}\left[\operatorname{Im}\Gamma_{ji}(\mathbf{r^{\prime},r};\omega)\operatorname{Im}\Gamma_{ij}(\mathbf{r,r^{\prime}};\omega)\right],$
(11b) and
$X_{AB}^{\vphantom{q}}(\omega)=\operatorname{Im}\chi_{A}^{\vphantom{q}}(\omega)\operatorname{Re}\chi_{B}^{\vphantom{q}}(\omega)-\operatorname{Re}\chi_{A}^{\vphantom{q}}(\omega)\operatorname{Im}\chi_{B}^{\vphantom{q}}(\omega).$
(11c)

Since the divergenceless part444The divergenceless part of the Green’s dyadic
is defined by $\bm{\Gamma^{\prime}}=\bm{\Gamma}+\bm{1}$, where $\bm{1}$
includes a spatial delta function. of the vacuum Green’s dyadic is

$\bm{\Gamma}^{(0)\prime}(\mathbf{r-r^{\prime}};\omega)=(\bm{\nabla\nabla}-\bm{1}\nabla^{2})\frac{e^{i\omega
R}}{4\pi R}=\left[\mathbf{\hat{R}\hat{R}}(3-3i\omega
R-\omega^{2}R^{2})-\bm{1}(1-i\omega R-\omega^{2}R^{2})\right]\frac{e^{i\omega
R}}{4\pi R^{3}},$ (12)

where $R=|\mathbf{R}|$ and $\mathbf{\hat{R}}=\mathbf{R}/R$, and the matrix
trace appearing in Eq. (11b) is

$\operatorname{Im}\Gamma_{ji}(-\mathbf{R};\omega)\operatorname{Im}\Gamma_{ij}(\mathbf{R};\omega)=\frac{2\Delta(\omega
R)}{(4\pi R^{3})^{2}},$ (13a) with $\Delta(\omega
R)=(3-2\omega^{2}R^{2}+\omega^{4}R^{4})\sin^{2}\\!\omega R-\omega
R(3-\omega^{2}R^{2})\sin 2\omega R+3\omega^{2}R^{2}\cos^{2}\\!\omega R.$ (13b)

Note that this trace is finite as $R\to 0$ since

$\Delta(u)\sim\frac{2}{3}u^{6}-\frac{2}{9}u^{8}+\cdots,\quad u\ll 1.$ (14a)
Although it therefore appears that the leading term, proportional to $u^{6}$,
never contributes to the geometrical integral (11b), this depends on the
convergence of the integrals—see Sec. VI.2. And, apart from subdominant
oscillating terms, the large-argument behavior is
$\Delta(u)\sim\frac{1}{2}u^{4}+\cdots,\quad u\gg 1.$ (14b)

An alternative, momentum-space formulation, of the force appears in Appendix
A.

##  III Inhomogeneous needle

This simplest example is a one-dimensional needle, lying along the $z$-axis,
with one side being made of uniform material $A$, and of length $a$; and the
other side of uniform material $B$, and of length $b$:

$\chi(\mathbf{r};\omega)=C\left[\delta(x)\delta(y)\theta(z)\theta(a-z)\chi_{A}^{\vphantom{q}}(\omega)+\delta(x)\delta(y)\theta(-z)\theta(b+z)\chi_{B}^{\vphantom{q}}(\omega)\right],$
(15)

for the needle having constant (small) cross-sectional area $C$. Then, the
double volume integral $I_{AB}$ just involves $R=z^{\prime}-z$
($z^{\prime}>0>z$), and can be explicitly evaluated:

$I_{AB}=\frac{C^{2}\omega^{5}}{(4\pi)^{2}}\left[f(\omega(a+b))-f(\omega
a)-f(\omega b)\right],$ (16)

where

$f(x)=\frac{1}{30x^{5}}g(x),$ (17a) with
$g(x)=-9-5x^{2}(1+3x^{2})+(9-13x^{2}+11x^{4})\cos 2x+2x(9-x^{2})\sin
2x+22x^{5}\operatorname{Si}(2x).$ (17b)

Here, the sine integral function is defined by

$\operatorname{Si}(x)=\int_{0}^{x}\frac{dt}{t}\sin t.$ (18)

The behavior of the function $g$ for small and large arguments is

$g(x)\sim 20x^{6}-\frac{20}{9}x^{8}\quad\mbox{as}\quad x\to 0,$ (19a) and
$g(x)\sim 11\pi x^{5}\quad\mbox{as}\quad x\to\infty.$ (19b)

Thus, the self-propulsive force (11a) on the needle can be written as

$F_{z}=\frac{C^{2}}{2\pi^{3}}\int_{0}^{\infty}d\omega\,\omega^{5}X_{AB}^{\vphantom{q}}(\omega)\left[\frac{1}{e^{\beta\omega}-1}-\frac{1}{e^{\beta^{\prime}\omega}-1}\right][f(\omega(a+b))-f(\omega
a)-f(\omega b)].$ (20)

One curious feature is immediately apparent: For high temperatures, or
$\beta$, $\beta^{\prime}\ll 1$, the frequency integral should be dominated by
large values of $\omega$, where $f(x)\sim 11/30$ is independent of $x$. Thus,
in this limit, the force is independent of the dimensions of the needle; that
is, only a relatively small region of the needle near the intersection of the
two halves contributes to the force. At room temperature, this independence of
the size of the needle should hold if $a,b\gg 10$ $\mu$m. For low
temperatures, or $\beta,\beta^{\prime}\gg 1$, low values of $\omega$ should
dominate, and in view of the behavior in Eq. (19a), the leading term in $g$
will give no contribution to the force, as already recognized above, and the
second term will yield bilinear behavior, that is $F_{z}\propto ab$.

Now let us compute the force for a situation with $\chi_{A}^{\vphantom{q}}$
being a real constant (because the frequencies involved should be small
compared to the atomic binding energies—but see the discussion at the end of
Sec. V), while $\chi_{B}^{\vphantom{q}}$ is given by the Drude model, which is
a good approximation for a metal:

$\chi_{B}^{\vphantom{q}}(\omega)=-\frac{\omega_{p}^{2}}{\omega^{2}+i\omega\nu},$
(21)

where $\omega_{p}$ is the plasma frequency and $\nu$ is the damping parameter,
about 9 eV and 0.035 eV, respectively, for gold. Let us express the
dimensional quantities in terms of reference scales, $\beta_{0}$ and $a_{0}$.
For numerical work we will take $\beta_{0}=40$ (eV)-1, corresponding to room
temperature, and a macroscopic size $a_{0}=1$ cm. For these values, the
dimensionless ratio is $a_{0}/\beta_{0}=1250$. Then, with

$\beta=\hat{\beta}\beta_{0},\quad\beta^{\prime}=\hat{\beta}^{\prime}\beta_{0},\quad\beta_{0}\omega=y,\quad\beta_{0}\nu=\lambda,$
(22)

we define

$h(u,\hat{\beta})=\frac{1}{u^{5}}\int_{0}^{\infty}\frac{dy}{y}\frac{1}{y^{2}+\lambda^{2}}g\left(\frac{a_{0}}{\beta_{0}}yu\right)\frac{1}{e^{\hat{\beta}y}-1},$
(23)

in terms of which the force is

$F=-\frac{C^{2}\omega_{p}^{2}\nu\chi_{A}^{\vphantom{q}}}{60\pi^{3}}\frac{\beta_{0}^{2}}{a_{0}^{5}}\hat{F},$
(24a) where
$\hat{F}=h\left(\frac{a+b}{a_{0}},\hat{\beta}\right)-h\left(\frac{a}{a_{0}},\hat{\beta}\right)-h\left(\frac{b}{a_{0}},\hat{\beta}\right)-h\left(\frac{a+b}{a_{0}},\hat{\beta}^{\prime}\right)+h\left(\frac{a}{a_{0}},\hat{\beta}^{\prime}\right)+h\left(\frac{b}{a_{0}},\hat{\beta}^{\prime}\right).$
(24b)

We now show numerical results. We take the above stated parameters so, for
gold, $\lambda=1.4$, and the magnitude of the prefactor in the force (24b) is
$3.85\chi_{A}\times 10^{-20}$ N if the cross-sectional radius of the needle is
1 mm. Figure 2 shows the saturation of the force at a value of about 0.06 N
for large $a$.

Figure 2: Dimensionless force $\hat{F}$ on an inhomogeneous needle with $a=b$
as a function of $a$ for $T=300$ K and $T^{\prime}=600$ K. The half-length of
the needle is given in cm.

The figure also shows how the force vanishes for small $a$. Finally, the
temperature dependence is sketched in Fig. 3.

Figure 3: Dimensionless force $\hat{F}$ on an inhomogeneous needle with $a=b$
as a function of $T^{\prime}$ for $T=300$ K. The half-length of the needle is
taken to be 1 cm.

However, this is likely to be a gross overestimate. Because the metal has a
small skin depth, which we estimate in terms of the conductivity $\sigma$ as
follows,

$\delta=(\omega\sigma/2)^{-1/2}=(\omega^{2}\operatorname{Im}\chi/2)^{-1/2}=\sqrt{\frac{2(\omega^{2}+\nu^{2})}{\omega\omega_{p}^{2}\nu}},$
(25)

which, for our nominal values and for $\omega\sim\beta_{0}^{-1}$, turns out to
be of order 50 nm. So, the force on a needle with this tiny cross section,
rather than one of 1 mm radius, would be reduced by a factor of $\sim
10^{-18}$. Interestingly, since $1/\beta_{0}$ and $\nu$ are comparable, the
skin depth is near its minimal value, achieved when $\omega=\nu$, $\delta_{\rm
min}=2/\omega_{p}$.

The reason the force is toward the metal end of the needle (negative in our
convention) might seem to be that emissivity of a metal is far lower than that
of an insulator, so there is more radiant energy emitted from the insulator
end of the needle. The corresponding momentum imbalance drives the object in
the $-z$ direction. However, this is not quite borne out by our perturbative
calculations, which show, for modest temperatures, more thermal radiation is
emitted by the metal than by the dielectric. However, the emission from the
dielectric is nearly entirely reflected by the metal, so the momentum
imbalance remains the same. An explicit example of this is detailed in
Appendix C.

##  IV Thin spherical shell

In view of the preceding remarks concerning skin depth, we next consider a
spherical shell, with thickness $t$, smaller than the skin depth, and much
smaller than the radius $a$ of the sphere. We imagine that the sphere is
inhomogeneous: in terms of the polar angle $\theta$,

$\chi(\mathbf{r};\omega)=\left\\{\begin{array}[]{cc}0<\theta<\frac{\pi}{2}:&\chi_{A}^{\vphantom{q}}(\omega),\\\
\frac{\pi}{2}<\theta<\pi:&\chi_{B}^{\vphantom{q}}(\omega).\end{array}\right.$
(26)

The square of the distance between two points on the two hemispheres is given
by

$R^{2}=2a^{2}(1-\cos\gamma)=2a^{2}[1-\cos\theta\cos\theta^{\prime}-\sin\theta\sin\theta^{\prime}\cos(\phi-\phi^{\prime})],$
(27)

where $\gamma$ is the angle between two points, one on the upper hemisphere
and one on the lower. Then, the geometrical integral is

$I_{AB}=\int_{A}(d\mathbf{r})\int_{B}(d\mathbf{r^{\prime}})\,\partial_{z}\left[\frac{\Delta(\omega
R)}{(4\pi
R^{3})^{2}}\right]=\frac{\omega^{8}a}{(4\pi)^{2}}\int_{A}(d\mathbf{r})\int_{B}(d\mathbf{r^{\prime}})\frac{\cos\theta-\cos\theta^{\prime}}{u}D(u),$
(28)

where $u=\omega R$. The $u$ derivative occurring here is explicitly

$D(u)=\frac{\partial}{\partial
u}\left[\frac{\Delta(u)}{u^{6}}\right]=\frac{1}{u^{7}}\left[-9-u^{2}(2+u^{2})+(9-16u^{2}+3u^{4})\cos
2u+u(18-8u^{2}+u^{4})\sin 2u\right],$ (29)

which behaves as

$\displaystyle D(u)$ $\displaystyle\sim$ $\displaystyle-\frac{4}{9}u,\quad
u\to 0,$ (30a) $\displaystyle D(u)$ $\displaystyle\sim$
$\displaystyle-\frac{1}{u^{3}}+\frac{\sin(2u)}{u^{2}},\quad u\to\infty.$ (30b)

The low-$u$ behavior follows from Eq. (14a), where again we note that the
$u^{6}$ term does not contribute. The last term for large $u$ rapidly
oscillates, and on the average gives a term of $O(u^{-4})$, so is subdominant.

For our inhomogeneous sphere, with vanishing thickness $t$, the geometrical
integral is a threefold one:

$I_{AB}=\frac{\omega^{8}a^{5}t^{2}}{8\pi}\mathscr{I}_{AB}(\omega a),$ (31a)
with $\mathscr{I}_{AB}(\omega
a)=\int_{0}^{\pi/2}d\theta\sin\theta\int_{0}^{2\pi}d\phi\int_{\pi/2}^{\pi}d\theta^{\prime}\sin\theta^{\prime}(\cos\theta-\cos\theta^{\prime})\frac{D(u)}{u}.$
(31b)

For a macroscopic object, one would think that the large $u$ behavior of $D$
is dominant, because for a 1 cm sphere at room temperature the characteristic
value of $\omega a\sim 1250$. But this might be misleading, since polar angles
close to the equator of the sphere correspond to $\gamma\to 0$. In Fig. 4 we
show the result of numerical integration, and compare with that arising from
the leading large-$u$ behavior.

Figure 4: The integral $-\mathscr{I}_{AB}$ in Eq. (31b), evaluated numerically
(blue dots), The line shows the result of replacing $D(u)$ by its asymptotic
value $-\frac{1}{u^{3}}$, which is exactly described by a power law $(\omega
a)^{-4}$. The coefficient of the fit to $\mathscr{I}_{AB}$ is
$\mathscr{N}=-26.88$. The “exact” results exhibit considerable scatter, only
part of which reflects the serious oscillations of the integrand; the
numerical integration becomes unstable for large values of $\omega a$.
Although those points lie below the asymptotic line, they seem to be
approaching it.

Using the result of the approximate fit for $\mathscr{I}_{AB}$, and the model
we used for the needle, that $\chi_{A}^{\vphantom{q}}(\omega)$ is constant and
$\chi_{B}^{\vphantom{q}}$ is given by the Drude model (21), we find for the
self-propulsive force (11a)

$F=-\frac{\omega_{p}^{2}t^{2}a}{16\pi^{2}}\chi_{A}^{\vphantom{q}}\nu^{3}\mathscr{N}\hat{F},$
(32a) where
$\hat{F}=\int_{0}^{\infty}dx\frac{x^{3}}{x^{2}+1}\left(\frac{1}{e^{\beta\nu
x}-1}-\frac{1}{e^{\beta^{\prime}x\nu}-1}\right),$ (32b)

the integrals being explicitly

$\int_{0}^{\infty}dx\frac{x^{3}}{x^{2}+1}\frac{1}{e^{xy}-1}=\frac{\pi^{2}}{6y^{2}}-\frac{1}{2}\left[\ln\left(\frac{y}{2\pi}\right)-\frac{\pi}{y}-\psi\left(\frac{y}{2\pi}\right)\right].$
(33)

The dimensionless force $\hat{F}$ is plotted in Fig. 5.

Figure 5: Dimensionless force on an inhomogeneous sphere with the northern
hemisphere having a constant susceptibility, and the southern hemisphere being
a Drude metal, with the environment being at room temperature, for various
values of the temperature of the sphere, $T^{\prime}$. The prefactor in the
force (32a), for our nominal values for gold, and the minimum value of the
skin depth, $t=2/\omega_{p}$, is $1.2\times 10^{-12}\chi_{A}^{\vphantom{q}}$
N, for a sphere of 1 cm radius. The force is computed in the approximation
that the geometric integral $I_{AB}$ is described by the fit to the large-$u$
asymptotics in Fig. 4.

##  V Janus Ball

We now turn to what has been called a Janus ball, a spherical volume with the
forward and rear halves composed of different homogeneous materials. In view
of our previous remarks about skin depth, we should restrict attention to a
nanometer-sized object. Because $I_{AB}$ is the integral of a gradient, the
double radial integral becomes a surface integral,

$I_{AB}=\int_{S_{A}}d\mathbf{S\cdot\hat{z}}\int_{B}(d\mathbf{r^{\prime}})\frac{\Delta(\omega
R)}{(4\pi R^{3})^{2}},$ (34)

where $S_{A}$ is the surface of the $A$ half-ball, that is, a hemisphere of
radius $a$, $0<\theta<\frac{\pi}{2}$ and a disk $\theta=\pi/2$, $0\leq r\leq
a$, while the second integral is over the volume of the lower B half-ball, for
polar angle $\frac{\pi}{2}>\theta>\pi$. Explicitly, the integral can be
written in two parts:

$I_{AB}=\frac{1}{8\pi a}\mathscr{I}_{AB},$ (35a) with
$\displaystyle\mathscr{I}_{AB}$ $\displaystyle=$
$\displaystyle\int_{0}^{\pi/2}d\theta\sin\theta\cos\theta\int_{0}^{1}d\rho^{\prime}\,\rho^{\prime
2}\int_{\pi/2}^{\pi}d\theta^{\prime}\,\sin\theta^{\prime}\int_{0}^{2\pi}d\phi^{\prime}\frac{\Delta(\omega
aP_{1})}{P_{1}^{6}}$ (35b)
$\displaystyle\quad\mbox{}-\int_{0}^{1}d\rho\,\rho\int_{0}^{1}d\rho^{\prime}\,\rho^{\prime
2}\int_{\pi/2}^{\pi}d\theta^{\prime}\,\sin\theta^{\prime}\int_{0}^{2\pi}d\phi^{\prime}\frac{\Delta(\omega
aP_{2})}{P_{2}^{6}},$

where $\rho=r/a$, $\rho^{\prime}=r^{\prime}/a$, and

$\displaystyle P_{1}^{2}$ $\displaystyle=$ $\displaystyle 1+\rho^{\prime
2}-2\rho^{\prime}(\cos\theta\cos\theta^{\prime}+\sin\theta\sin\theta^{\prime}\cos\phi^{\prime}),$
(36a) $\displaystyle P_{2}^{2}$ $\displaystyle=$
$\displaystyle\rho^{2}+\rho^{\prime
2}-2\rho\rho^{\prime}\sin\theta^{\prime}\cos\phi^{\prime},$ (36b)

because on the bisecting disk, $\theta=\pi/2$. Once these four-fold integrals
are done, the force, for the same model considered above, is obtained from

$F=-\frac{\omega_{p}^{2}\epsilon}{2\pi^{2}}\chi_{A}\int_{0}^{\infty}du\frac{1}{u(u^{2}+\epsilon^{2})}\mathscr{I}_{AB}(u)\left[\frac{1}{e^{\beta
u/a}-1}-\frac{1}{e^{\beta^{\prime}u/a}-1}\right],$ (37)

in terms of the abbreviations $\epsilon=\nu a$, $u=\omega a$.

Numerical integration of the two integrals in Eq. (35b) is shown in Fig. 6.

Figure 6: The dots show the results of numerical integration of the two four-
fold integrals in (35b). Shown is the magnitude of the difference of the two
integrals. The upper line (blue) is the large-$u$ asymptotic behavior
resulting from (14b), that is, $\mathscr{I}_{AB}\sim-0.927(\omega a)^{4}$. The
fit is very good for values of $\omega a>10$. Values of $\omega a$ smaller
than 1 are well described by $\mathscr{I}_{AB}\sim-\frac{2\pi}{27}(\omega
a)^{8}$, which results from the $u^{8}$ term in Eq. (14a), the $u^{6}$ term
never contributing.

The figure shows that for large values of $\omega a$, appropriate to a
macroscopic ball near room temperature, the linear fit from the large $u$
behavior of $\Delta(u)$ in Eq. (14b) is quite good, and results in the
following behavior of $I_{AB}$:

$I_{AB}^{\rm JB}\sim-\frac{1}{8\pi a}(0.927)(\omega a)^{4}.$ (38)

This has the same $\omega$ dependence as that for a thin spherical shell in
this limit, which, from Eq. (31a), is

$I_{AB}^{\rm HS}\sim-\frac{26.9}{8\pi}\omega^{4}at^{2}.$ (39)

This will be the same as $I_{AB}^{\rm JB}$ if $a=5.39t$. So apart from a
factor of $0.035(a/t)^{2}$, the forces will be the same in the large $a$
limit.

But as noted above, only the small sphere situation is likely to be physically
relevant. For small $a$, the forces will have a different behavior. The lower
(red) line in Fig. 6 is $\mathscr{I}_{AB}=-\frac{2\pi}{27}(\omega a)^{8}$,
which follows from performing the elementary integrations occurring when Eq.
(14a) holds: the $u^{6}$ term cancels between the disk and the hemisphere
surface integrals (as noted above, the $u^{6}$ term never contributes), and
the $u^{8}$ term gives the value stated. So for a small Janus ball the force
is given by

$F^{\rm JB}=\frac{1}{27\pi}\chi_{A}\omega_{p}^{2}(\nu a)^{7}\hat{F},$ (40a)
where the dimensionless force is $\displaystyle\hat{F}$ $\displaystyle=$
$\displaystyle\int_{0}^{\infty}dx\frac{x^{7}}{x^{2}+1}\left(\frac{1}{e^{\beta\nu
x}-1}-\frac{1}{e^{\beta^{\prime}\nu x}-1}\right)$ (40b) $\displaystyle=$
$\displaystyle-\frac{1}{2}\left[\ln\frac{T^{\prime}}{T}+\frac{\pi}{\nu}(T^{\prime}-T)+\psi\left(\frac{\nu}{2\pi
T^{\prime}}\right)-\psi\left(\frac{\nu}{2\pi T}\right)\right]$
$\displaystyle\quad\mbox{}+\frac{\pi^{2}}{6}\frac{1}{\nu^{2}}(T^{2}-T^{\prime
2})-\frac{\pi^{4}}{15}\frac{1}{\nu^{4}}(T^{4}-T^{\prime
4})+\frac{8\pi^{6}}{63}\frac{1}{\nu^{6}}(T^{6}-T^{\prime 6}).$

The absolute value of this function is plotted in Fig. 7, as a function of the
body temperature $T^{\prime}$, for the environment temperature $T=300$ K, and
$\nu=0.035$ eV, appropriate for gold. If further the radius of the sphere is 1
$\mu$m, the prefactor for a gold microsphere ($\omega_{p}=9$ eV) is
$3.84\times 10^{-18}\chi_{A}$ N, so this might be observable. Note that the
force is negative for $T^{\prime}>T$, so it is in the direction of the metal
end of the ball.

Figure 7: Dimensionless force $\hat{F}$ on a Janus ball composed of a northern
hemisphere made of a dispersionless dielectric, and the southern hemisphere
composed of a Drude metal. Parameters are as given in the text. The force,
given by Eq. (40b), is plotted as a function of the temperature of the sphere,
$T^{\prime}$, for the temperature of the environment fixed at room
temperature. The magnitude is shown in this log plot; the force is negative
for $T^{\prime}>T$.

These results are roughly consistent with those found in Ref. [44]. They
consider a Janus ball with one half SiO2 and the other half gold, and for
$T=0$ K and $T^{\prime}=300$ K find a force of about $4\times 10^{-18}$ N on a
1 $\mu$m ball. Here, for the same example, we find the prefactor in Eq. (40a)
to be $3\times 10^{-18}$ N, while $\hat{F}$ is about 15. A plot of $\hat{F}$
for a vacuum environment at 0 K is shown in Fig. 8, again qualitatively
similar to the results of Ref. [44]. However, the scaling with the radius of
the ball is quite different: we find a $a^{7}$ behavior, whereas they report
scaling roughly with the volume.

Figure 8: Dimensionless force $\hat{F}$ for a Janus ball composed of gold and
SiO2, for the vacuum at zero temperature, as a function of the temperature of
the ball.

For sufficiently high temperatures, the dispersion of the dielectric cannot be
ignored. Therefore, let use a more realistic Lorentzian model,

$\chi_{A}^{\vphantom{q}}(\omega)=\frac{\tilde{\omega}_{p}^{2}}{\omega_{0}^{2}-\omega^{2}-i\omega\gamma}.$
(41)

Then, the second-order susceptiblity factor that appears in the force is

$X(\omega)=-\frac{\omega_{p}^{2}\tilde{\omega}_{p}^{2}\left[\omega_{0}^{2}\nu+\omega^{2}(\gamma-\nu)\right]}{\left[(\omega_{0}^{2}-\omega)^{2}+\omega^{2}\gamma^{2}\right]\left[\omega(\omega^{2}+\nu^{2})\right]}.$
(42)

Of course, this agrees with what we had previously if $\omega\ll\omega_{0}$,
provided $\chi_{A}^{\vphantom{q}}=\tilde{\omega}_{p}^{2}/\omega_{0}^{2}$. The
propulsive force is now

$F=\frac{1}{27\pi}\omega_{p}^{2}\tilde{\omega}_{p}^{2}a^{7}T^{5}\hat{F},$
(43a) the dimensionless force being
$\hat{F}=\int_{0}^{\infty}dy\,y^{7}\frac{y_{0}^{2}\mu+y^{2}(\mu-\lambda)}{[(y_{0}^{2}-y^{2})^{2}+y^{2}\mu^{2}][y^{2}+\lambda^{2}]}\left[\frac{1}{e^{y/t}-1}-\frac{1}{e^{y/t^{\prime}}-1}\right],$
(43b) where in terms of the inverse of room temperature, $\beta_{0}$, the
dimensionless variables are
$\lambda=\beta_{0}\nu,\quad\mu=\beta_{0}\gamma,\quad
y_{0}=\beta_{0}\omega_{0},\quad y=\beta_{0}\omega,\quad t=\beta_{0}T,\quad
t^{\prime}=\beta_{0}T^{\prime}.$ (43c)

We illustrate the effect of this change on the force on a small Janus ball
where $I_{AB}=-\frac{(\omega a)^{8}}{108a}$ in Fig. 9.

Figure 9: Plot of the dimensionless force $\hat{F}$ for a Janus ball with
dielectric on the A side and metal on the B side, as a function of the
temperature of the ball. The metal is chosen to be gold, for which $\mu=1.4$,
the dielectric is polystyrene, for which, roughly [47] (taking the first term
in a four-term fit to the susceptibility) $y_{0}=240$ and $\lambda=26$, and we
consider a room-temperature background. The upper (red) curve assumes a
dispersionless dielectric, while the lower (black) curve includes the single
Lorentzian dispersion model. The effect of the latter is rather small, except
at very high temperature.

##  VI Blackbody/Metal Planar Surface

###  VI.1 Blackbody surface

We preface this section by considering the power absorbed by an object out of
thermal equilibrium. In first order, the power is classically

$P=\frac{\partial\mathscr{F}}{\partial
t}=-\int(d\mathbf{r})\mathbf{P}(\mathbf{r},t)\cdot\frac{\partial}{\partial
t}\mathbf{E}(\mathbf{r},t).$ (44)

We see here the reappearance of the (free) energy density (2). For an object
at temperature $T^{\prime}$ in vacuum at temperature $T$, the FDT implies

$P=\int(d\mathbf{r})\frac{d\omega}{2\pi}\omega\operatorname{Im}\chi_{ij}(\mathbf{r};\omega)\operatorname{Im}\Gamma_{ji}(\mathbf{r,r^{\prime}};\omega)\bigg{|}_{\mathbf{r^{\prime}\to
r}}\left[\coth\frac{\beta\omega}{2}-\coth\frac{\beta^{\prime}\omega}{2}\right].\
$ (45)

For an isotropic body,
$\chi_{ij}(\mathbf{r};\omega)=\chi(\mathbf{r};\omega)\delta_{ij}$, and a
vacuum environment, where
$\operatorname{tr}\operatorname{Im}\mathbf{\Gamma}(\mathbf{r,r^{\prime}};\omega)\big{|}_{\mathbf{r=r^{\prime}}}=\omega^{3}/(2\pi)$,
the power simplifies to

$P=\frac{1}{\pi^{2}}\int(d\mathbf{r})\int_{0}^{\infty}d\omega\,\omega^{4}\operatorname{Im}\chi(\mathbf{r};\omega)\left[\frac{1}{e^{\beta\omega}-1}-\frac{1}{e^{\beta^{\prime}\omega}-1}\right].$
(46)

Now the power absorbed by a blackbody is given by Stefan’s law,

$P=S\frac{\pi^{2}}{60}(T^{4}-T^{\prime 4}),$ (47)

where $S$ is the surface area of the blackbody. Since the radiation from a
blackbody arises entirely from the surface, we can write

$\int_{V}(d\mathbf{r})\chi(\mathbf{r};\omega)=\int_{S}dS\tilde{\chi}(\mathbf{r};\omega),$
(48)

so in order for Eq. (46) to describe a blackbody, the surface susceptibility
must satisfy

$\operatorname{Im}\tilde{\chi}(\omega)=\frac{1}{4\omega}\Rightarrow\tilde{\chi}(\omega)=\frac{i}{4}\frac{1}{\omega+i\epsilon}=\frac{\pi}{4}\delta(\omega)+\mathscr{P}\frac{i}{4\omega},$
(49)

$\mathscr{P}$ signifying the principal part, the latter following from the
Kramers-Kronig relation.

###  VI.2 Force on a blackbody/metal plate

Now we want to calculate the self-propulsive force on a composite plate
consisting of a blackbody surface on one side and a Drude metal on the other.
The force is given by Eq. (11a), in terms of the geometrical integral (11b),

$I_{AB}=\int_{A}(d\mathbf{r})\int_{B}(d\mathbf{r^{\prime}})\nabla_{z}\frac{\Delta(\omega|\mathbf{r-r^{\prime}}|)}{(4\pi|\mathbf{r-r^{\prime}}|^{3})^{2}}.$
(50)

Using polar coordinates for the transvserse spatial integrals, and integrating
out the $z$ derivative, we find for plates of (large) cross sectional area $S$
and thicknesses of the two sides $t_{A}$ and $t_{B}$, respectively,

$\displaystyle I_{AB}$ $\displaystyle=$
$\displaystyle\frac{S}{32\pi}\int_{0}^{\infty}\rho\,d\rho\int_{-t_{B}}^{0}dz^{\prime}\left[\frac{\Delta\left(\omega\sqrt{\rho^{2}+(t_{A}-z^{\prime})^{2}}\right)}{(\rho^{2}+(t_{A}-z^{\prime})^{2})^{3}}-\frac{\Delta(\omega\sqrt{\rho^{2}+z^{\prime
2}})}{(\rho^{2}+z^{\prime 2})^{3}}\right]$ (51) $\displaystyle=$
$\displaystyle\frac{S}{32\pi}\int_{-t_{B}}^{0}dz^{\prime}\int_{t_{A}-z^{\prime}}^{-z^{\prime}}du\frac{\Delta(\omega
u)}{u^{5}}.$

Because $\omega\sim T,\,T^{\prime}$ corresponds to a length scale $\sim
10\,\mu$m for temperatures around room temperature, for thicknesses much
smaller that that, we can expand $\Delta(\omega u)$ using only the first term
in Eq. (14a), and then we obtain immediately555 For this planar geometry, it
might seem more convenient to use a ($2+1$)-dimensional breakup,
$\Gamma_{ij}(\mathbf{r,r^{\prime}};\omega)=\int\frac{(d\mathbf{k}_{\perp})}{(2\pi)^{2}}e^{i\mathbf{k}_{\perp}\cdot(\mathbf{r-r^{\prime}})_{\perp}}g_{ij}(z,z^{\prime};\mathbf{k}_{\perp},\omega).$
(52) Carrying out the trace over the Green’s functions, and simplifying to the
case that the susceptibility is independent of $x$ and $y$, we readily obtain
$I_{AB}=-\frac{S}{32}\int_{|\mathbf{k}_{\perp}|<|\omega|}\frac{(d\mathbf{k}_{\perp})}{(2\pi)^{2}}\int_{A}dz\int_{B}dz^{\prime}\frac{\lambda^{4}+k^{4}}{\lambda}\sin
2\lambda(z-z^{\prime})$ (53) Now, if the two sides of the thin plate have
small thickness $t_{A}$ and $t_{B}$, respectively, we can expand the sine for
small $z$ and $z^{\prime}$ and find the same result as Eq. (54).

$I_{AB}=-\frac{S}{96\pi}t_{A}t_{B}(t_{A}+t_{B})\omega^{6}.$ (54)

Then the force is given by

$F_{z}=-\frac{S}{24\pi^{2}}t_{A}t_{B}(t_{A}+t_{B})\int_{0}^{\infty}d\omega\hat{\chi}_{AB}^{\vphantom{q}}(\omega)\omega^{6}\left[\frac{1}{e^{\beta\omega}-1}-\frac{1}{e^{\beta^{\prime}\omega}-1}\right].$
(55)

When $A$ is the blackbody,
$t_{A}\operatorname{Im}\chi_{A}^{\phantom{q}}=\frac{1}{4\omega}$, while $B$ is
the Drude metal, described by Eq. (21), the force is

$F_{z}=\frac{St_{B}(t_{A}+t_{B})}{96\pi^{2}}\omega_{p}^{2}\int_{0}^{\infty}d\omega\frac{\omega^{5}}{\omega^{2}+\nu^{2}}\left[\frac{1}{e^{\beta\omega}-1}-\frac{1}{e^{\beta^{\prime}\omega}-1}\right].$
(56)

Again, this is of the same class of functions explored in Appendix A of Ref.
[42],

$F_{z}=\frac{St_{B}(t_{A}+t_{B})\omega_{p}^{2}\nu^{4}}{96\pi^{2}}\hat{F},$
(57a) where $\displaystyle\hat{F}$ $\displaystyle=$
$\displaystyle\int_{0}^{\infty}dx\frac{x^{5}}{x^{2}+1}\left[\frac{1}{e^{\beta\nu
x}-1}-\frac{1}{e^{\beta^{\prime}\nu x}-1}\right]$ (57b) $\displaystyle=$
$\displaystyle\frac{\pi}{2\nu}(T^{\prime}-T)+\frac{1}{2}\ln\frac{T^{\prime}}{T}+\frac{1}{2}\psi\left(\frac{\nu}{2\pi
T^{\prime}}\right)-\frac{1}{2}\psi\left(\frac{\nu}{2\pi
T}\right)-\frac{\pi^{2}}{6\nu^{2}}(T^{2}-T^{\prime
2})+\frac{\pi^{4}}{15\nu^{4}}(T^{4}-T^{\prime 4}).$

This is plotted in Fig. 10.

Figure 10: Dimensionless force (57b) on a thin plate consisting of a blackbody
on the positive-$z$ side and a Drude metal on the negative-$z$ side.

The values there are to be multiplied by the prefactor in Eq. (57a), which for
the area of the plate being $S=1$ cm, $t=w=10$ nm, and, appropriate for gold,
$\omega_{p}=9$ eV, $\nu=0.035$ eV, is $\sim 10^{-13}$ N. Again, this force
should be observable.

##  VII Friction

Of course, once a body is put in motion by the self-propulsive force, it will
experience quantum friction in the vacuum. This friction was considered, in
first order, for example, in Ref. [28], but primarily in the condition of
nonequilibrium steady state (NESS). The general formula for the formula for
the frictional force on a particle in its rest frame is given by Eq. (5.7)
there, and shows the effects of field and dipole fluctuations,,

$F^{\rm P}=\frac{1}{3\pi^{2}\gamma
v}\int_{0}^{\infty}d\omega\,\omega^{4}\operatorname{Im}\alpha_{\rm
P}\int_{\gamma(1-v)}^{\gamma(1+v)}dy(y-\gamma)f^{\rm
P}(y)\left(\frac{1}{e^{\beta\omega
y}-1}-\frac{1}{e^{\beta^{\prime}\omega}-1}\right),$ (58)

where $\rm P$ designates the polarization state of the particle, with the
structure functions being [$\gamma=(1-v^{2})^{-1/2}$]

$f_{f}^{\rm P}(y)=\left\\{\begin{array}[]{cc}\frac{3}{2\gamma
v},&\mbox{ISO},\\\ \frac{3}{4\gamma
v}\left(1-\frac{(y-\gamma)^{2}}{\gamma^{2}v^{2}}\right),&\mbox{ZZ}.\end{array}\right.$
(59)

Here, the particle is described by a mean polarizability,

$\bm{\alpha}(\omega)=\int(d\mathbf{r})\bm{\chi}(\mathbf{r};\omega).$ (60)

ISO means a particle which is isotropically polarizable, and ZZ means one
which is only polarizable in the $z$ direction, the direction of motion. Due
to the symmetry of the $y$ integrand about $y=\gamma$, the second term here,
due to dipole fluctuations, vanishes, and hence there is no dependence on the
temperature of the particle. (See Appendix A for the second-order correction
to the force.)

For a needle moving in the direction of its axis, the ZZ polarization would
seem to be appropriate.666Actually, in this weak-susceptibility approximation
the susceptibility is assumed isotropic, so it might be more accurate to
multiply the friction by a factor of 5. When the force for this case is worked
out for low velocities, we obtain

$F_{f}^{\rm
ZZ}=-\frac{1}{5}\frac{v\beta}{12\pi^{2}}\int_{0}^{\infty}d\omega\,\omega^{5}\operatorname{Im}\alpha(\omega)\frac{1}{\sinh^{2}\beta\omega/2},$
(61)

which is just $1/5$ the Einstein-Hopf formula for an isotropic particle. The
essential point here is that $F_{f}$ is proportional to the velocity in the
low-velocity limit.

Due to the self-propulsive force, the body, out of equilibrium with its
environment, will begin to accelerate. It will reach a terminal velocity
$v^{\vphantom{q}}_{T}$ when the propulsive force exactly balances the
frictional force,

$F_{p}+v^{\vphantom{q}}_{T}F^{\prime}_{f}=0,\quad
v^{\vphantom{q}}_{T}=-\frac{F_{p}}{F^{\prime}_{f}}.$ (62)

Using the same model for the needle as in Sec. III, the derivative of the
frictional force is

$F^{\prime}_{f}=-\frac{2Ca\omega_{p}^{2}\nu}{15\pi^{2}\beta^{2}}\int_{0}^{\infty}dx\frac{x^{4}}{x^{2}+(\beta\nu/2)^{2}}\frac{1}{\sinh^{2}x},$
(63)

where for $\beta=40$ (eV)-1 ($T=$ room temperature) and a gold needle,
$\beta\nu/2=0.7$, and we find the $x$ integral here evaluates to 0.90. Then,
from the propulsive force (24a), we obtain the terminal velocity

$v^{\vphantom{q}}_{T}=-\frac{F_{p}}{F^{\prime}_{f}}=\frac{\chi_{A}^{\vphantom{q}}\beta_{0}^{2}\beta^{2}}{8}\frac{r_{0}^{2}}{a_{0}a^{5}}\frac{\hat{F}}{0.90}.$
(64)

(The velocity is in the $-z$ direction because $\hat{F}<0$ if $T^{\prime}>T$.)
For a needle of 1 cm half-length and radius 10 nm, at a temperature twice that
of the 300 K background, this equals $7.7\times 10^{-8}$ or about 20 m/s. This
would be seem to be very observable, if the nonequilibrium temperature
inbalance can be maintained.

However, it would take an extraordinarily long time to reach this terminal
velocity. From Newton’s law,

$v(t)=v^{\vphantom{q}}_{T}\left(1-e^{-t/t_{0}}\right),$ (65)

where the characteristic time is estimated from above to be

$t_{0}=\frac{m}{F_{f}^{\prime}}=\frac{15\pi^{2}\beta^{2}\rho}{2\omega_{p}^{2}\nu}\frac{1}{0.9}.$
(66)

where $\rho$ is the density of the metal part of the needle. Putting in
numbers appropriate for gold, we find something like $t_{0}\sim 2.7\times
10^{9}$ s, about 80 years. This long time is a reflection of the weakness of
the frictional force.

##  VIII Relaxation to thermal equilibrium

Quantum friction appears then to play a negligible role. However, unless the
body is equipped with a mechanism to maintain its elevated temperature, like a
real quantum thruster [46], is will rapidly cool to the temperature of the
environment. We estimate this effect as in Ref. [42]: The rate of heat loss

$\frac{dQ}{dt}=C_{V}(T^{\prime})\frac{dT^{\prime}}{dt}=P(T^{\prime},T),$ (67)

where $C_{V}(T^{\prime})$ is the specific heat of the body at temperature
$T^{\prime}$, and the power for a slowly moving body is given by Eq. (46)
[28], or its generalization (see Appendix B)

$P(T^{\prime},T)=\frac{1}{3\pi^{2}}\int_{0}^{\infty}d\omega\,\omega^{4}\operatorname{Im}\operatorname{tr}\bm{\alpha}(\omega)\left[\frac{1}{e^{\beta\omega}-1}-\frac{1}{e^{\beta^{\prime}\omega}-1}\right],$
(68)

but this transcends the weak susceptibility limit, in that the polarizability
is not simply $\alpha(\omega)=\int(d\mathbf{r})\chi(\mathbf{r},\omega)$. Then,
the time to cool the body from temperature $T^{\prime}_{0}$ to temperature
$T^{\prime}_{1}$ is

$t_{1}=\int_{T_{0}^{\prime}}^{T^{\prime}_{1}}dT^{\prime}\frac{C_{V}(T^{\prime})}{P(T^{\prime},T)}.$
(69)

We can solve this equation (numerically) for $T^{\prime}_{1}(t_{1},T)$
(assuming fixed environmental temperature), and then find the body’s
acceleration in terms of the propulsive force at that time ($m$ is the mass of
the body),

$m\frac{dv(t)}{dt}=F_{p}(T^{\prime}(t),T),$ (70)

which, when integrated over all time, gives the terminal velocity,

$v^{\vphantom{q}}_{T}=\frac{1}{m}\int_{0}^{\infty}dt\,F_{p}(T^{\prime}(t),T).$
(71)

This is precisely the scheme advocated in Ref. [44].

Let us specifically consider the example of the Janus ball discussed in Sec.
V. For the cooling, we consider the high-temperature approximation of the
Debye model treated in Ref. [42], together with the Lorenz-Lorentz model for
the polarizability, which gives

$t=t_{c}\int_{T^{\prime}_{0}/T}^{T^{\prime}/T}du\frac{1}{1-u^{6}},\quad
t_{c}=\frac{21}{8\pi^{4}}\frac{n\omega_{p}^{2}}{\nu T^{5}},$ (72)

$n$ being the number density. The time scale $t_{c}$ for gold at room
temperature is about 2,000 s. The numerical inverse function $T^{\prime}(t)$
is shown in Fig. 11.

Figure 11: The temperature $T^{\prime}$ of the body (relative to that of the
environment $T$) in terms of the cooling time relative to the scale $t_{c}$.
The two curves are for an initial temperature ratio of 10 (upper) and 2
(lower).

Now we insert this into the force formula for the propulsive force on a small
Janus ball, given by Eq. (40). Then we use the expression for the terminal
velocity (71) to give

$v^{\vphantom{q}}_{T}=\lim_{t\to\infty}v(t),\quad
v(t)=\frac{7}{72\pi^{5}}\chi_{A}^{\vphantom{q}}\omega_{p}^{4}\frac{n}{m}\frac{\nu^{6}a^{7}}{T^{5}}\int_{0}^{t/t_{c}}d\tau\hat{F}(\tau),\quad\tau=\frac{t^{\prime}}{t_{c}}.$
(73)

The prefactor here, for the metallic half-ball being gold, having a radius 100
nm, in an vacuum environment at room temperature, is about
$20\chi_{A}^{\vphantom{q}}$ $\mu$m/s. The dimensionless force,
$\hat{F}(t/t_{c})$ is shown in Fig. 12.

Figure 12: The dimensionless force $\hat{F}$ for a Janus ball, the metallic
half of which is gold, as a function of the time $t/t_{c}$, at a vacuum
temperature of $T=300$ K, and an initial temperature of the nanoball of
$T^{\prime}_{0}=600$ K.

The numerical integral of $\hat{F}$ for large $t/t_{0}$ when
$T_{0}^{\prime}/T=2$ gives about $-15$, yielding roughly
$v^{\vphantom{q}}_{T}=-300$ $\mu$m/s, the asymptotic value being achieved by
about $0.1\,t_{c}\approx 200$ s. If there had been no thermal cooling, in this
time the body would have reached a velocity of about 0.1 cm/s. Increasing the
intial temperature of the particle to $T_{0}^{\prime}=10T$ increases the
terminal velocity by about a factor of 6, while reducing the time to reach
equilibrium by about four orders of magnitude. (The discussion in this
paragraph assumes that the thermal radiation from the metal portion dominates
the energy loss mechanism, which is true, at least perturbatively.)

###  VIII.1 Weak-susceptibility model

The above treatment might seem inconsistent, since the perturbative
susceptibility is used for the propulsion but the dressed Lorenz-Lorentz
polarizability is used for the cooling. If, instead, we use the weak-coupling
susceptility for the metal,

$\operatorname{Im}\chi(\omega)=\frac{\omega_{p}^{2}\nu}{\omega(\omega^{2}+\nu^{2})},$
(74)

we obtain instead of Eq. (72) the time to cool from $T^{\prime}_{0}$ to
$T^{\prime}_{1}$

$t_{1}=-\frac{6\pi^{2}n}{\nu^{2}\omega_{p}^{2}}\int_{\beta^{\prime}_{0}\nu}^{\beta^{\prime}_{1}\nu}\frac{dx^{\prime}}{x^{\prime
2}}\left[\ln\frac{x^{\prime}}{x}+\psi\left(\frac{x}{2\pi}\right)-\psi\left(\frac{x^{\prime}}{2\pi}\right)+\pi\left(\frac{1}{x}-\frac{1}{x^{\prime}}\right)+\frac{\pi^{2}}{3}\left(\frac{1}{x^{2}}-\frac{1}{x^{\prime
2}}\right)\right]^{-1}.$ (75)

The prefactor here, $t_{c}$, for gold is about $2\times 10^{-4}$ s. This is
then inverted to find $T^{\prime}/T$ as a function of $\tau=t^{\prime}/t_{c}$
as above. Then the terminal velocity for a Janus ball, half of which is gold,
is found from Eq. (71) to be

$v^{\vphantom{q}}_{T}=\frac{1}{3}\frac{\chi_{A}^{\vphantom{q}}\nu^{5}a^{4}}{m_{\rm
Au}}\int_{0}^{\infty}d\tau\hat{F}(T^{\prime}(\tau),T),$ (76)

$m_{\rm Au}$ being the mass of a gold atom, where the prefactor is only
$5\times 10^{-1}$ nm/s, while the integral is only about 100, so this is
ridiculously small.777 Note that the ratio of the $t_{c}$ scales in Eqs. (72)
and (75) is $\frac{7}{16\pi^{6}}\frac{\omega_{p}^{4}\nu}{T^{5}}\sim 10^{7}$,
which is also the ratio of the forces, apart from the dimensionless integrals,
whch are in the ratio of 15/100.

One might think one could boost this substantially by considering a
macroscopic object. But, presumably it should be thin so that thicknesses do
not exceed the skin depth, otherwise the weak susceptibility approach to the
force will not be valid. So let’s consider the thin spherical shell model of
Sec. IV for which the propulsive force is given, approximately, by Eq. (32).
Since the cooling time is independent of the shape of the body, we can use Eq.
(75), and then find the terminal velocity from

$v^{\vphantom{q}}_{T}=\frac{3}{16\pi}\frac{|\mathscr{N}|\nu}{m_{\rm
Au}}\frac{t}{a}\chi_{A}^{\vphantom{q}}\int_{0}^{\infty}d\tau|\hat{F}(T^{\prime}(\tau),T)|$
(77)

The prefactor evaluates to $2\times 10^{-10}\chi_{A}^{\vphantom{q}}$ m/s,
where the integral over Eq. (32b) is about 0.4, so this still remains beyond
experimental reach.

##  IX Conclusions

In this paper we have tried to give a systematic treatment of self-propulsion
of an inhomogeneous body in vacuum, which occurs not in first order, but in
second order in the susceptibility (or polarizability) of the body. For a
force to occur, the body must have nonuniform electric susceptibility, be
dissipative, and be out of thermal equilibrium with the background. For a body
hotter than the vacuum environment and having two different halves, the force
is in the direction of the side having the lowest emissivity, the radiation
emitted being prodominantly from the opposite side. We consider four different
configurations: a thin needle, a thin spherical shell, a solid nanoball, and
finally a thin plate. Except in the last case, one side is regarded as non-
dissipative, while the other is modeled as a Drude metal; for the thin plate,
the nonmetal side is modelled as a total blackbody absorber. Presumably, in
all cases, the objects must have small thicknesses, so that the skin depth is
larger than the thickness.888 This may also validate the weak-susceptibility
approximation, in that we take the polarizability to be simply the volume
integral of the susceptibility. Therefore, the forces must be very small (not
sufficient for a quantum thruster [46]), but probably amenable to observation
if the temperature imbalance could be maintained. Quantum vacuum friction will
eventually cause the particle to acquire a terminal velocity. But if the
system is allowed to relax to thermal equilibrium, the body might reach an
observable velocity in a matter of seconds; however, this conclusion
sensitively depends on the model of heat loss.

###### Acknowledgements.

This work was supported in part by the US National Science Foundation, grant
number 2008417. GK thanks the Homer L. Dodge Department of Physics and
Astronomy at the University of Oklahoma for its hospitality. We thank Xin Guo,
Prachi Parashar, and Steve Fulling for collaborative assistance. This paper
reflects solely the authors’ personal opinions and does not represent the
opinions of the authors’ employers, present and past, in any way. For the
purpose of open access, the authors have applied a CCBY public copyright
license to any Author Accepted Manuscript version arising from this
submission.

##  Appendix A Force on moving body

In this appendix we derive the force on a body moving with velocity
$\mathbf{v}$ in the $z$ direction, which in general will include both a
propulsive force and a frictional force. For simplicity of presentation, we
will restrict attention to the case when the susceptibility has only a nonzero
$zz$ component, because that greatly simplifies the form of the Lorentz
boosts. We will work in a ($3+1$)-dimensional wave-vector representation.
Starting from Eq. (1), we have for the force on the dielectric body in its
rest frame

$F_{z}=\int(d\mathbf{r})\int\frac{d\omega\,d\nu}{(2\pi)^{2}}\int\frac{(d\mathbf{k})}{(2\pi)^{3}}\frac{(d\mathbf{q})}{(2\pi)^{3}}e^{-i(\omega+\nu)t}e^{i\mathbf{(k+q)\cdot
r}}\,\mathbf{P}(\mathbf{k},\omega)iq_{z}\mathbf{E}(\mathbf{q},\nu),$ (78)

where all quantities are in the rest frame of the body. Here, we have omitted
the first term in Eq. (1), because it will always vanish.

###  A.1 First-order friction

To begin, we examine this in first order, where in place of Eqs. (3) we have
in momentum space (leaving off the component indices, which are always $z$ in
our case)

$\displaystyle P(\mathbf{k},\omega)$ $\displaystyle=$
$\displaystyle\int\frac{(d\mathbf{k^{\prime}})}{(2\pi)^{3}}\chi(\mathbf{k-k^{\prime}})E(\mathbf{k}^{\prime},\omega),$
(79a) $\displaystyle\tilde{E}(\mathbf{k},\omega)$ $\displaystyle=$
$\displaystyle\Gamma({\mathbf{k},\omega)\tilde{P}(\mathbf{k},\omega}).$ (79b)

(Quantities without tildes are in the rest frame of the body, while those with
tildes are in the rest frame of the radiation.) Expanding $P$ to first order
in $E$, we have

$F_{z}^{(1,0)}=\int\frac{d\omega}{2\pi}\frac{d\nu}{2\pi}\frac{(d\mathbf{k})}{(2\pi)^{3}}\frac{(d\mathbf{k^{\prime}})}{(2\pi)^{3}}e^{-i(\omega+\nu)t}(-ik_{z})\chi(\mathbf{k-k^{\prime}},\omega)E(\mathbf{k}^{\prime},\omega)E(-\mathbf{k},\nu),$
(80)

where the body rest frame value of the electric field, $E$ is related to that
in the radiation rest frame, $\tilde{E}$, by

$E(\mathbf{k},\omega)=\tilde{E}(\tilde{\mathbf{k}},\tilde{\omega}),$ (81)

(this is where the simplification to have only $z$ polarization is crucial).
The boosted momentum and frequency variables are

$\tilde{k}_{z}=\gamma(k_{z}+\omega
v),\quad\tilde{k}_{x,y}=k_{x,y},\quad\tilde{\omega}=\gamma(\omega+k_{z}v),$
(82)

in terms of the relativistic dilation factor $\gamma=(1-v^{2})^{-1/2}$. Then,
we use the FDT in the form

$\langle\mathscr{S}\tilde{E}(\tilde{\mathbf{k}},\tilde{\omega})\tilde{E}(\tilde{\mathbf{q}},\tilde{\nu})\rangle=(2\pi)^{4}\delta(\tilde{\omega}+\tilde{\nu})\delta(\tilde{\mathbf{k}}+\tilde{\mathbf{q}})\operatorname{Im}\Gamma(\mathbf{\tilde{k}},\tilde{\omega})\coth\frac{\beta\tilde{\omega}}{2}.$
(83)

The product of $\delta$ functions is invariant under a Lorentz boost,

$\delta(\tilde{\omega}+\tilde{\nu})\delta(\tilde{\mathbf{k}}+\tilde{\mathbf{q}})=\delta(\omega+\nu)\delta(\mathbf{k}+\mathbf{q}),$
(84)

as is the $zz$ component of the vacuum Green’s dyadic:

$\tilde{\Gamma}(\tilde{\mathbf{k}},\tilde{\omega})=\Gamma(\mathbf{k},\omega)=\frac{k_{\perp}^{2}}{k^{2}-(\omega+i\epsilon)^{2}}.$
(85)

The imaginary part of the latter is

$\operatorname{Im}\Gamma(\mathbf{k},\omega)=k_{\perp}^{2}\pi\operatorname{sgn}(\omega)\delta(k^{2}-\omega^{2}).$
(86)

Putting this into the formula for the force (80) we obtain

$\displaystyle F^{(1,0)}_{z}$ $\displaystyle=$
$\displaystyle\int\frac{d\omega(d\mathbf{k})}{(2\pi)^{4}}(-ik_{z})\chi(\bm{0},\omega)\operatorname{Im}\Gamma(\mathbf{k},\omega)\coth\frac{\beta\tilde{\omega}}{2}$
(87) $\displaystyle=$
$\displaystyle\frac{1}{(2\pi)^{2}}\int_{0}^{\infty}d\omega\int_{-\omega}^{\omega}dk_{z}\operatorname{Im}\chi(\bm{0},\omega)k_{x}(\omega^{2}-k_{z}^{2})\frac{1}{e^{\beta\gamma(\omega+k_{z}v)}-1}.$

Introducing the variable $y$ through $\gamma(\omega+k_{z}v)=\omega y$, we
obtain exactly the result (58) for the $ZZ$ polarization: there is no
contribution at $v=0$ and the PP or the (0,1) contribution is zero, because
the corresponding integral is odd in $k_{z}$.

###  A.2 Second-order force and friction

To obtain a propulsive force, we have to go to second order in the
susceptibility. We begin by expanding the free energy so that $P$ is expanded
once, and $E$ is expanded twice, the term we refer to as (1,2). For simplicity
of notation, we leave off integration elements:

$P(\mathbf{k},\omega)E(\mathbf{k}^{\prime},\nu)=\chi(\mathbf{k-k^{\prime\prime}},\omega)\tilde{E}(\tilde{\mathbf{k}}^{\prime\prime},\tilde{\omega})\Gamma(\mathbf{\tilde{k}^{\prime}},\tilde{\nu})\chi(\mathbf{k^{\prime}-k^{\prime\prime\prime}},\nu)\tilde{E}(\tilde{\mathbf{k}}^{\prime\prime\prime},\tilde{\nu}),$
(88)

doing successive Lorentz boosts back and forth between the two natural frames,
and then use the FDT (83). The result for this force component is

$F_{z}^{(1,2)}=\int\frac{d\omega\,(d\mathbf{k})\,(d\mathbf{k}^{\prime})}{(2\pi)^{7}}k^{\prime}_{z}|\chi(\mathbf{k-k^{\prime}},\omega)|^{2}\operatorname{Im}\Gamma(\mathbf{k},\omega)\operatorname{Im}\Gamma(\mathbf{k^{\prime}},\omega)\coth\frac{\beta\tilde{\omega}}{2},\quad\tilde{\omega}=\gamma(\omega+k_{z}v),$
(89)

where the second imaginary part of a Green’s dyadic comes from the symmetry
$(\omega,\mathbf{k,k^{\prime}})\to-(\omega,\mathbf{k,k^{\prime}})$, and we
have noted that

$\chi(\mathbf{k},\omega)^{*}=\chi(-\mathbf{k},-\omega),$ (90)

because the underlying coordinate-space response function is real.

Following the same procedure, expanding $P$ out to third order, we find

$F^{(3,0)}_{z}=\int\frac{d\omega\,(d\mathbf{k})\,(d\mathbf{k}^{\prime})}{(2\pi)^{7}}(-ik_{z})\chi(\mathbf{k-k^{\prime}},\omega)\chi(\mathbf{k^{\prime}-k},\omega)\Gamma(\mathbf{k^{\prime}},\omega)\operatorname{Im}\Gamma(\mathbf{k},\omega)\coth\frac{\beta\tilde{\omega}}{2}.$
(91)

The two contributions to the EE force correspond to what we called I [Eq.
(91)] and II [Eq. (89)] in Ref. [27], We note that the (3,0) contribution
vanishes at $v=0$ because it is then antisymmetric under the reflection
$(\mathbf{k,k^{\prime}})\to-(\mathbf{k,k^{\prime}})$. It is possible then to
show that the $v=0$ contribution of $F^{(1,2)}$ coincides with that found from
Eq. (6a). namely,

$F^{\rm
EE}_{z}\bigg{|}_{v=0}=\int(d\mathbf{r})(d\mathbf{r^{\prime}})\frac{d\omega}{2\pi}\operatorname{Im}[\chi(\mathbf{r};\omega)\chi(\mathbf{r}^{\prime};\omega)^{*}]\operatorname{Im}\Gamma(\mathbf{r,r^{\prime}};\omega)\nabla_{z}\operatorname{Im}\Gamma(\mathbf{r,r^{\prime}};\omega)\coth\frac{\beta\omega}{2},$
(92)

having the same form as Eq. (9).

In general, however, because the product of susceptibilities in
$F_{z}^{(3,0)}$ is not real, the real part of
$\Gamma(\mathbf{k^{\prime}},\omega)$ occurs, which leads to divergences when
the integration over $\mathbf{k}^{\prime}$ is performed. The solution to this
difficulty was provided in Ref. [27]: when $F_{z}^{(1,0)}$ [Eq. (87)] and
$F_{z}^{(3,0)}$ [Eq. (91)] are combined, we have

$F_{z}^{(1,0)+(3,0)}=\int\frac{d\omega\,(d\mathbf{k})\,(d\mathbf{k}^{\prime})}{(2\pi)^{7}}(-ik_{z})\hat{\chi}(\mathbf{k-k^{\prime}},\omega)\operatorname{Im}\Gamma(\mathbf{k},\omega)\coth\frac{\beta\tilde{\omega}}{2},$
(93)

where

$\hat{\chi}(\mathbf{k-k^{\prime}},\omega)=\chi(\mathbf{k-k^{\prime}},\omega)\left[(2\pi)^{3}\delta(\mathbf{k-k^{\prime}})+\Gamma(\mathbf{k}^{\prime},\omega)\chi(\mathbf{k^{\prime}-k},\omega)\right].$
(94)

These are the first two terms of an infinite series, written in symbolic
notation:

$\hat{\chi}=\chi^{\vphantom{q}}_{0}\left(1-\Gamma\chi^{\vphantom{q}}_{0}\right)^{-1},$
(95)

where we have denoted by $\chi^{\vphantom{q}}_{0}$, the bare intrinsic
susceptibility, what we previously called $\chi$. Then, breaking up $\Gamma$
into its real and imaginary parts, we have for the effective susceptibility,
just as in Eq. (4.31) of Ref. [27],

$\hat{\chi}=\chi(1-i\operatorname{Im}\Gamma\chi)^{-1},$ (96a) in terms of the
renormalized intrinsic susceptibility
$\chi=\chi^{\vphantom{q}}_{0}(1-\operatorname{Re}\Gamma\chi^{\vphantom{q}}_{0})^{-1}.$
(96b)

This is how the bare intrinsic susceptibilty is dressed by the radiation
field. (After all, the momentum-space Fourier transform of $\chi$ is the
generalization of the polarizability discussed in Ref. [27].) Therefore, we
conclude that the renormalized version of $F^{(3,0)}$ should read

$F^{(3,0)}_{z,R}=\int\frac{d\omega\,(d\mathbf{k})\,(d\mathbf{k}^{\prime})}{(2\pi)^{7}}k_{z}\operatorname{Re}[\chi(\mathbf{k-k^{\prime}},\omega)\chi(\mathbf{k^{\prime}-k},\omega)]\operatorname{Im}\Gamma(\mathbf{k^{\prime}},\omega)\operatorname{Im}\Gamma(\mathbf{k},\omega)\coth\frac{\beta\tilde{\omega}}{2},$
(97)

which is free of all difficulties.

For the PP fluctuation contributions, the calculation proceeds similarly. Now
we either expand the free energy out to third order in $E$, giving a (0,3)
contribution, or expand $P$ out to second order and $E$ out to first, the
(2,1) term. The FDT (4b), in the Fourier domain, reads

$\langle\mathscr{S}P(\mathbf{k},\omega)P(\mathbf{k^{\prime}},\nu)\rangle=2\pi\delta(\omega+\nu)\Im\chi(\mathbf{k+k^{\prime}},\omega)\coth\frac{\beta^{\prime}\omega}{2},$
(98a) with
$\Im\chi(\mathbf{k+k^{\prime}},\omega)=\frac{1}{2i}\left[\chi(\mathbf{k+k^{\prime}},\omega)-\chi(\mathbf{k+k^{\prime}},-\omega)\right],$
(98b)

neither the imaginary nor the anti-Hermitian part. The two contributions to
the force are, respectively,999Note that no renormalization of $F_{z}^{(0,3)}$
is necessary, because $F_{z}^{(0,1)}$ vanishes.

$\displaystyle F_{z}^{(0,3)}$ $\displaystyle=$ $\displaystyle
i\int\frac{d\omega(d\mathbf{k})(d\mathbf{k^{\prime}})}{(2\pi)^{7}}k_{z}\Gamma(\mathbf{k},-\omega)\Gamma(\mathbf{k}^{\prime},-\omega)\chi(\mathbf{k-k^{\prime}},-\omega)\Im\chi(\mathbf{k^{\prime}-k},-\omega)\coth\frac{\beta^{\prime}\omega}{2},$
(99a) $\displaystyle F_{z}^{(2,1)}$ $\displaystyle=$
$\displaystyle-i\int\frac{d\omega(d\mathbf{k})(d\mathbf{k^{\prime}})}{(2\pi)^{7}}k_{z}\Gamma(\mathbf{k},-\omega)\Gamma(\mathbf{k^{\prime}},\omega)\chi(\mathbf{k-k^{\prime}},\omega)\Im\chi(\mathbf{k^{\prime}-k},\omega)\coth\frac{\beta^{\prime}\omega}{2}.$
(99b)

Using the symmetries $(\mathbf{k,k^{\prime}})\to-(\mathbf{k,k^{\prime}})$ and
$\mathbf{k\leftrightarrow k^{\prime}}$, we obtain the total PP contribution to
the force

$F^{(0,3)+(2,1)}=-\frac{1}{4}\int\frac{d\omega(d\mathbf{k})(d\mathbf{k^{\prime}})}{(2\pi)^{7}}(k_{z}-k_{z}^{\prime})\operatorname{Im}\Gamma(\mathbf{k},\omega)\operatorname{Im}\Gamma(\mathbf{k^{\prime}},\omega)\left[|\chi(\mathbf{k^{\prime}-k},\omega)|^{2}-|\chi(\mathbf{k-k^{\prime}},\omega)|^{2}\right]\coth\frac{\beta^{\prime}\omega}{2}.$
(100)

and the total $v=0$ propulsive force is

$\displaystyle F^{EE+PP}$ $\displaystyle=$
$\displaystyle-\int_{0}^{\infty}\frac{d\omega}{2\pi}\int\frac{(d\mathbf{k})(d\mathbf{k^{\prime}})}{(2\pi)^{6}}(k_{z}-k_{z}^{\prime})\operatorname{Im}\Gamma(\mathbf{k},\omega)\operatorname{Im}\Gamma(\mathbf{k^{\prime}},\omega)\left[|\chi(\mathbf{k-k^{\prime}},\omega)|^{2}-|\chi(\mathbf{k^{\prime}-k},\omega)|^{2}\right]$
(101)
$\displaystyle\qquad\times\left[\frac{1}{e^{\beta\omega}-1}-\frac{1}{e^{\beta^{\prime}\omega}-1}\right].$

This agrees with the propulsive force given in Sec. II.

As for the $v\neq 0$ frictional force, we can compare with the second-order
results of Ref. [27], where we considered a point particle with a real
polarizability. Then there are only EE fluctuations, and the spatial Fourier
transform of the susceptibility is the polarizability of the particle,
$\chi(\mathbf{k-k^{\prime}},\omega)=\alpha(\omega)$. Then it is
straightforward to show that the EE force is entirely frictional and coincides
with Eq. (3.16) of Ref. [27]. This is because

$\int dk_{x}\operatorname{Im}\Gamma(\mathbf{k},\omega)=\int
dk_{x}(k_{x}^{2}+k_{y}^{2})\operatorname{Im}\frac{1}{k_{x}^{2}+k_{y}^{2}+k_{z}^{2}-(\omega-i\epsilon)^{2}}=\pi\operatorname{sgn}(\omega)\frac{\omega^{2}-k_{z}^{2}}{\sqrt{\omega^{2}-k_{y}^{2}-k_{z}^{2}}}\eta(\omega^{2}-k_{y}^{2}-k_{z}^{2}).$
(102)

That is, the frictional force in this case is

$F^{(3,0)+(1,2)}=\frac{1}{32\pi^{3}}\int_{0}^{\infty}d\omega\int_{0}^{\omega}dk_{z}\int_{0}^{\omega}dk_{z}^{\prime}\,\alpha^{2}(\omega)(k_{z}-k_{z}^{\prime})(\omega^{2}-k_{z}^{2})(\omega^{2}-k_{z}^{\prime
2})\frac{1}{e^{\beta\gamma(\omega+k_{z}v)}-1}.$ (103)

If $F^{(1,2)}$ is expanded to first order in $v$, the frictional force so
arising, proportional to $k_{z}k_{z}^{\prime}$, vanishes by
$k_{z}\leftrightarrow k^{\prime}_{z}$ symmetry. This statement is true for any
object with uniform susceptibility; that is, the frictional force only arises
from $F^{(3,0)}$, while the propulsive force on a nonuniform object has an EE
contribution coming only from $F^{(1,2)}$. (We already recognized the latter
in Sec. II, while the former was observed in Ref. [27].) For a small object,
with polarizability proportional to its volume, this second-order friction
would be expected to be much smaller than the first-order friction given in
Eq. (58). Consider the thin needle as an example. The friction comes only from
$F^{(3,0)}$, and, in the approximation that the friction arises from a region
where the susceptibility is real and uniform, reads, for low velocity (the
Einstein-Hopf limit),

$F=16C^{2}vT^{6}\int_{0}^{\infty}dz\,z^{6}\chi(2Tz)^{2}\frac{r(zbT)}{\sinh^{2}z},$
(104)

where

$r(t)=\int_{0}^{1}dx\int_{0}^{1}dy\frac{x^{2}(1-x^{2})(1-y^{2})}{(x-y)^{2}}\sin^{2}[(x-y)t].$
(105)

The integral $r(t)$ may be given in closed form in terms of the sine-integral
function. but only the large-$t$ behavior is relevant for a 1 cm object at
room temperature because typical values of $t$ are of order $bT=625$ at room
temperature:

$r(t)\sim\frac{8\pi t}{105},\quad t\gg 1.$ (106)

For a representative value of susceptibility $\chi_{0}$, the force (104) is
smaller than the frictional force found in Eq. (58) by a factor of $5\times
10^{-7}\chi_{0}^{2}$.

##  Appendix B Radiated energy

In Ref. [28] we derived the energy lost by a polarizable particle out of
thermal equilibrium by looking at the rate with which the particle does work
on the field. Alternatively, we can examine the rate at which the energy flows
across a surface surrounding a body, which follows from energy conservation,

$\frac{\partial}{\partial t}u+\bm{\nabla}\cdot\mathbf{S}+\mathbf{j\cdot E}=0,$
(107)

provided, as is the case here, that the situation is stationary. Here the
energy flux vector or Poynting vecctor is

$\mathbf{S(r,t)}=\int\frac{d\omega}{2\pi}\frac{d\nu}{2\pi}e^{-i(\omega+\nu)t}\mathbf{E(r;\omega)}\times\left[\frac{1}{i\nu}\bm{\nabla}\times\mathbf{E(r;\nu)}\right].$
(108)

If the electric field is expanded out to first order with the polarization as
given in Eq. (3a), and we use the FDT (4b) for the polarization field product,
we find for the PP fluctuation part of the Poynting vector

$S^{\rm
PP}_{k}(\mathbf{r})=-\int(d\mathbf{r^{\prime}})\frac{d\omega}{2\pi}\frac{1}{i\omega}\Gamma_{ij}(\mathbf{r-r^{\prime}};\omega)\operatorname{Im}\chi_{jl}^{\vphantom{q}}(\mathbf{r^{\prime}};\omega)\coth\frac{\beta^{\prime}\omega}{2}\left[\nabla_{k}\Gamma_{il}(\mathbf{r-r^{\prime}};-\omega)-\nabla_{i}\Gamma_{kl}(\mathbf{r-r^{\prime}};-\omega)\right].$
(109)

Now we imagine surrounding the body with a sphere of radius $R$, large
compared to the size of the body and centered on it. Then the Green’s dyadic
appearing here is asymptotically

$\bm{\Gamma}_{ij}(\mathbf{R};\omega)\sim\omega^{2}(\bm{1}-\hat{\mathbf{R}}\hat{\mathbf{R}})\frac{e^{i\omega
R}}{4\pi R},\quad R\gg a,$ (110)

where $a$ is a typical size of the object. The corresponding expectation value
of the flux vector is

$\displaystyle S^{\rm PP}_{k}(\mathbf{r})$ $\displaystyle=$
$\displaystyle\int\frac{d\omega}{2\pi}(d\mathbf{r^{\prime}})\frac{\omega^{4}}{(4\pi
R)^{2}}\hat{R}_{k}(\delta_{ij}-\hat{R}_{i}\hat{R}_{j})\operatorname{Im}\chi_{ij}^{\vphantom{q}}(\mathbf{r^{\prime}};\omega)\coth\frac{\beta^{\prime}\omega}{2}$
(111a) $\displaystyle\to$ $\displaystyle
2\int\frac{d\omega}{2\pi}(d\mathbf{r^{\prime}})\frac{\omega^{4}}{(4\pi
R)^{2}}\hat{R}_{k}\operatorname{Im}\chi(\mathbf{r^{\prime}};\omega)\coth\frac{\beta^{\prime}\omega}{2},$
(111b)

where the second form applies for isotropic susceptibility. Integrating this
over a large sphere surrounding the body gives the total energy emitted per
unit time:

$\int d\sigma\hat{\mathbf{R}}\cdot\mathbf{S}^{\rm
PP}(\mathbf{r})=\frac{1}{4\pi^{2}}\int
d\omega\,\omega^{4}\operatorname{Im}\alpha(\omega)\coth\frac{\beta^{\prime}\omega}{2}.$
(112)

Here, we use the lowest-order susceptibility,
$\alpha(\omega)=\int(d\mathbf{r^{\prime}})\chi(\mathbf{r^{\prime}};\omega)$.

What we did in the preceding paragraph is what we denote as the (1,1) term in
the expansion. To obtain the EE fluctuation contribution, we compute, in the
same way, the (2,0) and (0,2) expansions. The result is as expected; the total
flux vector for large $R$ being for arbitrary polarizability

$S_{k}=S_{k}^{\rm PP}+S_{k}^{\rm EE}=\frac{\hat{R}_{k}}{(4\pi
R)^{2}}\int\frac{d\omega}{2\pi}\omega^{4}(\delta_{ij}-\hat{R}_{i}\hat{R}_{j})\operatorname{Im}\alpha_{ij}(\omega)\left(\coth\frac{\beta^{\prime}\omega}{2}-\coth\frac{\beta\omega}{2}\right).$
(113)

##  Appendix C Dielectic in front of a mirror

To understand the sense of the force, which in all the examples we have
considered is in the direction of the metal side of the composite object, we
consider a simple example of a perfectly conducting plane, a perfect mirror,
located at $z=0$, attached to which is a dilute dielctric body, as shown in
Fig. 13.

Figure 13: Perfectly reflecting mirror located in the plane $z=0$ to which a
dilute dielectric is attached. The radiation fields are observed on the
distant screen at $z=Z$. The screen to the left of the mirror sees nothing but
the background field modified by the mirror. The dielectric is at temperature
$T^{\prime}$ while the background has temperature $T$.

It is evident from the diagram that, because energy is radiated to the right
by the composite assembly, there will be a force on that assembly to the left,
in the $-z$ direction. The purpose of this appendix is to make that picture
quantitative.

We treat the effect of the dielectric perturbatively, to first order in its
susceptibility, $\chi(\mathbf{r^{\prime}};\omega)$. The effect of the mirror
is to replace the free electromagnetic Green’s dyadic
$\bm{\Gamma}^{(0)}(\mathbf{r-r^{\prime}};\omega)$ by the Green’s dyadic [48]

$\bm{\Gamma}(\mathbf{r,r^{\prime}};\omega)=\bm{\Gamma}^{(0)}(\mathbf{r-r^{\prime}};\omega)-\bm{\Gamma}^{(0)}(\mathbf{r-r^{\prime}}+2z^{\prime}\mathbf{\hat{z};\omega})\cdot(\bm{1}-2\mathbf{\hat{z}\hat{z}}),$
(114)

which has an obvious interpretation in terms of an image charge. We are
interested in fields far from the mirror/dielectric assembly, $r\gg
r^{\prime}$, where $\mathbf{r^{\prime}}$ locates a point within the
dielectric. In the radiation zone, then,

$\bm{\Gamma}(\mathbf{r,r^{\prime}};\omega)\sim
2\bm{\Gamma}^{(0)}(\mathbf{r};\omega)\cdot\mathbf{\hat{z}\hat{z}}.$ (115)

For large $r$, the only surviving components of the Green’s dyadic are from
Eq. (12),

$\Gamma_{xz}\sim-\frac{2\omega^{2}}{4\pi r}\frac{xz}{r^{2}}e^{i\omega
r},\quad\Gamma_{yz}\sim-\frac{2\omega^{2}}{4\pi r}\frac{yz}{r^{2}}e^{i\omega
r}\quad\Gamma_{xz}\sim\frac{2\omega^{2}}{4\pi
r}\left(1-\frac{z^{2}}{r^{2}}\right)e^{i\omega r}.$ (116)

Now we calculate the first-order energy flux as in Appendix B, see Eq. (109):

$S_{z}=\int\frac{d\omega}{2\pi}(d\mathbf{r^{\prime}})\frac{1}{\omega}\operatorname{Im}\chi(\mathbf{r^{\prime}};\omega)\left(\coth\frac{\beta\omega}{2}-\coth\frac{\beta^{\prime}\omega}{2}\right)\operatorname{Im}\left\\{\Gamma_{ij}(\mathbf{r,r^{\prime}};\omega)\left[\partial_{z}\Gamma_{ij}(\mathbf{r,r^{\prime}};-\omega)-\partial_{i}\Gamma_{zj}(\mathbf{r,r^{\prime}};-\omega)\right]\right\\},$
(117)

for isotropic polarizability. When the asymptotic forms of the Green’s dyadic
are inserted into this, we find

$S_{z}\sim-\frac{1}{4\pi^{2}}\frac{z(r^{2}-z^{2})}{r^{5}}\int\frac{d\omega}{2\pi}\omega^{4}\operatorname{Im}\alpha(\omega)\left(\coth\frac{\beta\omega}{2}-\coth\frac{\beta^{\prime}\omega}{2}\right).$
(118)

Here we have introduced again
$\alpha(\omega)=\int(d\mathbf{r^{\prime}})\chi(\mathbf{r^{\prime}};\omega)$.
Now we integrate $S_{z}$ over the infinite screen at $z=Z$ by introducing
polar coordinates on that plane:

$r=Z\sec\theta,\quad\rho=\sqrt{x^{2}+y^{2}}=Z\tan\theta.$ (119)

Therefore, the total power radiated is

$P=-\frac{2}{3\pi^{2}}\int_{0}^{\infty}d\omega\,\omega^{4}\operatorname{Im}\alpha(\omega)\left(\frac{1}{e^{\beta\omega}-1}-\frac{1}{e^{\beta^{\prime}\omega}-1}\right).$
(120)

The sign means that, if $T^{\prime}>T$, the energy flux is, as is obvious, in
the $+z$ direction.

For the force of the composite mirror/dielectric object, we compute the
surface integral of the $zz$ component of the electromagnetic stress tensor,
$T_{zz}$, over the same surface, as follows from the local statement of
momentum conservation,

$\frac{\partial}{\partial
t}\mathbf{G}+\bm{\nabla}\cdot\mathbf{T}+\mathbf{f}=\bm{0},$ (121)

where $\mathbf{f}$ is the force density on all the free and bound charges,
$\mathbf{T}$ is the stress tensor, and $\mathbf{G}$ is the momentum density of
the field. [The contributions to $S_{z}$ and $T_{zz}$ from the mirror alone in
vacuum is zero.] The force is, in our static situation,

$F_{z}=-\int_{z=Z}d\sigma\,T_{zz}=-\frac{1}{2}\int_{z=Z}d\sigma\left(E_{x}^{2}+E_{y}^{2}-E_{z}^{2}+H_{x}^{2}+H_{y}^{2}-H_{z}^{2}\right).$
(122)

The FDT gives PP contributions [that we referred to as (1,1) expansion terms]
and EE contributions [(2,0) and (0,2) contributions]. A straightforward
calculation for the former leads to the following radiation-zone result:

$F^{\rm
PP}_{z}=-2\int_{z=Z}d\sigma\frac{z^{2}(x^{2}+y^{2})}{r^{6}}\int\frac{d\omega}{2\pi}\operatorname{Im}\alpha(\omega)\frac{\omega^{4}}{4\pi^{2}}\coth\frac{\beta^{\prime}\omega}{2},$
(123)

which exhibits the same positional dependence on the screen as the energy
flux. The corresponding calculation for the EE contributions involves
recognizing the average

$\operatorname{Im}\left(e^{i\omega R}\right)e^{-i\omega
R}\to-\frac{i}{2},\quad\operatorname{Im}\left(i\,e^{i\omega
R}\right)i\,e^{i\omega R}\to+\frac{i}{2};$ (124)

the result of including the EE and PP contributions is

$F_{z}=\frac{1}{8\pi^{2}}\int_{0}^{\infty}d\omega\,\omega^{4}\operatorname{Im}\alpha(\omega)\left(\coth\frac{\beta\omega}{2}-\coth\frac{\beta^{\prime}\omega}{2}\right).$
(125)

Indeed, the force is directed in the $-z$ direction, but the precise
connection between the energy radiated (120) and the force on the composite
object (125) is obscure.

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