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Maxwell’s Radiation Pressure

**53.** While in the preceding part the phenomena of radiation have been presented
with the assumption of only well known elementary laws of optics summarized in
Sec. 2, which are common to all optical theories, we shall hereafter make use of
the electromagnetic theory of light and shall begin by deducing a consequence
characteristic of that theory. We shall, namely, calculate the magnitude
of the mechanical force, which is exerted by a light or heat ray passing
through a vacuum on striking a reﬂecting (Sec. 10) surface assumed to be at
rest.

For this purpose we begin by stating Maxwell’s general equations for an electromagnetic process in a vacuum. Let the vector $E$ denote the electric ﬁeld-strength (intensity of the electric ﬁeld) in electric units and the vector $H$ the magnetic ﬁeld-strength in magnetic units. Then the equations are, in the abbreviated notation of the vector calculus,

$$\begin{array}{ccccc}\hfill \stackrel{\u0307}{E}& =ccurlH\hfill & \hfill \stackrel{\u0307}{H}& =-ccurlE\hfill & \hfill \\ \hfill div.E& =0\hfill & \hfill div.H& =0.\hfill \end{array}$$ | (52) |

Should the reader be unfamiliar with the symbols of this notation, he may
readily deduce their meaning by working backward from the subsequent
equations (53).
**
**

** 54.** In order to pass to the case of a plane wave in any direction we
assume that all the quantities that ﬁx the state depend only on the
time $t$ and on one of the coordinates
$x\prime $, $y\prime $, $z\prime $,
of an orthogonal right-handed system of coordinates, say
on $x\prime $.
Then the equations (52) reduce to

Hence the most general expression for a plane wave passing through a vacuum in the direction of the positive $x\prime $-axis is

where $f$ and $g$
represent two arbitrary functions of the same argument.
**
**

** 55.** Suppose now that this wave strikes a reﬂecting surface, *e.g.*, the surface of
an absolute conductor (metal) of inﬁnitely

large conductivity. In such a conductor even an inﬁnitely small electric
ﬁeld-strength produces a ﬁnite conduction current; hence the electric
ﬁeld-strength $E$
in it must be always and everywhere inﬁnitely small. For simplicity we also
suppose the conductor to be non-magnetizable, *i.e.*, we assume the magnetic
induction $B$ in it to be equal to
the magnetic ﬁeld-strength $H$,
just as is the case in a vacuum.

If we place the $x$-axis of a right-handed coordinate system $(xyz)$ along the normal of the surface directed toward the interior of the conductor, the $x$-axis is the normal of incidence. We place the $(x\prime y\prime )$ plane in the plane of incidence and take this as the plane of the ﬁgure (Fig. 4). Moreover, we can also, without any restriction of generality, place the $y$-axis in the plane of the ﬁgure, so that the $z$-axis coincides with the $z\prime $-axis (directed from the ﬁgure toward the observer). Let the common origin $O$ of the two coordinate systems lie in the surface. If ﬁnally $\theta $ represents the angle of incidence, the coordinates with and without accent are related to each other by the following equations:

$$\begin{array}{llllllll}\hfill x& =x\prime cos\theta -y\prime sin\theta \phantom{\rule{2em}{0ex}}& \hfill x\prime & =xcos\theta +ysin\theta \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill y& =x\prime sin\theta +y\prime cos\theta \phantom{\rule{2em}{0ex}}& \hfill y\prime & =-xsin\theta +ycos\theta \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill z& =z\prime \phantom{\rule{2em}{0ex}}& \hfill z\prime & =z.\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$$By the same transformation we may pass from the components of the electric or magnetic ﬁeld-strength in the ﬁrst coordinate system to their components in the second system. Performing this transformation the following values are obtained from (54) for the components of the electric and magnetic ﬁeld-strengths of the incident wave in the coordinate system without accent,

Herein the argument of the functions $f$ and $g$ is

$$t-\frac{x\prime}{c}=t-\frac{xcos\theta +ysin\theta}{c}.$$ | (56) |

** 56.** In the surface of separation of the two media
$x=0$.
According to the general electromagnetic boundary conditions the components of
the ﬁeld-strengths in the surface of separation, *i.e.*, the four quantities
${E}_{y}$, ${E}_{z}$,
${H}_{y}$, ${H}_{z}$
must be equal to each other on the two sides of the surface of separation for this value
of $x$. In the conductor the
electric ﬁeld-strength $E$
is inﬁnitely small in accordance with the assumption made above. Hence
${E}_{y}$ and ${E}_{z}$
must vanish also in the vacuum for $x=0$.
This condition cannot be satisﬁed unless we assume in the vacuum, besides
the incident, also a reﬂected wave superposed on the former in such a
way that the components of the electric ﬁeld of the two waves in the
$y$ and $z$
direction just cancel at every instant and at every point in the surface of
separation. By this assumption and the condition that the reﬂected wave is a
plane wave returning into the interior of the vacuum, the other four components
of the reﬂected wave are also completely determined. They are all functions of the
single argument

$$t-\frac{-xcos\theta +ysin\theta}{c}.$$ | (57) |

The actual calculation yields as components of the total electromagnetic ﬁeld produced in the vacuum by the superposition of the two waves, the following expressions valid for points of the surface of separation $x=0$,

In these equations the argument of the functions $f$ and $g$ is, according to (56) and (57),

$$t-\frac{ysin\theta}{c}.$$

From these values the electric and magnetic ﬁeld-strength within the
conductor in the immediate neighborhood of the separating surface
$x=0$ is
obtained: where again the argument $t-\frac{ysin\theta}{c}$ is to be substituted in the functions $f$ and $g$. For the components of $E$ all vanish in an absolute conductor and the components ${H}_{x}$, ${H}_{y}$, ${H}_{z}$ are all continuous at the separating surface, the two latter since they are tangential components of the ﬁeld-strength, the former since it is the normal component of the magnetic induction $B$ (Sec. 55), which likewise remains continuous on passing through any surface of separation.

On the other hand, the normal component of the electric ﬁeld-strength ${E}_{x}$ is seen to be discontinuous; the discontinuity shows the existence of an electric charge on the surface, the surface density of which is given in magnitude and sign as follows:

$$\frac{1}{4\pi}2sin\theta \cdot f=\frac{1}{2\pi}sin\theta \cdot f.$$ | (60) |

In the interior of the conductor at a ﬁnite distance from the bounding surface,
*i.e.*, for $x>0$,
all six ﬁeld components are inﬁnitely small. Hence, on
increasing $x$, the values of
${H}_{y}$ and ${H}_{z}$,
which are ﬁnite for $x=0$,
approach the value $0$
at an inﬁnitely rapid rate.
**
**

** 57.** A certain mechanical force is exerted on the substance of
the conductor by the electromagnetic ﬁeld considered. We shall
calculate the component of this force normal to the surface. It is
partly of electric, partly of magnetic, origin. Let us ﬁrst consider the
former, ${F}_{e}$.
Since the electric charge existing on the surface of the conductor is in an electric ﬁeld,
a mechanical force equal to the product of the charge and the ﬁeld-strength is
exerted on it. Since, however, the ﬁeld-strength is discontinuous, having the value
$-2sin\theta f$ on the side of the
vacuum and $0$ on the
side of the conductor, from a well-known law of electrostatics the magnitude of the mechanical
force ${F}_{e}$ acting on an
element of surface $d\sigma $
of the conductor is obtained by multiplying the electric charge of the element of area
calculated in (60) by the arithmetic mean of the electric ﬁeld-strength on the two sides.
Hence

$${F}_{e}=\frac{sin\theta}{2\pi}f\phantom{\rule{0.3em}{0ex}}d\sigma \phantom{\rule{0.3em}{0ex}}(-sin\theta f)=-\frac{{sin}^{2}\theta}{2\pi}{f}^{2}\phantom{\rule{0.3em}{0ex}}d\sigma .$$

This force acts in the direction toward the vacuum and therefore exerts a
tension.
** 58.** We shall now calculate the mechanical force of magnetic
origin ${F}_{m}$.
In the interior of the conducting substance there are certain conduction
currents, whose intensity and direction are determined by the
vector $I$
of the current density

$$I=\frac{c}{4\pi}curlH.$$ | (61) |

A mechanical force acts on every element of space $d\tau $ of the conductor through which a conduction current ﬂows, and is given by the vector product

$$\frac{d\tau}{c}[I\times H].$$ | (62) |

Hence the component of this force normal to the surface of the conductor $x=0$ is equal to

$$\frac{d\tau}{c}({I}_{y}{H}_{z}-{I}_{z}{H}_{y}).$$

On substituting the values of
${I}_{y}$ and ${I}_{z}$
from (61) we obtain $$\frac{d\tau}{4\pi}\left[{H}_{z}\left(\frac{\partial {H}_{x}}{\partial z}-\frac{\partial {H}_{z}}{\partial x}\right)-{H}_{y}\left(\frac{\partial {H}_{y}}{\partial x}-\frac{\partial {H}_{x}}{\partial y}\right)\right].$$

In this expression the diﬀerential coeﬃcients with respect to
$y$ and $z$
are negligibly small in comparison to those with respect
to $x$,
according to the remark at the end of Sec. 56; hence the expression reduces to
$$-\frac{d\tau}{4\pi}\left({H}_{y}\frac{\partial {H}_{y}}{\partial x}+{H}_{z}\frac{\partial {H}_{z}}{\partial x}\right).$$

Let us
now consider a cylinder cut out of the conductor perpendicular to the surface with the
cross-section $d\sigma $, and
extending from $x=0$
to $x=\infty $. The
entire mechanical force of magnetic origin acting on this cylinder in the direction of the
$x$-axis, since
$d\tau =d\sigma \phantom{\rule{0.3em}{0ex}}x$, is given by
$${F}_{m}=-\frac{d\sigma}{4\pi}{\int}_{0}^{\infty}dx\left({H}_{y}\frac{\partial {H}_{y}}{\partial x}+{H}_{z}\frac{\partial {H}_{z}}{\partial x}\right).$$

On integration,
since $H$ vanishes
for $x=\infty $, we
obtain $${F}_{m}=\frac{d\sigma}{8\pi}{\left({H}_{y}^{2}+{H}_{z}^{2}\right)}_{x=0}$$

or by
equation (59) $${F}_{m}=\frac{d\sigma}{2\pi}({cos}^{2}\theta \cdot {g}^{2}+{f}^{2}).$$

By adding ${F}_{e}$ and ${F}_{m}$ the total mechanical force acting on the cylinder in question in the direction of the $x$-axis is found to be

$$F=\frac{d\sigma}{2\pi}{cos}^{2}\theta ({f}^{2}+{g}^{2}).$$ | (63) |

This force exerts on the surface of the conductor a pressure, which acts in a
direction normal to the surface toward the interior and is called “*Maxwell’s*
radiation pressure.” The existence and the magnitude of the radiation pressure as
predicted by the theory was ﬁrst found by delicate measurements with the
radiometer by *P. Lebedew.*^{16}
**
**

** 59.** We shall now establish a relation between the radiation pressure and the energy of
radiation $I\phantom{\rule{0.3em}{0ex}}dt$ falling on the
surface element $d\sigma $ of the
conductor in a time element $dt$.
The latter from *Poynting’s* law of energy ﬂow is

$$I\phantom{\rule{0.3em}{0ex}}dt=\frac{c}{4\pi}({E}_{y}{H}_{z}-{E}_{z}{H}_{y})\phantom{\rule{0.3em}{0ex}}d\sigma \phantom{\rule{0.3em}{0ex}}dt,$$

hence
from (55) $$I\phantom{\rule{0.3em}{0ex}}dt=\frac{c}{4\pi}cos\theta ({f}^{2}+{g}^{2})\phantom{\rule{0.3em}{0ex}}d\sigma \phantom{\rule{0.3em}{0ex}}dt.$$

By comparison with (63) we obtain $$F=\frac{2cos\theta}{c}I.$$ | (64) |

From this we ﬁnally calculate the total
pressure $p$,
*i.e.*, that mechanical force, which an arbitrary radiation proceeding from the vacuum
and totally reﬂected upon incidence on the conductor exerts in a normal direction
on a unit surface of the conductor. The energy radiated in the conical element

$$d\Omega =sin\theta \phantom{\rule{0.3em}{0ex}}d\theta \phantom{\rule{0.3em}{0ex}}d\phi $$

in the
time $dt$ on the
element of area $d\sigma $ is,
according to (6), $$I\phantom{\rule{0.3em}{0ex}}dt=Kcos\theta \phantom{\rule{0.3em}{0ex}}d\Omega \phantom{\rule{0.3em}{0ex}}d\sigma \phantom{\rule{0.3em}{0ex}}dt,$$

where $K$ represents
the speciﬁc intensity of the radiation in the
direction $d\Omega $
toward the reﬂector. On substituting this in (64) and integrating
over $d\Omega $
we obtain for the total pressure of all pencils which fall on the surface and are
reﬂected by it $$p=\frac{2}{c}\int K{cos}^{2}\theta \phantom{\rule{0.3em}{0ex}}d\Omega ,$$ | (65) |

the integration with respect to $\phi $ extending from $0$ to $2\pi $ and with respect to $\theta $ from $0$ to $\frac{\pi}{2}$.

In case $K$ is independent of direction as in the case of black radiation, we obtain for the radiation pressure

$$p=\frac{2K}{c}{\int}_{0}^{2\pi}d\phi {\int}_{0}^{\frac{\pi}{2}}d\theta {cos}^{2}\theta sin\theta =\frac{4\pi K}{3c}$$

or, if we introduce
instead of $K$ the volume
density of radiation $u$
from (21) $$p=\frac{u}{3}.$$ | (66) |

This value of the radiation pressure holds only when the reﬂection of the
radiation occurs at the surface of an absolute non-magnetizable conductor.
Therefore we shall in the thermodynamic deductions of the next chapter
make use of it only in such cases. Nevertheless it will be shown later on
(Sec. 66) that equation (66) gives the pressure of uniform radiation against
any totally reﬂecting surface, no matter whether it reﬂects uniformly or
diﬀusely.
**
**

** 60.** In view of the extraordinarily simple and close relation between the
radiation pressure and the energy of radiation, the question might be raised
whether this relation is really a special consequence of the electromagnetic theory,
or whether it might not, perhaps, be founded on more general energetic or
thermodynamic considerations. To decide this question we shall calculate the
radiation pressure that would follow by Newtonian mechanics from *Newton’s*
(emission) theory of light, a theory which, in itself, is quite consistent with
the energy principle. According to it the energy radiated onto a surface
by a light ray passing through a vacuum is equal to the kinetic energy
of the light particles striking the surface, all moving with the constant
velocity $c$.
The decrease in intensity of the energy radiation with the distance is
then explained simply by the decrease of the volume density of the light
particles.

Let us denote by $n$ the number of the light particles contained in a unit volume and by $m$ the mass of a particle. Then for a beam of parallel light the number of particles impinging in unit time on the element $d\sigma $ of a reﬂecting surface at the angle of incidence $\theta $ is

$$nccos\theta \phantom{\rule{0.3em}{0ex}}d\sigma .$$ | (67) |

Their kinetic energy is given according to Newtonian mechanics by

$$I=nccos\theta \phantom{\rule{0.3em}{0ex}}d\sigma \phantom{\rule{0.3em}{0ex}}\frac{m{c}^{2}}{2}=nmcos\theta \frac{{c}^{3}}{2}\phantom{\rule{0.3em}{0ex}}d\sigma .$$ | (68) |

Now, in order to determine the normal pressure of these particles on the surface, we may note that the normal component of the velocity $ccos\theta $ of every particle is changed on reﬂection into a component of opposite direction. Hence the normal component of the momentum of every particle (impulse-coordinate) is changed through reﬂection by $-2mccos\theta $. Then the change in momentum for all particles considered will be, according to (67),

$$-2nm{cos}^{2}\theta \phantom{\rule{0.3em}{0ex}}{c}^{2}\phantom{\rule{0.3em}{0ex}}d\sigma .$$ | (69) |

Should the reﬂecting body be free to move in the direction of the
normal of the reﬂecting surface and should there be no force acting on it
except the impact of the light particles, it would be set into motion by the
impacts. According to the law of action and reaction the ensuing motion
would be such that the momentum acquired in a certain interval of time
would be equal and opposite to the change in momentum of all the light
particles reﬂected from it in the same time interval. But if we allow a
separate constant force to act from outside on the reﬂector, there is to be
added to the change in momenta of the light particles the impulse of
the external force, *i.e.*, the product of the force and the time interval in
question.

Therefore the reﬂector will remain continuously at rest, whenever the constant external force exerted on it is so chosen that its impulse for any time is just equal to the change in momentum of all the particles reﬂected from the reﬂector in the same time. Thus it follows that the force $F$ itself which the particles exert by their impact on the surface element $d\sigma $ is equal and opposite to the change of their momentum in unit time as expressed in (69)

$$F=2nm{cos}^{2}\theta \phantom{\rule{0.3em}{0ex}}{c}^{2}\phantom{\rule{0.3em}{0ex}}d\sigma $$

and by making
use of (68), $$F=\frac{4cos\theta}{c}I.$$

On comparing this relation with equation (64) in which all symbols have the
same physical signiﬁcance, it is seen that *Newton’s* radiation pressure is twice as
large as *Maxwell’s* for the same energy radiation. A necessary consequence of
this is that the magnitude of *Maxwell’s* radiation pressure cannot be
deduced from general energetic considerations, but is a special feature of the
electromagnetic theory and hence all deductions from *Maxwell’s* radiation
pressure are to be regarded as consequences of the electromagnetic theory
of light and all conﬁrmations of them are conﬁrmations of this special
theory.

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