Chapter I

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,

(52)

$\begin{matrix} \dot{E} & = c curl H & \dot{H} & = - c curl E \\ div. E & = 0 & div. H & = 0 . \end{matrix}$

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 $x'$ , $y'$ , $z'$ , of an orthogonal right-handed system of coordinates, say on $x'$ . Then the equations (52) reduce to

(53)

$\begin{matrix} \frac{\partial E_{x'}}{\partial t} & = 0 & \frac{\partial H_{x'}}{\partial t} & = 0 \\ \frac{\partial E_{y'}}{\partial t} & = - c \frac{\partial H_{z'}}{\partial x'} & \frac{\partial H_{y'}}{\partial t} & = c \frac{\partial E_{z'}}{\partial x'} \\ \frac{\partial E_{x'}}{\partial t} & = c \frac{\partial H_{y'}}{\partial x'} & \frac{\partial H_{z'}}{\partial t} & = - c \frac{\partial E_{y'}}{\partial x'} \\ \frac{\partial E_{x'}}{\partial x'} & = 0 & \frac{\partial H_{x'}}{\partial x'} & = 0 . \end{matrix}$

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

(54)

$\begin{matrix} E_{x'} & = 0 & H_{x'} & = 0 \\ E_{y'} & = f (t - \frac{x'}{c}) & H_{y'} & = - g (t - \frac{x'}{c}) \\ E_{z'} & = g (t - \frac{x'}{c}) & H_{z'} & = f (t - \frac{x'}{c}) \end{matrix}$

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

Fig. 4

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 $(x y z)$ along the normal of the surface directed toward the interior of the conductor, the $x$ -axis is the normal of incidence. We place the $(x' y')$ 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 $z'$ -axis (directed from the figure toward the observer). Let the common origin $O$ of the two coordinate systems lie in the surface. If finally $θ$ represents the angle of incidence, the coordinates with and without accent are related to each other by the following equations:

$\begin{array}{l} x & = x' cos θ - y' sin θ & x' & = x cos θ + y sin θ \\ y & = x' sin θ + y' cos θ & y' & = - x sin θ + y cos θ \\ z & = z' & z' & = z . \end{array}$

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,

(55)

$\begin{matrix} E_{x} & = - sin θ \cdot f & H_{x} & = sin θ \cdot g \\ E_{y} & = cos θ \cdot f & H_{y} & = - cos θ \cdot g \\ E_{z} & = g & H_{z} & = f . \end{matrix}$

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

(56)

$t - \frac{x'}{c} = t - \frac{x cos θ + y sin θ}{c} .$

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

(57)

$t - \frac{- x cos θ + y sin θ}{c} .$

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 $x = 0$ ,

(58)

$\begin{matrix} E_{x} & = - sin θ \cdot f - sin θ \cdot f = - 2 sin θ \cdot f \\ E_{y} & = cos θ \cdot f - cos θ \cdot f = 0 \\ E_{z} & = g - g = 0 \\ H_{x} & = sin θ \cdot g - sin θ \cdot g = 0 \\ H_{y} & = - cos θ \cdot g - cos θ \cdot g = - 2 cos θ \cdot g \\ H_{z} & = f + f = 2 f . \end{matrix}$

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

$t - \frac{y sin θ}{c} .$

From these values the electric and magnetic field-strength within the conductor in the immediate neighborhood of the separating surface $x = 0$ is obtained:

(59)

$\begin{matrix} E_{x} & = 0 & H_{x} & = 0 \\ E_{y} & = 0 & H_{y} & = - 2 cos θ \cdot g \\ E_{z} & = 0 & H_{z} & = 2 f \end{matrix}$

where again the argument $t - \frac{y sin θ}{c}$ 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:

(60)

$\frac{1}{4 π} 2 sin θ \cdot f = \frac{1}{2 π} sin θ \cdot f .$

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 $- 2 sin θ f$ 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 $d σ$ 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

$F_{e} = \frac{sin θ}{2 π} f d σ (- sin θ f) = - \frac{{sin}^{2} θ}{2 π} f^{2} d σ .$

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

(61)

$I = \frac{c}{4 π} curl H .$

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

(62)

$\frac{d τ}{c} [I \times H] .$

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

$\frac{d τ}{c} (I_{y} H_{z} - I_{z} H_{y}) .$

On substituting the values of $I_{y}$ and $I_{z}$ from (61) we obtain

$\frac{d τ}{4 π} [H_{z} (\frac{\partial H_{x}}{\partial z} - \frac{\partial H_{z}}{\partial x}) - H_{y} (\frac{\partial H_{y}}{\partial x} - \frac{\partial H_{x}}{\partial y})] .$

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

$- \frac{d τ}{4 π} (H_{y} \frac{\partial H_{y}}{\partial x} + H_{z} \frac{\partial H_{z}}{\partial x}) .$

Let us now consider a cylinder cut out of the conductor perpendicular to the surface with the cross-section $d σ$ , 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 $d τ = d σ x$ , is given by

$F_{m} = - \frac{d σ}{4 π} \int_{0}^{\infty} d x (H_{y} \frac{\partial H_{y}}{\partial x} + H_{z} \frac{\partial H_{z}}{\partial x}) .$

On integration, since $H$ vanishes for $x = \infty$ , we obtain

$F_{m} = \frac{d σ}{8 π} {(H_{y}^{2} + H_{z}^{2})}_{x = 0}$

or by equation (59)

$F_{m} = \frac{d σ}{2 π} ({cos}^{2} θ \cdot g^{2} + f^{2}) .$

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

(63)

$F = \frac{d σ}{2 π} {cos}^{2} θ (f^{2} + g^{2}) .$

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 $I d t$ falling on the surface element $d σ$ of the conductor in a time element $d t$ . The latter from Poynting’s law of energy flow is

$I d t = \frac{c}{4 π} (E_{y} H_{z} - E_{z} H_{y}) d σ d t,$

hence from (55)

$I d t = \frac{c}{4 π} cos θ (f^{2} + g^{2}) d σ d t .$

By comparison with (63) we obtain

(64)

$F = \frac{2 cos θ}{c} I .$

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

$d Ω = sin θ d θ d φ$

in the time $d t$ on the element of area $d σ$ is, according to (6),

$I d t = K cos θ d Ω d σ d t,$

where $K$ represents the specific intensity of the radiation in the direction $d Ω$ toward the reflector. On substituting this in (64) and integrating over $d Ω$ we obtain for the total pressure of all pencils which fall on the surface and are reflected by it

(65)

$p = \frac{2}{c} \int K {cos}^{2} θ d Ω,$

the integration with respect to $φ$ extending from $0$ to $2 π$ and with respect to $θ$ from $0$ to $\frac{π}{2}$ .

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

$p = \frac{2 K}{c} \int_{0}^{2 π} d φ \int_{0}^{\frac{π}{2}} d θ {cos}^{2} θ sin θ = \frac{4 π K}{3 c}$

or, if we introduce instead of $K$ the volume density of radiation $u$ from (21)

(66)

$p = \frac{u}{3} .$

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 $d σ$ of a reflecting surface at the angle of incidence $θ$ is

(67)

$n c cos θ d σ .$

Their kinetic energy is given according to Newtonian mechanics by

(68)

$I = n c cos θ d σ \frac{m c^{2}}{2} = n m cos θ \frac{c^{3}}{2} d σ .$

Now, in order to determine the normal pressure of these particles on the surface, we may note that the normal component of the velocity $c cos θ$ 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 $- 2 m c cos θ$ . Then the change in momentum for all particles considered will be, according to (67),

(69)

$- 2 n m {cos}^{2} θ c^{2} d σ .$

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 $d σ$ is equal and opposite to the change of their momentum in unit time as expressed in (69)

$F = 2 n m {cos}^{2} θ c^{2} d σ$

and by making use of (68),

$F = \frac{4 cos θ}{c} I .$

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.

Chapter I

16P. Lebedew, Annalen d. Phys. 6, p. 433, 1901. See also E. F. Nichols and G. F. Hull, Annalen d. Phys. 12, p. 225, 1903.