On the Motion of Particles Suspended in Liquids at Rest, as Required by the Molecular-Kinetic Theory of Heat

English

On the Motion of Particles Suspended in Liquids at Rest, as Required by the Molecular-Kinetic Theory of HeatShow the German source

English

by A. EinsteinShow the German source

English

In this paper it is to be shown that, according to the molecular-kinetic theory of heat, bodies of microscopically visible size suspended in liquids must, as a consequence of the molecular motion of heat, carry out motions of such magnitude that these motions can easily be detected with the microscope.Show the German source It is possible that the motions to be treated here are identical with the so-called “Brownian molecular motion”; the information about the latter that is available to me is, however, so imprecise that I could form no judgment on the matter.Show the German source

English

If the motion to be treated here can actually be observed, together with the regularities to be expected for it, then classical thermodynamics can no longer be regarded as exactly valid even for spaces distinguishable under the microscope, and an exact determination of the true size of atoms is then possible.Show the German source If, conversely, the prediction of this motion should prove incorrect, a weighty argument against the molecular-kinetic conception of heat would thereby be furnished.Show the German source

English

§ 1. On the osmotic pressure to be ascribed to the suspended particles.Show the German source

English

In the partial volume $V^*$ of a liquid of total volume $V$, let $z$ gram-molecules of a non-electrolyte be dissolved.Show the German source If the volume $V^*$ is separated from the pure solvent by a wall that is permeable to the solvent but not to the dissolved substance, then this wall is acted on by the so-called osmotic pressure, which, for sufficiently large values of $V^*/z$, satisfies the equation:Show the German source $$p V^* = R T z.$$$p$Osmotic pressure$V^{\ast }$Volume$R$Gas constant$T$Absolute temperature$z$Solute amountShow the German source

English

If, on the other hand, instead of the dissolved substance, small suspended bodies are present in the partial volume $V^*$ of the liquid, bodies that likewise cannot pass through the wall permeable to the solvent, then according to the classical theory of thermodynamics one is not to expect — at least when gravity, which does not interest us here, is neglected — that a force acts on the wall; for on the usual view the “free energy” of the system appears to depend not on the position of the wall and of the suspended bodies, but only on the total masses and the qualities of the suspended substance, of the liquid and of the wall, and on pressure and temperature.Show the German source It is true that the energy and entropy of the boundary surfaces (capillary forces) would also come into consideration in calculating the free energy; we can, however, disregard these, since in the changes of position of the wall and of the suspended bodies that are to be considered, changes in the size and nature of the surfaces of contact are supposed not to occur.Show the German source

English

From the standpoint of the molecular-kinetic theory of heat, however, one arrives at a different view.Show the German source According to this theory a dissolved molecule differs from a suspended body only in size, and one does not see why a number of suspended bodies should not correspond to the same osmotic pressure as the same number of dissolved molecules.Show the German source One will have to assume that the suspended bodies, as a consequence of the molecular motion of the liquid, carry out an irregular motion in the liquid, even if a very slow one; if they are prevented by the wall from leaving the volume $V^*$, they will exert forces on the wall, just as dissolved molecules do.Show the German source If, therefore, $n$ suspended bodies are present in the volume $V^*$, that is, $n/V^* = \nu$ in unit volume, and if neighboring ones among them are sufficiently far apart, then there will correspond to them an osmotic pressure $p$ of magnitude:Show the German source $$p = \frac{R T}{V^*} \frac{n}{N} = \frac{R T}{N} \cdot \nu,$$$p$Osmotic pressure$R$Gas constant$T$Absolute temperature$V^{\ast }$Volume$n$Particle count$N$Avogadro's number$\nu$Number densityShow the German source where $N$ denotes the number of actual molecules contained in one gram-molecule.Show the German source In the next section it is to be shown that the molecular-kinetic theory of heat does in fact lead to this extended view of osmotic pressure.Show the German source

English

§ 2. The osmotic pressure from the standpoint of the molecular-kinetic theory of heat.^[1)]Show the German source

English

If $p_1 p_2 \ldots p_l$ are variables of state of a physical system that completely determine its instantaneous state (for example, the coordinates and velocity components of all the atoms of the system), and if the complete system of the equations of change of these variables of state is given in the formShow the German source $$\frac{\partial p_\nu}{\partial t} = \varphi_\nu(p_1 \ldots p_l) \quad (\nu = 1, 2 \ldots l)$$$p_{\nu }$State variable$t$Elapsed time${\phi }_{\nu }$State variable rate$p_1$State variable$p_l$State variableShow the German source where $\sum \frac{\partial \varphi_\nu}{\partial p_\nu} = 0$, then the entropy of the system is given by the expression:Show the German source $$S = \frac{\bar{E}}{T} + 2 \varkappa \lg \int e^{-\frac{E}{2 \varkappa T}} d p_1 \ldots d p_l.$$$S$Entropy$\bar{E}$System energy$T$Absolute temperature$\varkappa$Boltzmann constant$E$Microstate energy$p_1$State variable$p_l$State variableShow the German source Here $T$ denotes the absolute temperature, $\bar{E}$ the energy of the system, and $E$ the energy as a function of the $p_\nu$.Show the German source The integral is to be extended over all combinations of values of the $p_\nu$ that are compatible with the conditions of the problem.Show the German source $\varkappa$ is connected with the constant $N$ mentioned above by the relation $2 \varkappa N = R$.Show the German source For the free energy $F$ we therefore obtain:Show the German source $$F = -\frac{R}{N} T \lg \int e^{-\frac{E N}{R T}} d p_1 \ldots d p_l = -\frac{R T}{N} \lg B.$$$F$Free energy$R$Gas constant$N$Avogadro's number$T$Absolute temperature$E$Microstate energy$p_1$State variable$p_l$State variable$B$Configuration integralShow the German source

English

We now imagine a liquid enclosed in the volume $V$; in the partial volume $V^*$ of $V$ let there be $n$ dissolved molecules or suspended bodies, which are held in the volume $V^*$ by a semipermeable wall; the limits of integration of the integral $B$ that occurs in the expressions for $S$ and $F$ are thereby affected.Show the German source Let the total volume of the dissolved molecules or suspended bodies be small compared with $V^*$.Show the German source Let this system be completely described, in the sense of the theory mentioned, by the variables of state $p_1 \ldots p_l$.Show the German source

English

Even if the molecular picture were fixed down to every detail, the evaluation of the integral $B$ would still present such difficulties that an exact calculation of $F$ could hardly be contemplated.Show the German source Here, however, we need only know how $F$ depends on the size of the volume $V^*$ in which all the dissolved molecules or suspended bodies (called “particles” for short in what follows) are contained.Show the German source

English

We call $x_1, y_1, z_1$ the rectangular coordinates of the center of gravity of the first particle, $x_2, y_2, z_2$ those of the second, etc., $x_n, y_n, z_n$ those of the last particle, and we assign to the centers of gravity of the particles the infinitely small parallelepiped-shaped regions $d x_1 d y_1 d z_1, d x_2 d y_2 d z_2 \ldots d x_n d y_n d z_n$, all of which are to lie in $V^*$.Show the German source Let the value be sought of the integral occurring in the expression for $F$, with the restriction that the centers of gravity of the particles lie in the regions just assigned to them.Show the German source This integral can in any case be brought into the formShow the German source $$d B = d x_1 d y_1 \ldots d z_n \cdot J$$$B$Configuration integral$x_1$Latent position (one dimension)$y_1$Latent position (one dimension)$z_n$Latent position (one dimension)$J$Configuration integral factorShow the German source where $J$ is independent of $d x_1 d y_1$ etc., and of $V^*$, that is, of the position of the semipermeable wall.Show the German source But $J$ is also independent of the particular choice of the positions of the regions for the centers of gravity, and of the value of $V^*$, as is now to be shown.Show the German source For let a second system of infinitely small regions for the centers of gravity of the particles be given, denoted by $d x'_1 d y'_1 d z'_2, d x'_2 d y'_2 d z'_2 \ldots d x'_n d y'_n d z'_n$, regions that are to differ from those originally given only in their position and not in their size, and that are likewise all to be contained in $V^*$; then, analogously,Show the German source $$d B' = d x'_1 d y'_1 \ldots d z'_n \cdot J',$$$B^{\prime }$Configuration integral$x_1^{\prime }$Latent position (one dimension)$y_1^{\prime }$Latent position (one dimension)$z_n^{\prime }$Latent position (one dimension)$J^{\prime }$Configuration integral factorShow the German source whereShow the German source $$d x_1 d y_1 \ldots d z_n = d x'_1 d y'_1 \ldots d z'_n.$$$x_1$Latent position (one dimension)$y_1$Latent position (one dimension)$z_n$Latent position (one dimension)$x_1^{\prime }$Latent position (one dimension)$y_1^{\prime }$Latent position (one dimension)$z_n^{\prime }$Latent position (one dimension)Show the German source We therefore have:Show the German source $$\frac{d B}{d B'} = \frac{J}{J'}.$$$B$Configuration integral$B^{\prime }$Configuration integral$J$Configuration integral factor$J^{\prime }$Configuration integral factorShow the German source

English

From the molecular theory of heat given in the papers cited, however, it can easily be inferred^[1)] that $d B/B$ (respectively $d B'/B$) is equal to the probability that at an arbitrarily chosen instant the centers of gravity of the particles lie in the regions $(d x_1 \ldots d z_n)$ (respectively in the regions $(d x'_1 \ldots d z'_n)$).Show the German source Now if the motions of the individual particles are independent of one another (to a sufficient approximation), if the liquid is homogeneous, and if no forces act on the particles, then for regions of equal size the probabilities belonging to the two systems of regions must be equal to each other, so that:Show the German source $$\frac{d B}{B} = \frac{d B'}{B}.$$$B$Configuration integral$B^{\prime }$Configuration integralShow the German source But from this equation and the one found last it follows thatShow the German source $$J = J'.$$$J$Configuration integral factor$J^{\prime }$Configuration integral factorShow the German source

English

It is thus proved that $J$ depends neither on $V^*$ nor on $x_1, y_1 \ldots z_n$.Show the German source By integration one obtainsShow the German source $$B = \int J d x_1 \ldots d z_n = J V^{*n}$$$B$Configuration integral$J$Configuration integral factor$x_1$Latent position (one dimension)$z_n$Latent position (one dimension)$V$Volume$n$Particle countShow the German source and from thisShow the German source $$F = -\frac{R T}{N} \{ \lg J + n \lg V^* \}$$$F$Free energy$R$Gas constant$T$Absolute temperature$N$Avogadro's number$J$Configuration integral factor$n$Particle count$V^{\ast }$VolumeShow the German source andShow the German source $$p = -\frac{\partial F}{\partial V^*} = \frac{R T}{V^*} \frac{n}{N} = \frac{R T}{N} \nu.$$$p$Osmotic pressure$F$Free energy$V^{\ast }$Volume$R$Gas constant$T$Absolute temperature$n$Particle count$N$Avogadro's number$\nu$Number densityShow the German source

English

This consideration shows that the existence of osmotic pressure is a consequence of the molecular-kinetic theory of heat, and that according to this theory dissolved molecules and suspended bodies present in equal numbers behave in exactly the same way with respect to osmotic pressure at great dilution.Show the German source

English

§ 3. Theory of the diffusion of small suspended spheres.Show the German source

English

Let suspended particles be distributed irregularly in a liquid.Show the German source We shall investigate their state of dynamic equilibrium under the assumption that on the individual particles there acts a force $K$ that depends on position but not on time.Show the German source For the sake of simplicity let it be assumed that the force everywhere has the direction of the $X$-axis.Show the German source

English

Let $\nu$ be the number of suspended particles per unit volume; then in the case of thermodynamic equilibrium $\nu$ is a function of $x$ such that, for an arbitrary virtual displacement $\delta x$ of the suspended substance, the variation of the free energy vanishes.Show the German source We therefore have:Show the German source $$\delta F = \delta E - T \delta S = 0.$$$F$Free energy$E$System energy$T$Absolute temperature$S$EntropyShow the German source Let it be assumed that the liquid has cross-section 1 perpendicular to the $X$-axis and is bounded by the planes $x = 0$ and $x = l$.Show the German source We then have:Show the German source $$\delta E = -\int_0^l K \nu \delta x d x$$$E$System energy$l$Column length$K$External force per particle$\nu$Number densityShow the German source andShow the German source $$\delta S = \int_0^l R \frac{\nu}{N} \frac{\partial \delta x}{\partial x} d x = -\frac{R}{N} \int_0^l \frac{\partial \nu}{\partial x} \delta x d x.$$$S$Entropy$l$Column length$R$Gas constant$\nu$Number density$N$Avogadro's numberShow the German source The equilibrium condition sought is therefore:Show the German source $$-K \nu + \frac{R T}{N} \frac{\partial \nu}{\partial x} = 0$$$K$External force per particle$\nu$Number density$R$Gas constant$T$Absolute temperature$N$Avogadro's numberShow the German source orShow the German source $$K \nu - \frac{\partial p}{\partial x} = 0.$$$K$External force per particle$\nu$Number density$p$Osmotic pressureShow the German source The last equation states that the force $K$ is held in equilibrium by forces of osmotic pressure.Show the German source

English

We use equation (1) to determine the diffusion coefficient of the suspended substance.Show the German source We can regard the state of dynamic equilibrium just considered as the superposition of two processes running in opposite directions, namely
1. a motion of the suspended substance under the action of the force $K$ acting on each individual suspended particle,
2. a process of diffusion, which is to be regarded as a consequence of the irregular motions of the particles due to the molecular motion of heat.Show the German source

English

If the suspended particles are spherical (sphere radius $P$) and the liquid has the coefficient of friction $k$, the force $K$ imparts to the individual particle the velocity^[1)]Show the German source $$\frac{K}{6 \pi k P},$$$K$External force per particle$k$Dynamic viscosity$P$Particle radiusShow the German source and through the unit of cross-section there pass per unit timeShow the German source $$\frac{\nu K}{6 \pi k P}$$$\nu$Number density$K$External force per particle$k$Dynamic viscosity$P$Particle radiusShow the German source particles.Show the German source

English

If, further, $D$ denotes the diffusion coefficient of the suspended substance and $\mu$ the mass of a particle, then as a result of diffusion there pass through the unit of cross-section per unit timeShow the German source $$-D \frac{\partial (\mu \nu)}{\partial x} \quad \text{Gramm}$$$D$Diffusion coefficient$\mu$Particle mass$\nu$Number densityShow the German source orShow the German source $$-D \frac{\partial \nu}{\partial x}$$$D$Diffusion coefficient$\nu$Number densityShow the German source particles.Show the German source Since dynamic equilibrium is to prevail, we must have:Show the German source $$\frac{\nu K}{6 \pi k P} - D \frac{\partial \nu}{\partial x} = 0.$$$\nu$Number density$K$External force per particle$k$Dynamic viscosity$P$Particle radius$D$Diffusion coefficientShow the German source

English

From the two conditions (1) and (2) found for dynamic equilibrium, the diffusion coefficient can be calculated.Show the German source One obtains:Show the German source $$D = \frac{R T}{N} \frac{1}{6 \pi k P}.$$$D$Diffusion coefficient$R$Gas constant$T$Absolute temperature$N$Avogadro's number$k$Dynamic viscosity$P$Particle radiusShow the German source The diffusion coefficient of the suspended substance therefore depends, apart from universal constants and the absolute temperature, only on the coefficient of friction of the liquid and on the size of the suspended particles.Show the German source

English

§ 4. On the irregular motion of particles suspended in a liquid and its relation to diffusion.Show the German source

English

We now turn to a closer investigation of the irregular motions which, produced by the molecular motion of heat, give rise to the diffusion investigated in the last section.Show the German source

English

It must evidently be assumed that each individual particle carries out a motion that is independent of the motion of all the other particles; the motions of one and the same particle in different time intervals are also to be regarded as mutually independent processes, as long as we think of these time intervals as not chosen too small.Show the German source

English

We introduce into the discussion a time interval $\tau$, which is to be very small compared with the observable time intervals, but still so large that the motions carried out by a particle in two successive time intervals $\tau$ are to be regarded as mutually independent events.Show the German source

English

Now let a total of $n$ suspended particles be present in a liquid.Show the German source In a time interval $\tau$ the $X$-coordinates of the individual particles will increase by $\Delta$, where $\Delta$ has a different (positive or negative) value for each particle.Show the German source A certain frequency law will hold for $\Delta$; the number $d n$ of particles that in the time interval $\tau$ undergo a displacement lying between $\Delta$ and $\Delta + d \Delta$ will be expressible by an equation of the formShow the German source $$d n = n \varphi(\Delta) d \Delta$$$n$Particle count$\phi$Jump density$\Delta$Displacement in one intervalShow the German source whereShow the German source $$\int_{-\infty}^{+\infty} \varphi(\Delta) d \Delta = 1$$$\phi$Jump density$\Delta$Displacement in one intervalShow the German source and $\varphi$ differs from zero only for very small values of $\Delta$ and satisfies the conditionShow the German source $$\varphi(\Delta) = \varphi(-\Delta)$$$\phi$Jump density$\Delta$Displacement in one intervalShow the German source

English

We now investigate how the diffusion coefficient depends on $\varphi$, again restricting ourselves to the case in which the number $\nu$ of particles per unit volume depends only on $x$ and $t$.Show the German source

English

Let $\nu = f(x, t)$ be the number of particles per unit volume; we calculate the distribution of the particles at the time $t + \tau$ from their distribution at the time $t$.Show the German source From the definition of the function $\varphi(\Delta)$ one easily obtains the number of particles that at the time $t + \tau$ are located between two planes perpendicular to the $X$-axis with the abscissas $x$ and $x + d x$.Show the German source One obtains:Show the German source $$f(x, t + \tau) d x = d x \cdot \int_{\Delta = -\infty}^{\Delta = +\infty} f(x + \Delta) \varphi(\Delta) d \Delta.$$$f$Number density$t$Time$\tau$Step interval$\Delta$Displacement in one interval$\phi$Jump densityShow the German source But since $\tau$ is very small, we can now set:Show the German source $$f(x, t + \tau) = f(x, t) + \tau \frac{\partial f}{\partial t}.$$$f$Number density$t$Time$\tau$Step intervalShow the German source Further, we expand $f(x + \Delta, t)$ in powers of $\Delta$:Show the German source $$f(x + \Delta, t) = f(x, t) + \Delta \frac{\partial f(x, t)}{\partial x} + \frac{\Delta^2}{2!} \frac{\partial^2 f(x, t)}{\partial x^2} \ldots \text{ in inf.}$$$f$Number density$\Delta$Displacement in one interval$t$TimeShow the German source We can carry out this expansion under the integral, since only very small values of $\Delta$ contribute anything to the latter.Show the German source We obtain:Show the German source $$\begin{aligned} f + \frac{\partial f}{\partial t} \cdot \tau = f \cdot \int_{-\infty}^{+\infty} \varphi(\Delta) d \Delta &+ \frac{\partial f}{\partial x} \int_{-\infty}^{+\infty} \Delta \varphi(\Delta) d \Delta \\ &+ \frac{\partial^2 f}{\partial x^2} \int_{-\infty}^{+\infty} \frac{\Delta^2}{2} \varphi(\Delta) d \Delta \ldots \end{aligned}$$$f$Number density$t$Time$\tau$Step interval$\phi$Jump density$\Delta$Displacement in one intervalShow the German source On the right-hand side, because $\varphi(x) = \varphi(-x)$, the second, fourth, etc. terms vanish, while of the first, third, fifth, etc. terms each successive one is very small compared with the one before.Show the German source From this equation we obtain, taking into account thatShow the German source $$\int_{-\infty}^{+\infty} \varphi(\Delta) d \Delta = 1,$$$\phi$Jump density$\Delta$Displacement in one intervalShow the German source and settingShow the German source $$\frac{1}{\tau} \int_{-\infty}^{+\infty} \frac{\Delta^2}{2} \varphi(\Delta) d \Delta = D$$$\tau$Step interval$\Delta$Displacement in one interval$\phi$Jump density$D$Diffusion coefficientShow the German source and taking into account only the first and third terms on the right-hand side:Show the German source $$\frac{\partial f}{\partial t} = D \frac{\partial^2 f}{\partial x^2}.$$$f$Number density$t$Time$D$Diffusion coefficientShow the German source

English

This is the well-known differential equation of diffusion, and one sees that $D$ is the diffusion coefficient.Show the German source

English

Another important consideration can be attached to this development.Show the German source We have assumed that the individual particles were all referred to the same coordinate system.Show the German source This is not necessary, however, since the motions of the individual particles are independent of one another.Show the German source We shall now refer the motion of each particle to a coordinate system whose origin coincides with the position of the center of gravity of the particle in question at the time $t = 0$, with the difference that $f(x, t) d x$ now denotes the number of particles whose $X$-coordinate has increased from the time $t = 0$ to the time $t = t$ by an amount lying between $x$ and $x + d x$.Show the German source In this case too, then, the function $f$ changes according to equation (1).Show the German source Further, for $x \gtrless 0$ and $t = 0$ we must evidently haveShow the German source $$f(x, t) = 0 \quad \text{und} \quad \int_{-\infty}^{+\infty} f(x, t) d x = n$$$f$Number density$t$Time$n$Particle countShow the German source The problem, which coincides with the problem of diffusion from a single point (neglecting the interaction of the diffusing particles), is now mathematically completely determined; its solution is:Show the German source $$f(x, t) = \frac{n}{\sqrt{4 \pi D}} \frac{e^{-\frac{x^2}{4 D t}}}{\sqrt{t}}.$$$f$Number density$t$Time$n$Particle count$D$Diffusion coefficientShow the German source

English

The frequency distribution of the changes of position that have taken place in an arbitrary time $t$ is therefore the same as that of random errors, as was to be expected.Show the German source Of importance, however, is how the constant in the exponent is connected with the diffusion coefficient.Show the German source With the help of this equation we now calculate the displacement $\lambda_x$ in the direction of the $X$-axis that a particle undergoes on average, or — more precisely expressed — the square root of the arithmetic mean of the squares of the displacements in the direction of the $X$-axis; it is:Show the German source $$\lambda_x = \sqrt{\overline{x^2}} = \sqrt{2 D t}.$$${\lambda }_x$Coordinate RMS displacement$\overline{x^2}$Mean-square displacement (one dimension)$D$Diffusion coefficient$t$Observation intervalShow the German source

English

The mean displacement is therefore proportional to the square root of the time.Show the German source It can easily be shown that the square root of the mean of the squares of the total displacements of the particles has the value $\lambda_x \sqrt{3}$.Show the German source

English

§ 5. Formula for the mean displacement of suspended particles. A new method for determining the true size of atoms.Show the German source

English

In § 3 we found, for the diffusion coefficient $D$ of a substance suspended in a liquid in the form of small spheres of radius $P$, the value:Show the German source $$D = \frac{R T}{N} \frac{1}{6 \pi k P}.$$$D$Diffusion coefficient$R$Gas constant$T$Absolute temperature$N$Avogadro's number$k$Dynamic viscosity$P$Particle radiusShow the German source Further, in § 4 we found, for the mean value of the displacements of the particles in the direction of the $X$-axis in the time $t$:Show the German source $$\lambda_x = \sqrt{2 D t}.$$${\lambda }_x$Coordinate RMS displacement$D$Diffusion coefficient$t$Observation intervalShow the German source By eliminating $D$ we obtain:Show the German source $$\lambda_x = \sqrt{t} \cdot \sqrt{\frac{R T}{N} \frac{1}{3 \pi k P}}.$$${\lambda }_x$Coordinate RMS displacement$t$Observation interval$R$Gas constant$T$Absolute temperature$N$Avogadro's number$k$Dynamic viscosity$P$Particle radiusShow the German source This equation shows how $\lambda_x$ must depend on $T$, $k$ and $P$.Show the German source

English

We shall calculate how large $\lambda_x$ is for one second if $N$ is set equal to $6 \cdot 10^{23}$ in accordance with the results of the kinetic theory of gases; let the liquid chosen be water at 17° C ($k = 1{,}35 \cdot 10^{-2}$), and let the diameter of the particles be 0,001 mm.Show the German source One obtains:Show the German source $$\lambda_x = 8 \cdot 10^{-5} \, \text{cm} = 0{,}8 \, \text{Mikron}.$$${\lambda }_x$Coordinate RMS displacementShow the German source The mean displacement in 1 minute would therefore be about 6 microns.Show the German source

English

Conversely, the relation found can be used to determine $N$.Show the German source One obtains:Show the German source $$N = \frac{t}{\lambda_x^2} \cdot \frac{R T}{3 \pi k P}.$$$N$Estimated Avogadro number$t$Observation interval${\lambda }_x$Coordinate RMS displacement$R$Gas constant$T$Absolute temperature$k$Dynamic viscosity$P$Particle radiusShow the German source

English

May a researcher soon succeed in deciding the question raised here, which is important for the theory of heat!Show the German source

English

Bern, May 1905.Show the German source

English

(Received 11 May 1905.)Show the German source