Transformation of Matter into Radiant Energy

Ya. K. Syrkin, Ivanovo-Voznesensk.

Mass and energy are essentially identical; they represent different manifestations of one and the same thing.
—Einstein.

§ 1. Introduction.

Classical physics made use of the law of conservation of energy as the most general law, fulfilled in every phenomenon of nature. Within the limits of experimental error, in the case of slowly moving bodies, the conservation or equivalence of different forms of energy is an empirical regularity.

When the object of physical investigation was the study of the transition of a system from a given initial state to some final state, the question of energy was posed precisely with respect to the change in energy, independently of its absolute value. In this connection the concept of energy acquired great significance, for the total energy is a function of state and does not depend on the path of transition (the differential of the total energy is a complete differential).

The law of conservation of weight in chemical reactions plays in chemistry a role analogous to that of the law of conservation of energy in physics. The equality of the masses of the initial and resulting bodies, realized in every chemical process, makes possible strict accounting, analysis, and verification of the reaction. Thus physicochemical phenomena, accompanied by energetic and structural-molecular changes,

can be described in two ways, in the language of conservation of matter and energy.

It happens as though, for a possible complete description of a phenomenon, one has to speak in two languages. On the one hand, we encompass the process by the equation of the chemical reaction, whereby the law of conservation of matter is satisfied; and on the other, we compose an energy equation expressing the balance of mutually transforming energies in the transition from the initial to the final state.

There naturally arises the question of the connection between these two fundamental laws. Can they not be combined into one more general law? The physical answer to this question was given by Einstein in the special principle of relativity and is expressed by the now well-known relation connecting energy with mass. For the energy of a moving body (velocity \(v\)) we have

\[ E=\frac{mc^2}{\sqrt{1-\frac{v^2}{c^2}}}. \tag{1} \]

For \(v=0\), eq. (1) becomes

\[ E_0=mc^2, \tag{2} \]

where \(m\) is the rest mass of the body, and \(c\) is the speed of light in vacuum—\(3\cdot 10^{10}\) cm/sec.

We do not give the derivation of eq. [1], found in manuals on the principle of relativity, nor do we attempt to obtain eq. [1] by other methods. Such attempts have been made up to the very recent time (for example, Haas’s derivation \((^1)\) in connection with wave mechanics).

Einstein’s fundamental equation unites both conservation laws. The mass of a body is a measure of its energy. A change in energy corresponds to an equivalent change in mass. Already in his first paper Einstein indicated that the possibility of an experimental verification of eqs. [1] and [2] is not excluded for processes accompanied by a large release of energy, for example, in the case of radioactive transformations. The formation of atomic nuclei, which possess tremendous stability

in a large interval of variation of the conditions, is accompanied by a noticeable loss of mass. In fact, the improvement of the method for determining the atomic weights of isotopes made it possible for Aston (²) to discover deviations from whole numbers. In the present case one may speak of a loss, a defect of mass, if it is assumed that nuclei are constructed from protons and electrons.

Equation [1] signifies an extension of our views in another direction as well. In our energy equations we speak only of the change of energy accompanying a process. Therefore, for example, the thermodynamic equalities of the first law contain an integration constant. The character of these uncertainties is revealed by Einstein’s formula.

Furthermore, equation [1] shows that the usual expression for kinetic energy \(\frac{mv^2}{2}\) is valid only for \(v\) small in comparison with \(c\). In fact, the excess of the energy of a moving body over that of a body at rest is equal to

\[ E_{kin}=\frac{mc^2}{\sqrt{1-\frac{v^2}{c^2}}}-mc^2=\frac{mv^2}{2}+\frac{3mv^4}{8c^2}+\ldots \tag{3} \]

Thus \(E_{kin}\) is equal to \(\frac{mv^2}{2}\) only in the first approximation.

It is clear that in the case of ordinary chemical reactions with a heat effect of the order of \(100\ \mathrm{cal}\), there is no need to replace the old equations by Einsteinian ones. Indeed, the liberation of \(100\ \mathrm{cal}\) means a decrease in the mass of the products, as compared with the initial substances, by \(4.6\cdot 10^{-9}\ \mathrm{g}\). Such a correction lies far beyond the limits of experiment. But it is necessary to resort to Einstein’s equation when the processes are accompanied by an enormous liberation of energy. Such a case we have in the radiation of the sun and stars, and also in cosmic penetrating rays of wavelength of the order of \(10^{-12}\ \mathrm{cm}\).

§ 2. Sources of Stellar Energy.

The Sun radiates per second \(3.79\cdot 10^{33}\) erg. This amounts to \(2.9\cdot 10^{33}\) cal per year. Of this energy the Earth absorbs one \(4.5\cdot 10^{-10}\) part. Life and energy transformations on Earth take place at the expense of this energy, so that the question of the source of solar energy is, essentially, the question of the source of terrestrial energy, since the latter is solar radiation accumulated by plants and transformed into other forms.

Usually, under terrestrial conditions, we obtain energy from the combustion of fuel. However, the assumption that solar energy is obtained as a result of similar combustion is completely incorrect. A Sun made of burning coal could cover the above-mentioned expenditure of energy for only \(5000\) years. This completely excludes the possibility of ordinary chemical reactions as sources of stellar energy.

The well-known meteoric theory of Mayer (1848) is also clearly insufficient. If an infinitely distant meteor of mass \(m\), having zero initial velocity, falls onto the Sun (of mass \(M\) and radius \(r\)), then the amount of energy obtained is equal to

\[ g\frac{Mm}{r\cdot 4.19\cdot 10^7}\, cal. \]

Substituting the corresponding values, we obtain \(4.44\cdot 10^7 m\) cal. Since the Sun radiates annually \(2.9\cdot 10^{33}\) cal, it is therefore necessary, according to the meteoric theory, that at least \(6.5\cdot 10^{25}\) g fall upon it. This amounts to \(3.25\cdot 10^{-8}\) of the entire solar mass. Such a possibility is apparently excluded. If one assumes that the meteors fly uniformly from the surrounding space, then a considerable part of them would have to fall upon the Earth. Calculation shows that the Earth, owing to the infalling meteors, would in that case have to receive \(1/213\) of the energy which it receives from the Sun. The increase in the mass of the Earth would have to cause a lengthening of the year by \(2.2\) sec. Moreover, the falling meteors could not in any case prolong the life of the Sun, since the interior of the latter would remain unchanged by this fall. All this compels us to reject

THE TRANSFORMATION OF MATTER INTO RADIANT ENERGY

the theory mentioned, which had adherents until quite recently (³).

Another theory, reducing the source of stellar energy to gravitational forces—the contraction hypothesis of Helmholtz–Kelvin—likewise cannot be maintained at present. The formation of a body of radius $r$ and mass $M$ from rarefied matter entails the release of energy equal to

$$ \frac{a g M^{2}}{r \cdot 4.19 \cdot 10^{7}} \, cal. $$

Here $a$ is a numerical coefficient depending on the distribution of density along the radius. For constant density $a = {}^{3}/_{5}$. If the Sun were formed in this way, an energy of $5.17 \cdot 10^{40}\, cal.$ could be obtained. At the present rate of expenditure this energy would suffice for approximately 18 million years. Further condensation, with a decrease of the radius by $\Delta r$, gives

$$ \frac{a g M^{2} \Delta r}{r^{2}} \, erg. $$

Depending on the nature of the gases and their heat capacities, only a part of this energy can be emitted outward, while the remaining energy must stay in the star itself (see more fully in Emden). But a time of the order of several tens of millions of years is insufficient. Nernst estimates the age of the Sun at about $3 \cdot 10^{9}$ years. According to Jeans, this figure must be reduced to $3 \cdot 10^{8}$ years. Geology operates with a time of the order of $10^{8}$ years. Thus the contraction hypothesis cannot ensure a sufficiently long expenditure of solar energy. The question arises whether solar energy might not appear as the result of radioactive decay. One gram of radium gives, on decay, $10^{9}\, cal.$ To cover the Sun’s radiation, about $3 \cdot 10^{24}\, g$ of radium would have to decay annually. This quantity amounts to $1.5 \cdot 10^{-9}$ of the solar mass. But from the presence of radium on the Sun at present one must conclude that uranium, the progenitor of radium, is present. As is known, $1\,g$ of uranium in equilibrium with its decay products gives $10^{-4}\, cal$ per hour. The Sun, if composed entirely of uranium, would give $1.75 \cdot 10^{33}\, cal.$ per year. This quantity amounts to $0.6$ of the actual radiation. Moreover, it is very probable that formerly the Sun radiated more than it does now. Consequently, this attempt at explanation too must be rejected as insufficient.

Apparently only one way out remains. The source of energy must be considered to be processes in which a noticeable decrease of mass occurs. Each gram of mass, according to the formula \(E=mc^2\), corresponds to \(9\cdot10^{20}\) ergs. To maintain the solar expenditure in a year, a mass of \(1.3\cdot10^{20}\) g must be “radiated.” Under such conditions the sun could disappear entirely in \(1.5\cdot10^{13}\) years. Since the duration of a star’s life is measured by a time of the order of \(10^{11}\) years, and, on the other hand, the masses of stars fluctuate within small intervals compared with the mass of the sun, Einstein’s equation provides a star with a full life at the cost of losing approximately one percent of its mass. With respect to the sun, an approximate calculation shows a loss of \(0.31\%\) over the elapsed time of its existence.

Attempts to detect a decrease of mass are made from observations of binary stars. If the smaller component of a binary star has mass \(m\), and the larger one \(M\), then, taking into account the predominance of the radiation of the larger component, the mass of the latter must decrease faster than the mass of the former, and the ratio \(m/M\) must approach unity as the binary star exists (according to Russell’s diagram). From observations of 85 objects, Vogt \((^4)\) considers this proven. According to Shajn \((^5)\), other factors also play a role here. Investigations of star clusters also provide grounds for the conclusion that mass is lost \((^6)\).

At present the question reduces to a detailed examination and to an attempt to elucidate the mechanism by which the mass defect occurs with the appearance of radiation. Thermodynamics leads to interesting results in the question of the equilibrium of matter and radiant energy.

First of all, what processes with a noticeable mass defect are conceivable?

1. The formation of helium nuclei is the most probable. We imagine atomic nuclei as consisting ultimately of electrons and protons (hydrogen nuclei). The artificial disintegration of a nucleus makes it possible to detect H-particles, i.e., protons of nuclear origin. The well-known stability of the helium nucleus is apparently connected with the enormous energy necessary for its decomposition. Since

if the helium nucleus has an atomic weight equal to 4 and two positive charges, then its formation is possible from 4 protons and 2 electrons. Mass-spectrographic determinations of atomic weights give very precise values (to 4 figures). It is easy to calculate the decrease in mass in the formation of a helium nucleus. The atomic weight of hydrogen according to Aston is 1.00778, of the electron—0.00054, of helium—4.00216. Thus, in the formation of a gram-atom of helium nuclei we have a loss of

\[ 4 \times 1.00778 + 2 \times 0.00054 - 4.00216 = 0.03 \text{ g}. \]

This corresponds to the liberation of \(27 \cdot 10^{18}\) ergs. Dividing this by Avogadro’s number \(6.06 \cdot 10^{23}\), we find the energy in the formation of one helium nucleus, equal to \(4.45 \cdot 10^{-5}\) ergs. If the energy in the elementary process is emitted in the form of a quantum of light \(h\nu\), then, knowing \(h = 6.55 \cdot 10^{-27}\), we find the frequency of the emitted light \(\nu = 68 \cdot 10^{20}\), to which corresponds a wave of length \(\lambda = \frac{c}{\nu} = 4.4 \cdot 10^{-12}\) cm. Such waves, as well as still shorter ones, are found in penetrating cosmic radiation. This may serve as a justification for the fact that the process of helium formation in the universe proceeds according to the indicated scheme with a mass defect.

1. Alongside the energy liberated in the formation of helium nuclei, there may also be a mass defect in the occurrence of nuclei of other elements. According to Aston’s new data\((^2)\), for a number of isotopes we have slight deviations from integral values. When converted into energy, however, this gives a considerable effect. We present the table:

Ya. K. Syrkin

3. It would seem that the formation of helium represents a sufficiently powerful source of stellar energy. However, it is doubtful whether this source can cover the full expenditure of radiation. If at the beginning of its development a star consisted entirely of protons and electrons, then complete transformation into helium would cause a decrease in mass of 0.75%. If one also takes into account the further mass defect in the formation of heavier nuclei, the total change will hardly exceed 1%. Thus the following conclusion is obtained: if a star during the course of its full development loses more than 1% of its mass, then the radiation requires another, more powerful source than the formation of helium. In addition, it is doubtful that a star, even at an early stage of life, consisted only of protons and electrons. Eddington estimates their content at approximately 10%.

Cosmic radiation, coming chiefly from nebulae, i.e. from comparatively cold regions of the universe, shows that the formation of helium apparently takes place already in the preliminary stage of accumulation of stellar masses.

Eddington attempted to cut this Gordian knot by a radical hypothesis. He suggested the possibility of the complete disappearance of mass when a proton is neutralized by an electron. Matter is known to us as an aggregate of positive and negative electric charges. The combination of a proton with an electron in a “merged” form ought to yield some new unknown kind of substance. Eddington’s assumption amounts to saying that such a “merger” is in essence complete self-radiation. The question of the mechanism of this is not yet clear. Jeans and Yus (⁷) represent this elementary process as the collision of two electrons with one proton. In this case one quantum of radiation is formed at the expense of the proton and electron, while the second electron flies away in free form.

4. In order to avoid the solution given above, another way out was proposed, namely, the existence in stars of radioactive bodies of special power, yielding more energy upon decay than the radium known on earth

and others. It cannot be thought that these are terrestrial radioactive bodies merely under different conditions. Even the temperatures inside stars, of the order of \(10^7\)—\(10^8\) degrees, are too low to alter radioactive transformation. The hypothesis of such super-radioactive bodies was put forward by Nernst and Jeans. These, in their opinion, are nuclei of very heavy elements unknown to us (heavier than uranium), formed under stellar conditions and extremely unstable. In this connection Nernst\({}^{(8)}\) admits their emergence from the “zero energy of the ether.” Unfortunately, these postulated radioactive bodies cannot be known to us, since they are formed under conditions unattainable on earth.

In reality the two theories do not contradict one another. The disappearance of the proton and the electron is not conceived by Eddington as the result of a simple collision. If this were so, then the energy arising, corresponding to the number of “neutralized” particles, would be proportional to the number of collisions, which is not in fact the case. It is therefore admitted that the transformation of mass into radiation is a process taking place in the nucleus, as a result, perhaps, of the metastability of the latter. On the other hand, the hypothesis of super-radioactivity is nothing other than the assumption of the formation of unstable nuclei, rapidly disintegrating with a large mass defect. Thus, in principle, the two theories do not contradict one another. But the condition of a large weight of the nucleus is by no means a necessary prerequisite for a powerful release of energy during disintegration. It is possible that, alongside the stable nuclei that are formed, there is the formation of unstable ones, rapidly disintegrating, with an insignificant lifetime. These metastable formations may also consist of a small number of protons and electrons, which makes their appearance more probable. These are isomers of certain isotopes.

Until very recently, all the possibilities cited were merely statements of a general character, so to speak, qualitative, preliminary hypotheses. At the present time, after the work of Stern, Lenz, and others, a more detailed, quantitative approach to the indicated pro-

problems. The thermodynamic and kinetic consideration of the question has led to very interesting results, which will be presented below.

§ 3. EQUILIBRIUM BETWEEN MATTER AND BLACK RADIATION.

The equivalence of mass and energy compels one to think that, if a process of loss of mass with transition into radiation is possible, then there must also take place the reverse process of the transformation of radiation into mass, so to speak, the materialization of radiant energy.

If this phenomenon is reversible, then it is natural to make an attempt to clarify the conditions of thermodynamic equilibrium, which is characterized by a maximum of the entropy of the system. Stern’s work is devoted to the solution of this question (9).

Let us introduce the following notation: \(U\) is the total energy of the space under consideration, \(V\) its volume, \(m\) the mass of an atom, \(u_g\) the energy of an atom, \(N\) the number of atoms in the volume \(V\), \(T\) the temperature, \(S\) the total entropy of the volume, \(s_g\) the mean entropy of one atom, \(u_s\) the energy of \(1\ \mathrm{cm}^3\) of black radiation, \(k\) Boltzmann’s constant, \(s_s\) the entropy of \(1\ \mathrm{cm}^3\) of radiation.

The total energy of the given volume is the sum of the energies of the atoms and of the black radiation.

\[ U = N u_g + V u_s \tag{4} \]

In exactly the same way, the total entropy is composed of the entropy of the masses and the entropy of the radiation

\[ S = N s_g + V s_s \tag{5} \]

The condition of equilibrium consists in a maximum of the entropy \(\delta S = 0\), at constant energy \(\delta U = 0\) and constant volume \(\delta V = 0\). Let us consider the conditions for the formation of one atom. From eqs. [4] and [5] we easily obtain

$$ \delta U=u_g+N\delta u_g+V\delta u_s=0 \tag{6} $$

$$ \delta S=s_g+N\delta s_g+V\delta s_s=0 \tag{7} $$

Thermodynamics gives the known relation

$$ \delta u_s=T\delta s_s \tag{8} $$

Multiplying eq. [7] by \(T\) and subtracting it from [6], we obtain, observing condition [8],

$$ u_g-Ts_g+N(\delta u_g-T\delta s_g)=0 \tag{9} $$

The energy of an atom of mass \(m\) is equal to

$$ U_g=\frac{mc^2}{\sqrt{1-\frac{v^2}{c^2}}}=mc^2+\frac{3}{2}kT \tag{10} $$

For the entropy of an ideal gas, classical thermodynamics gives

$$ s_g=\frac{3}{2}k\ln T+k\ln\frac{V}{N}+S_0 \tag{11} $$

\(S_0\) is an undetermined entropy constant. Quantum statistics removes this indeterminacy and gives the value of the absolute entropy

$$ s_g=k\ln\frac{(2\pi mkT)^{3/2}Ve^{5/2}}{Nh^3} \tag{12} $$

Thus we obtain

$$ \delta u_g=\frac{3}{2}k\,\delta T \tag{13} $$

$$ s\,\delta_g=-\frac{k}{N}+\frac{3k}{2T}\,\delta T \tag{14} $$

Substituting [13] and [14] into eq. [9], we find

$$ u_g-Ts_g+kT=0 \tag{15} $$

Substituting further, in place of \(u_g\) and \(s_g\), their values from equations [10] and [12], we obtain for the number of atoms \(n\) per unit volume in equilibrium with black radiation

\[ n=\frac{N}{V}=\frac{(2\pi m kT)^{3/2}}{h^3}\,e^{-\frac{mc^2}{kT}} \tag{16} \]

In the case of a gas that is not sufficiently rarefied, the dielectric constant of the medium must also be taken into account.

If in equation 16 we substitute for \(m\) the mass of the electron, equal to \(9\cdot 10^{-28}\), and further \(k=1.37\cdot 10^{-16}\); \(h=6.55\cdot 10^{-27}\), then we find that \(n\) is equal to unity at approximately one hundred million degrees. In other words, even at so high a temperature only one electron is in equilibrium with \(1\ \mathrm{cm}^3\) of black radiation. As for protons, whose mass is 1,844 times greater, the results are still less favorable. As Jordan has shown, Stern’s formula (10) [16] is consistent both with Bose–Einstein statistics and with Fermi statistics.

Stern’s conclusion leads to a number of difficulties. If the radiation is in equilibrium with some quantity of electrons, then it must also coexist with a much smaller quantity of protons, i.e. the condition of total electroneutrality is not fulfilled—a condition that appears most probable. Furthermore, thermodynamic equilibrium leads to the requirement of very insignificant masses alongside a large energy of black radiation—whereas in the stars we have precisely the opposite, namely, an enormous accumulation of masses in a small region of the universe. The highest astronomical temperatures, of the order of \(10^8\) degrees, prove too low for the condition of comparable quantities of mass and radiation.

If the process of the decay of masses and their transition into the energy of radiation is required by Stern’s formula, then the reverse process of the formation and accumulation of matter, for the purpose of saving the universe from heat death, becomes impossible. Nernst’s assumption concerning the emergence of heavy radioactive atoms at the expense of the “zero energy of the ether” is a specially devised hypothesis, not connected with other phenomena.

Apparently, Stern’s equation needs correction. Were certain factors omitted in his derivation that would introduce a substantial change? In fact, the picture changes if the reasoning is applied to a closed universe in connection with the general principle of relativity.

§ 4. Matter and Radiation in a Closed Universe.

Is it legitimate to apply the law of increasing entropy to the universe? Clausius, who in 1865 formulated the second principle of thermodynamics in the words: “the entropy of the universe tends toward a maximum,” omitted this passage in the second edition of the mechanical theory of heat (1876). The basis of these difficulties was the assumption of an infinite, unbounded universe, whereas the principle of entropy in the physical sense is applicable to a closed system.

The question appears in another aspect if one takes into account the closedness of the universe—an idea expressed by Einstein (¹¹). According to the general principle of relativity, the geometrical properties of bodies are not independent, but are determined by the distribution of masses. The curvature of space depends on the distribution of matter within it. For a uniform density of matter in the universe \(\rho\), the value obtained for the radius of the latter is

\[ r=\sqrt{\frac{2}{\varkappa\rho}} \tag{17} \]

Here \(\varkappa\) is a universal constant, connected, according to the theory, with the gravitational constant \(g\) and with the speed of light \(c\) by the relation

\[ \varkappa=\frac{8\pi g}{c^{2}}=\frac{8\pi\cdot 6.67\cdot 10^{-8}}{9\cdot 10^{20}}=1.86\cdot 10^{-27} \]

A space of constant positive curvature is possible in the form of Riemann’s spherical space or Newcomb’s elliptical space. The volume of the universe, conceived as a spherical space, is equal to:

\[ V=2\pi^{2}r^{3} \tag{18} \]

(for Newcomb’s elliptical space \(V=\pi^{2}r^{3}\)),

One must not, of course, imagine this volume as the interior of some sphere in three-dimensional space. This volume bounds a hypersphere in a space of 4 dimensions, in the same way as a spherical surface bounds a ball.

From [17] and [18] we obtain

\[ V=\frac{2^{5/2}\pi^2}{\chi^{5/2}\rho^{3/2}} \tag{19} \]

If the total mass of the universe is equal to \(M\), then the mean density will be \(M/V\). This gives, with equation [19],

\[ V=\frac{\chi^3 M^3}{32\pi^4} \tag{20} \]

For an average particle mass equal to \(m\) and a number of particles \(N\), equation [20] takes the form

\[ V=\frac{\chi^3 m^3 N^3}{32\pi^4} \tag{21} \]

If space is closed, then there is no reason not to apply the second principle of thermodynamics to an equilibrium universe. However, Stern’s conclusions must be modified, since there the discussion concerned a process at constant volume \((\delta V=0)\). In reality, however, the volume of the universe depends on the masses. An increase of mass, the appearance of one new particle, entails an increase of the volume of the universe by an appreciable amount, which can readily be calculated from equation [21]:

\[ \delta V=\frac{3N^2\chi^3 m^3\delta N}{32\pi^4}=\frac{3V\delta N}{N} \tag{22} \]

When the mass of the universe is increased by one atom, its volume increases by

\[ \frac{3V}{N} \]

or, since

\[ V=\frac{mN}{\rho} \]

the increase of volume is equal to

\[ \frac{3m}{\rho}. \]

The average density of masses in the world according to de Sitter is about \(10^{-26}\, g/cm^{3}\). According to Hubble it is equal to \(\sim 1.5\cdot 10^{-31}\). Thus, with the appearance of one atom of hydrogen \((m=1.66\cdot 10^{-24})\), the volume of the universe increases by \(3.3\cdot 10^{7}\, cm^{3}\), i.e. by a quantity of about \(33\, m^{3}\). This compels us to consider the process of the appearance and disappearance of matter taking into account the variable volume of the pulsating universe. Stern’s equations thereby undergo, as Lenz \((^{12})\) showed, a substantial change. Since \(\delta V\) is not equal to zero, from equations [4] and [5] we obtain

\[ u_g - Ts_g + N(\delta u_g - T\delta s_g) + \delta V(u_s - Ts_s)=0 \tag{23} \]

Now \(U\), \(V\), and \(N\) denote the total energy, volume, and number of atoms in the world.

Further, \(\delta s_g\) is now equal to (see equation [12])

\[ \delta s_g=-\frac{3\varkappa\delta T}{2T}+\frac{\varkappa\delta V}{V}-\frac{\varkappa}{N} \tag{24} \]

According to the Stefan–Boltzmann law, \(u_s=aT^4\). The entropy of radiation is obtained from the relation

\[ ds_s=\frac{du_s}{T}=4aT^2dT \]

Whence

\[ s_s=\frac{4aT^3}{3}=\frac{4u_s}{3T} \tag{25} \]

Introducing equations [24], [25], and [13] into [23], we obtain

\[ u_g - Ts_g - \frac{kNT\delta V}{V} + kT - \frac{u_s\delta V}{3}=0 \tag{26} \]

If now, instead of \(\delta V\), we substitute its value \(\frac{3V}{N}\), then equation [26] becomes

\[ u_g - Ts_g - 2kT - \frac{u_sV}{N}=0 \tag{27} \]

This expression can be simplified. The first two terms in it are small compared with the last. Indeed, \(u_s\)—the density of black radiation—increases proportionally

of the fourth power of the temperature; further, \(V/N\), as we have seen, is a significant quantity; the constants entering are \(a = 7.64\cdot 10^{-15}\), and \(k = 1.37\cdot 10^{-16}\); in \(s_g\) the constant \(k\) is multiplied by the number standing under the logarithm sign (see eq. [12]). Neglecting the two middle terms, we obtain

\[ Nu_\varphi = Vu_s \tag{28} \]

In this expression, on the right stands the total radiant energy of the universe, and on the left—the total energy of all masses. The result obtained, as is evident, is very simple: in the state of equilibrium the total energy of the masses is equal to the total energy of radiation. Consequently, there exists a uniform distribution of energy into “mass” and radiant energy. Relation [28] makes it possible to calculate the mean temperature of the universe1. Let us rewrite eq. [28] in the form

\[ Mc^2 = VaT^4 \]

or

\[ \rho c^2 = aT^4; \]

whence, for \(\rho = 1.5\cdot 10^{-31}\) and \(a = 7.64\cdot 10^{-15}\), we find \(T = 11.5\) degrees. The difficulty concerning the electroneutrality of the universe is resolved by eq. [28] very simply, for the condition of equilibrium requires that in one place just as many particles of one kind or another be formed as disappear in another. Thus, for a given electroneutrality the latter is not disturbed by the conditions of equilibrium.

A universe in which relation [28] has been attained is a dead world of icy cold, a world of maximum entropy, the realm of heat death. For us, however, the processes in the present universe are of special interest: a universe living far from thermodynamic equilibrium, a universe where the sun shines, the stars shine, and radiant energy is formed at the expense of loss of mass. One of the sources of this energy is the formation of helium nuclei. This phenomenon we can

to interpret it not only by the phenomenological method of thermodynamics, but also in the sense of elucidating its mechanism and the course of the process in time.

§ 5. Thermodynamics of the formation of helium nuclei.

The formation of helium nuclei proceeds according to the equation

\[ 4\overset{+}{\mathrm{H}}+2\mathrm{E}\rightleftarrows \overset{++}{\mathrm{He}} \]

Here the symbol \(\mathrm{E}\) denotes electrons.

To compute the equilibrium state one must know the energy effect of the reaction, as well as the thermal data of all the substances involved. The protons, electrons, and helium nuclei may be treated as ideal gases. Eddington \({}^{(13)}\) showed that this assumption is quite legitimate, taking into account the temperature inside the stars. Corrections were introduced for the action of electrostatic forces (analogous to Debye forces in electrolyte solutions), and the deviations from the gas laws are apparently insignificant.

For the case of the formation of a helium atom from four hydrogens, the equilibrium was calculated by Taylor \({}^{(14)}\). But under conditions of stellar temperatures there can be no question of electrically neutral helium and hydrogen atoms. In the present case complete ionization of the atom takes place. Therefore we shall consider the formation of nuclei. Applying ordinary thermodynamics to the indicated reaction, we say that at a certain temperature \(T\) an equilibrium is established between \(\overset{+}{\mathrm{H}}\), \(\mathrm{E}\), and \(\overset{++}{\mathrm{He}}\), upon reaching which the concentrations of the components cease to change. Let, in the equilibrium state, the concentrations of protons, electrons, and helium nuclei be \(C_{\mathrm{H}}, C_e, C_{\mathrm{He}}\); then we have the relation

\[ \frac{C_{\mathrm{H}}^{4} C_e^{2}}{C_{\mathrm{He}}}=K \tag{29} \]

\(K\) is the equilibrium constant, dependent on the temperature. Let the energy effect of the reaction at constant volume

and temperature \(T\) is equal to \(q_v\). If 4 protons and 2 electrons give a helium nucleus, then the change in entropy in this case is equal to

\[ \Delta S=-\frac{q_v+5kT}{T} \tag{30} \]

In equation [30] there appears \(5kT\), since out of 6 particles one is formed. According to the first principle of thermodynamics, the change of the heat effect with temperature is equal to the difference of the heat capacities \(C_v\) of the initial and final substances. Assuming that \(\mathrm{H}^{+}\), \(\mathrm{E}\), and \(\mathrm{He}^{++}\) behave as ideal gases with \(C_v=\frac{3}{2}k\), we obtain

\[ q_v=q_0+\frac{15}{2}kT \tag{31} \]

where \(q_0\) is the effect at absolute zero. In our case

\[ q_0=4.45\cdot10^{-5}\ \mathrm{erg}. \]

On the other hand, the change in entropy is equal to the entropy of the nucleus that has arisen minus the entropy of the 4 vanished protons and 2 electrons.

\[ \Delta S=S_{\mathrm{He}}-4S_{\mathrm{H}}-2S_e \tag{32} \]

The values of \(S\) are known from equation [12]. The concentrations are quantities inverse to the volumes, i.e.

\[ C_{\mathrm{H}}=\frac{1}{V_{\mathrm{H}}};\quad C_e=\frac{1}{V_e};\quad C_{\mathrm{He}}=\frac{1}{V_{\mathrm{He}}} \tag{33} \]

From equations [29], [30], [31], [32], [33], and [12] we find

\[ K=\frac{C_{\mathrm{H}}^{4}C_e^{2}}{C_{\mathrm{He}}} =\frac{(2\pi kT)^{15/2}m_{\mathrm{H}}^{6}m_e^{3}}{N^{5}h^{15}m_{\mathrm{He}}^{3/2}}e^{-\frac{q_0}{kT}} \tag{34} \]

Here \(m_{\mathrm{He}}\), \(m_{\mathrm{H}}\), and \(m_e\) are the masses of helium nuclei, protons, and electrons, equal respectively to \(6.64\cdot10^{-24}\), \(1.66\cdot10^{-24}\), and \(9\cdot10^{-28}\ \mathrm{g}\). Substituting the remaining values, we find

\[ K=T^{15/2}\,6.4\cdot10^{-29}\,e^{-\frac{3.2\cdot10^{11}}{T}} \tag{35} \]

By this formula we can compute the equilibrium constant at any temperature. As can be seen, even inside stars, at \(T \leq 10^7—10^8\) degrees K, it must be an extremely small quantity. This means that, in the fraction \(\frac{C_{\mathrm H}\cdot C_e^2}{C_{\mathrm{He}}}\), the denominator is colossally large in comparison with the numerator. In other words, even at hundreds of millions of degrees (and these are the maximum temperatures known in nature) helium is practically undecomposed, i.e. it is in equilibrium with an infinitely small quantity of the products of its decay. Formula [35] speaks of the enormous stability of the helium nucleus. This is caused by the large energy effect of the reaction of its formation. Thus, in stars the decomposition of helium does not occur. But we are more interested in another question. Does its formation occur?

§ 6. Kinetics of the formation of helium nuclei.

To explain the sources of stellar energy one has to resort to processes with a noticeable mass defect and with a powerful liberation of energy. But precisely this leads to certain difficulties. In such processes a flexible, mobile equilibrium with measurable quantities of all the constituent parts can occur only at a very high temperature. For the decomposition of helium in a considerable quantity about \(10^{11}\) degrees are needed. Radioactive processes, as is sometimes indicated, come under the influence of temperature at \(10^{11}—10^{12}\) degrees. The formation of a hydrogen atom requires an energy quantum of \(0.0015\) ergs. For this a temperature of about \(75\cdot 10^{11}\) degrees is needed. (15). Meanwhile, inside stars, oscillations occur at about \(10^8\) degrees. Sometimes, in this connection, it is pointed out that stellar temperatures are not capable of changing and influencing subatomic processes. If we pay attention to the way the questions were posed in the last three paragraphs, we shall notice that up to now the discussion has been of an equilibrium state. Meanwhile, in stars the attainment of such equilibrium is still very far off. If a small mass can exist with a considerable quantity of radiation energy, then in a star this has evidently not been attained, and there, consequently, a process with loss of mass is закономерен.

Equilibrium has not yet been reached, and the process is therefore predominantly one-sided. The question is how, in what manner, the phenomenon proceeds in time, tending toward the predicted and thermodynamically required end. The same is true with respect to the formation of helium nuclei. Here, too, equilibrium has not been reached, since in the Sun and in the stars there are quite a large number of protons and electrons. The attainment of equilibrium would mean that in a unit of time as many helium nuclei are formed as decay. The energy effect would be equal to zero. But the Sun and the stars before our eyes radiate energy; more helium nuclei are formed than decay. This, of course, is to be expected, taking into account the very large concentration, compared with the equilibrium one, of protons and electrons. The question is whether the rate of this process is comprehensible to us, whether, according to our ideas, at the indicated temperatures it should proceed more slowly. Here we see that the thermodynamic treatment is insufficient. It would be more correct to regard it as thermostatic, since the conclusions refer to the final equilibrium state. The question, consequently, must be posed differently. Can processes with a noticeable mass defect proceed at stellar temperatures with sufficient speed? The problem lies not in the statics, but in the kinetics of the process. An attempt at such a consideration was made by the author of the present article \((^{16})\). The constant (equation [29]) obtained thermodynamically denotes a definite ratio of concentrations in the equilibrium state. The same constant can also be obtained from kinetic considerations. In the present case we assume that equilibrium occurs when the rates of the forward and reverse reactions are equal. The rate depends on the concentrations, since some fraction of the available number of particles reacts. If at some moment the concentration of helium nuclei is equal to \([\mathrm{He}]\), then the rate of their decay, i.e. the decrease of the concentration with time, is equal to:

\[ V_2=-\frac{d[\mathrm{He}]}{dt}=K_2[\mathrm{He}] \tag{36} \]

\(K_2\) is a certain quantity, the rate constant, which can practically be defined as follows: \(K_2\) indicates what fraction

of the available quantity of helium nuclei decays in one second. The rate of the reverse reaction of formation of helium nuclei is equal to:

\[ V_2 = K_1[\mathrm{H}]^4[\mathrm{E}]^2 \tag{37} \]

where \(K_1\) is, correspondingly, the rate constant for the combination of protons and electrons. After some time the concentrations reach the values \(C_{\mathrm{He}}, C_{\mathrm{H}}\), and \(C_e\), with \(V_1 = V_2\). Then we have:

\[ \frac{C_{\mathrm{H}}^{4}\cdot C_e^{2}}{C_{\mathrm{He}}}=K=\frac{K_2}{K_1}. \tag{38} \]

From the point of view of chemical kinetics, the equilibrium constant is nothing other than the ratio of the rate constants of mutually reverse reactions. We have seen that \(K\) is a small quantity; this means that \(K_2\) is much smaller than \(K_1\), i.e., that the rate constant for the formation of helium nuclei is much greater than the constant for their decay. The question, therefore, is to estimate the value of \(K_1\). According to the theory of chemical kinetics, the rate constants can be represented in the following form:

\[ K_1 = ae^{-\frac{q_1}{kT}} \tag{39} \]

\[ K_2 = be^{-\frac{q_2}{kT}}, \tag{40} \]

where \(a\) and \(b\) are certain quantities connected with the number of collisions and, in general, with the mechanism by which the reaction occurs; \(q_1\) and \(q_2\) denote certain energies, with

\[ q_2-q_1=q_0, \tag{41} \]

The expressions [39] and [40] are based on the assumption that not all particles react, but only those which have passed through an excited active state with a high energy level. In this case \(q_1\) and \(q_2\) are, respectively, the activation energies.

For the rate constant of a monomolecular reaction there exist a number of expressions for the factor \(b\), depending on the theoretical premises. We give here the equation of Dushman\({}^{(17)}\)

\[ K_2=\nu e^{-\frac{h\nu}{kT}}. \tag{42} \]

Here \(b\) is related to \(q_2\) by the relation \(b=\dfrac{q_2}{h}\). For some reactions this expression satisfactorily gives the order of \(K_2\); in other cases the experimental \(b\) is greater than the calculated \(b\). Expression [42] may be obtained, as Lewis showed\({}^{(18)}\), if one assumes that the reaction is caused by activation with the aid of a light quantum. We cannot enter here into a detailed consideration of these questions. Let us merely indicate that they have been discussed in this journal\({}^{(19)}\).

The first and simplest assumption that can be made for determining the \(K_1\) of interest to us is to take \(q_1=0\).

Using equation [42], we obtain \(b=\dfrac{q_2}{h}=\dfrac{q_0}{h}=\)

\(68\cdot10^{20}\); hence

\[ K_1=\frac{1.05\cdot10^{50}}{T^{15/2}}. \]

Such a solution is apparently unsatisfactory. The rate constant in this case would increase rapidly with decreasing temperature. By taking \(q_1=0\), we would in effect be assuming that all protons and electrons are capable of combining, i.e. that no activation energy is required for them and everything is determined by collisions. Therefore one might point out that the equation given for \(K_1\) is not fulfilled at low temperatures, since here one cannot admit the approximation of protons to electrons to distances of subatomic order. The incidence of protons upon electrons at low temperatures would lead to the formation of hydrogen atoms.

In the case of ordinary chemical reactions \(q_1\) and \(q_2\) are quantities of one and the same order. If this is also assumed for the case of the formation of helium nuclei, i.e. if one sets \(q_1\sim q_2\), then the rate constant \(K_1\) proves to be extremely small, from which it would follow that helium nuclei are formed extremely slowly. Meanwhile, this process proceeds at a noticeable rate. Helium nuclei in stars must form, if only because, as Eddington wittily observes, there is no hotter place in nature. If both solutions, \(q_1=0\) and \(q_1\sim q_2\), are inadequate, then a third way out remains. One may assume that \(q_1\) is small in comparison with \(q_2\). This way out means the following. From protons and electrons there are formed—

CONVERSION OF MATTER INTO RADIANT ENERGY

an unstable helium nucleus is first formed. Some of these metastable nuclei again decompose into \(\overset{+}{\mathrm H}\) and \(\mathrm E\), while another part spontaneously, with the release of the corresponding energy, passes into stable helium nuclei. The process may be represented thus:

\[ 4\overset{+}{\mathrm H} + 2\mathrm E \rightleftarrows \overset{++}{\mathrm{He}}\;(\text{unstable}) \longrightarrow \overset{++}{\mathrm{He}}\;(\text{stable}). \]

Soon after Bohr’s first works there were attempts to represent the structure of the helium nucleus. Lenz\(^{(20)}\) proposed for this purpose a model of an inverted atom. If in the latter the electrons revolve around the nucleus, then in the nucleus the protons revolve around the electrons. The electron was taken as the cementing principle in the nucleus. When the quantization condition for the angular momentum of the proton is fulfilled \(\left(mvr = \dfrac{nh}{2\pi}\right)\), owing to its large mass (1844 times greater than that of the electron), radii of the order of \(10^{-12}\) cm are obtained, i.e. close to subatomic dimensions. Lenz’s model—undoubtedly inaccurate—is as follows: around two fixed electrons 4 protons revolve in one orbit in a plane perpendicular to the line joining the electrons and passing through its middle. Such a model is unacceptable in view of its dynamical instability even with respect to X-rays. The total energy of such a nucleus is equal to \(-5.37 \cdot 10^{-8}\) erg, whereas from the mass defect it should be \(-4.45 \cdot 10^{-5}\). If Lenz’s model is plainly insufficient for a stable helium nucleus, then it may be assumed that it, or something similar to it, with respect to the order of magnitude of the energy and dimensions, represents unstable, excited states of the helium nucleus. The kinetic energy of the protons in this metastable nucleus is equal to \(+5.37 \cdot 10^{-8}\) (the total energy with the opposite sign). It may be admitted that free protons having energy of a similar order can form a temporary, unstable, intermediate nucleus of size of the order of \(10^{-12}\) cm (in Lenz’s model the radius of the orbit of the protons is \(5 \cdot 10^{-12}\) cm), from which, spontaneously, with the release of energy, the final stable nucleus is formed. Energy

of the order of \(10^{-8}\) erg, and is precisely intermediate between the magnitude of the atomic energy of hydrogen \(\sim 10^{-11}\) and that of the helium nucleus \(\sim 10^{-5}\). If in equation [40] \(\nu\) does not depend on temperature (the experience of monomolecular reactions shows that the temperature dependence of the rate constant is determined by the exponential term), and \(q_1\) denotes the kinetic energy of the protons of the unstable nucleus, \(5.37 \cdot 10^{-8}\), then from equations [41], [39], [40], [55], and [38] we obtain, for the rate constant of formation of helium nuclei,

\[ K_1=\frac{A}{T^{15/2}} e^{-\frac{3.9\cdot 10^8}{T}}, \tag{43} \]

where \(A\) is a certain quantity depending on the coefficient \(b\) and on what fraction of the metastable nuclei is transformed into stable ones. From eq. [43] it is seen that \(K_1\) at first increases with temperature, then reaches a maximum at \(T = 52\,000\,000\) degrees, and then begins to fall with further increase of \(T\). This result has astronomical significance. Usually inside stars the temperature varies within comparatively small limits.

For the Sun this temperature lies near \(50\,000\,000\) degrees; in some stars it fluctuates around \(10^8\) degrees. If strongly exothermic processes are taken into account, then heating must be expected, which, as it were, may accelerate the reaction. Astronomers have to assume the action of some regulating mechanism that keeps the temperature within a certain interval. Formula [43] shows that above a certain maximum \(T\) the rate constant begins to decrease. Too great a kinetic energy of protons and electrons is an obstacle to the formation of a metastable nucleus. The latter is apparently not formed because of the large magnitude of the centrifugal forces, which prevent protons and electrons from assembling into an unstable, intermediate nucleus. For the formation of the latter, the energy of motion of the free protons must be of the same order of magnitude as the energy of rotation in the temporarily arising nucleus. In this way the quantity \(q_1\) can be interpreted. Further, the higher the temperature, the greater the probability of dis-

the decay of the unstable nucleus into primary protons and electrons. The hypothesis of an intermediate formation with an energy of the order of \(10^{-8}\) erg, corresponding to a grouping over an extent of \(\sim 10^{-12}\) cm, leads to a measurable rate of formation of helium nuclei and, consequently, to a rather powerful radiation of the star over a long period of time.

We have already indicated that there is reason to suppose that the radiation is also the result of other processes with a large mass defect. These are probably processes in other metastable nuclei. The hypothesis of especially radioactive bodies has already been mentioned. It is possible that under the conditions of stellar temperatures and pressures other, weakly stable nuclei are also formed. The kinetic method presented above can also be applied to such intermediate formations, in which the transition to the final stable nucleus is connected with the complete conversion of one proton and one electron into radiation energy.

The impossibility of carrying out the processes cited above under our laboratory conditions, where the attainable temperatures vary by only a few thousand degrees, and the certain unusualness, the absence, so to speak, of a laboratory habit for such phenomena, should not compel us to regard the indicated hypotheses as unnecessary. On this occasion we shall quote Nernst’s words: “Cosmic physics is not ordinary physics. That which in the latter calls forth censure as unprovable speculation may here become a logical necessity, with irresistible force taking its place among investigations.” Indeed, one way or another, processes with a significant mass defect are taking place. They occur near us, in the Sun. Moreover, we live at the expense of the energy released in these processes. Under such conditions, hypotheses, even if not directly connected with the totality of our everyday experiences, have a right to exist, provided only that they do not contradict basic physical views.

An attempt at a kinetic solution of the question requires:

1) the admission of intermediate nuclear formations. This admission is justified by atoms and molecules known in an excited state with a high level of energy.

2) a spontaneous transition to a final, stable state with the liberation of the corresponding energy. And this admission is justified for atoms and molecules returning from an excited state to a stable one.

Kinetics does not contradict thermodynamics. The latter speaks of an equilibrium stable state from the standpoint of the second principle. Kinetics merely indicates the path to this final state. This path is very long and ensures a long life for the sun and the stars. In explaining the present state, kinetics gives a long reprieve before the thermodynamic, thermal death of the universe. Whether a new accumulation of energy occurs in the atoms being formed at the expense of the “zero energy of the ether,” or by some other route, we do not yet know. At present one can make only general statements of a qualitative character concerning the reverse transition (21), while trying to bring to bear here the method of fluctuations. In the light of new achievements of experiment and theory, future science will have to overcome thermal death and explain the eternal circulation of mass and energy in the universe.

Literature

Works of general significance: A. Eddington — The Internal Constitution of the Stars, 1926; Emden — Thermodynamik der Himmelskörper, Encykl. der Mathem. Wiss., 1926.

1) A. Haas, Phys. ZS. 28, 632, 1927.
2) Aston, Proc. of Royal. Soc. A. 115, 487, 1927.
3) M. MacMillan, Astropn. J. 48, 55, 1918; Scientia 17, 3, 103, 1923.
4) H. Vogt, ZS. f. Phys. 26, 139, 1925.
5) G. Shajn, Month. Not. 85, 245, 1925.
6) P. ten Bruggencate, Naturwiss. 13, 261, 1925.
7) G. Jauncey and A. Hughes, Proc. Nat. Acad. Amer. 12, 169, 1926.
8) Herbst, “Uspekhi fizich. nauk”, 3, 151, 1923.
9) O. Stern, ZS. f. phys. Chem. 120, 60, 1926; Zs. f. Elektrochem., 31, 448, 1929.
10) P. Jordan, ZS. f. Phys. 41, 711, 1927.
11) A. Einstein, Sitz. d. Preuss. Akad. 6, 142, 1917.
12) W. Lenz, Phys. ZS. 27, 642, 1926.
13) Eddington, Month. Not. 88, 352, 1928.

14) Taylor, J. Amer. Chem. Soc. 44, 1902, 1922.
15) Jeans, Nature, 121, 463, 1928.
16) Ya. K. Syrkin, Journal of the Russian Physico-Chemical Society, physical part, 1928.
17) S. Dushman, J. Amer. Chem. Soc. 43, 397, 1921; Daniels, Chem. Rev. 5, 39, 1928.
18) Lewis and Smith, J. Amer. Chem. Soc. 47, 1508, 1925.
19) Gimmelvud, “Advances in the Physical Sciences,” 7, 407, 1927; Ya. Syrkin, Communications of Scientific and Technical Works of the Republic, 23, 61, 1927.
20) Lenz, München. Akad. 355, 1913; Harkins and Wilson, Zs. f. anorg. Chem. 95, 1, 1916.
21) I. Ghosh, Naturwiss. 15, 445, 1927.

1. From the given \(\rho\) we find the radius of the universe \(8.4\cdot 10^{28}\) cm and the volume \(1.16\cdot 10^{88}\ \text{cm}^3\); \(M = 1.74\cdot 10^{57}\) g. ↩