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I.—ON THE HEAT GIVEN OUT IN THE CONTRACTION OF THE NEBULA.

§ 1. Fundamental Theorems in the Attraction of Gravitation.

The first theorem to be proved is as follows:—

The attraction of a thin homogeneous spherical shell on any point in its interior vanishes.

Take any point P within the sphere. Let this be the vertex of a cone produced both ways, but with a very small vertical angle, so that the small areas S and S´, in which the two parts of the cone cut the sphere, may be regarded as planes. Draw the tangent planes at S and S´. Let the plane of the paper pass through P and be perpendicular to both these tangent planes. Let O P O´ be one of the generators of the cone, and let fall P Q perpendicular to the tangent plane at O, and P Q´ perpendicular to the tangent plane at O´. The volume of the cone with the vertex at P and the base S is ? P Q × S, and the other part of the cone has the volume ? P Q´ × S´.

As the vertical angles of the cones are small, their volumes will, in the limit, be in the ratio of O P³ to O´ P³, and accordingly ? P Q · S ÷ ? P Q´ · S´ = P O³ ÷ O´ P³. But from the figure P Q ÷ P Q´ = P O ÷ P O´, and hence S ÷ O P² = S´ ÷ O´ P².

As the shell is uniform, the masses of the parts cut out by the cones are respectively proportional to S and S´. Hence we see that the attractions of S and S´ on P will neutralise. The same must be true for every such cone through P, and accordingly the total attraction of the shell on a particle inside is zero.

The second fundamental theorem is as follows:—

A thin spherical homogeneous shell produces the same attraction at an external point as if its entire mass were concentrated at the centre of the sphere.

This is another famous theorem due to Newton. He gives a beautiful geometrical proof in Section XII. of the first book of the “Principia.” We shall here take it for granted, and we shall consequently assume that—

The attraction by the law of gravitation of a homogeneous sphere on an external point is the same as if the entire mass of the sphere were concentrated at its centre.

§ 2. On the Energy between Two Attracting Masses.

Let m and m´ be two attracting bodies supposed to be small in comparison with their distance x. Let the force between them be e m m´ ÷ x² when e is the force between two unit masses at unit distance. It is required to find the energy necessary to separate them to infinity, it being supposed that they start from an initial distance a. The energy required is obtained by integrating between the limits infinity and a, and is consequently e m m´ ÷ a.

§ 3. On the Energy Given Out in the Contraction of the Nebula.

We assume that the nebula is contracting symmetrically, so that at any moment it is a homogeneous sphere. We shall consider the shell which lies between the two spheres of radii, r + dr and r respectively.

Let M´ be the mass of the nebula contained within the sphere of radius r, and let dM´ be the mass of the shell just defined. Then it follows from § 1 that the condensation of the shell will have been effected by the attraction of the mass M´ solely. The exterior parts of the nebula can have had no effect, for the outer part has always been in symmetrical spherical shells exterior to dM´, and the attraction of these is zero. We see from § 2 that the contraction of dM´ from infinity, until it forms a shell with radius r, represents a quantity of energy,

for it is obvious that the energy involved in the contraction of the whole shell is the sum of the energies corresponding to its several parts.

If M be the total mass and a the radius of the nebula always supposed homogeneous

and therefore

Hence the work done in the contraction is

Integrating, therefore, the total work of contraction is

At the present moment a mass of 1 lb. at the surface of the sun would weigh 27 lbs. if tested by a spring balance. Hence

With this substitution we find the expression for the foot-pounds of work corresponding to the contraction of the nebula from infinity to a sphere of radius a to be,

Hence we have the following fundamental theorem due to Helmholtz, which is the basis of the theory of sun heat.

If the sun he regarded as a homogeneous sphere of mass M pounds and radius a feet, then the foot-pounds of energy rendered available for sun heat by the contraction of the solar material from, an infinite distance is 16 a M.

§ 4. Evaluation of the Sun Heat Given Out in Contraction.

The number of foot-pounds of work given out in the contraction from infinity is 16 a M. As 772 foot-pounds are equal to one unit of heat, i.e. to the quantity of heat necessary to raise 1 lb. of water 1° Fahrenheit, we see that 772 M is the work required to raise a mass of water equal to the mass of the sun through 1° Fahrenheit. Hence the number of globes of water, each equal to the sun in mass, which would be raised 1° Fahrenheit by the total heat arising from the contraction, is

but a, the radius of the sun in feet, is 2,280,000,000, and hence we have the following theorem:—

The energy liberated in the contraction of the sun from infinity to its present dimensions would, if turned into heat, suffice to raise 47,000,000 globes of water, each having the same mass as the sun, through 1° Fahr.

It is found by experiment that 1 lb. of good coal may develop 14,000 units of heat, and is therefore equivalent to 14,000 × 772 foot-pounds of work. A mass of coal equal to the sun would therefore (granted oxygen enough) be equivalent to 14,000 × 772 × M foot-pounds of work. But we have

Hence we see that

The energy liberated in the contraction of the sun from infinity to its present dimensions, is as great as could be produced by the combustion of 3,400 globes of coal, each as heavy as the sun.

We may speak of 3,400 in this case as the coal equivalent.

§ 5. On the Further Contraction of the Sun and the Heat that may thus be Given Out.

Let us suppose the sun contracts to the radius r, and then, as already proved, § 3, the energy it gives out is

but we have

whence on contraction to the radius r the total energy given out from the commencement is

The average density of the sun at present is 1.4. Let us suppose it condenses until it has a density ?.

whence the energy becomes

but the coal equivalent of 16 a M has been found in § 4 to be 3,400, and hence the coal equivalent in this case is

If we take ? to be the density of platinum (21.5), we get a coal equivalent 8,300. This, therefore, seems to represent a major limit to the quantity of heat which can be obtained from the condensation of the nebula from infinity into a sun of the utmost density.

§ 6. On the Present Emission of Sun Heat.

According to Scheiner, “Strahlung und Temperatur der Sonne, Leipzig, 1899,” the value of the solar constant, i.e. the number of cubic centimetres of water which would be raised 1° Centigrade by the quantity of sun heat which, if there were no atmospheric absorption, would fall perpendicularly on a square centimetre, at the earth’s mean distance from the sun, is between 3.5 and 4.0. If we take the mean value, we have (translated into British units), the following statement:—

If at a point in space, distant from the sun by the earth’s mean distance, one square foot was exposed perpendicularly to the solar rays, then the sun heat that would fall upon it in one minute would raise one pound of water 14° Fahr.

This shows that the solar energy emitted daily amounts to

§ 7. On the Daily Contraction of the Sun Necessary to Supply the Present Expenditure of Heat.

We have seen that at the radius r the energy is

Hence for a change dr it is

At its present size, accordingly, the energy given out by a shrinkage dr is

One cubic foot of the sun averages 87 pounds, so that

We have to equate this to the expression in the last article, and we get

This is the shrinkage of the sun’s radius expressed in feet. Hence the daily reduction of the sun’s diameter is 16 inches.

One coal equivalent possesses energy represented by M × 14,000 × 772. Hence we can calculate that one coal equivalent would supply the solar radiation at its present rate for about 2,800 years.

II.—THE CONSERVATION OF MOMENT OF MOMENTUM.

We give here an elementary investigation of the fundamental dynamical principle which has been of such importance throughout this volume.

§ 8. Case where there are no forces.

Newton’s first law of motion tells us that a particle in motion if unacted upon by force, will move continuously in a straight line without change of velocity.

Let A₀, Fig. 60, be the position of the particle at any moment. Let A₁ be its position after the time t; A₂ be the position at the time 2t; A₃ be the position at the time 3t, and so on.

Then the first law of motion tells us that the distances A₀ A₁, A₁ A₂, A₂ A₃, A₃ A₄, must form parts of the same straight line and must be all equal.

If lines O A₀, O A₁, O A₂, etc., be drawn from any fixed point 0, then the areas of the triangles O A₀ A₁, O A₁ A₂, O A₂ A₃, 0 A₃ A₄, will be all equal. For each area is one-half the product of the base of the triangle into the perpendicular O T from O on A₀ A₁, and, as the bases of all the triangles are equal, it follows that their areas are equal.

Thus we learn that a particle moving without the action of force will describe around any fixed point O equal areas in equal times.

The product of the mass of the particle and its velocity is termed the momentum. If the momentum be multiplied by O T the product is termed the moment of momentum around O. We have in this case the simplest example of the important principle known as the conservation of moment of momentum.

The moment of momentum of a system of particles moving in a plane is defined to be the excess of the sum of the moments of momentum of those particles which tend round O in one direction, over the sum of the moments of momentum of those particles which tend round O in the opposite direction.

If we deem those moments in one direction round O as positive, and those in the other direction as negative, then we may say that the moment of momentum of a system of particles moving in a plane is the algebraical sum of the several moments of momentum of each of the particles.

§ 9. A Geometrical Proposition.

The following theorem in elementary geometry will be required:—

Let A B and A C be adjacent sides of a parallelogram, Fig. 61, of which A D is the diagonal, and let O be any point in its plane. Then the area O A C is the difference of the areas O A D and O A B.

Draw D Q and C P parallel to O A. Then O A D = O A Q, whence O A D – O A B = O B Q = O A P = O A C.

§ 10. Relation Between the Change of Moment of Momentum and the Force Acting on the Particle.

Let A₁ and A₂, Fig. 62, be two adjacent points on the path of the particle, and let A₁ Q and A₂ R be the tangents at those points. Let S Q represent the velocity of the particle at A₁, and SR the velocity of the particle at A₂. Then Q R represents both in magnitude and direction the change in velocity due to the force F, which we suppose constant both in magnitude and direction, while the particle moves from A₁ to A₂ in the small time t; we have also Q R = F t ÷ m.

Complete the parallelogram S Q R U, and let fall O P₁, O P₂, O T perpendiculars from O on S Q, S R, S U respectively. Since S Q is the velocity of the particle when at A₁ the moment of momentum is m O P₁ × S Q; when the particle is at A₂ the moment of momentum is m O P₂ × S R. Whence the difference of the moments of momentum at A₁ and A₂ is m (O P₂ × S R - O P₁ × S Q) = 2 m (O S R - O S Q) = 2 m O S U = m O T × S U = m O T. Q R = F t × O T. But in the limit S coincides with A₁ and A₂, and we see that the gain in moment of momentum is t times the moment of the force around O. Hence we deduce the following fundamental theorem, in which, by the expression acceleration of moment of momentum, we mean the rate at which the moment of momentum increases:—

If a particle under the action of force describes a plane orbit, then the acceleration of the moment of momentum around any point in the plane is equal to the moment of the force around the point.

If the force is constantly directed to a fixed point, then the moment of the force about this point is always zero. Hence the acceleration of the moment of momentum around this point is zero, and the moment of momentum is constant. Thus we have Kepler’s law of the description of equal areas in equal times, and we learn that the velocity is inversely proportional to the perpendicular on the tangent.

§ 11. If Two or More Forces Act on a Point, then the Acceleration of the Moment of Momentum, due to the Resultant of these Forces, is Equal to the Algebraic Sum of the Moments of Momentum due to the Action of the Several Components.

Let A D, Fig. 61, be a force, and A C and A B its two components. Then, since O A D = O A B + O A C, we see that the moment of A D around O is equal to the sum of the moments of its components. Hence we easily infer that if a force be resolved into several components the moment of that force around a point is equal to the algebraical sum of the moments of its several components.

The acceleration of the moment of momentum around O, due to the resultant of a number of forces, is equal to the moment of that resultant around O. But, as we have just shown, this is equal to the sum of the moments of the separate forces, and hence the theorem is proved.

This important theorem is deduced from the fact stated in the third law of motion, that action and reaction are equal and opposite. Let us take any two particles; then, the acceleration of the moment of momentum of one of them, A, by the action of the other, B, will be the moment of the force between them. The acceleration of the moment of momentum of B by the action of A will be the same moment, but with an opposite sign. Hence the total acceleration of the moment of momentum of the system by the mutual action of A and B is zero. In like manner we dispose of every other pair of actions, and thus, as the total acceleration of the moment of momentum is zero, it follows that the moment of momentum of the system itself must be constant.

This fundamental principle is also known as the doctrine of the conservation of areas. It may be stated in the following manner:—

If a system of particles are moving in a plane under the influence of their mutual actions only, the algebraic sum of the areas swept out around a point, each multiplied by the mass of the particle, is directly proportional to the time.

§ 13. If a Particle of Mass m, is Moving in Space under the Action of any Force F, then the Projection of that Particle on any Fixed Plane will Move as if it were a Particle of Mass m Acted upon by that Component of F which is Parallel to the Plane.

This is evident from the consideration that the acceleration of the particle parallel to the plane must be proportional to this component of F.

Let us now suppose a system of particles moving in space under their mutual actions. The projections of these particles on a plane will move as if they were the particles themselves subjected to the action of forces which are the projections of the actual forces on the same plane, and as the reactions between any two particles are equal and opposite, the projections of those reactions on the plane are equal and opposite. Hence the proof already given of the constancy of the moments of momentum of a plane system, will apply equally to prove the constancy of the moments of momentum of the projections of the particles on the plane. Hence we have the following important theorem:—

Let a system of particles be moving in space under the action of forces internal to the system only. Let any plane be taken, and any point in that plane, and let the momentum of each particle be projected into the plane, then the algebraic sum of the moments of these projections around the point is constant.

§ 14. On the Principal Plane of a System.

Let us suppose a system of particles moving under the influence of their mutual actions. Let O be any point, and draw any plane L through O. Then the moment of momentum of the system around the point O and projected into the plane L is constant. Let us call it S. If another plane, L´, had been drawn through O, the similar moment with regard to L´ is S´. Thus for each plane through O there will be a corresponding value of S. We have now to show that one plane can be drawn through O, such that the value of S is greater than it is for any other plane. This is the principal plane of the system.

If v be the velocity of a particle, then in a small time t it moves over the distance v t. If p be the perpendicular from O on the tangent to the motion, then the area of the triangle swept round O in the time t is ½ p v t, and we see that the momentum is proportional to the mass of the particle multiplied into the area swept over in the time t. The quantity S will, therefore, be proportional to the sum of the projections of the areas in L, swept over in the time t, each increased in the proportion of the mass of the particle. It is easily seen that the projection of an area in one plane on another is obtained by multiplying the original area by the cosine of the angle between the two planes. For if the area be divided into thin strips by lines parallel to the line of intersection of the planes, then in the projection of these strips the lengths are unchanged, while the breadths are altered by being multiplied by the cosine of the angle between the two planes. If, therefore, we mark off on the normal to a plane L a length h proportional to any area in that plane, then the projection of this area on any other plane L´ may be measured by the projection of h on the normal to L´.

To determine the moment of momentum resolved in any plane we therefore proceed as follows: Draw a plane through O, and the tangent to the path of one of the particles, and mark off on the normal drawn through O to this plane a length l proportional to the moment of momentum. Repeat the same process for each of the other particles with lengths l´, l, etc., on their several normals. Suppose that l, l´, l represent forces acting at O, and determine their resultant R. Then R, resolved along any other direction, will give the component of moment of momentum in the plane to which that direction is normal. In any plane which passes through R the component of moment of momentum is zero. The plane perpendicular to R contains the maximum projection of moment of momentum. This is the principal plane of the system which we have seen to be of such importance in connection with the nebular theory.

§ 15. Collisions.

The conservation of moment of momentum remains true in a system, even though there may have been actual collisions between the several parts. This is included in the proof already given, for collisions are among the mutual actions referred to. It may, however, be instructive to give a direct proof of a particular case.

Let two particles collide when meeting in the directions A P and B P (Fig. 63) respectively. Whether the particles be elastic or inelastic is quite immaterial, for in both cases the action and reaction must be equal and opposite, and take place along some line P Q. The action on the particle moving along A P will give to it an acceleration of moment of momentum which is equal to the moment of the action around O. The acceleration of the moment of momentum coming along B P will be equal and opposite. Thus the total acceleration of the moment of momentum is zero. Hence the collision has no effect on the total moment of momentum.

§ 16. Friction and Tides.

We have shown that such actions as collisions cannot affect the moment of momentum of the system, neither can it be affected by friction of one body on another. Here, as in the former case, the actions and reactions are equal and opposite, and consequently the accelerations of moment of momentum are zero. Nor is it possible for any tidal action to affect the total moment of momentum of the system. Every such action must be composed of the effects of one particle in the system on another, and as this must invariably produce an equal and opposite reaction the total moment of momentum is unaltered.