The Case of Continuous Particle Density

The consideration given in part 5. leads toward the solution for continuous distributions of the particle density. We divide the interval 0 ≤rro into an infinite number of equal parts dr. We imagine that there is constructed in the center of each partition dr a shell of a two dimensional character of the type discussed in part 4.

The shells may be chosen so that they are equivalent to a continuous distribution of mass. Between any two subsequent shells we shall have a gravitational field of the Schwarzschild type

…

where A, B, and are constants which differ only infinitesimally for two neigh- boring regions. Then the sum total of all these partial solutions constitutes the gravitational field inside the cluster. Our task is to determine A, B, and 7 as functions of r.

We consider two neighboring Schwarzschild solutions which belong to the radius intervals rdr to r+dr and r+dr tor + dr. In the first region the values of A, B, and 7 belong to the value r of the radius, in the second to the valuer + dr. If we use the quantities introduced by (2) then the two local solutions are given by and

where a, b are functions of r in accordance with (17). assume the same values for a and b in the point r +

These 2 solutions are to dr because these quantities must not change when we pass through a shell occupied by particles. It follows, up to quantities of the first order

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or, in accordance with

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where σ is written for 7/2r.

These equations determine A, B as functions of r when 7 or σ is given as func- tion of r. It turns out that a, 8, computed from the solutions A, B of (18), are the solutions of (13), represented with the help of the “parameter” function σ. 7 is arbitrary within certain limits because it is closely connected with the mass distribution.

On the other hand, A, B, and r have to satisfy the condition that (17) makes possible circular particle paths for all values of r, i.e. a and b have to satisfy the inequality (6).

In connection with (17) we obtain the inequality

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(18) and (19)

together completely determine the problem within the cluster; ☛ is arbitrary save for the only restriction that, together with the values of A and B, calculated from (18), it has to satisfy (19).

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For r > ro we have AB = 1, with r = const.. By using (18) we may write (19) thus:

or, with some transformations:

This inequality has to hold within as well as outside the cluster. For infinite values of r, vanishes. Further has to be positive, as negative masses are excluded. Because of the denominator, σ can nowhere be greater than 1. Therefore the numerator of the left hand side has to be positive. As the second factor of the numerator is always negative the first factor has to be negative, too. We therefore obtain

This is a generalization of (6a) as (6a) was only proven to hold for the outside boundary of the cluster. 7 represents the mass enclosed by the spherical surface of the radius r. In order that negative masses should be ruled out it is necessary that everywhere dr (20) dr = 0.

It is further necessary that vanishes for r = 0. Save for this condition may be chosen arbitrarily if only a satisfies (19b). When and therefore is given then the problem of determining the gravitational field of the form (17) is reduced to the carrying out of two integrations, according to (18).

The equations (18) give us the integration of (13) with arbitrary mass density distribution, where the latter is expressed by 7 or σ. (14) gives the correspond- ign particle density n. We shall express n in terms of σ. We have

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together with the relations

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Therefore, when σ is given as a function of r we obtain n by carrying out one integration only. is positive and stays below the limit 2 3. The square root of the denominator of the third term in (21) therefore is always positive. We further

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For rro we have, of course, AB = 1, with r = const.. By using (18) we may write (19) thus:

…

or, with some transformations:

…

This inequality has to hold within as well as outside the cluster. For infinite values of r, vanishes. Further has to be positive, as negative masses are excluded. Because of the denominator, σ can nowhere be greater than 1. Therefore the numerator of the left hand side has to be positive. As the second factor of the numerator is always negative the first factor has to be negative, too. We therefore obtain

This is a generalization of (6a) as (6a) was only proven to hold for the outside boundary of the cluster. 7 represents the mass enclosed by the spherical surface of the radius r. In order that negative masses should be ruled out it is necessary that everywhere dr (20) dr = 0.

It is further necessary that vanishes for r = 0. Save for this condition may be chosen arbitrarily if only a satisfies (19b). When and therefore is given then the problem of determining the gravitational field of the form (17) is reduced to the carrying out of two integrations, according to (18).

The equations (18) give us the integration of (13) with arbitrary mass density distribution, where the latter is expressed by 7 or σ. (14) gives the correspond- ign particle density n. We shall express n in terms of σ. We have

…

together with the relations

…

Therefore, when σ is given as a function of r we obtain n by carrying out one integration only.

…

is positive and stays below the limit 2 3. The square root of the denominator of the third term in (21) therefore is always positive. We further