_{1}

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In this paper, we investigate orbiting of two stars having equal masses. We consider two models: with a circular orbit and with two elliptical orbits having a common center of a mass located in a common focal point. In the case of the circular orbit, we applied the notion of the instantaneous complex frequency. The paper is illustrated with numerous formulas, derivations and discussion of results.

Three years ago, the author presented a paper describing the gravitational forces as a result of anisotropic energy exchange between baryonic matter and quantum vacuum [

• MKS system of units is applied.

• Ellipse: a―semi-major axis, b = semi-minor axis, ε―eccentricity;

• s ( t ) = α ( t ) + j ω ( t ) ―instantaneous complex frequency;

• G = 6.67384 × 10 − 11 [ m 3 / ( kg ⋅ s 2 ) ] ―gravitational constant;

• (ρ, j)―polar coordinates of the ellipse centered at the focus;

• ρ_{c}―curvature radius;

• P―power;

• E―energy;

• F―force;

• T = orbital period.

_{0}. Both stars are separated by the distance l 0 = 2 ρ 0 . The angular position of the first star is defined by the phasor ψ 1 ( t ) = ρ 0 exp ( j ω 0 t ) ; φ 1 ( t ) = ω 0 t , and the second by ψ 2 ( t ) = ρ 0 exp ( j ω 0 t + π ) ; φ 2 ( t ) = ω 0 t + π _{.}_{ }This system is described by the equality of two forces: the gravitational force of attraction and the centrifugal inertial force.

For the circular orbit of

F g = G m 2 4 ρ 0 2 = F c = m ω 0 2 ρ 0 . (1)

We get the following time-independent relations between the angular velocity ω_{0} and the radius ρ_{0}:

ρ 0 = G m 4 ω 0 2 3 or ω 0 = G m 4 ρ 0 3 . (2)

The orbital tangential velocity is v 0 = ω 0 ρ 0 . Therefore, the kinetic energy of the system is E k = m v 0 2 and the potential energy E p = − 2 ρ 0 | F g | . The potential energy is negative. Its value equals twice the kinetic energy. Therefore, the total energy of the system is negative and time independent. Several authors derived formulae for calculation of the power of gravitational waves emitted by the system of

Valeria Ferrari [

P = 2 5 G 4 c 5 m 5 ρ 0 5 [ W ] . (3)

In the book of Gasperini [

defined as one half of the orbital separation. Deleting this error and insertion

the formula (2), we get again (3). The above presented well known theory does not explain the phenomenon of inspiral. The famous observations by Taylor and Hulse [

We have to show that due to the emission of gravitational waves, the trajectory of the stars is not circular since the instantaneous radius decrease in time and the angular velocity and tangential velocity increase in time. The stars are orbiting along spirals (

ψ 1 ( t ) = ρ 0 exp [ ∫ 0 t s ( t ) d t + j φ ( 0 ) ] ; s ( t ) = α ( t ) + j ω ( t ) (4)

and the second one

ψ 2 ( t ) = ρ 0 exp [ ∫ 0 t s ( t ) d t + j [ φ ( 0 ) + π ] ] , (5)

α (t) is called instantaneous radial frequency and ω(t)―instantaneous angular frequency. The instantaneous radius is ρ ( t ) = ρ 0 exp [ ∫ 0 t α ( t ) d t ] . In the

simplest description of the properties of inspiral, the instantaneous complex frequency is time independent: s ( t ) = − α 0 + j ω 0 . In this case, the Cartesian coordinates of the orbit are:

• For the first star x ( t ) = ρ 0 exp ( − α 0 t ) cos ( ω 0 t ) ; y ( t ) = ρ 0 exp ( − α 0 t ) sin ( ω 0 t ) ;

• For the second x ( t ) = − ρ 0 exp ( − α 0 t ) cos ( ω 0 t ) ; y ( t ) = − ρ 0 exp ( − α 0 t ) sin ( ω 0 t ) .

We have to explain why the stars accelerate by orbiting along the inspiral orbit.

Let us show that the gravitational force F g = G m 2 4 ρ 2 ( t ) and the centrifugal force

F c = m ω 0 2 ρ c differ by magnitude and direction. The orbital distance 2 ρ ( t ) = 2 ρ 0 exp [ − ∫ 0 t α ( t ) d t ] is a line connecting the mass centers through the origin (0, 0) and defines the direction of gravitation force. Differently, the direction of the centrifugal force F_{c} is defined by the curvature radius ρ_{c}. The geometry of the addition of the two forces is presented in

tan ( γ ( t = 0 ) ) = − α 0 ω 0 ,sin ( γ ( t = 0 ) ) = − α 0 / ω 0 1 + ( α 0 / ω 0 ) 2 , cos ( γ ( t = 0 ) ) = 1 1 + ( α 0 / ω 0 ) 2 . (6)

The inspiral orbit is defined by the equation

Gravitationalforce × cos [ γ ( t ) ] = Centrifugalforce . (7)

However, there is a tangential force (see

Tangentialforce = F tan = Gravitationalforce × sin [ γ ( t ) ] . (8)

However, for the quasi-circular orbit, the tangential force is extremely small. This force induces acceleration of the mass m given by

a ( t ) = m F tan . (9)

In consequence, the instantaneous angular frequency is increasing in time

ω ( t ) = ω 0 + ∫ 0 t a ( t ) d t = 2 π T ( t ) (10)

where T(t) is a decreasing instantaneous period. The energies of the system also increase in time. The instantaneous tangential velocity of the stars is

v ( t ) = ω ( t ) ρ c ( t ) . (11)

The curvature radius is (see Appendix 2)

ρ c ( t ) = ρ 0 e − α 0 t ( 1 + ( α 0 / ω 0 ) 2 ) 3 / 2 1 − ( α 0 / ω 0 ) 2 . (12)

The instantaneous kinetic energy of both stars is E k ( t ) = m v 2 ( t ) and the

instantaneous negative potential energy is E p ( t ) = F g ρ ( t ) = − G m 2 4 ρ ( t ) . We start

the investigations with Equation (2). We can define the value of the radius r_{0} or of ω_{0} but not of both. Our choice is the value ω 0 = 2 π / T 0 = 2 .259554 × 10 -4 [ rad / s ] ; T 0 = 2.788720 × 10 4 [ s ] measured by Taylor and Hulse [_{k}) is E p = − G m 2 / ( 2 ρ 0 ) = − 2.692268 × 10 41 [ J ] = − 2 E k . The total energy of the system is negative. The power of the gravitational waves emitted by the system given by Equation (3) is P = 6.523698 × 10 23 [ W ] .

1) Estimation of the value of the radial frequency α_{0}

The decrease of the radius of the circular model in one period T_{0} is

ρ 1 = ρ 0 e − α 0 T 0 = ρ 0 e − 2 π ( α 0 / ω 0 ) . (13)

We have an increase of the negative value of the potential energy

E p 1 = − G m 2 4 ρ 1 ≈ − G m 4 ρ 0 ( 1 + 2 π α 0 / ω 0 ) . (14)

Therefore, we get the increase

∇ E p 1 ≈ − G m 4 ρ 0 ( 2 π α 0 / ω 0 ) . (15)

Assuming arbitrary that this increase should be equal to the energy emitted by gravitational waves during one period we get

∇ E p 1 ≈ − G m 4 ρ 0 ( 2 π α 0 / ω 0 ) = P T 0 = P 2 π / ω 0 . (16)

Therefore,

α 0 ≈ − P E p [ s − 1 ] = − 6.526980 × 10 23 / 2.692268 × 10 41 = − 2.424343 × 10 − 18 . (17)

2) The increase of the angular frequency (or decrease of the period T_{0}) during the inspiral

Taylor and Hulse have measured that the period of the PSR system decreases by 76.5 μs per year [_{0} getting

ω ( t ) = G m 4 ρ 0 4 e − 3 α 0 t = ω 0 e 3 2 α 0 t → T ( t ) = T 0 e − 3 2 α 0 t . (18)

The decrease of the period per year is

∇ T year = T 0 − T ( t year ) = T 0 [ 1 − e − 3 2 α 0 t year ] ≈ − 3 2 T 0 α 0 t year = 3.18022 × 10 − 6 [ s ] . (19)

It is more than one order of magnitude smaller in comparison to 76, 5 μs per year of the PSR system. Therefore, the circular model cannot be applied to describe the properties of the PSR elliptical system.

3) The increase of the negative value of the potential energy

The potential energy at the moment t = 0 is

E p ( 0 ) = − G m 2 2 ρ 0 = − 2.692268 × 10 41 [ J ] . (20)

The value after one year is

E p ( t year ) = − G m 2 2 ρ 0 e − α 0 t year [ J ] . (21)

The increase is

∇ E p = E p ( 0 ) ( e α 0 t year − 1 ) = − 2.052708 × 10 31 [ J ] . (22)

The division by t_{year} yields the power

P = ∇ E p t year = − 6.526917 × 10 23 [ W ] , (21)

i.e., exactly the value defined by Equation (3) which represents the power of the emitted gravitational waves. This result validates the correctness of Equation (17) defining α_{0} and Equation (19) defining the delay per year. The negative sign of this power is applied in the book of Gasperini [

4) The inspiral time

The main goal of this paper is to validate the explanation of the nature of gravity presented in [

process of inspiral. During each period the radius ρ ( t ) = ρ 0 exp ( − ∫ 0 t α ( t ) d t ) is

a bit shorter, each next period is shorter corresponding to an increase of the angular frequency which is a function of time. As well, the instantaneous radial frequency increases with time. The angular frequency is

ω ( t ) = G m 2 4 ρ 3 ( t ) = ω 0 e 3 2 α 0 t ; α ( t ) = α 0 . (22)

We get

T ( t ) = T 0 e − 3 2 α 0 t and ∇ T ( t ) = T 0 − T ( t ) = T 0 [ 1 − e − 3 2 α 0 t ] ≈ − 3 2 T 0 α 0 t . (23)

The instantaneous radius is

ρ ( t ) = ρ 0 exp [ − ∫ 0 t α ( t ) d t ] ≈ ρ 0 e − α 0 t ; α ( t ) ≈ α 0 . (24)

The decrease of the radius during a year is

∇ ρ year = ρ 0 ( 1 − e − α 0 t year ) . (25)

The overestimated number of years of the total inspiral (overestimated since calculated using the constant value α_{0}) is

Numberofyears = ρ 0 ∇ ρ year = 1 1 − e − α 0 t year ≈ ~ 1 α 0 t year = 1.098 × 10 10 [ years ] . (27)

5) Concluding remarks about the circular system

During a single revolution, the emitted power may be classified as time independent since the increase is negligible. The directional pattern σ ( Ω ) [ W / steradian ] is circular symmetric with maximum radiation in the plane of the circle. The total energy of the system during a single revolution has a negligible time dependence. The radiation is emitted in twice the orbital frequency with a circular polarization [

The PSR system differs considerably from the above described circular system. The two stars are orbiting along elliptical orbits (see _{max} = 450 [km/s] to 120 [km/s]. These changes overshadow by several orders the inspiral change due to the emission of gravitational waves.

1) The description of the elliptical orbits in polar coordinates centered at the focus

The elliptical orbit of the first star can be defined in polar coordinates (ρ, φ) centered at the right focus of the left ellipse (

ρ ( φ ) = a 1 − ε 2 1 + ε cos ( φ ) (28)

and for the second one centered in the left focus of the right ellipse is

ρ ( φ ) = a 1 − ε 2 1 − ε cos ( φ + π ) (29)

Note that the angle φ is a function of time. However, our derivations have the form of functions of φ. The inverse function t(φ) has no closed form. Our goals do not require the presentation of these relations.

For φ = 0, we get periastrone separation = 2 a [ 1 − ε ] and for j = π, the apastrone separation = 2 a [ 1 + ε ] . The insertion of the semi-major axis a = 9.7506 × 10 8 [ m ] and eccentricity ε = 0.617733 yields (see [

Periastronseparation = 7 .45466 × 10 8 [ m ] and Apastronseparation = 3 .154477 × 10 8 [ m ] .

2) Why the stars accelerate and decelerate?

In the previous section, we explained why the stars on a circular orbit accelerate. For a circular orbit, this acceleration is extremely small. Differently, in the

case of the two elliptical orbits we have large accelerations and decelerations during each period. Again, the gravitational attraction force is given by

F g ( φ ) = G m 2 4 ρ 2 ( φ ) [ N ] (30)

and the centrifugal force is given by

F c ( t ) = m ω 2 ( φ ) ρ c ( φ ) [ N ] (31)

where ρ_{c} is the curvature radius of the ellipse and the angular velocity is defined by the rotation of the curvature radius. Again, the force vectors have different direction defined by the angle γ. The gravitational force can be represented by a vector sum of two perpendicular terms (

F g ( φ ) cos ( γ ) = G m 2 4 ρ 2 ( φ ) cos ( γ ) [ N ] (32)

has the direction of (30) perpendicular to the tangent of the ellipse and the term

F g ( φ ) sin ( γ ) = G m 2 4 ρ 2 ( φ ) sin ( γ ) [ N ] (33)

represents a tangential force. Equating the terms (31) and (32) yields the following formula for the local angular velocity (local means the function of j).

ω ( φ ) = G m cos ( γ ) 4 ρ ( φ ) ρ c ( φ ) [ rad / s ] . (34)

The local tangential acceleration is

a ( φ ) = F g ( φ ) sin ( γ ) / m = G m 4 ρ 2 ( φ ) sin ( γ ) [ m / s 2 ] . (35)

The local velocity of the stars by orbiting from periastrone to apastron is (deceleration)

v ( φ ) = v max − ∫ 0 π a ( φ ) d φ (36)

and in opposite direction (acceleration)

v ( φ ) = v min + ∫ π 2π a ( φ ) d φ . (37)

The mean velocity (in terms of φ) is the same for both directions

v mean = 1 π ∫ 0 π v ( φ ) d φ . (38)

The local tangential velocity is alternatively defined as

v ( φ ) = ω ( φ ) ρ c ( φ ) (39)

Where ρ_{c} is the local curvature radius (see

Note that in this model, the maxima and minima of the velocity are located near the periastrone and apastrone (not exactly at these locations). The mean value in terms of φ (as in Equation (38)) is 269.782 [km/s]. The time average is

v ¯ ( t ) = circumferenceofellipse timeofasinglerevolution = 181.757 [ km / s ] . (40)

The local average differs from the time average by the factor 1.319. The local angular velocity is shown in

3) Kinetic and potential energies in the PSR system

The local kinetic energy is (

E k ( φ ) = m v 2 ( φ ) [ J ] (41)

and the local negative potential energy is

E P ( φ ) = − G m 2 2 ρ ( φ ) [ J ] . (42)

The corresponding local averages are E ¯ k = 2.38257 × 10 41 [ J ] and E ¯ p = − 4.369225 × 10 41 [ J ] . The ratio is | E p | / E k = 2.02435 ≈ 2 , i.e., almost the value defined by the circular system. It is reasonable to assume that the same ratio is valid for time averages.

4) Concluding remarks about the PSR system

Differently to the circular system the power emitted during a single revolution is a function of time and the radiation pattern σ ( Ω ) [ W / steradian ] is a periodic function of time. Therefore, the reported by Taylor and Hulse power of the emitted gravitational waves P = 7.35 × 10^{24} [W] should be classified as a mean value.

• Both orbits, the circular and the elliptical, are defined by two forces of opposite directions: the centrifugal force and a term of the gravitational force (see

• In the case of a circular orbit, the tangential acceleration is extremely small. We have shown that the notion of the instantaneous complex frequency is a convenient tool to study the process of inspiral of circular systems in the range of a linear gravitational force.

Arguments in favor of the radiation recoil nature of gravity presented in [1 m]

In 2015, the author presented a paper “Gravitational Forces Explained as the Result of Exchange between Baryonic Matter and the Quantum Vacuum” [

• The orbit is defined by two forces: the gravitational force directed towards inside of the orbit and the centrifugal force directed towards the outside of the orbit. The forces are not collinear.

• The centripetal force is perpendicular to the tangent of the orbit.

• The gravitational force except some points is not perpendicular to the tangent and can be decomposed in two terms: the perpendicular compensates the centripetal force. The two forces cancel.

• The tangent force is responsible for acceleration or deceleration of the star.

• In the case of the common nearly circular orbit of the stars, we have an extremely small acceleration. In the case of a double-elliptical orbit, we have large accelerations and deceleration.

• The cancelation of the two forces shows that gravity and inertia have the same physical origin. They are recoil forces of radiation. The radiation pattern should be symmetric w.r.t. the tangent of the orbit. Differently, the pattern is asymmetric w.r.t. the line perpendicular to the orbit resulting in a recoil force of radiation.

In a word, it is logical to assume that all described here forces are recoil forces of radiation. The radiation pattern is symmetric w.r.t. the tangent of the orbit (cancellation of gravitation and inertia) and asymmetric w.r.t. the line perpendicular to the orbit, i.e., the direction of the curvature radius.

The authors declare no conflicts of interest regarding the publication of this paper.

Hahn, S.L. (2018) Gravity in View of the Theory of Orbiting Binary Stars. Journal of Modern Physics, 9, 1954-1969. https://doi.org/10.4236/jmp.2018.910124

This paper is illustrated by the properties of the binary pulsar PSR1913+16, a system of two binary neutron stars discovered and measured during many years by Taylor and Hulse [

Mass of detected pulsar m 1 = 1.441 × solarmass = 2 .8764 × 10 30 [ kg ] and of its companion m 2 = 1.387 × s .m . = 2 .7205 × 10 30 [ kg ] .

•rbital period T 0 = 7.7511939106 [ hr ] = 27807.19557 [ s ] .

Eccentricity of the elliptical orbits ε = 0. 617131 .

Semi-major axis 2 a = 1950100 [ km ] (remark of this author: the name semi-major axis should be replaced by major axis. Semi-major axis is equal not 2a but a. Periastronseparation = 746600 [ km ] , Apastronseparation = 3153600 [ km ]

•rbital velocity of stars relative to the center of mass: at periastrone 450 [km/s], at apastrone: 110 [km/s].

Let us define the in spiral orbit by the equation

ψ ( t ) = ρ 0 e ∫ 0 t s ( t ) d t (A1)

s ( t ) = − ( α + Δ α ∗ t ) + j ( ω 0 + Δ ω ∗ t ) (A2)

We assume that the binary stars have equal mass and rotate synchronously around common center of mass located at x = y = 0. The Cartesian coordinates of the first star are defined by the complex function ψ ( t ) = x ( t ) + j y ( t ) with

x ( t ) = ρ 0 e − ( α t + 0.5 Δ α ∗ t 2 ) cos ( ω 0 t + 0.5 Δ ω ∗ t 2 ) (A3)

y ( t ) = ρ 0 e − ( α t + 0.5 Δ α ∗ t 2 ) sin ( ω 0 t + 0.5 Δ ω ∗ t 2 ) (A4)

and for the second star

x ( t ) = − ρ 0 e − ( α t + 0.5 Δ α ∗ t 2 ) cos ( ω 0 t + 0.5 Δ ω ∗ t 2 ) (A5)

y ( t ) = − ρ 0 e − ( α t + 0.5 Δ α ∗ t 2 ) sin ( ω 0 t + 0.5 Δ ω ∗ t 2 ) (A6)

The curvature radius at the point defined by t = t_{0} is

ρ c = [ ( x ˙ ( t 0 ) 2 ) + ( y ˙ ( t 0 ) 2 ) ] | y ¨ ( t 0 ) x ˙ ( t 0 ) + x ¨ ( t 0 ) y ˙ ( t 0 ) | (A7)

Let us calculate the derivatives beginning with zero values of Da and Dω. We have

x ˙ ( t ) = ρ 0 [ − α 0 e − α 0 t cos ( ω 0 t ) − ω 0 e − α 0 t sin ( ω 0 t ) ] (A8)

y ˙ ( t ) = ρ 0 [ − α 0 e − α 0 t sin ( ω 0 t ) + e − α 0 t ω 0 cos ( ω 0 t ) ] (A9)

x ¨ ( t ) = ρ 0 [ α 0 2 e − α 0 t cos ( ω 0 t ) + α 0 ω 0 e − α t sin ( ω 0 t ) + α 0 ω 0 e − α t sin ( ω 0 t ) − ω 0 2 e − α 0 t cos ( ω 0 t ) ]

y ¨ ( t ) = R 0 [ α 0 2 e − α 0 t sin ( ω 0 t ) − α 0 ω 0 e − α 0 t cos ( ω 0 t ) − α 0 ω 0 e − α 0 t cos ( ω 0 t ) − ω 0 2 e − α t sin ( ω 0 t ) ]

The insertion using t = 0 yields

ρ c ( t ) = ρ 0 e − α 0 t ( α 0 2 + ω 0 2 ) 3 2 | ω 0 ( ω 0 2 − α 0 2 ) | = ρ 0 e − α 0 t ( 1 + ( α 0 / ω 0 ) 2 ) 3 2 1 − ( α 0 / ω 0 ) 2 (A10)

•f course, for a circular orbit a_{0} = 0 and ρ_{c} = ρ_{0}. Otherwise, ρ_{c} > ρ_{0}. The center of the curvature radius is located at

x c = x ( t 0 ) − y ˙ ( t 0 ) ( ( x ˙ ( t 0 ) ) ) 2 + ( y ˙ ( t 0 ) ) 2 y ¨ ( t 0 ) x ˙ ( t 0 ) − y ˙ ( t 0 ) x ¨ ( t 0 ) ; y c = y ( t 0 ) + x ˙ ( t 0 ) ( ( x ˙ ( t 0 ) ) ) 2 + ( y ˙ ( t 0 ) ) 2 y ¨ ( t 0 ) x ˙ ( t 0 ) − y ˙ ( t 0 ) x ¨ ( t 0 ) . (A11)

If t = 0 , x ( 0 ) = ρ 0 , y ( 0 ) = 0 , x ˙ ( 0 ) = − α 0 ρ 0 , y ˙ ( 0 ) = ω 0 ρ 0 , x ¨ ( 0 ) = ρ 0 ( α 0 2 − ω 0 2 ) , y ¨ ( 0 ) = − 2 α 0 ω 0 R 0 .

The insertion yields x c ( 0 ) = 0 , y c ( 0 ) = − α 0 ω 0 ρ 0 . The angle between ρ 0 and ρ c is

tan ( φ ( t = 0 ) ) = − α 0 ω 0 ,sin ( φ ( t = 0 ) ) = α 0 / ω 0 1 + ( α 0 / ω 0 ) 2 , cos ( φ ( t = 0 ) ) = 1 1 + ( α 0 / ω 0 ) 2

The curvature radius of an ellipse in Cartesian coordinates (x, y) is ρ c = a 2 b 2 ( x 2 a 4 + y 2 b 4 ) 3 2 .