the two body problem. almost complete-1.pdf

QUEEN MARY UNIVERSITY OF LONDON

The Two Body Problem MTH717U: MSci Project

Antonis Iaponas

4/28/2010

The Two Body Problem April 28, 2010

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Contents The Two Body Problem ........................................................................................................................... 3

I. Abstract ...................................................................................................................................... 3

II. Introduction ............................................................................................................................... 3

Part 1 ....................................................................................................................................................... 4

III. The orbital equation and periods of the two-body problem ...................................................... 4

IV. Reducing the two-body problem into a one-body problem ....................................................... 4

V. Proving that the system moves in a plane ................................................................................. 6

VI. Proving that angular momentum is conserved .......................................................................... 7

VII. Orbits and the use of Laplace-Runge-Lenz vector ...................................................................... 8

VIII. Proving that Laplace-Runge-Lenz vector is conserved for an inverse square law ..................... 8

IX. Deriving the general equation of the orbit using Laplace-Runge-Lenz vector ......................... 11

X. Relating conserved quantities with Laplace-Runge-Lenz vector .............................................. 13

XI. Conic sections ........................................................................................................................... 14

XII. Period of the orbit .................................................................................................................... 15

XIII. Circle ........................................................................................................................................ 15

XIV. Ellipse ....................................................................................................................................... 16

Part 2 ..................................................................................................................................................... 20

XV. Special relativity in a central force field ................................................................................... 20

XVI. Binet’s equation using classical mechanics .............................................................................. 21

XVII. Deriving first order differential equation of the orbit .............................................................. 22

XVIII. Different approach for deriving Binet’s equation .................................................................... 23

XIX. Relativistic equation of the orbit of a one-body system in an arbitrary central force ............. 23

XX. Deriving the relativistic non-linear first order differential equation ........................................ 23

XXI. Deriving relativistic Binet’s equation ....................................................................................... 26

XXII. Solving classical and relativistic Binet’s equation .................................................................... 27

XXIII. Classical .................................................................................................................................... 27

XXIV. Relativistic ................................................................................................................................ 28

XXV. Conclusion ................................................................................................................................ 30

References ............................................................................................................................................. 31


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The Two Body Problem

Abstract

This project investigates how two masses interact with each other in free space. The first part of the project describes the way we reduce the two-body problem into a one-body problem and explains the way the central force equation gets modified. We subsequently prove that the motion lies on a plane with the help of angular momentum vector. Then we derive the orbits and the dependence of eccentricity on {𝐸𝑡𝑜𝑡𝑎𝑙 } with the aid of another conserve quantity called the Runge-Lenz-vector and we finally obtain the period of the orbits for a circle and ellipse. In the second part of the project we obtain Binet’s differential equation for the orbit. We then illustrate the way we compute the relativistic differential equation for the orbit and by solving the differential equations, we give a description in what ways the orbits differ.

Introduction

Modelling the motion, (for example a solar system with ‘n’ masses) can be portrayed as the n-body problem. Each planet can be represented as a point particle. Hence planetary motion can be given in terms of differential equations. The masses obey an inverse square law (in other words the force between them is proportional to the { 𝑟2}). Furthermore, in the case of the two-body problem it can be considered to be as a system with only two point masses that attract each other and move under their common gravitational force. Thus reducing the two-body problem is fairly easy compared to the n-body problem or even the three- body problem. Sir Isaac Newton solved the two-body problem by considering only the gravitational force between them, so by excluding any other force he gave a detailed solution for the problem and he derived the same laws found by Kepler. While an explicit solution was given for the two-body problem, a solution for systems with additional masses { 𝑛 > 2 } was much more complicated. Even Hilbert placed the solution of only a three-body system into the same category as Fermat’s Last Theorem. However, Sir Isaac Newton believed that an exact solution to the three-body problem was feasible. The complex nature of the three-body problem made Poincare believe in 1890 that a solution was not possible without new mathematics. However, after twenty years an astronomer by the name of Karl Sundman derived a mathematical solution for the problem by the means of uniform convergent infinite series. Even with Sundman’s solution many questions concerning the three-body problem were left unanswered [1]. An answer to a two-body system though can also be given using the theory of Stuckelberg relativistic dynamics, the result is given by means of Lorentz-invariant work function [2]. However in this case we will try to give a complete solution of the two-body problem using Newtonian mechanics.

Figure 1 (A vector representation diagram of the displacements between the mass { 𝑚1 } and { 𝑚2 })

Origin

𝑚2

𝑚1 𝒘

𝒓 1 𝑹 𝒓 2

𝒓


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Part 1

The orbital equation and periods of the two-body problem

Reducing the two-body problem into a one-body problem

𝑟 = 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑏𝑒𝑡𝑤𝑒𝑒𝑛 𝑡𝑕𝑒 𝑡𝑤𝑜 𝑚𝑎𝑠𝑠𝑒𝑠

𝒓 = 𝑑𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡 𝑏𝑒𝑡𝑤𝑒𝑒𝑛 𝑡𝑕𝑒 𝑡𝑤𝑜 𝑚𝑎𝑠𝑠𝑒𝑠

𝒓 1 & 𝒓 2 = 𝑝𝑜𝑠𝑖𝑡𝑖𝑜𝑛 𝑣𝑒𝑐𝑡𝑜𝑟𝑠

𝑚1 & 𝑚2 = 𝑝𝑜𝑖𝑛𝑡 𝑚𝑎𝑠𝑠𝑒𝑠

𝑹 = 𝑝𝑜𝑠𝑖𝑡𝑖𝑜𝑛 𝑣𝑒𝑐𝑡𝑜𝑟(𝑏𝑎𝑟𝑦𝑐𝑒𝑛𝑡𝑒𝑟)

𝑮 = 𝑔𝑟𝑎𝑣𝑖𝑡𝑎𝑡𝑖𝑜𝑛𝑎𝑙 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡

𝒘 = 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛 𝑣𝑒𝑐𝑜𝑟

𝜇 = 𝑟𝑒𝑑𝑢𝑐𝑒𝑑 𝑚𝑎𝑠𝑠

In order to solve the problem, we consider two point particles { 𝑚1} and { 𝑚2} where according to the law of Isaac Newton they exert equal and opposite force onto each other (Newton’s third law of motion). We can also see that the force obeys an inverse square law (meaning that the strength of the force is inversely proportional to the square of the distance) thus the force is also given by this formula

𝑭 = −𝑮𝑚1𝑚2

𝑟3 𝒓 (1)

This is the force exerted by gravitational pull which is Newton’s gravitational law. Hence the forces that particle { 𝑚1} exerts on { 𝑚2} and vice versa are

𝑚1𝒓 1 = −𝑭 (2)

𝑚2𝒓 2 = 𝑭 (3)

The equation (2) and (3) are two coupled second order differential equations where each equation has 3 degrees of freedom. Consequently both of them have 6 degrees of freedom. Therefore, we have a 6 dimensional system where we can select our coordinates. In order to reduce this coupled differential equation we introduce the idea of centre of mass. We introduce the equation of the centre of mass (barycentre) which is given by

𝑚1 + 𝑚2 𝑹 = 𝑚1𝒓 1 + 𝑚2𝒓 2 (4)

When we make { 𝑹 } the subject of the formula the equation (4) (barycentre) between the two point masses become

𝑹 =𝑚1𝒓 1+𝑚2𝒓 2

𝑚1+𝑚2 (5)

By differentiate equation (5) twice in respect to time, we find its second derivative and we show that

𝑹 = (𝑚1𝒓

1+𝑚2𝒓

2

𝑚1+𝑚2) (6)

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Then we substitute equations (2) and (3) into equation (6) resulting in

𝑹 =𝑭 −𝑭

𝑚1+𝑚2= 0 ⇒ 𝑹 = 0 (7)

Thus it implies that

𝑹 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 (8)

The result from equation (8) shows that the centre of mass is moving with a constant velocity so it does not accelerate and no forces are present. According to Newton First law, if we have a mass that is moving with a constant velocity and a second mass that is stationary, then both have the same inertial of frame. Since velocity of the barycentre does not change with time throughout the whole motion of the masses, this can help us to relate equation (2) and (3). Now we continue to try to relate equation (2) and (3) into a new equation which satisfies both. In other words we want to achieve a reduction of the two-body system into one-body system. When we equate equation (3) and (1) yields

𝑚2𝒓 2 = 𝑭 = −𝑮

𝑚1𝑚2

𝑟3 𝒓 (9)

In Figure 1 we observe that the vectors {𝒓 2 & 𝒓 1} is given by

𝒓 2 = 𝑹 + 𝒘 (10)

And

𝒓 1 = 𝑹 − 𝒓 + 𝒘 (11)

Using the equation of the barycentre (4) and after we substitute equation (10) and (11) it becomes

𝑚1 + 𝑚2 𝑹 = 𝑚1𝑹 − 𝑚1𝒓 + 𝑚1𝒘 + 𝑚2𝑹 + 𝑚2𝒘 (12)

This is then simplified to

−𝒘 𝑚1 + 𝑚2 = −𝑚1𝒓 (13)

Where equation (13) can also be written as

𝒘 =𝑚1

𝑚1+𝑚2𝒓 (14)

We continue and substitute equation (14) into equation (10) and differentiate this equation two times with respect to time to find its second derivative

𝒓 2 = 𝑹 +𝑚1

𝑚1+𝑚2𝒓 (15)

However we know from equation (7) that { 𝑹 = 0}. Consequently when we substitute equation (15) into equation (9) it becomes

𝑚2 0 +𝑚1

𝑚1+𝑚2 𝒓 =

𝑚1𝑚2

𝑚1+𝑚2𝒓 = 𝑭 = −𝑮

𝑚1𝑚2

𝑟3 𝒓

𝒓 = −𝑮𝑚1+𝑚2

𝑟3 𝒓 (16)


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Equation (16) shows the acceleration of the one-body problem and how the masses have been

modified. Then we multiply both sides of equation (16) again by { 𝑚1𝑚2

𝑚1+𝑚2 } and we find that it turns

into the equation

𝜇𝒓 = −𝑮𝑚1𝑚2

𝑟𝟑 𝒓 (17)

Such equation corresponds to a system of one-body with a reduced mass {𝜇} which is given by

1

𝜇=

1

𝑚1+

1

𝑚2 (18)

With slight rearrangements we have

𝜇 =𝑚1𝑚2

𝑚1+𝑚2 (19)

As a result by introducing the centre of mass we reduce 3 out of the 6 degrees of freedom. In other words we have reduced our two-body problem into one-body problem.

Proving that the system moves in a plane

𝒓 = 𝑑𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡 𝑟𝑒𝑙𝑎𝑡𝑖𝑣𝑒 𝑡𝑜 𝑡𝑕𝑒 𝑜𝑟𝑖𝑔𝑖𝑛

𝒑 = 𝑙𝑖𝑛𝑒𝑎𝑟 𝑚𝑜𝑚𝑒𝑛𝑡𝑢𝑚

The previous section reveals that our analysis is now restricted to a one-body system with a central force derived from a potential. Now that we have reduced our system we are interested to check if the motion lies in a plane. If we succeed we show that the system has only 2 degrees of freedom and it therefore becomes easier to work with. In order to do this we introduce the idea of angular momentum which is the cross product between displacement and linear momentum and is given by this formula

𝑳 = 𝒓 × 𝒑 (20)

We know that the one-body particle with reduce mass {𝜇} is moving in rotations along a fixed force which we consider it as the origin of the coordinate system. Thus by proving that angular momentum is conserved we restrict the rotations to lie in a plane. By definition the cross product is the operation between two vectors which lie in a plane in a three dimensional Euclidean space and the cross product between them results into the formation of a new vector which is perpendicular to the plane. Therefore if we find that angular momentum is constant, then we are showing that the orbits of the one point particle lie in a plane, perpendicular to the angular momentum vector. By definition of linear momentum we know that

1. 𝒑 = 𝜇𝒓 And when we differentiate once in respect of time we have

2. 𝒑 = 𝜇𝒓 (21)


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Proving that angular momentum is conserved

In order to show that the angular momentum is conserved, we prove that { 𝑳 = 0}. To achieve this we differentiate equation (20) in respect of time once and results

𝑳 = 𝒓 × 𝒑 + 𝒓 × 𝒑

Then we substitute (21) part 1 & 2 accordingly and yields

𝑳 = 𝒓 × 𝜇𝒓 + 𝒓 × 𝜇 𝒓 (22)

After that we substitute equation (17) and we have

𝑳 = 𝒓 × 𝜇𝒓 + 𝒓 × −𝑮𝑚1𝑚2

𝑟𝟑 𝒓 (23)

The vector { 𝑳 } from equation (23) can be divided into two sections (a) and (b) { 𝑳 = a × b}

(a) 𝒓 × 𝜇𝒓

(b) 𝒓 × −𝑮𝑚1𝑚2

𝑟3 𝒓

By definition we know that

𝒂 × 𝒂 = 𝒂𝟐 sin𝜃 𝒏 (Where { 𝜃} is the angle between the vectors { 𝒂 })

From vector geometry we know that the angle between identical vectors is zero thus

𝒂 × 𝒂 = 𝒂𝟐 sin(0) 𝒏 = 0

𝜶 × 𝜶 = 0 (24)

Therefore using (24) we can see that in part (a) and (b) the cross product is between the same vectors consequently both parts are equal to zero thus equation (23) becomes

𝑳 = 𝒓 × 𝜇𝒓 + 𝒓 × −𝑮𝑚1+𝑚2

𝑟3 𝒓 ⇒ 𝑳 = 0 + 0 = 0 ⇒ 𝑳 = 0 (25)

This implies the fact that

𝑳 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 (26)

Hence we have proved that the orbit of the particle lies in a plane perpendicular to the angular momentum vector. This is due to the fact that angular momentum is a vector and when is conserved, it means that remains unchanged (constant) within the plane that is found. So if it lies in different planes logically it will have different direction which implies that our angular momentum is not conserved (i.e. a different vector). So we can safely claim that the rotation of the one-body particle lies in a plane. As a result we have reduced our 3 degrees of freedom of the one-body particle into 2, or putting it differently, we have reduced our 3 dimensional system into 2 dimensions.


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Orbits and the use of Laplace-Runge-Lenz vector

The Laplace-Runge-Lenz Vector is vector not so well-known but it can be shown that it remains conserved for any gravitational force that is inversely proportional to 𝑟2. This is a vector which is very useful in describing the movements of celestial bodies and more generally the motion between two particles that interact with each other. Several great mathematicians have discovered the vector independently. The first mathematician that found that the vector is conserved for inverse square central force was Jakob Herman. Even though the vector after his discovery did not become so famous among the physicists thus it was forgotten. However it was rediscovered, by Pierre Simon de Laplace followed by William Rowan Hamilton. The vector lies in the plane of the orbit with a constant magnitude and has the property of being conserved throughout the whole motion of the particles hence it remains the same everywhere on the orbit [3][4].

Proving that Laplace-Runge-Lenz vector is conserved for an inverse square law

𝑚 = 𝑚𝑎𝑠𝑠

𝑘 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡

𝒓 = 𝑢𝑛𝑖𝑡 𝑣𝑒𝑐𝑡𝑜𝑟

𝒓 = 𝑖𝑠 𝑡𝑕𝑒 𝑑𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡 𝑓𝑟𝑜𝑚 𝑡𝑕𝑒 𝑓𝑜𝑐𝑢𝑠 𝑡𝑜 𝑡𝑕𝑒 𝑝𝑒𝑟𝑖𝑚𝑒𝑡𝑒𝑟 𝑜𝑓 𝑡𝑕𝑒 𝑜𝑟𝑏𝑖𝑡

Now Laplace-Runge-Lenz vector for a single particle is defined as

𝑨 = 𝒑 × 𝑳 − 𝑚𝑘𝒓 (27)

At first we have to show first that vector { 𝑨 } is conserved for a gravitational force that is inversely

proportional to 𝑟2. So in order to do this we have to show that { 𝑨 = 0}. The definition of the force is also given by

𝑭 = 𝒑 (28)

We can always rewrite 𝑭 vectorially as a function of { 𝑓(𝑟) } hence

1. 𝑭 = −𝑮𝑚1𝑚2

𝑟3 𝒓

2. 𝑭 =−𝑘

𝑟2 𝒓

3. 𝑭 = 𝑓(𝑟)𝒓 (29)

Where

𝒓 = 𝒓

𝑟 (30)

And

−𝑘 = −𝑮𝑚1𝑚2 (31)

Also when we equate (28) and part 3 of equation (29) we have

𝑓(𝑟)𝒓 = 𝒑 (32)

Next for convenience we equate part 2 and 3 of equation (29) resulting in

−𝑘

𝑟2 𝒓 = 𝑓(𝑟)𝒓


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−𝑘 = 𝑓 𝑟 𝑟2 (33)

Previously when we have reduced our two-body problem into one-body (single particle) problem we have replaced our masses using a single mass which we have called reduced mass{ 𝜇 }. Consequently because our analysis is restricted to a one-body particle with reduce mass { 𝜇 } the equation (27) is given by

𝑨 = 𝒑 × 𝑳 − 𝜇𝑘𝒓 (34)

Note: From now on when we refer to vector { 𝑨 } we will imply expression (34)

We continue and differentiate equation (34) once in respect of the time and it can be written as

𝑨 = (𝒑 × 𝑳 ). − (𝜇𝑘𝒓 )∙ (35)

Hence using the product rule, equation (35) becomes

𝑨 = 𝒑 × 𝑳 + 𝒑 × 𝑳 − (𝜇𝑘𝒓 )∙ (36)

From equation (25) we have shown that the angular momentum is conserved so { 𝑳 = 0} thus equation (36) can be expressed as

𝑨 = 𝒑 × 𝑳 + 𝒑 × 0 − 𝜇𝑘𝒓 ∙

From vector geometry we know that { 𝒂 × 0 = 0} so

𝑨 = 𝒑 × 𝑳 − 𝜇𝑘𝒓 ∙

Then using equation (32), we can carry out the following substitution into expression (36) and we have

𝑨 = 𝑓(𝑟)𝒓 × 𝑳 −(𝜇𝑘𝒓 )∙

We then replace { 𝑳 } by equation (20) after we have substitute into (20) part 1 of equation (21) and we have

𝑨 = 𝑓(𝑟)𝒓 × 𝒓 × 𝜇 𝒓 −(𝜇𝑘𝒓 )∙ (37)

Afterwards using the Lagrange’s formula

𝒂 × 𝒃 × 𝒄 = 𝒃 𝒂 ∙ 𝒄 − 𝒄 (𝒂 ∙ 𝒃 ) (38)

Equation (37) can be expanded as

𝑨 = 𝒓 𝑓 𝑟 𝒓 ∙ 𝜇𝒓 − 𝜇𝒓 𝑓 𝑟 𝒓 ∙ 𝒓 −(𝜇𝑘𝒓 )∙ (39)

Next we take out 𝜇𝑓 𝑟

𝑟 as a common factor and yields

𝑨 =𝜇𝑓 𝑟

𝑟[𝒓 𝒓 ∙ 𝒓 − 𝒓 (𝒓 ∙ 𝒓 )]−(𝜇𝑘𝒓 )∙ (40)

The equation (40) can be further simplified by noting that

𝒓 ∙ 𝒓 = 𝑟2 (41)


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Then when we differentiate equation (41) once in respect to time and we get

(𝒓 ∙ 𝒓 )∙ = 𝒓 ∙ 𝒓 + 𝒓 ∙ 𝒓 = 2𝒓 ∙ 𝒓

We can also write it differently

(𝒓 ∙ 𝒓 )∙ = 𝑟2 ∙ = 2𝑟𝑟

Therefore we find that this relation holds

2𝒓 ∙ 𝒓 = 2𝑟𝑟

𝒓 ∙ 𝒓 = 𝑟𝑟 (42)

Moreover by differentiating expression (30) once in respect to time we demonstrate that

𝒓 = 𝒓 𝑟 + 𝒓 𝑟 (43)

Then with slight rearrangements

𝒓 =𝒓

𝑟+

𝒓 𝑟

𝑟2 (44)

Using equation (41) and (42) we can re-write equation (40) as

𝑨 =𝜇𝑓 𝑟

𝑟 𝒓 𝑟𝑟 − 𝒓 (𝑟2 ]− 𝜇𝑘𝒓 ∙ (45)

Then we take {−𝑟3} out of the equation as a common factor and we obtain the following result

𝑨 = −𝜇𝑓 𝑟 𝑟2[𝒓

𝑟+

𝒓 𝑟

𝑟2]− 𝜇𝑘𝒓 ∙ (46)

We expand { 𝜇𝑘𝒓 ∙ = 𝜇𝑘𝒓 } and next using equation (33) we can write { 𝜇𝑘𝒓 = 𝜇𝑓 𝑟 𝑟2𝒓 }. Thus we have

𝑨 = −𝜇𝑓 𝑟 𝑟2[𝒓

𝑟+

𝒓 𝑟

𝑟2] + 𝜇𝑓 𝑟 𝑟2𝒓 (47)

Then using equation (44) we can write that

𝑨 = −𝜇𝑓 𝑟 𝑟2𝒓 + 𝜇𝑓 𝑟 𝑟2𝒓 = 0 ⇒ 𝑨 = 0 (48)

Hence this implies that

𝑨 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 (49)

The conservation of the Laplace-Runge-Lenz vector is a rather subtle symmetry and it is related to

the dependence of the force on 𝑟. So we have proved that vector { 𝑨 } is conserved for an inverse square central force.


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Deriving the general equation of the orbit using Laplace-Runge-Lenz vector

We then continue and by means of vector { 𝑨 } we can derive the possible orbits of the one-body system.

Figure 2 (shows that { 𝒑 × 𝑳 } is orthogonal to { 𝑳 } and { 𝒓 × 𝒑 } is orthogonal to { 𝒓 })

The general definition of dot product where { 𝜗 } is the angle between the two vectors { 𝒂 1} and { 𝒂 2} is given by

𝒂 1 ∙ 𝒂 2 = 𝑎1𝑎2 cos 𝜗 (50)

In other words, it is a binary operation between two vectors on the same plane in which they produce a scalar number. Then using equation (29) part 2 and substituting equation (30) we can

write {𝑭 } as

𝑭 = −𝑘

𝑟𝟐

𝒓

𝑟 (51)

By means of Figure 2 it implies the fact that Laplace-Runge-Lenz vector { 𝑨 } is orthogonal to the

angular momentum { 𝑳 } so { 𝑨 ∙ 𝑳 = 0}. Therefore this suggests that { 𝑨 } lies in the plane of the

orbit and is fixed. However the displacement { 𝒓 } changes, thus the angle {𝜃} that {𝒓 } makes with

{𝑨 } also changes, so by using dot product, we can relate the angle with {𝑨 ∙ 𝑳 }.

Figure 3 (shows the angle { 𝜃 } between { 𝑨 } and the vector { 𝒓 })

Hence using definition (50) we can write the following

𝒓

𝑨

𝜃

𝒑

𝒓

𝒓 × 𝒑 𝒑 × 𝑳

𝒑 𝑳


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𝑨 ∙ 𝒓 = 𝐴𝑟 cos 𝜃 = (𝒑 × 𝑳 − 𝜇𝑘𝒓 ) ∙ 𝒓

Where { 𝜃 } is the angle between { 𝑨 } and { 𝒓 } as shown in the Figure 3, furthermore dot product is distributive thus we can write

𝐴𝑟 cos𝜃 = (𝒑 × 𝑳 ) ∙ 𝒓 − 𝜇𝑘 𝒓 ∙ 𝒓 (52)

Then we continue and use the triple scalar product which is given by

𝒂 × 𝒃 ∙ 𝒄 = 𝒄 ∙ 𝒂 × 𝒃 = 𝒃 ∙ 𝒄 × 𝒂 (53)

Therefore using (53) we can reformulate (52) and yield

𝐴𝑟 cos 𝜃 = 𝑳 ∙ 𝒓 × 𝒑 − 𝜇𝑘 𝒓 ∙ 𝒓 (54)

After that using equation (20) we can then write that

𝒓 ∙ 𝒑 × 𝑳 = 𝑳 ∙ 𝒓 × 𝒑 = 𝑳 ∙ 𝑳 = 𝐿2 (55)

Also it can be noted that

𝒓 ∙ 𝒓 =1

𝑟 𝒓 ∙ 𝒓 =

𝑟2

𝑟= 𝑟 (56)

So by substituting (55) and (56) back into equation (54), yields

𝐴𝑟 cos 𝜃 = 𝐿𝟐 − 𝜇𝑘𝑟

𝑟 𝐴 cos 𝜃 + 𝜇𝑘 = 𝐿𝟐

𝑟 =𝐿𝟐

𝐴 cos 𝜃+𝜇𝑘

1

𝑟=

𝐴 cos 𝜃+𝜇𝑘

𝐿𝟐 ⇒ 1

𝑟=

𝜇𝑘

𝐿2 (1 +𝐴

𝜇𝑘cos 𝜃) (57)

Finally we have simplified our equations into an equation which by comparison looks like the general conic equation.

1

𝑟=

1

𝑙(1 + 𝑒 cos 𝜃) (By definition) (58)

So by comparing result (57) and equation (58) we can write the following

1

𝑙=

1

𝑎(1−𝑒2) (By definition)

1

𝑙=

𝐿2

𝜇𝑘

𝐿2

𝜇𝑘=

1

𝑎(1−𝑒2) (59)

Plus eccentricity (conic section parameter) is given by

𝑒 =𝐴

𝜇𝑘 (Eccentricity defines the shape of the orbit) (60)

Thus we have derived an equation that defines the possible orbits of the on-body system.


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Relating conserved quantities with Laplace-Runge-Lenz vector

Then we continue with our original equation of Laplace-Runge-Lenz vector and we use the dot product with the vector itself in order to find how conserved quantities are related with each other, we do this because we know that relations exist among symmetries.

𝑨 ∙ 𝑨 = 𝒑 × 𝑳 − 𝜇𝑘𝒓 ∙ (𝒑 × 𝑳 − 𝜇𝑘𝒓 ) (61)

As a result using the distributive property of dot product we can write equation (61) as

𝑨 ∙ 𝑨 = 𝒑 × 𝑳 2− 2(𝒑 × 𝑳 ) ∙ 𝜇𝑘𝒓 + 𝜇2𝑘2𝒓 2 (62)

We can simplify the above equation further using Lagrange’s identity

𝒂 × 𝒃 2

= 𝒂 ∙ 𝒂 𝒃 ∙ 𝒃 − 𝒂 ∙ 𝒃 (𝒂 ∙ 𝒃 ) (63)

Where it can also be written as

𝐴2 = 𝒑 ∙ 𝒑 𝑳 ∙ 𝑳 − 𝒑 ∙ 𝑳 (𝒑 ∙ 𝑳 ) − 2(𝒑 × 𝑳 ) ∙ 𝜇𝑘𝒓 + 𝜇2𝑘2 (64)

But we know that the vector { 𝒑 ∙ 𝑳 = 0 } because { 𝒑 } is perpendicular to { 𝑳 } as a result equation

(64) can be also stated as

𝐴2 = 𝑝2𝐿2 − 2(𝒑 × 𝑳 ) ∙ 𝜇𝑘𝒓 + 𝜇2𝑘2 (65)

When we dot product { 𝑨 } with itself and do some manipulations we can see that a relation of the total energy with eccentricity can be derived. So by definition total energy is related to potential and kinetic energy in this way

𝐸𝑡𝑜𝑡𝑎𝑙 = 𝐾𝑘𝑖𝑛 + 𝑈𝑝𝑜𝑡 (66)

Where

𝐾𝑘𝑖𝑛 =𝜇𝑟 2

2 (67)

And

𝑈𝑝𝑜𝑡 = −𝑘

𝑟 (68)

Consequently we can write 𝐸𝑡𝑜𝑡𝑎𝑙 as

𝐸𝑡𝑜𝑡𝑎𝑙 =𝜇𝑟 2

2−

𝑘

𝑟 (69)

We also know that kinetic energy is related to linear momentum therefore by definition we have

𝐾𝑘𝑖𝑛 =𝑝2

2𝜇 (70)

Thus substituting (68) and (70) we can write into (60) it becomes

𝐸𝑡𝑜𝑡𝑎𝑙 =𝑝2

2𝜇−

𝑘

𝑟 (71)

We then continue with equation (65) and equation (53), then with slight manipulation we can simplify (65) and write that


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𝐴2 = 𝑝2𝐿2 − 2𝒓 ∙ 𝒑 × 𝑳 𝜇𝑘

𝑟+ 𝜇2𝑘2 ⇒ 𝐴2 = 𝑝2𝐿2 + 2𝐿2𝜇 −

𝑘

𝑟 + 𝜇2𝑘2 (72)

Then we take { 2𝐿2𝜇 } out as a common factor and substituting equation (71) we have

𝐴2 = 2𝐿2𝜇 𝑝2

2𝜇−

𝑘

𝒓 + 𝜇2𝑘2 ⇒ 𝐴2 = 2𝐿2𝜇𝐸𝑡𝑜𝑡𝑎𝑙 + 𝜇2𝑘2 (73)

We can simplify (73) using the eccentricity given by equation (60). Then with slight rearrangements we have

𝐴2

𝜇2𝑘2 − 1 =2𝑙2𝐸𝑡𝑜𝑡𝑎𝑙

𝜇𝑘2 ⇒ 𝑒2 − 1 =2𝑙2𝐸𝑡𝑜𝑡𝑎𝑙

𝜇𝑘2

𝑒 = 1 +2𝐸𝑡𝑜𝑡𝑎𝑙 𝐿2

𝜇𝑘2 (74)

Consequently when we have

1. 𝑒 > 1 𝐸𝑡𝑜𝑡𝑎𝑙 > 0: 𝐻𝑦𝑝𝑒𝑟𝑏𝑜𝑙𝑎 2. 𝑒 = 1 𝐸𝑡𝑜𝑡𝑎𝑙 = 0: 𝑃𝑎𝑟𝑎𝑏𝑜𝑙𝑎 3. 0 < 𝑒 < 1 𝐸𝑡𝑜𝑡𝑎𝑙 < 0: 𝐸𝑙𝑙𝑖𝑝𝑠𝑒

4. 𝑒 = 0 𝐸𝑡𝑜𝑡𝑎𝑙 = −𝜇𝑘2

2𝐿2 ∶ 𝑐𝑖𝑟𝑐𝑙𝑒 (75)

From this equation we can determine the nature of the orbit since it only depends on the parameter {𝑒} and {𝐸𝑡𝑜𝑡𝑎𝑙 }. So our orbits can be bounded or unbounded and each time forming an ellipse, a parabola, a hyperbola or a circle. This depends if the eccentricity is larger smaller or equal with one, or equal with zero which in this case we have a circle [5].

Conic sections

All these orbits are geometrical shapes found in the Euclidean geometry and are grouped under the category of conic sections, in view of the fact that they are all curves intersecting a cone. Conic sections can be categorized into three type’s, ellipse, (which circle is a special case of ellipse), hyperbolas and parabolas. All conic section have foci(s), a semi-major axes denoted by the letter {𝑎} (for the circle the semi-major axes is its radius) as well as a semi-minor axes. All shapes have analogous properties as they share similar parameters of describing them. Therefore using polar coordinates we can produce a common equation for all three of them, this equation is given in (58). So the common parameters are

(I) Eccentricity is denote by the letter {𝑒}. This parameter basically measures how much the conic section shape changes from being a circle.

(II) Semi-latus {𝑙} which is a chord parallel to directrix

(III) Focal parameter {𝜌} which is the distance from foci to the directrix

Focal parameter, eccentricity and semi-latus are related by this formula

𝜌 = 𝑒𝑙 (76)

The directrix is a line parallel to the line of symmetry of each conic section and is used in describing the shapes.


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Figure 4 (shows the 3 types of conic sections, hyperbola, parabola and ellipse circle is a special case of ellipse having the two foci’s at the centre)

Period of the orbit

Circle

The period of the circular orbit can be determined purely using the idea of conservation of energy. Since we know that in this two-body system energy is conserved we can equate { 𝐸𝑡𝑜𝑡𝑎𝑙 } result found for the circle from equation (75) with equation (66).

𝐸𝑡𝑜𝑡𝑎𝑙 = −𝜇𝑘2

2𝑙2 = 𝐾𝑘𝑖𝑛 + 𝑈𝑝𝑜𝑡 (77)

First we are going to derive the period for the circle and then we will derive the period for an ellipse. We do the simple case first and when we find the result for the ellipse we will check how they are related. Previously we have found that the general equation for the orbit is given by equation (57) and that of the eccentricity by equation (60). Therefore substituting the result from equation (75) part 4, that eccentricity of the circle is { 𝑒 = 0} we have

1

𝑟=

𝜇𝑘

𝐿2 (1 +𝐴

𝜇𝑘cos 𝜃)

1

𝑟=

𝜇𝑘

𝐿2 (1 + 0)

1

𝑟=

𝜇𝑘

𝐿2 (78)

Therefore using result from equation (75) part 4, and substituting into (69), becomes

−𝜇𝑘2

2𝐿2 =𝜇𝑟 2

2−

𝑘

𝑟 (79)

In addition we know that the tangential velocity of a body that rotates can be given as

𝒓 = 𝑟𝝎 (𝜔 =2𝜋

𝑇) (80)

Hence by substituting equation (80) into (79) then multiply by 2, we have

Parabola

Hyperbola

Ellipse

Circle


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−𝜇𝑘2

2𝐿2 =𝜇𝑟2𝜔2

2−

𝑘

𝑟 . 2

−𝜇𝑘2

𝐿2 +2𝑘

𝑟= 𝜇𝑟2𝜔2

And then taking { 𝑘} out as a common factor we obtain

𝑘 2

𝑟−

𝜇𝑘

𝐿2 = 𝜇𝑟2𝜔2 (81)

Using equation (78) we can substitute into (81) and find that

𝑘 2

𝑟−

1

𝑟 = 𝜇𝑟2𝜔2 ⇒

𝑘

𝑟= 𝜇𝑟2𝜔2 (82)

Then by substituting equation (31), { 𝜔 =2𝜋

𝑇} and (19) into (82) we get

𝑮 𝑚1𝑚2 =𝑚1𝑚2

𝑚1+𝑚2𝑟3 4𝜋

𝑇2

𝑇 = 4𝜋𝑟3

𝑮(𝑚1+𝑚2) (83)

Ellipse

In the case of an elliptic orbit, we know that again { 𝐸𝑡𝑜𝑡𝑎𝑙 } is conserved, consequently we can

always write { 𝐸𝑡𝑜𝑡𝑎𝑙 =𝜇𝑟 2

2−

𝑘

𝑟 } given by equation (69). Therefore we will try to express conserved

quantities using elliptic parameters, consequently to help us express time in terms of the parameters. So we start and relate { 𝐸𝑡𝑜𝑡𝑎𝑙 } with angular momentum. But before we do that, we first express angular momentum into its polar form given by

Figure 5 (shows the relation of angular momentum with { 𝑑𝜃

𝑑𝑡})

𝐿 = 𝜇𝑟2𝜃 (84)

We can then write equation (84) as

𝜃 =𝐿

𝜇𝑟2 (85)

In addition when we square it we get

𝑟

𝜔


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𝜃 2 =𝐿2

𝜇2𝑟4 (86)

Then we can re-arrange it and we can write

𝜃 2𝑟2 =𝐿2

𝜇2𝑟2 (87)

Thus, we continue using { 𝒓 = 𝜽 𝜃 } (a result that we are going to prove later) and then squaring equation (43) we obtain

𝑟 2 = 𝑟2𝜃 2 + 𝑟 2 (88)

As a result we have derived the speed of the kinetic energy in polar form, this in turn it will help us to relate the angular momentum with { 𝐸𝑡𝑜𝑡𝑎𝑙 } thus we substitute equation (88) into equation (69) and we get the following

𝐸𝑡𝑜𝑡𝑎𝑙 =𝜇𝑟2𝜃 2

2+

𝜇𝑟 2

2−

𝑘

𝑟 (89)

Then substituting equation (87) into (89) we get

𝐸𝑡𝑜𝑡𝑎𝑙 = 𝐿2

2𝜇𝑟2 +𝜇𝑟 2

2−

𝑘

𝑟 (90)

Then making speed (writing it using Leibniz notation) the subject of the formula we can write

𝐸𝑡𝑜𝑡𝑎𝑙 +𝑘

𝑟−

𝐿2

2𝜇𝑟2 2

𝜇=

𝑑𝑟

𝑑𝑡 (91)

So we find velocity in respect of conserved quantities. Then we can integrate {𝑑𝑡} and we find the time {𝑡}

𝑡 = 𝑑𝑟


𝑟 –

𝐿2

2𝜇 𝑟2 2

𝜇

Then at {𝑡 = 0} we let {𝑟} to have the initial value { 𝑟0}, the closest distance to the centre of force (perihelion) and the final value of {𝑟′ } thus the integral becomes

𝑡 = 𝜇

2

𝑑𝑟


𝑟 −

𝐿2

2𝜇 𝑟2

𝑟 ′

𝑟0 (92)

Next by means of equation (59) we can state that

𝐿2

𝜇𝑘=

1

𝑎 1−𝑒2 ⇒ 𝐿2 = 𝜇𝑘𝑎(1 − 𝑒2) (93)

Also using (74) we derive the following result

𝐸𝑡𝑜𝑡𝑎𝑙 = (𝑒2 − 1)𝜇𝑘2

2𝐿2 (94)

We can then substitute equation (93) into (94) and becomes

𝐸𝑡𝑜𝑡𝑎𝑙 = (𝑒2 − 1)𝜇𝑘2

−2𝜇𝑘𝑎 (𝑒2−1)

𝐸𝑡𝑜𝑡𝑎𝑙 = −𝑘

2𝑎 (95)


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Therefore when we substitute (93) and (95) back into our integral given by equation (92) and we get

𝑡 = 𝜇

2

𝑑𝑟

− 𝑘

2𝑎+

𝑘

𝑟 −

𝑘𝑎 (1−𝑒2)

2𝑟2

𝑟 ′

𝑟0 ⇒ 𝑡 =

𝜇

2𝑘

𝑑𝑟

− 1

2𝑎+

1

𝑟 −

𝑎(1−𝑒2)

2𝑟2

𝑟 ′

𝑟0

𝑡 = 𝜇

2𝑘

𝑑𝑟

– 𝑟2

2𝑎+

1

𝑟 −

𝑎 1−𝑒2

2𝑟2

𝑟 ′

𝑟0

𝑡 = 𝜇

2𝑘

𝑟𝑑𝑟

− 𝑟2

2𝑎+𝑟 −

𝑎(1−𝑒2)

2

𝑟 ′

𝑟0 (96)

This is an elliptic integral and in order to solve (96) we use the substitution

𝑟 = 𝑎 1 − 𝑒 cos 𝛿 (97)

The equation (97) is a relation of { 𝑟 } with the semi-major axes { 𝑎 }, eccentricity {𝑒} and the angle { 𝛿 }, it is also called the eccentric anomaly.

Figure 6 (shows how eccentric anomaly is related with the radius { 𝑎 } that makes an angle { 𝛿 } with the horizontal of the auxiliary circle and the distance { 𝑟} from the focus of the ellipse)

Next we differentiate (97) in respect to { 𝛿 } and we get

𝑑𝑟 = 𝑎𝑒 sin𝛿 𝑑𝛿 (98)

Then we substitute equation (97) and (98) into (96) along with the limits of integration (by letting {𝛿 = 0} we find the perihelion {𝑟0}) then replacing {𝑟} in terms of {𝛿} we have

𝛼 𝛿

𝛼

𝑟


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𝑡 = 𝜇

2𝑘

𝑎 1−𝑒 cos 𝛿 𝑎𝑒 sin 𝛿 𝑑𝛿

𝑎 1−𝑒 cos 𝛿 − 𝑎2

2𝑎 1−2𝑒 cos 𝛿+𝑒2 cos 2 𝛿 −

𝑎(1−𝑒2)

2

𝛿 ′

0

𝑡 = 𝜇𝑎4

2𝑘

1−𝑒 cos 𝛿 𝑒 sin 𝛿 𝑑𝛿

𝑎−𝑒𝑎 cos 𝛿− 𝑎

2+𝑎𝑒 cos 𝛿−

𝑎𝑒 2 cos 2 𝛿

2−

𝑎

2+

𝑎𝑒2

2

𝛿 ′

0

Then with slight rearrangements leads to the following result

𝑡 = 𝜇𝑎4

2𝑘

1−𝑒 cos 𝛿 𝑒 sin 𝛿 𝑑𝛿

𝑎𝑒2

2 1−𝑒 cos 2 𝛿

𝛿 ′

0

𝑡 = 𝜇𝑎3

𝑘 1 − 𝑒 cos 𝛿 𝑑𝛿

𝛿 ′

0 (99)

So when we integrate { 𝑑𝛿 } we have

𝑡 = 𝜇𝑎3

𝑘𝛿 − 𝑒

𝜇𝑎3

𝑘sin 𝛿

0

𝛿 ′

(100)

Figure 7 (when the auxiliary angle does one complete revolution of 2𝜋)

Then, when the auxiliary angle makes one complete revolution we have the time taken for one cycle therefore we have the period. Thus if we use the limits from { 0 → 𝛿 ′ = 2𝜋 } we have the time for one complete revolution which we can call period denoted by the letter { 𝑇}. Finally after we substitute the limits, yields

2𝜋


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𝑇 = 2𝜋 𝜇𝑎3

𝑘− 𝑒

𝜇𝑎3

𝑘sin 2𝜋

𝑇 = 2𝜋 𝜇𝑎3

𝑘 (101)

Hence we have obtained the general formula for the period of the ellipse. In addition we see that is a similar result we have found for the circle, thus this must be true since the circle is a general case of the ellipse [5].

Part 2

Special relativity in a central force field

In this section we are going to investigate how a relativistic point particle (moving with high speed) is

moving in a central force field with a { 1

𝑟 } potential. We are going to try to find its second order

deferential equation of the orbit in terms of how { 𝑢 =1

𝑟 } changes with angle { 𝜃 } and how it differs

from a non relativistic particle in a central force field. In special relativity, more specifically when we have high speed particles we know that there is no absolute space and time thus everything is relative to each other, therefore we are expecting to find a modified differential equation for the relativistic orbiting particle.

In 1905 Albert Einstein was the one who proposed the special relativity principles

(i) The speed of light {𝑐} is constant in an empty space and hence is the same for everyone and everything, no matter of their inertial frame of reference.

(ii) The laws of nature are the same in all inertial frames including the homogeneity and isotropy of space.

This led Einstein to discover that time is another dimension of space and is not constant, thus time changes when a body is moving and hence it can be treated as a vector quantity. This phenomenon becomes considerably observable when the body is moving with high speeds. This is special theory of relativity and using the theory we can easily calculate time dilation and length contraction. One of the main reasons that led Einstein to the discovery of the theory was due to the chaos that was created after the derivation of Maxwell equations for electromagnetism where Galilean transformation did not remain invariant. As a result new ideas were brought into the light to try and resolve the problem. Einstein believed that a relativity principle existed in the whole physics therefore Galilean transformations needed modification, this was resolved using Lorentz transformation were Einstein re-derived from special relativity and gave an answer to the question. In addition to that he had expanded the understanding of nature. The equation { 𝐸 = 𝑚0𝑐

2 } relates rest mass and energy with the speed of light, (basically it implies that mass is energy). This in turn implies that bodies that are moving with high speeds are becoming heavier due to the fact that more kinetic energy is needed to make them move faster therefore more mass. So for that reason momentum and kinetic energy need modification in special relativity. In classical mechanics masses that are not moving very fast, are almost equal to the rest mass consequently equal to Newtonian masses but when they are moving near the speed of light masses become relativistic. Putting it more mathematically, letting { 𝑙𝑖𝑚𝑖𝑡 𝑐 → ∞} in relativistic equation, then the equation starts to act as classical one. We can always do this because in special relativity { 𝑐 } is a constant and contributes to the results, where in classical mechanics { 𝑐 } does not contribute to the results therefore we take the limit to infinity in order to omit the modified relativistic parts.


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Binet’s equation using classical mechanics

So we start by equation (17) where it relates the one-body problem with reduced mass and a central field force. Thus using equation (17) we are going to translate it into polar coordinates, therefore the polar coordinates in an inertial frame of reference can be written as

𝒓 = 𝑥, 𝑦 = (𝑟 cos 𝜃 , 𝑟 sin 𝜃)

We can then write the unit vectors in this way, where we can easily obtain using Figure 8

𝒓 = cos𝜃 , sin𝜃

𝜽 = (− sin 𝜃 , cos 𝜃) (102)

Figure 8 (shows the unit vectors of polar coordinates)

Then we differentiate the unit vectors and we obtain

𝒓 = − sin 𝜃 , cos 𝜃 𝑑𝜃

𝑑𝑡 ⇒ 𝒓 = 𝜽 𝜃 (103)

𝜽 = − cos 𝜃 , − sin𝜃 𝑑𝜃

𝑑𝑡 ⇒ 𝜽 = −𝒓 𝜃 (104)

Then we observe that we can relate the unit vectors of polar coordinates with their differential functions. So we continue using equation (103), and after we substitute into equation (43) we have

𝒓 = 𝜽 𝜃 𝑟 + 𝒓 𝑟 (105)

Next we differentiate again to find acceleration and we have

𝒓 = 𝒓 𝑟 + 𝑟 𝒓 + 𝑟 𝜃 𝜽 + 𝑟𝜃 𝜽 + 𝑟𝜃 𝜽 ⇒ 𝒓 = 𝒓 𝑟 + 𝑟 𝜃 𝜽 + 𝑟 𝜃 𝜽 + 𝑟𝜃 𝜽 − 𝑟𝒓 𝜃 2

𝒓 = 𝑟 𝒓 + 2𝑟 𝜃 𝜽 + 𝑟𝜃 𝜽 − 𝑟𝜃 2𝒓 ⇒ 𝒓 = 𝑟 − 𝑟𝜃 2 𝒓 + (2𝑟 𝜃 + 𝑟𝜃 )𝜽 (106)

Then we can substitute the results found in (106) into equation (17) accordingly to get

𝜇 𝑟 − 𝑟𝜃 2 𝒓 + 𝜇(2𝑟 𝜃 + 𝑟𝜃 )𝜽 = −𝑮𝑚1𝑚2

𝑟2 𝒓

After we equate the equations we obtain

𝜇 𝑟 − 𝑟𝜃 2 = −𝑮𝑚1𝑚2

𝑟2 (107)

𝜃 𝑟


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Here we have expressed the magnitude of the force of gravitational pull between two masses in polar coordinates. Next we will try to modify this differential equation by introducing a new variable

𝑢 =1

𝑟 (108)

Equation (108) gives the potential. Then angular momentum of the orbit can be defined using polar coordinates given by equation (84). Thus by using different notation (Leibniz notation) we can write

(85) as { 𝑑𝜃

𝑑𝑡=

𝐿

𝑟2𝜇 }. What we want to do is to reduce our independent variable {𝑡} and introduce

the new variable {𝑢}, we do this in order to find how potential changes when our angle changes. For that reason we differentiate (108) with respect of time and we have

𝑑𝑢

𝑑𝑡= −

1

𝑟2

𝑑𝑟

𝑑𝑡 (109)

Then we write equation (85) using the chain rule 𝑑𝜃

𝑑𝑡∙𝑑𝑡

𝑑𝑢=

𝑑𝜃

𝑑𝑢 and yields

𝑑𝜃

𝑑𝑢

𝑑𝑢

𝑑𝑡=

𝐿

𝑟2𝜇 (110)

Then we substitute (109) into (110) and we have

𝑑𝜃

𝑑𝑢 −

1

𝑟2

𝑑𝑟

𝑑𝑡 =

𝐿

𝑟2𝜇 ⇒

𝑑𝑢

𝑑𝜃= −

𝑟 𝜇

𝐿 (111)

Now we differentiate (111) for a second time in respect of {𝜃} and we find that

𝑑2𝑢

𝑑𝜃2 = −𝜇

𝐿

𝑑

𝑑𝑡 𝑑𝑟

𝑑𝑡

𝑑𝑡

𝑑𝜃 (112)

Then substituting back equation (85) we get

𝑑2𝑢

𝑑𝜃2 = −𝜇

𝐿

𝑟 𝑟2𝜇

𝐿 ⇒

𝑑2𝑢

𝑑𝜃2 = −𝜇2𝑟2𝑟

𝐿2

Plus when we rearrange it becomes

𝑟 = −𝐿2

𝑟2𝜇2

𝑑2𝑢

𝑑𝜃2 (113)

Then we substitute equation (113) and (86) into (107) and we obtain

𝜇 −𝐿2

𝑟2𝜇2

𝑑2𝑢

𝑑𝜃2 −r

r2 𝐿2

𝑟2𝜇2 = −𝑮𝑚1𝑚2

𝑟2 ⇒ −𝐿2

𝜇𝑟2 𝑑2𝑢

𝑑𝜃2 + 𝑢 = −𝑮𝑚1𝑚2

𝑟2

This in turn simplifies, to

𝑑2𝑢

𝑑𝜃2 + 𝑢 = 𝜇𝑮𝑚1𝑚2

𝐿2 (114)

Here we have obtained Binet’s equation, we have demonstrated the derivation of the linear second order non-homogeneous differential equation of the orbit into its polar form. We have reduced the variable { 𝑡} and we introduce { 𝑢} because in this way we get a better picture of what is happening when a force is directed to a single fixed point [6].

Deriving first order differential equation of the orbit

Binet’s equation can also be derived differently. This time we start with equation (69) and when we substitute (88) with slights rearrangements we obtain


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𝐸𝑡𝑜𝑡𝑎𝑙 =𝜇 𝜃 2𝑟2+𝑟 2

2−

𝑘

𝑟 ⇒ 𝐸𝑡𝑜𝑡𝑎𝑙 +

𝑘

𝑟−

𝜃 2𝑟2𝜇

2

2

𝜇= 𝑟 2 (115)

After that, we substitute (85) into (115) and we have


𝑟−

𝐿

𝑟2𝜇

2 𝑟2𝜇

2

2

𝜇= 𝑟 2 ⇒ 𝐸𝑡𝑜𝑡𝑎𝑙 +

𝑘

𝑟−

𝐿2

2𝜇𝑟

2

𝜇= 𝑟 2 (116)

When we square equation (111) and make speed the subject of the formula, yields

𝑟 2 = 𝐿

𝜇

2 𝑑𝑢

𝑑𝜃

2 (117)

Then we substitute equation (117) into equation (116) and we obtain


𝑟−

𝐿2

2𝜇𝑟

2

𝜇=

𝐿

𝜇

2 𝑑𝑢

𝑑𝜃

2

By further simplification we get

2𝐸𝑡𝑜𝑡𝑎𝑙

𝜇+

2𝑘

𝑟𝜇=

𝐿

𝜇

2 𝑑𝑢

𝑑𝜃

2+

𝐿2

𝜇2𝑟 ⇒

2

𝜇 𝐸𝑡𝑜𝑡𝑎𝑙 +

𝑘

𝑟 =

𝐿2

𝜇2 𝑑𝑢

𝑑𝜃

2+ 𝑢2

𝐸𝑡𝑜𝑡𝑎𝑙 + 𝑘𝑢 =𝐿2

2𝜇

𝑑𝑢

𝑑𝜃

2+ 𝑢2 (118)

Here we have found that adding the energies, kinetic and potential yields a first order non linear differential equation in respect of the variables { 𝑢} and { 𝜃}.

Different approach for deriving Binet’s equation

Now by differentiating equation (118) in respect of { 𝑢} leads to the result we have derived previously in equation (114). Hence

2𝜇𝑘

𝐿2 = 2𝑑𝑢

𝑑𝜃

𝑑2𝑢

𝑑𝜃2

𝑑𝜃

𝑑𝑢+ 2𝑢

𝑑2𝑢

𝑑𝜃2 + 𝑢 = 𝜇𝑘

𝐿2 (119)

As a result we see that by differentiating the non linear differential equation (118) in respect of { 𝑢} we can derive Binet’s equation.

Relativistic equation of the orbit of a one-body system in an arbitrary central force

This time we want to find the equation of the orbit when we have a relativistic particle moving in a

{ 1

𝑟 } potential and how this differs from the classical mechanics equation (114) or (119)

Deriving the relativistic non-linear first order differential equation

We start by the definition of linear momentum for a relativistic particle which is given by

𝒑 = 𝛾𝜇𝒓 (120)

This is only the spatial piece of the 4-momentum vector. Where { 𝛾} is a factor given as


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𝛾 =1

1−𝑟 2

𝑐2

(121)

We also want to define the kinetic energy in terms of a relativistic particle. This can be done as

𝐾𝑘𝑖𝑛 = 𝑟 ∙ 𝑑𝒑 (122)

Then we differentiate (120) in respect of time and becomes

𝑑𝒑

𝑑𝑡= 𝜇𝛾𝒓 + 𝜇𝛾 𝒓

𝑑𝒑 = 𝜇𝛾𝒓 𝑑𝑡 + 𝜇𝛾 𝒓 𝑑𝑡 (123)

Then we substitute into equation (122) and we obtain

𝐾𝑘𝑖𝑛 = 𝜇𝛾𝑑𝒓

𝑑𝑡𝑟 𝑑𝑡 + 𝜇𝑟 2

𝑑𝛾

𝑑𝑡𝑑𝑡 ⇒ 𝐾𝑘𝑖𝑛 = 𝜇 𝛾𝑟 𝑑𝒓 + 𝜇 𝑟 2𝑑𝛾 (124)

Then using equation (121) we write

1

𝛾2 =1−𝑟 2

𝑐2 (125)

Plus

𝑟 2 = 𝑐2(1 −1

𝛾2) (126)

Then we differentiate (125) in respect of velocity and we have

−2𝛾−3 𝑑𝛾

𝑑𝒓 = −2

𝑟

𝑐2 ⇒ 𝑑𝒓 =𝑐2

𝛾3𝑟 (127)

Then we substitute equation (126) and (127) into (124) and we obtain

𝐾𝑘𝑖𝑛 = 𝜇𝑐2 1

𝛾2 𝑑𝛾 + 𝜇𝑐2 𝑑𝛾 − 𝜇𝑐2 1

𝛾2 𝑑𝛾

𝐾𝑘𝑖𝑛 = 𝛾𝜇𝑐2 + 𝐶 (128)

Here 𝐶 is the constant of integration. Then in order to find the value of { 𝐶 } we let speed { 𝑟 = 0}, then { 𝐾𝑘𝑖𝑛 = 0 } because is not moving as a result { 𝛾 = 1 } as a result { 𝐶 = −𝜇𝑐2}. Consequently we have define the relativistic kinetic energy and is given by the following equation

𝐾𝑘𝑖𝑛 = 𝜇𝑐2 𝛾 − 1 (129)

We have identified that kinetic energy of a high speed particle is given by equation (129). Then we choose and let the particle moving in a central force field, described by the potential energy in equation (68), in this way we keep the symmetries { 𝐸𝑡𝑜𝑡𝑎𝑙 } and {𝐿} in our system. Then { 𝑈𝑝𝑜𝑡 } is

given by the equation (68) and { 𝐸𝑡𝑜𝑡𝑎𝑙 } by equation (66), we also know that angular momentum of a relativistic particle in polar form can be given by

𝐿 = 𝛾𝜇𝑟2𝜃 (130)

In this equation the only difference from classical one (84) is the { 𝛾 } factor. Now using equation (121) we substitute equation (88) and we find that { 𝛾 } in its polar form is given by


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𝛾 =1

1−𝜃 2𝑟2+𝑟 2

𝑐2

(131)

Then using Leibniz notation (130) can be written as

𝑑𝜃

𝑑𝑡=

𝐿

𝜇𝛾 𝑟2 (132)

We then introduce the variable given by (108) and using equation (109) and chain rule

{ 𝑑𝜃

𝑑𝑡∙𝑑𝑡

𝑑𝑢=

𝑑𝜃

𝑑𝑢 } we have

𝐿

𝜇𝛾 𝑟2 ∙−𝑟2

𝑟 =

𝐿

𝜇𝛾 𝑟 =

𝑑𝜃

𝑑𝑢

So

𝑟 =𝑑𝑢

𝑑𝜃

𝐿

𝜇𝛾 (133)

Now using (131) we can rearrange it and write that

1

𝛾2 = 1 −𝑟 2

𝑐2 −𝑟2𝜃 2

𝑐2 (134)

Then we substitute (133) in (134) results in

1

𝛾2 = 1 − 𝑑𝑢

𝑑𝜃

2 𝐿2

𝑐2𝜇2𝛾2 − 𝑟2 𝐿2

𝜇2𝛾2𝑟4

Then taking {−𝐿2

𝑐2𝜇2𝛾2 } as a common factor we have

1

𝛾2 = 1 −𝐿2

𝑐2𝜇2𝛾2 𝑑𝑢

𝑑𝜃

2+ 𝑢2

1 = 𝛾2 −𝐿2

𝑐2𝜇2 𝑑𝑢

𝑑𝜃

2+ 𝑢2

𝛾2 = 1 +𝐿2

𝑐2𝜇2 𝑑𝑢

𝑑𝜃

2+ 𝑢2 (135)

We have found factor { 𝛾2 } in respect of a non-linear first order differential equation. By means of equation (66) and (129) we can write that

𝐸𝑡𝑜𝑡𝑎𝑙 = 𝜇𝑐2 𝛾 − 1 + 𝑈𝑝𝑜𝑡 ⇒ 𝐸𝑡𝑜𝑡𝑎𝑙 = 𝜇𝑐2𝛾 − 𝜇𝑐2 + 𝑈𝑝𝑜𝑡

𝐸𝑡𝑜𝑡𝑎𝑙 −𝑈𝑝𝑜𝑡

𝜇𝑐2 + 1 = 𝛾 (136)

Now we square 𝛾 and we find { 𝛾2} in respect of energies thus we have

𝛾2 = 1 +2 𝐸𝑡𝑜 𝑡𝑎𝑙 −𝑈𝑝𝑜𝑡

𝜇𝑐2 + 𝐸𝑡𝑜𝑡𝑎𝑙 −𝑈𝑝𝑜𝑡

2

𝜇𝑐2 (137)

Now we equate (135) with (137) and we obtain

1 +2 𝐸𝑡𝑜𝑡𝑎𝑙 −𝑈𝑝𝑜𝑡


2

𝜇𝑐2 = 1 +𝐿2

𝑐2𝜇2 𝑑𝑢

𝑑𝜃

2+ 𝑢2


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2 𝐸𝑡𝑜𝑡𝑎𝑙 −𝑈𝑝𝑜𝑡


2

𝜇𝑐2 =𝐿2

𝑐2𝜇2 𝑑𝑢

𝑑𝜃

2+ 𝑢2

2 𝐸𝑡𝑜𝑡𝑎𝑙 −𝑈𝑝𝑜𝑡

𝜇𝑐2 +𝐸𝑡𝑜𝑡𝑎𝑙

2

𝜇2𝑐4 −2𝐸𝑡𝑜𝑡𝑎𝑙 𝑈𝑝𝑜𝑡

𝜇2𝑐4 +𝑈𝑝𝑜𝑡

2

𝜇2𝑐4 =𝐿2

𝑐2𝜇2 𝑑𝑢

𝑑𝜃

2+ 𝑢2

2𝐸𝑡𝑜𝑡𝑎𝑙

𝜇𝑐2 −2𝑈𝑝𝑜𝑡

𝜇𝑐2 +𝐸𝑡𝑜𝑡𝑎𝑙

2

𝜇2𝑐4 −2𝐸𝑡𝑜𝑡𝑎𝑙 𝑈𝑝𝑜𝑡

𝜇2𝑐4 +𝑈𝑝𝑜𝑡

2

𝜇2𝑐4 =𝐿2

𝑐2𝜇2 𝑑𝑢

𝑑𝜃

2+ 𝑢2

Then simplifying further we get

2

𝑐2𝜇 𝐸𝑡𝑜𝑡𝑎𝑙 − 𝑈𝑝𝑜𝑡 +

𝐸𝑡𝑜𝑡𝑎𝑙2

2𝜇𝑐2 −2𝐸𝑡𝑜𝑡𝑎𝑙 𝑈𝑝𝑜𝑡

𝜇𝑐2 +𝑈𝑝𝑜𝑡

2

2𝜇𝑐2 =𝐿2

𝑐2𝜇2 𝑑𝑢

𝑑𝜃

2+ 𝑢2

𝐸𝑡𝑜𝑡𝑎𝑙 1 +𝐸𝑡𝑜𝑡𝑎𝑙

2𝜇𝑐2 − 𝑈𝑝𝑜𝑡 1 +𝐸𝑡𝑜𝑡𝑎𝑙

𝜇𝑐2 +𝑈𝑝𝑜𝑡

2

2𝜇𝑐2 =𝐿2

2𝜇

𝑑𝑢

𝑑𝜃

2+ 𝑢2

𝐸𝑡𝑜𝑡𝑎𝑙 1 +𝐸𝑡𝑜𝑡𝑎𝑙

2𝜇𝑐2 − 𝑈𝑝𝑜𝑡 1 +𝐸𝑡𝑜𝑡𝑎𝑙

𝜇𝑐2 −𝑈𝑝𝑜𝑡

2𝜇𝑐2 =𝐿2

2𝜇

𝑑𝑢

𝑑𝜃

2+ 𝑢2 (138)

Now by letting

𝑈𝑝𝑜𝑡′ = −𝑈𝑝𝑜𝑡 1 +

𝐸𝑡𝑜𝑡𝑎𝑙

𝜇𝑐2 −𝑈𝑝𝑜𝑡

2𝜇𝑐2 (139)

And

𝐸𝑡𝑜𝑡𝑎𝑙′ = 𝐸𝑡𝑜𝑡𝑎𝑙 1 +


2𝜇𝑐2 (140)

Which here equation (138) takes on the form

𝐿2

2𝜇

𝑑𝑢

𝑑𝜃

2+ 𝑢2 = 𝑈𝑝𝑜𝑡

′ + 𝐸𝑡𝑜𝑡𝑎𝑙′ (141)

Deriving relativistic Binet’s equation

Subsequently we can differentiate (141) in respect of { 𝑢} to find the second order differential equation of the orbit for a relativistic particle. We can always do this, similar to what we did for Binet’s equation (classical case), the left hand side will transform exactly like Binet’s equation and the right hand side because it is in terms of modified potential and kinetic energy it will also get

modified. Next we know that potential energy can be written like this {𝑈𝑝𝑜𝑡 = −𝑘

𝑟= −𝑘𝑢 } hence

when we substitute into equation (139) we have

𝑈𝑝𝑜𝑡′ = 𝑘𝑢 1 +


𝜇𝑐2 +𝑘𝑢

2𝜇𝑐2 (142)

We the substitute equation (142) and (140) back into (141) and we obtain

𝐿2

2𝜇

𝑑𝑢

𝑑𝜃

2+ 𝑢2 = 𝑘𝑢 1 +


𝜇𝑐2 +𝑘𝑢

2𝜇𝑐2 + 𝐸𝑡𝑜𝑡𝑎𝑙 1 +𝐸𝑡𝑜 𝑡𝑎𝑙

2𝜇𝑐2 (143)

Now when we differentiate equation (143) in respect of { 𝑢 } we obtain the second order differential equation of the orbit for a relativistic particle

𝑑2𝑢

𝑑𝜃2 + 𝑢 = 𝑘 +𝑘𝐸𝑡𝑜𝑡𝑎𝑙

𝜇𝑐2 +2𝑘2𝑢

𝜇𝑐2 𝜇

𝐿2 (144)


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In order to check if this equation satisfies the classical one (when the particle is not relativistic) we

let { 𝑙𝑖𝑚𝑖𝑡 𝑐 → ∞ } when we do this we can observe that we are left with { 𝑑2𝑢


𝐿2 } which is

Binet’s equation [7].

Solving classical and relativistic Binet’s equation

Here we will try to solve the differential equations (classical and relativistic) and attempt to discover the orbital equation for a relativistic particle differs from the classical one. First we will show how classical equation can be solved and by applying the same method how one can solve the relativistic equation.

Classical

We start by equation (114) or (119) and we see that is a second order non-homogeneous differential equation. Therefore in order to solve the differential equation we do the following substitution

𝑢 = 𝒆𝑚𝜃 (145)

Then we differentiate equation (145) twice in respect of { 𝜃} and yields

𝑑𝑢

𝑑𝜃= 𝑚𝒆𝑚𝜃

𝑑2𝑢

𝑑𝜃2 = 𝑚2𝒆𝑚𝜃 (146)

When we substitute (146) equation (114) or (119) takes the form

𝑚2𝒆𝑚𝜃 + 𝒆𝑚𝜃 =𝜇𝑘

𝐿2

𝒆𝑚𝜃 𝑚2 + 1 =𝜇𝑘

𝐿2 (147)

The solution of this type of differential equation is given in this form

𝑢 = 𝑐𝑜𝑚𝑝𝑙𝑒𝑚𝑒𝑛𝑡𝑎𝑟𝑦 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛 + 𝑝𝑎𝑟𝑖𝑔𝑢𝑙𝑎𝑟 𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑙 (𝑄) (148)

As a result in order to find the complementary function we let the right hand side of equation (147) equals with zero

𝒆𝑚𝜃 𝑚2 + 1 = 0 (149)

Therefore is either { 𝑒𝑚𝜃 = 0 } or the auxiliary polynomial { 𝑚2 + 1 = 0 } that turns equation into

zero, but { 𝒆𝑚𝜃 > 0 } as a result

𝑚2 + 1 = 0

𝑚 = ±𝑖 (150)

Thus the particular solution for equation (150) is given by

𝑢 = 𝐹 cos 𝜃 + 𝐺 sin 𝜃 (151)

Then we can change solution (151) using the trigonometric identity

𝑅 cos 𝜃 − 𝜃0 = 𝑅 cos 𝜃0 cos 𝜃 − 𝑅 sin𝜃0 sin 𝜃 (152)


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Where

𝑅 cos 𝜃0 = 𝐹

And

𝑅 sin 𝜃0 = 𝐺 And 𝜃0 = 0 (153)

Hence (151) becomes

𝑢 = 𝑅 cos 𝜃 (154)

Then we continue, to find the particular integral, this is done by letting

𝑢 = 𝑄 (155)

We next differentiate equation (155) twice in respect of { 𝜃 } and we have

𝑑𝑢

𝑑𝜃= 0

𝑑2𝑢

𝑑𝜃2 = 0 (156)

After that we substitute equation (156) into (114) or (119) and we obtain the particular integral which is given by

𝑄 =𝜇𝑘

𝐿2 (157)

Now the complete solution can be written as

𝑢 = 𝑅 cos 𝜃 +𝜇𝑘

𝐿2

𝑢 = 𝜇𝑘

𝐿2 𝑅𝐿2

𝜇𝑘cos 𝜃 + 1

1

𝑟=

𝜇𝑘

𝐿2 𝑅𝐿2

𝜇𝑘cos 𝜃 + 1 (158)

Which is the same as equation (57)[8]. By means of equation (57) we observe that { 𝑅𝐿2 = 𝐴 } so ({𝑅} is a constant), hence we have

𝑅 =𝐴

𝐿2 (159)

Relativistic

Now using the same procedure we are going to solve equation (144) in order to find the differences from the classical one. Thus with slight rearrangements equation (144) becomes

𝑑2𝑢

𝑑𝜃2 + 𝑢 = 𝑘 +𝑘𝐸𝑡𝑜𝑡𝑎𝑙

𝜇𝑐2 +2𝑘2𝑢

𝜇𝑐2 𝜇

𝐿2

𝑑2𝑢


𝐿2 +𝑘𝐸𝑡𝑜𝑡𝑎𝑙

𝐿2𝑐2 +2𝑘2𝑢

𝐿2𝑐2


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𝑑2𝑢

𝑑𝜃2 + 𝑢 −2𝑘2𝑢

𝐿2𝑐2 = 𝜇𝑘


𝐿2𝑐2

𝑑2𝑢

𝑑𝜃2 + 𝑢 1 −2𝑘2

𝐿2𝑐2 = 𝜇𝑘


𝐿2𝑐2

𝑑2𝑢

𝑑𝜃2 + 𝑢𝑞2 = 𝑊 (163)

Where

1 −2𝑘2

𝐿2𝑐2 = 𝑞2

And

𝜇𝑘


𝐿2𝑐2 = 𝑊 (161)

Therefore the solution of differential equation (161) can be obtained using the same procedure demonstrated before for the classical equation and is given as

𝑢 = 𝑅 cos𝑞2𝜃 + 𝑊

𝑢 = 𝑊 𝑅

𝑊cos 𝑞2𝜃 + 1 (162)

Now using equation (159) and (161) we can rewrite equation (162) in this form

𝑢 =𝜇𝑘


𝐿2𝑐2 𝐴

𝜇𝑘 +𝑘𝐸𝑡𝑜𝑡𝑎𝑙

𝑐2

cos 1 −2𝑘2

𝐿2𝑐2 𝜃 + 1

1

𝑟=

𝜇𝑘


𝐿2𝑐2 𝐴

𝜇𝑘 +𝑘𝐸𝑡𝑜𝑡𝑎𝑙

𝑐2

cos 1 −2𝑘2

𝐿2𝑐2 𝜃 + 1

By the means of equation (74) we can show that

1

𝑟=

𝜇𝑘


𝐿2𝑐2

𝜇2 1+2𝐸𝑡𝑜𝑡𝑎𝑙 𝐿2

𝜇 𝑘2

𝜇+𝐸𝑡𝑜𝑡𝑎𝑙

𝑐2 cos 1 −

2𝑘2

𝐿2𝑐2 𝜃 + 1

(163)

Here we can observe the modified eccentricity and the semi-latus.

𝑒 ′ =

𝜇2 1+2𝐸𝑡𝑜𝑡𝑎𝑙 𝐿2

𝜇 𝑘2

𝜇+𝐸𝑡𝑜𝑡𝑎𝑙

𝑐2

1

𝑙

′=

𝜇𝑘


𝐿2𝑐2 (164)

Now in order to check if (163) satisfies the classical one (when the particle is not relativistic) we let

{ 𝑙𝑖𝑚𝑖𝑡 𝑐 → ∞ } and it turns out to be { 1

𝑟=

𝜇𝑘

𝐿2 (1 +𝐴

𝜇𝑘cos𝜃)} which is the classical equation of the


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orbit given by (57). Consequently the equation (163) is the orbital equation for a relativistic particle (in this case our one-body system) in arbitrary central force.

Figure 9 (Shows the relativistic orbits of shape with constant {𝑒 ′ = 0.5} and we vary only the {𝑞2})

Now keeping the modified eccentricity the same all the time, we check how the shape with {𝑒 ′ =0.5} changes when we vary only the constant { 𝑞2} in front of { 𝜃}. Hence the following orbits shown in Figure 9 take place

1. 𝑒 ′ = 0.5 1 −2𝑘2

𝐿2𝑐2 = 0.9

2. 𝑒 ′ = 0.5 1 −2𝑘2

𝐿2𝑐2 = 2

3. 𝑒 ′ = 0.5 1 −2𝑘2

𝐿2𝑐2 = 3

4. 𝑒 ′ = 0.5 1 −2𝑘2

𝐿2𝑐2 = 4 (165)

Conclusion

In part 1, we have reduced the two-body system into a one-body. The solution of this problem in most of the text is derived using Lagrangian dynamics, but we have derived a solution by means of Newtonian dynamics and then using symmetries (angular momentum and Laplace-Runge-Lenz vector) we have found the orbital equation. One can expand and talk more about the symmetries in such a system and prove them using Noether’s theorem. In Part 2 at first using classical mechanics we derived the second order differential (Binet’s) equation of an orbiting particle with a reduced mass {𝜇}. Then we derived the relativistic orbital second order differential equation using the special theory of relativity. We then solve it and find the relativistic equation of the orbit and compare in what ways does the relativistic orbit equation differs from the classical one. When we compare the 2 equations we examine that the eccentricity and the semi-latus are different but the most important thing that we observe is that in the relativistic equation a constant { 𝑞2} is multiplied by { 𝜃}. This constant multiplied by { 𝜃} is what makes the orbit to shift. The derivation of the problem using the General relativity is a lot better approach where it gives more explicit result but at the

1

2

3

4


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same time is a really daring one because of the many difficulties one can face. Using General relativity is considered a complex approach because you don’t have symmetries in the system and many singularities appear.

References

[1] Gowers,T. Barrow-Green, J. & Leader, I.2008. The Princeton Companion to Mathematics. Oxfordshire: Princeton University Press, pp.47-49,pp.179-181.

[2] Trump,M.A. &Schieve,W.C.,1998. Perihelion Precession in the Special Relativistic Two-body Problem. Foundations of Physics,28(9), pp1407-1416.

[3] Goldstein, H., 1975.Prehistory of the “Runge-Lenz” vector. America Journal of Physics, 43(8), pp. 737-738.

[4] Goldstein, H., 1976 more on the prehistory of the Laplace or Runge-Lenz vector. America Journal of Physics, 44(11), pp. 1123-1124.

[5] Goldstein, H. Poole, C & Safko, J., 2002.Classical Mechanics.3rd ed. San Francisco: Addison Wesley. Ch. 3.9.

[6] Dyke, P. Whitworth, R., 2001. Guide to Mechanics. Hampshire: Palgrave. Ch.9.

[7] Aaron, F. D., 2005. Relativistic equation of the orbit of a particle in an arbitrary central force field. Romanian Journal of Physics, 50(5-6), p.615-619.

Available at: http://www.nipne.ro/rjp/2005_50_5-6/0615_0620.pdf

[Accessed 15 Feb 2010].

[8] Mannall, G. & Kenwood M., 2004. Further Pure Mathematics 1. Oxford: Heinemann Educational.Ch.6.

http://www.nipne.ro/rjp/2005_50_5-6/0615_0620.pdf


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