the two body problem. almost complete-1.pdf
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Two body problemTRANSCRIPT
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. Binetโs equation using classical mechanics .............................................................................. 21
XVII. Deriving first order differential equation of the orbit .............................................................. 22
XVIII. Different approach for deriving Binetโs 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 Binetโs equation ....................................................................................... 26
XXII. Solving classical and relativistic Binetโs 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 {๐ธ๐ก๐๐ก๐๐ } 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 Binetโs 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 โnโ 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 { ๐2}). 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 { ๐ > 2 } was much more complicated. Even Hilbert placed the solution of only a three-body system into the same category as Fermatโs 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 Sundmanโs 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 { ๐1 } and { ๐2 })
Origin
๐2
๐1 ๐
๐ 1 ๐น ๐ 2
๐
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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
๐ = ๐๐๐ ๐ก๐๐๐๐ ๐๐๐ก๐ค๐๐๐ ๐ก๐๐ ๐ก๐ค๐ ๐๐๐ ๐ ๐๐
๐ = ๐๐๐ ๐๐๐๐๐๐๐๐๐ก ๐๐๐ก๐ค๐๐๐ ๐ก๐๐ ๐ก๐ค๐ ๐๐๐ ๐ ๐๐
๐ 1 & ๐ 2 = ๐๐๐ ๐๐ก๐๐๐ ๐ฃ๐๐๐ก๐๐๐
๐1 & ๐2 = ๐๐๐๐๐ก ๐๐๐ ๐ ๐๐
๐น = ๐๐๐ ๐๐ก๐๐๐ ๐ฃ๐๐๐ก๐๐(๐๐๐๐ฆ๐๐๐๐ก๐๐)
๐ฎ = ๐๐๐๐ฃ๐๐ก๐๐ก๐๐๐๐๐ ๐๐๐๐ ๐ก๐๐๐ก
๐ = ๐๐๐๐๐๐ก๐๐๐ ๐ฃ๐๐๐๐
๐ = ๐๐๐๐ข๐๐๐ ๐๐๐ ๐
In order to solve the problem, we consider two point particles { ๐1} and { ๐2} where according to the law of Isaac Newton they exert equal and opposite force onto each other (Newtonโs 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
๐ญ = โ๐ฎ๐1๐2
๐3 ๐ (1)
This is the force exerted by gravitational pull which is Newtonโs gravitational law. Hence the forces that particle { ๐1} exerts on { ๐2} and vice versa are
๐1๐ 1 = โ๐ญ (2)
๐2๐ 2 = ๐ญ (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
๐1 + ๐2 ๐น = ๐1๐ 1 + ๐2๐ 2 (4)
When we make { ๐น } the subject of the formula the equation (4) (barycentre) between the two point masses become
๐น =๐1๐ 1+๐2๐ 2
๐1+๐2 (5)
By differentiate equation (5) twice in respect to time, we find its second derivative and we show that
๐น = (๐1๐
1+๐2๐
2
๐1+๐2) (6)
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Then we substitute equations (2) and (3) into equation (6) resulting in
๐น =๐ญ โ๐ญ
๐1+๐2= 0 โ ๐น = 0 (7)
Thus it implies that
๐น = ๐๐๐๐ ๐ก๐๐๐ก (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
๐2๐ 2 = ๐ญ = โ๐ฎ
๐1๐2
๐3 ๐ (9)
In Figure 1 we observe that the vectors {๐ 2 & ๐ 1} is given by
๐ 2 = ๐น + ๐ (10)
And
๐ 1 = ๐น โ ๐ + ๐ (11)
Using the equation of the barycentre (4) and after we substitute equation (10) and (11) it becomes
๐1 + ๐2 ๐น = ๐1๐น โ ๐1๐ + ๐1๐ + ๐2๐น + ๐2๐ (12)
This is then simplified to
โ๐ ๐1 + ๐2 = โ๐1๐ (13)
Where equation (13) can also be written as
๐ =๐1
๐1+๐2๐ (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
๐ 2 = ๐น +๐1
๐1+๐2๐ (15)
However we know from equation (7) that { ๐น = 0}. Consequently when we substitute equation (15) into equation (9) it becomes
๐2 0 +๐1
๐1+๐2 ๐ =
๐1๐2
๐1+๐2๐ = ๐ญ = โ๐ฎ
๐1๐2
๐3 ๐
๐ = โ๐ฎ๐1+๐2
๐3 ๐ (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 { ๐1๐2
๐1+๐2 } and we find that it turns
into the equation
๐๐ = โ๐ฎ๐1๐2
๐๐ ๐ (17)
Such equation corresponds to a system of one-body with a reduced mass {๐} which is given by
1
๐=
1
๐1+
1
๐2 (18)
With slight rearrangements we have
๐ =๐1๐2
๐1+๐2 (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
๐ = ๐๐๐ ๐๐๐๐๐๐๐๐๐ก ๐๐๐๐๐ก๐๐ฃ๐ ๐ก๐ ๐ก๐๐ ๐๐๐๐๐๐
๐ = ๐๐๐๐๐๐ ๐๐๐๐๐๐ก๐ข๐
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
๐ณ = ๐ ร ๐ (20)
We know that the one-body particle with reduce mass {๐} 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. ๐ = ๐๐ And when we differentiate once in respect of time we have
2. ๐ = ๐๐ (21)
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Proving that angular momentum is conserved
In order to show that the angular momentum is conserved, we prove that { ๐ณ = 0}. To achieve this we differentiate equation (20) in respect of time once and results
๐ณ = ๐ ร ๐ + ๐ ร ๐
Then we substitute (21) part 1 & 2 accordingly and yields
๐ณ = ๐ ร ๐๐ + ๐ ร ๐ ๐ (22)
After that we substitute equation (17) and we have
๐ณ = ๐ ร ๐๐ + ๐ ร โ๐ฎ๐1๐2
๐๐ ๐ (23)
The vector { ๐ณ } from equation (23) can be divided into two sections (a) and (b) { ๐ณ = a ร b}
(a) ๐ ร ๐๐
(b) ๐ ร โ๐ฎ๐1๐2
๐3 ๐
By definition we know that
๐ ร ๐ = ๐๐ sin๐ ๐ (Where { ๐} is the angle between the vectors { ๐ })
From vector geometry we know that the angle between identical vectors is zero thus
๐ ร ๐ = ๐๐ sin(0) ๐ = 0
๐ถ ร ๐ถ = 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
๐ณ = ๐ ร ๐๐ + ๐ ร โ๐ฎ๐1+๐2
๐3 ๐ โ ๐ณ = 0 + 0 = 0 โ ๐ณ = 0 (25)
This implies the fact that
๐ณ = ๐๐๐๐ ๐ก๐๐๐ก (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 ๐2. 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
๐ = ๐๐๐ ๐
๐ = ๐๐๐๐ ๐ก๐๐๐ก
๐ = ๐ข๐๐๐ก ๐ฃ๐๐๐ก๐๐
๐ = ๐๐ ๐ก๐๐ ๐๐๐ ๐๐๐๐๐๐๐๐๐ก ๐๐๐๐ ๐ก๐๐ ๐๐๐๐ข๐ ๐ก๐ ๐ก๐๐ ๐๐๐๐๐๐๐ก๐๐ ๐๐ ๐ก๐๐ ๐๐๐๐๐ก
Now Laplace-Runge-Lenz vector for a single particle is defined as
๐จ = ๐ ร ๐ณ โ ๐๐๐ (27)
At first we have to show first that vector { ๐จ } is conserved for a gravitational force that is inversely
proportional to ๐2. So in order to do this we have to show that { ๐จ = 0}. The definition of the force is also given by
๐ญ = ๐ (28)
We can always rewrite ๐ญ vectorially as a function of { ๐(๐) } hence
1. ๐ญ = โ๐ฎ๐1๐2
๐3 ๐
2. ๐ญ =โ๐
๐2 ๐
3. ๐ญ = ๐(๐)๐ (29)
Where
๐ = ๐
๐ (30)
And
โ๐ = โ๐ฎ๐1๐2 (31)
Also when we equate (28) and part 3 of equation (29) we have
๐(๐)๐ = ๐ (32)
Next for convenience we equate part 2 and 3 of equation (29) resulting in
โ๐
๐2 ๐ = ๐(๐)๐
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โ๐ = ๐ ๐ ๐2 (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{ ๐ }. Consequently because our analysis is restricted to a one-body particle with reduce mass { ๐ } the equation (27) is given by
๐จ = ๐ ร ๐ณ โ ๐๐๐ (34)
Note: From now on when we refer to vector { ๐จ } we will imply expression (34)
We continue and differentiate equation (34) once in respect of the time and it can be written as
๐จ = (๐ ร ๐ณ ). โ (๐๐๐ )โ (35)
Hence using the product rule, equation (35) becomes
๐จ = ๐ ร ๐ณ + ๐ ร ๐ณ โ (๐๐๐ )โ (36)
From equation (25) we have shown that the angular momentum is conserved so { ๐ณ = 0} thus equation (36) can be expressed as
๐จ = ๐ ร ๐ณ + ๐ ร 0 โ ๐๐๐ โ
From vector geometry we know that { ๐ ร 0 = 0} so
๐จ = ๐ ร ๐ณ โ ๐๐๐ โ
Then using equation (32), we can carry out the following substitution into expression (36) and we have
๐จ = ๐(๐)๐ ร ๐ณ โ(๐๐๐ )โ
We then replace { ๐ณ } by equation (20) after we have substitute into (20) part 1 of equation (21) and we have
๐จ = ๐(๐)๐ ร ๐ ร ๐ ๐ โ(๐๐๐ )โ (37)
Afterwards using the Lagrangeโs formula
๐ ร ๐ ร ๐ = ๐ ๐ โ ๐ โ ๐ (๐ โ ๐ ) (38)
Equation (37) can be expanded as
๐จ = ๐ ๐ ๐ ๐ โ ๐๐ โ ๐๐ ๐ ๐ ๐ โ ๐ โ(๐๐๐ )โ (39)
Next we take out ๐๐ ๐
๐ as a common factor and yields
๐จ =๐๐ ๐
๐[๐ ๐ โ ๐ โ ๐ (๐ โ ๐ )]โ(๐๐๐ )โ (40)
The equation (40) can be further simplified by noting that
๐ โ ๐ = ๐2 (41)
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Then when we differentiate equation (41) once in respect to time and we get
(๐ โ ๐ )โ = ๐ โ ๐ + ๐ โ ๐ = 2๐ โ ๐
We can also write it differently
(๐ โ ๐ )โ = ๐2 โ = 2๐๐
Therefore we find that this relation holds
2๐ โ ๐ = 2๐๐
๐ โ ๐ = ๐๐ (42)
Moreover by differentiating expression (30) once in respect to time we demonstrate that
๐ = ๐ ๐ + ๐ ๐ (43)
Then with slight rearrangements
๐ =๐
๐+
๐ ๐
๐2 (44)
Using equation (41) and (42) we can re-write equation (40) as
๐จ =๐๐ ๐
๐ ๐ ๐๐ โ ๐ (๐2 ]โ ๐๐๐ โ (45)
Then we take {โ๐3} out of the equation as a common factor and we obtain the following result
๐จ = โ๐๐ ๐ ๐2[๐
๐+
๐ ๐
๐2]โ ๐๐๐ โ (46)
We expand { ๐๐๐ โ = ๐๐๐ } and next using equation (33) we can write { ๐๐๐ = ๐๐ ๐ ๐2๐ }. Thus we have
๐จ = โ๐๐ ๐ ๐2[๐
๐+
๐ ๐
๐2] + ๐๐ ๐ ๐2๐ (47)
Then using equation (44) we can write that
๐จ = โ๐๐ ๐ ๐2๐ + ๐๐ ๐ ๐2๐ = 0 โ ๐จ = 0 (48)
Hence this implies that
๐จ = ๐๐๐๐ ๐ก๐๐๐ก (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 ๐. So we have proved that vector { ๐จ } 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 { ๐จ } we can derive the possible orbits of the one-body system.
Figure 2 (shows that { ๐ ร ๐ณ } is orthogonal to { ๐ณ } and { ๐ ร ๐ } is orthogonal to { ๐ })
The general definition of dot product where { ๐ } is the angle between the two vectors { ๐ 1} and { ๐ 2} is given by
๐ 1 โ ๐ 2 = ๐1๐2 cos ๐ (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 {๐ญ } as
๐ญ = โ๐
๐๐
๐
๐ (51)
By means of Figure 2 it implies the fact that Laplace-Runge-Lenz vector { ๐จ } is orthogonal to the
angular momentum { ๐ณ } so { ๐จ โ ๐ณ = 0}. Therefore this suggests that { ๐จ } lies in the plane of the
orbit and is fixed. However the displacement { ๐ } changes, thus the angle {๐} that {๐ } makes with
{๐จ } also changes, so by using dot product, we can relate the angle with {๐จ โ ๐ณ }.
Figure 3 (shows the angle { ๐ } between { ๐จ } and the vector { ๐ })
Hence using definition (50) we can write the following
๐
๐จ
๐
๐
๐
๐ ร ๐ ๐ ร ๐ณ
๐ ๐ณ
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๐จ โ ๐ = ๐ด๐ cos ๐ = (๐ ร ๐ณ โ ๐๐๐ ) โ ๐
Where { ๐ } is the angle between { ๐จ } and { ๐ } as shown in the Figure 3, furthermore dot product is distributive thus we can write
๐ด๐ cos๐ = (๐ ร ๐ณ ) โ ๐ โ ๐๐ ๐ โ ๐ (52)
Then we continue and use the triple scalar product which is given by
๐ ร ๐ โ ๐ = ๐ โ ๐ ร ๐ = ๐ โ ๐ ร ๐ (53)
Therefore using (53) we can reformulate (52) and yield
๐ด๐ cos ๐ = ๐ณ โ ๐ ร ๐ โ ๐๐ ๐ โ ๐ (54)
After that using equation (20) we can then write that
๐ โ ๐ ร ๐ณ = ๐ณ โ ๐ ร ๐ = ๐ณ โ ๐ณ = ๐ฟ2 (55)
Also it can be noted that
๐ โ ๐ =1
๐ ๐ โ ๐ =
๐2
๐= ๐ (56)
So by substituting (55) and (56) back into equation (54), yields
๐ด๐ cos ๐ = ๐ฟ๐ โ ๐๐๐
๐ ๐ด cos ๐ + ๐๐ = ๐ฟ๐
๐ =๐ฟ๐
๐ด cos ๐+๐๐
1
๐=
๐ด cos ๐+๐๐
๐ฟ๐ โ 1
๐=
๐๐
๐ฟ2 (1 +๐ด
๐๐cos ๐) (57)
Finally we have simplified our equations into an equation which by comparison looks like the general conic equation.
1
๐=
1
๐(1 + ๐ cos ๐) (By definition) (58)
So by comparing result (57) and equation (58) we can write the following
1
๐=
1
๐(1โ๐2) (By definition)
1
๐=
๐ฟ2
๐๐
๐ฟ2
๐๐=
1
๐(1โ๐2) (59)
Plus eccentricity (conic section parameter) is given by
๐ =๐ด
๐๐ (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.
๐จ โ ๐จ = ๐ ร ๐ณ โ ๐๐๐ โ (๐ ร ๐ณ โ ๐๐๐ ) (61)
As a result using the distributive property of dot product we can write equation (61) as
๐จ โ ๐จ = ๐ ร ๐ณ 2โ 2(๐ ร ๐ณ ) โ ๐๐๐ + ๐2๐2๐ 2 (62)
We can simplify the above equation further using Lagrangeโs identity
๐ ร ๐ 2
= ๐ โ ๐ ๐ โ ๐ โ ๐ โ ๐ (๐ โ ๐ ) (63)
Where it can also be written as
๐ด2 = ๐ โ ๐ ๐ณ โ ๐ณ โ ๐ โ ๐ณ (๐ โ ๐ณ ) โ 2(๐ ร ๐ณ ) โ ๐๐๐ + ๐2๐2 (64)
But we know that the vector { ๐ โ ๐ณ = 0 } because { ๐ } is perpendicular to { ๐ณ } as a result equation
(64) can be also stated as
๐ด2 = ๐2๐ฟ2 โ 2(๐ ร ๐ณ ) โ ๐๐๐ + ๐2๐2 (65)
When we dot product { ๐จ } 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
๐ธ๐ก๐๐ก๐๐ = ๐พ๐๐๐ + ๐๐๐๐ก (66)
Where
๐พ๐๐๐ =๐๐ 2
2 (67)
And
๐๐๐๐ก = โ๐
๐ (68)
Consequently we can write ๐ธ๐ก๐๐ก๐๐ as
๐ธ๐ก๐๐ก๐๐ =๐๐ 2
2โ
๐
๐ (69)
We also know that kinetic energy is related to linear momentum therefore by definition we have
๐พ๐๐๐ =๐2
2๐ (70)
Thus substituting (68) and (70) we can write into (60) it becomes
๐ธ๐ก๐๐ก๐๐ =๐2
2๐โ
๐
๐ (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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๐ด2 = ๐2๐ฟ2 โ 2๐ โ ๐ ร ๐ณ ๐๐
๐+ ๐2๐2 โ ๐ด2 = ๐2๐ฟ2 + 2๐ฟ2๐ โ
๐
๐ + ๐2๐2 (72)
Then we take { 2๐ฟ2๐ } out as a common factor and substituting equation (71) we have
๐ด2 = 2๐ฟ2๐ ๐2
2๐โ
๐
๐ + ๐2๐2 โ ๐ด2 = 2๐ฟ2๐๐ธ๐ก๐๐ก๐๐ + ๐2๐2 (73)
We can simplify (73) using the eccentricity given by equation (60). Then with slight rearrangements we have
๐ด2
๐2๐2 โ 1 =2๐2๐ธ๐ก๐๐ก๐๐
๐๐2 โ ๐2 โ 1 =2๐2๐ธ๐ก๐๐ก๐๐
๐๐2
๐ = 1 +2๐ธ๐ก๐๐ก๐๐ ๐ฟ2
๐๐2 (74)
Consequently when we have
1. ๐ > 1 ๐ธ๐ก๐๐ก๐๐ > 0: ๐ป๐ฆ๐๐๐๐๐๐๐ 2. ๐ = 1 ๐ธ๐ก๐๐ก๐๐ = 0: ๐๐๐๐๐๐๐๐ 3. 0 < ๐ < 1 ๐ธ๐ก๐๐ก๐๐ < 0: ๐ธ๐๐๐๐๐ ๐
4. ๐ = 0 ๐ธ๐ก๐๐ก๐๐ = โ๐๐2
2๐ฟ2 โถ ๐๐๐๐๐๐ (75)
From this equation we can determine the nature of the orbit since it only depends on the parameter {๐} and {๐ธ๐ก๐๐ก๐๐ }. 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 typeโs, 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 {๐} (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 {๐}. This parameter basically measures how much the conic section shape changes from being a circle.
(II) Semi-latus {๐} which is a chord parallel to directrix
(III) Focal parameter {๐} which is the distance from foci to the directrix
Focal parameter, eccentricity and semi-latus are related by this formula
๐ = ๐๐ (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 fociโs 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 { ๐ธ๐ก๐๐ก๐๐ } result found for the circle from equation (75) with equation (66).
๐ธ๐ก๐๐ก๐๐ = โ๐๐2
2๐2 = ๐พ๐๐๐ + ๐๐๐๐ก (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 { ๐ = 0} we have
1
๐=
๐๐
๐ฟ2 (1 +๐ด
๐๐cos ๐)
1
๐=
๐๐
๐ฟ2 (1 + 0)
1
๐=
๐๐
๐ฟ2 (78)
Therefore using result from equation (75) part 4, and substituting into (69), becomes
โ๐๐2
2๐ฟ2 =๐๐ 2
2โ
๐
๐ (79)
In addition we know that the tangential velocity of a body that rotates can be given as
๐ = ๐๐ (๐ =2๐
๐) (80)
Hence by substituting equation (80) into (79) then multiply by 2, we have
Parabola
Hyperbola
Ellipse
Circle
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โ๐๐2
2๐ฟ2 =๐๐2๐2
2โ
๐
๐ . 2
โ๐๐2
๐ฟ2 +2๐
๐= ๐๐2๐2
And then taking { ๐} out as a common factor we obtain
๐ 2
๐โ
๐๐
๐ฟ2 = ๐๐2๐2 (81)
Using equation (78) we can substitute into (81) and find that
๐ 2
๐โ
1
๐ = ๐๐2๐2 โ
๐
๐= ๐๐2๐2 (82)
Then by substituting equation (31), { ๐ =2๐
๐} and (19) into (82) we get
๐ฎ ๐1๐2 =๐1๐2
๐1+๐2๐3 4๐
๐2
๐ = 4๐๐3
๐ฎ(๐1+๐2) (83)
Ellipse
In the case of an elliptic orbit, we know that again { ๐ธ๐ก๐๐ก๐๐ } is conserved, consequently we can
always write { ๐ธ๐ก๐๐ก๐๐ =๐๐ 2
2โ
๐
๐ } 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 { ๐ธ๐ก๐๐ก๐๐ } 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 { ๐๐
๐๐ก})
๐ฟ = ๐๐2๐ (84)
We can then write equation (84) as
๐ =๐ฟ
๐๐2 (85)
In addition when we square it we get
๐
๐
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๐ 2 =๐ฟ2
๐2๐4 (86)
Then we can re-arrange it and we can write
๐ 2๐2 =๐ฟ2
๐2๐2 (87)
Thus, we continue using { ๐ = ๐ฝ ๐ } (a result that we are going to prove later) and then squaring equation (43) we obtain
๐ 2 = ๐2๐ 2 + ๐ 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 { ๐ธ๐ก๐๐ก๐๐ } thus we substitute equation (88) into equation (69) and we get the following
๐ธ๐ก๐๐ก๐๐ =๐๐2๐ 2
2+
๐๐ 2
2โ
๐
๐ (89)
Then substituting equation (87) into (89) we get
๐ธ๐ก๐๐ก๐๐ = ๐ฟ2
2๐๐2 +๐๐ 2
2โ
๐
๐ (90)
Then making speed (writing it using Leibniz notation) the subject of the formula we can write
๐ธ๐ก๐๐ก๐๐ +๐
๐โ
๐ฟ2
2๐๐2 2
๐=
๐๐
๐๐ก (91)
So we find velocity in respect of conserved quantities. Then we can integrate {๐๐ก} and we find the time {๐ก}
๐ก = ๐๐
๐ธ๐ก๐๐ก๐๐ +๐
๐ โ
๐ฟ2
2๐ ๐2 2
๐
Then at {๐ก = 0} we let {๐} to have the initial value { ๐0}, the closest distance to the centre of force (perihelion) and the final value of {๐โฒ } thus the integral becomes
๐ก = ๐
2
๐๐
๐ธ๐ก๐๐ก๐๐ +๐
๐ โ
๐ฟ2
2๐ ๐2
๐ โฒ
๐0 (92)
Next by means of equation (59) we can state that
๐ฟ2
๐๐=
1
๐ 1โ๐2 โ ๐ฟ2 = ๐๐๐(1 โ ๐2) (93)
Also using (74) we derive the following result
๐ธ๐ก๐๐ก๐๐ = (๐2 โ 1)๐๐2
2๐ฟ2 (94)
We can then substitute equation (93) into (94) and becomes
๐ธ๐ก๐๐ก๐๐ = (๐2 โ 1)๐๐2
โ2๐๐๐ (๐2โ1)
๐ธ๐ก๐๐ก๐๐ = โ๐
2๐ (95)
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Therefore when we substitute (93) and (95) back into our integral given by equation (92) and we get
๐ก = ๐
2
๐๐
โ ๐
2๐+
๐
๐ โ
๐๐ (1โ๐2)
2๐2
๐ โฒ
๐0 โ ๐ก =
๐
2๐
๐๐
โ 1
2๐+
1
๐ โ
๐(1โ๐2)
2๐2
๐ โฒ
๐0
๐ก = ๐
2๐
๐๐
โ ๐2
2๐+
1
๐ โ
๐ 1โ๐2
2๐2
๐ โฒ
๐0
๐ก = ๐
2๐
๐๐๐
โ ๐2
2๐+๐ โ
๐(1โ๐2)
2
๐ โฒ
๐0 (96)
This is an elliptic integral and in order to solve (96) we use the substitution
๐ = ๐ 1 โ ๐ cos ๐ฟ (97)
The equation (97) is a relation of { ๐ } with the semi-major axes { ๐ }, eccentricity {๐} and the angle { ๐ฟ }, it is also called the eccentric anomaly.
Figure 6 (shows how eccentric anomaly is related with the radius { ๐ } that makes an angle { ๐ฟ } with the horizontal of the auxiliary circle and the distance { ๐} from the focus of the ellipse)
Next we differentiate (97) in respect to { ๐ฟ } and we get
๐๐ = ๐๐ sin๐ฟ ๐๐ฟ (98)
Then we substitute equation (97) and (98) into (96) along with the limits of integration (by letting {๐ฟ = 0} we find the perihelion {๐0}) then replacing {๐} in terms of {๐ฟ} we have
๐ผ ๐ฟ
๐ผ
๐
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๐ก = ๐
2๐
๐ 1โ๐ cos ๐ฟ ๐๐ sin ๐ฟ ๐๐ฟ
๐ 1โ๐ cos ๐ฟ โ ๐2
2๐ 1โ2๐ cos ๐ฟ+๐2 cos 2 ๐ฟ โ
๐(1โ๐2)
2
๐ฟ โฒ
0
๐ก = ๐๐4
2๐
1โ๐ cos ๐ฟ ๐ sin ๐ฟ ๐๐ฟ
๐โ๐๐ cos ๐ฟโ ๐
2+๐๐ cos ๐ฟโ
๐๐ 2 cos 2 ๐ฟ
2โ
๐
2+
๐๐2
2
๐ฟ โฒ
0
Then with slight rearrangements leads to the following result
๐ก = ๐๐4
2๐
1โ๐ cos ๐ฟ ๐ sin ๐ฟ ๐๐ฟ
๐๐2
2 1โ๐ cos 2 ๐ฟ
๐ฟ โฒ
0
๐ก = ๐๐3
๐ 1 โ ๐ cos ๐ฟ ๐๐ฟ
๐ฟ โฒ
0 (99)
So when we integrate { ๐๐ฟ } we have
๐ก = ๐๐3
๐๐ฟ โ ๐
๐๐3
๐sin ๐ฟ
0
๐ฟ โฒ
(100)
Figure 7 (when the auxiliary angle does one complete revolution of 2๐)
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 โ ๐ฟ โฒ = 2๐ } we have the time for one complete revolution which we can call period denoted by the letter { ๐}. Finally after we substitute the limits, yields
2๐
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๐ = 2๐ ๐๐3
๐โ ๐
๐๐3
๐sin 2๐
๐ = 2๐ ๐๐3
๐ (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
๐ } potential. We are going to try to find its second order
deferential equation of the orbit in terms of how { ๐ข =1
๐ } changes with angle { ๐ } 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 {๐} 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 { ๐ธ = ๐0๐
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 { ๐๐๐๐๐ก ๐ โ โ} in relativistic equation, then the equation starts to act as classical one. We can always do this because in special relativity { ๐ } is a constant and contributes to the results, where in classical mechanics { ๐ } 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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Binetโs 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
๐ = ๐ฅ, ๐ฆ = (๐ cos ๐ , ๐ sin ๐)
We can then write the unit vectors in this way, where we can easily obtain using Figure 8
๐ = cos๐ , sin๐
๐ฝ = (โ sin ๐ , cos ๐) (102)
Figure 8 (shows the unit vectors of polar coordinates)
Then we differentiate the unit vectors and we obtain
๐ = โ sin ๐ , cos ๐ ๐๐
๐๐ก โ ๐ = ๐ฝ ๐ (103)
๐ฝ = โ cos ๐ , โ sin๐ ๐๐
๐๐ก โ ๐ฝ = โ๐ ๐ (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
๐ = ๐ฝ ๐ ๐ + ๐ ๐ (105)
Next we differentiate again to find acceleration and we have
๐ = ๐ ๐ + ๐ ๐ + ๐ ๐ ๐ฝ + ๐๐ ๐ฝ + ๐๐ ๐ฝ โ ๐ = ๐ ๐ + ๐ ๐ ๐ฝ + ๐ ๐ ๐ฝ + ๐๐ ๐ฝ โ ๐๐ ๐ 2
๐ = ๐ ๐ + 2๐ ๐ ๐ฝ + ๐๐ ๐ฝ โ ๐๐ 2๐ โ ๐ = ๐ โ ๐๐ 2 ๐ + (2๐ ๐ + ๐๐ )๐ฝ (106)
Then we can substitute the results found in (106) into equation (17) accordingly to get
๐ ๐ โ ๐๐ 2 ๐ + ๐(2๐ ๐ + ๐๐ )๐ฝ = โ๐ฎ๐1๐2
๐2 ๐
After we equate the equations we obtain
๐ ๐ โ ๐๐ 2 = โ๐ฎ๐1๐2
๐2 (107)
๐ ๐
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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
๐ข =1
๐ (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 { ๐๐
๐๐ก=
๐ฟ
๐2๐ }. What we want to do is to reduce our independent variable {๐ก} and introduce
the new variable {๐ข}, 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
๐๐ข
๐๐ก= โ
1
๐2
๐๐
๐๐ก (109)
Then we write equation (85) using the chain rule ๐๐
๐๐กโ๐๐ก
๐๐ข=
๐๐
๐๐ข and yields
๐๐
๐๐ข
๐๐ข
๐๐ก=
๐ฟ
๐2๐ (110)
Then we substitute (109) into (110) and we have
๐๐
๐๐ข โ
1
๐2
๐๐
๐๐ก =
๐ฟ
๐2๐ โ
๐๐ข
๐๐= โ
๐ ๐
๐ฟ (111)
Now we differentiate (111) for a second time in respect of {๐} and we find that
๐2๐ข
๐๐2 = โ๐
๐ฟ
๐
๐๐ก ๐๐
๐๐ก
๐๐ก
๐๐ (112)
Then substituting back equation (85) we get
๐2๐ข
๐๐2 = โ๐
๐ฟ
๐ ๐2๐
๐ฟ โ
๐2๐ข
๐๐2 = โ๐2๐2๐
๐ฟ2
Plus when we rearrange it becomes
๐ = โ๐ฟ2
๐2๐2
๐2๐ข
๐๐2 (113)
Then we substitute equation (113) and (86) into (107) and we obtain
๐ โ๐ฟ2
๐2๐2
๐2๐ข
๐๐2 โr
r2 ๐ฟ2
๐2๐2 = โ๐ฎ๐1๐2
๐2 โ โ๐ฟ2
๐๐2 ๐2๐ข
๐๐2 + ๐ข = โ๐ฎ๐1๐2
๐2
This in turn simplifies, to
๐2๐ข
๐๐2 + ๐ข = ๐๐ฎ๐1๐2
๐ฟ2 (114)
Here we have obtained Binetโs 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 { ๐ก} and we introduce { ๐ข} 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
Binetโs 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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๐ธ๐ก๐๐ก๐๐ =๐ ๐ 2๐2+๐ 2
2โ
๐
๐ โ ๐ธ๐ก๐๐ก๐๐ +
๐
๐โ
๐ 2๐2๐
2
2
๐= ๐ 2 (115)
After that, we substitute (85) into (115) and we have
๐ธ๐ก๐๐ก๐๐ +๐
๐โ
๐ฟ
๐2๐
2 ๐2๐
2
2
๐= ๐ 2 โ ๐ธ๐ก๐๐ก๐๐ +
๐
๐โ
๐ฟ2
2๐๐
2
๐= ๐ 2 (116)
When we square equation (111) and make speed the subject of the formula, yields
๐ 2 = ๐ฟ
๐
2 ๐๐ข
๐๐
2 (117)
Then we substitute equation (117) into equation (116) and we obtain
๐ธ๐ก๐๐ก๐๐ +๐
๐โ
๐ฟ2
2๐๐
2
๐=
๐ฟ
๐
2 ๐๐ข
๐๐
2
By further simplification we get
2๐ธ๐ก๐๐ก๐๐
๐+
2๐
๐๐=
๐ฟ
๐
2 ๐๐ข
๐๐
2+
๐ฟ2
๐2๐ โ
2
๐ ๐ธ๐ก๐๐ก๐๐ +
๐
๐ =
๐ฟ2
๐2 ๐๐ข
๐๐
2+ ๐ข2
๐ธ๐ก๐๐ก๐๐ + ๐๐ข =๐ฟ2
2๐
๐๐ข
๐๐
2+ ๐ข2 (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 { ๐ข} and { ๐}.
Different approach for deriving Binetโs equation
Now by differentiating equation (118) in respect of { ๐ข} leads to the result we have derived previously in equation (114). Hence
2๐๐
๐ฟ2 = 2๐๐ข
๐๐
๐2๐ข
๐๐2
๐๐
๐๐ข+ 2๐ข
๐2๐ข
๐๐2 + ๐ข = ๐๐
๐ฟ2 (119)
As a result we see that by differentiating the non linear differential equation (118) in respect of { ๐ข} we can derive Binetโs 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
๐ } 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
๐ = ๐พ๐๐ (120)
This is only the spatial piece of the 4-momentum vector. Where { ๐พ} is a factor given as
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๐พ =1
1โ๐ 2
๐2
(121)
We also want to define the kinetic energy in terms of a relativistic particle. This can be done as
๐พ๐๐๐ = ๐ โ ๐๐ (122)
Then we differentiate (120) in respect of time and becomes
๐๐
๐๐ก= ๐๐พ๐ + ๐๐พ ๐
๐๐ = ๐๐พ๐ ๐๐ก + ๐๐พ ๐ ๐๐ก (123)
Then we substitute into equation (122) and we obtain
๐พ๐๐๐ = ๐๐พ๐๐
๐๐ก๐ ๐๐ก + ๐๐ 2
๐๐พ
๐๐ก๐๐ก โ ๐พ๐๐๐ = ๐ ๐พ๐ ๐๐ + ๐ ๐ 2๐๐พ (124)
Then using equation (121) we write
1
๐พ2 =1โ๐ 2
๐2 (125)
Plus
๐ 2 = ๐2(1 โ1
๐พ2) (126)
Then we differentiate (125) in respect of velocity and we have
โ2๐พโ3 ๐๐พ
๐๐ = โ2
๐
๐2 โ ๐๐ =๐2
๐พ3๐ (127)
Then we substitute equation (126) and (127) into (124) and we obtain
๐พ๐๐๐ = ๐๐2 1
๐พ2 ๐๐พ + ๐๐2 ๐๐พ โ ๐๐2 1
๐พ2 ๐๐พ
๐พ๐๐๐ = ๐พ๐๐2 + ๐ถ (128)
Here ๐ถ is the constant of integration. Then in order to find the value of { ๐ถ } we let speed { ๐ = 0}, then { ๐พ๐๐๐ = 0 } because is not moving as a result { ๐พ = 1 } as a result { ๐ถ = โ๐๐2}. Consequently we have define the relativistic kinetic energy and is given by the following equation
๐พ๐๐๐ = ๐๐2 ๐พ โ 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 { ๐ธ๐ก๐๐ก๐๐ } and {๐ฟ} in our system. Then { ๐๐๐๐ก } is
given by the equation (68) and { ๐ธ๐ก๐๐ก๐๐ } by equation (66), we also know that angular momentum of a relativistic particle in polar form can be given by
๐ฟ = ๐พ๐๐2๐ (130)
In this equation the only difference from classical one (84) is the { ๐พ } factor. Now using equation (121) we substitute equation (88) and we find that { ๐พ } in its polar form is given by
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๐พ =1
1โ๐ 2๐2+๐ 2
๐2
(131)
Then using Leibniz notation (130) can be written as
๐๐
๐๐ก=
๐ฟ
๐๐พ ๐2 (132)
We then introduce the variable given by (108) and using equation (109) and chain rule
{ ๐๐
๐๐กโ๐๐ก
๐๐ข=
๐๐
๐๐ข } we have
๐ฟ
๐๐พ ๐2 โโ๐2
๐ =
๐ฟ
๐๐พ ๐ =
๐๐
๐๐ข
So
๐ =๐๐ข
๐๐
๐ฟ
๐๐พ (133)
Now using (131) we can rearrange it and write that
1
๐พ2 = 1 โ๐ 2
๐2 โ๐2๐ 2
๐2 (134)
Then we substitute (133) in (134) results in
1
๐พ2 = 1 โ ๐๐ข
๐๐
2 ๐ฟ2
๐2๐2๐พ2 โ ๐2 ๐ฟ2
๐2๐พ2๐4
Then taking {โ๐ฟ2
๐2๐2๐พ2 } as a common factor we have
1
๐พ2 = 1 โ๐ฟ2
๐2๐2๐พ2 ๐๐ข
๐๐
2+ ๐ข2
1 = ๐พ2 โ๐ฟ2
๐2๐2 ๐๐ข
๐๐
2+ ๐ข2
๐พ2 = 1 +๐ฟ2
๐2๐2 ๐๐ข
๐๐
2+ ๐ข2 (135)
We have found factor { ๐พ2 } in respect of a non-linear first order differential equation. By means of equation (66) and (129) we can write that
๐ธ๐ก๐๐ก๐๐ = ๐๐2 ๐พ โ 1 + ๐๐๐๐ก โ ๐ธ๐ก๐๐ก๐๐ = ๐๐2๐พ โ ๐๐2 + ๐๐๐๐ก
๐ธ๐ก๐๐ก๐๐ โ๐๐๐๐ก
๐๐2 + 1 = ๐พ (136)
Now we square ๐พ and we find { ๐พ2} in respect of energies thus we have
๐พ2 = 1 +2 ๐ธ๐ก๐ ๐ก๐๐ โ๐๐๐๐ก
๐๐2 + ๐ธ๐ก๐๐ก๐๐ โ๐๐๐๐ก
2
๐๐2 (137)
Now we equate (135) with (137) and we obtain
1 +2 ๐ธ๐ก๐๐ก๐๐ โ๐๐๐๐ก
๐๐2 + ๐ธ๐ก๐๐ก๐๐ โ๐๐๐๐ก
2
๐๐2 = 1 +๐ฟ2
๐2๐2 ๐๐ข
๐๐
2+ ๐ข2
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2 ๐ธ๐ก๐๐ก๐๐ โ๐๐๐๐ก
๐๐2 + ๐ธ๐ก๐๐ก๐๐ โ๐๐๐๐ก
2
๐๐2 =๐ฟ2
๐2๐2 ๐๐ข
๐๐
2+ ๐ข2
2 ๐ธ๐ก๐๐ก๐๐ โ๐๐๐๐ก
๐๐2 +๐ธ๐ก๐๐ก๐๐
2
๐2๐4 โ2๐ธ๐ก๐๐ก๐๐ ๐๐๐๐ก
๐2๐4 +๐๐๐๐ก
2
๐2๐4 =๐ฟ2
๐2๐2 ๐๐ข
๐๐
2+ ๐ข2
2๐ธ๐ก๐๐ก๐๐
๐๐2 โ2๐๐๐๐ก
๐๐2 +๐ธ๐ก๐๐ก๐๐
2
๐2๐4 โ2๐ธ๐ก๐๐ก๐๐ ๐๐๐๐ก
๐2๐4 +๐๐๐๐ก
2
๐2๐4 =๐ฟ2
๐2๐2 ๐๐ข
๐๐
2+ ๐ข2
Then simplifying further we get
2
๐2๐ ๐ธ๐ก๐๐ก๐๐ โ ๐๐๐๐ก +
๐ธ๐ก๐๐ก๐๐2
2๐๐2 โ2๐ธ๐ก๐๐ก๐๐ ๐๐๐๐ก
๐๐2 +๐๐๐๐ก
2
2๐๐2 =๐ฟ2
๐2๐2 ๐๐ข
๐๐
2+ ๐ข2
๐ธ๐ก๐๐ก๐๐ 1 +๐ธ๐ก๐๐ก๐๐
2๐๐2 โ ๐๐๐๐ก 1 +๐ธ๐ก๐๐ก๐๐
๐๐2 +๐๐๐๐ก
2
2๐๐2 =๐ฟ2
2๐
๐๐ข
๐๐
2+ ๐ข2
๐ธ๐ก๐๐ก๐๐ 1 +๐ธ๐ก๐๐ก๐๐
2๐๐2 โ ๐๐๐๐ก 1 +๐ธ๐ก๐๐ก๐๐
๐๐2 โ๐๐๐๐ก
2๐๐2 =๐ฟ2
2๐
๐๐ข
๐๐
2+ ๐ข2 (138)
Now by letting
๐๐๐๐กโฒ = โ๐๐๐๐ก 1 +
๐ธ๐ก๐๐ก๐๐
๐๐2 โ๐๐๐๐ก
2๐๐2 (139)
And
๐ธ๐ก๐๐ก๐๐โฒ = ๐ธ๐ก๐๐ก๐๐ 1 +
๐ธ๐ก๐๐ก๐๐
2๐๐2 (140)
Which here equation (138) takes on the form
๐ฟ2
2๐
๐๐ข
๐๐
2+ ๐ข2 = ๐๐๐๐ก
โฒ + ๐ธ๐ก๐๐ก๐๐โฒ (141)
Deriving relativistic Binetโs equation
Subsequently we can differentiate (141) in respect of { ๐ข} 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 Binetโs equation (classical case), the left hand side will transform exactly like Binetโs 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 {๐๐๐๐ก = โ๐
๐= โ๐๐ข } hence
when we substitute into equation (139) we have
๐๐๐๐กโฒ = ๐๐ข 1 +
๐ธ๐ก๐๐ก๐๐
๐๐2 +๐๐ข
2๐๐2 (142)
We the substitute equation (142) and (140) back into (141) and we obtain
๐ฟ2
2๐
๐๐ข
๐๐
2+ ๐ข2 = ๐๐ข 1 +
๐ธ๐ก๐๐ก๐๐
๐๐2 +๐๐ข
2๐๐2 + ๐ธ๐ก๐๐ก๐๐ 1 +๐ธ๐ก๐ ๐ก๐๐
2๐๐2 (143)
Now when we differentiate equation (143) in respect of { ๐ข } we obtain the second order differential equation of the orbit for a relativistic particle
๐2๐ข
๐๐2 + ๐ข = ๐ +๐๐ธ๐ก๐๐ก๐๐
๐๐2 +2๐2๐ข
๐๐2 ๐
๐ฟ2 (144)
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In order to check if this equation satisfies the classical one (when the particle is not relativistic) we
let { ๐๐๐๐๐ก ๐ โ โ } when we do this we can observe that we are left with { ๐2๐ข
๐๐2 + ๐ข = ๐๐
๐ฟ2 } which is
Binetโs equation [7].
Solving classical and relativistic Binetโs 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
๐ข = ๐๐๐ (145)
Then we differentiate equation (145) twice in respect of { ๐} and yields
๐๐ข
๐๐= ๐๐๐๐
๐2๐ข
๐๐2 = ๐2๐๐๐ (146)
When we substitute (146) equation (114) or (119) takes the form
๐2๐๐๐ + ๐๐๐ =๐๐
๐ฟ2
๐๐๐ ๐2 + 1 =๐๐
๐ฟ2 (147)
The solution of this type of differential equation is given in this form
๐ข = ๐๐๐๐๐๐๐๐๐๐ก๐๐๐ฆ ๐๐ข๐๐๐ก๐๐๐ + ๐๐๐๐๐๐ข๐๐๐ ๐๐๐ก๐๐๐๐๐ (๐) (148)
As a result in order to find the complementary function we let the right hand side of equation (147) equals with zero
๐๐๐ ๐2 + 1 = 0 (149)
Therefore is either { ๐๐๐ = 0 } or the auxiliary polynomial { ๐2 + 1 = 0 } that turns equation into
zero, but { ๐๐๐ > 0 } as a result
๐2 + 1 = 0
๐ = ยฑ๐ (150)
Thus the particular solution for equation (150) is given by
๐ข = ๐น cos ๐ + ๐บ sin ๐ (151)
Then we can change solution (151) using the trigonometric identity
๐ cos ๐ โ ๐0 = ๐ cos ๐0 cos ๐ โ ๐ sin๐0 sin ๐ (152)
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Where
๐ cos ๐0 = ๐น
And
๐ sin ๐0 = ๐บ And ๐0 = 0 (153)
Hence (151) becomes
๐ข = ๐ cos ๐ (154)
Then we continue, to find the particular integral, this is done by letting
๐ข = ๐ (155)
We next differentiate equation (155) twice in respect of { ๐ } and we have
๐๐ข
๐๐= 0
๐2๐ข
๐๐2 = 0 (156)
After that we substitute equation (156) into (114) or (119) and we obtain the particular integral which is given by
๐ =๐๐
๐ฟ2 (157)
Now the complete solution can be written as
๐ข = ๐ cos ๐ +๐๐
๐ฟ2
๐ข = ๐๐
๐ฟ2 ๐ ๐ฟ2
๐๐cos ๐ + 1
1
๐=
๐๐
๐ฟ2 ๐ ๐ฟ2
๐๐cos ๐ + 1 (158)
Which is the same as equation (57)[8]. By means of equation (57) we observe that { ๐ ๐ฟ2 = ๐ด } so ({๐ } is a constant), hence we have
๐ =๐ด
๐ฟ2 (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
๐2๐ข
๐๐2 + ๐ข = ๐ +๐๐ธ๐ก๐๐ก๐๐
๐๐2 +2๐2๐ข
๐๐2 ๐
๐ฟ2
๐2๐ข
๐๐2 + ๐ข = ๐๐
๐ฟ2 +๐๐ธ๐ก๐๐ก๐๐
๐ฟ2๐2 +2๐2๐ข
๐ฟ2๐2
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๐2๐ข
๐๐2 + ๐ข โ2๐2๐ข
๐ฟ2๐2 = ๐๐
๐ฟ2 +๐๐ธ๐ก๐๐ก๐๐
๐ฟ2๐2
๐2๐ข
๐๐2 + ๐ข 1 โ2๐2
๐ฟ2๐2 = ๐๐
๐ฟ2 +๐๐ธ๐ก๐๐ก๐๐
๐ฟ2๐2
๐2๐ข
๐๐2 + ๐ข๐2 = ๐ (163)
Where
1 โ2๐2
๐ฟ2๐2 = ๐2
And
๐๐
๐ฟ2 +๐๐ธ๐ก๐๐ก๐๐
๐ฟ2๐2 = ๐ (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
๐ข = ๐ cos๐2๐ + ๐
๐ข = ๐ ๐
๐cos ๐2๐ + 1 (162)
Now using equation (159) and (161) we can rewrite equation (162) in this form
๐ข =๐๐
๐ฟ2 +๐๐ธ๐ก๐๐ก๐๐
๐ฟ2๐2 ๐ด
๐๐ +๐๐ธ๐ก๐๐ก๐๐
๐2
cos 1 โ2๐2
๐ฟ2๐2 ๐ + 1
1
๐=
๐๐
๐ฟ2 +๐๐ธ๐ก๐๐ก๐๐
๐ฟ2๐2 ๐ด
๐๐ +๐๐ธ๐ก๐๐ก๐๐
๐2
cos 1 โ2๐2
๐ฟ2๐2 ๐ + 1
By the means of equation (74) we can show that
1
๐=
๐๐
๐ฟ2 +๐๐ธ๐ก๐๐ก๐๐
๐ฟ2๐2
๐2 1+2๐ธ๐ก๐๐ก๐๐ ๐ฟ2
๐ ๐2
๐+๐ธ๐ก๐๐ก๐๐
๐2 cos 1 โ
2๐2
๐ฟ2๐2 ๐ + 1
(163)
Here we can observe the modified eccentricity and the semi-latus.
๐ โฒ =
๐2 1+2๐ธ๐ก๐๐ก๐๐ ๐ฟ2
๐ ๐2
๐+๐ธ๐ก๐๐ก๐๐
๐2
1
๐
โฒ=
๐๐
๐ฟ2 +๐๐ธ๐ก๐๐ก๐๐
๐ฟ2๐2 (164)
Now in order to check if (163) satisfies the classical one (when the particle is not relativistic) we let
{ ๐๐๐๐๐ก ๐ โ โ } and it turns out to be { 1
๐=
๐๐
๐ฟ2 (1 +๐ด
๐๐cos๐)} 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 {๐ โฒ = 0.5} and we vary only the {๐2})
Now keeping the modified eccentricity the same all the time, we check how the shape with {๐ โฒ =0.5} changes when we vary only the constant { ๐2} in front of { ๐}. Hence the following orbits shown in Figure 9 take place
1. ๐ โฒ = 0.5 1 โ2๐2
๐ฟ2๐2 = 0.9
2. ๐ โฒ = 0.5 1 โ2๐2
๐ฟ2๐2 = 2
3. ๐ โฒ = 0.5 1 โ2๐2
๐ฟ2๐2 = 3
4. ๐ โฒ = 0.5 1 โ2๐2
๐ฟ2๐2 = 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 Noetherโs theorem. In Part 2 at first using classical mechanics we derived the second order differential (Binetโs) equation of an orbiting particle with a reduced mass {๐}. 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 { ๐2} is multiplied by { ๐}. This constant multiplied by { ๐} 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 donโt 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 โRunge-Lenzโ 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.
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