# General equation for Space-Time geodesics and orbit equation in relativistic gravity

• Orbit equation and orbital precession

General Relativity explains gravity as Space-Time curvature and orbits of planets as geodesics of curved Space-Time. However, this concept is extremely hard to understand and geodesics hard to compute. If we can find an analytical orbit equation for planets like Newtonian orbit equation, relativistic gravity will become intuitive and straightforward so that most people can understand.

From gravitational force and acceleration, I have derived the analytical orbit equation for relativistic gravity which is equation (1). Below I will explain the derivation of this equation. Albert Einstein had correctly predicted the orbital precession of planet Mercury which had definitively validated General Relativity. Equation (2) is the angle of orbital precession that this orbit equation gives, which is identical to the one Albert Einstein had given [1][2].

If this orbit equation gave the same result than Space-Time geodesics, then everyone can compute the orbit of any object in gravitational field which obeys General Relativity using personal computer rather than big or super computer. Also, everyone can see how gravity leads to Space-Time curvature without the need of knowing Einstein tensor.

The derivation of the orbit equation is rather tedious and lengthy. So, for clarity of the reasoning and explanation, I have collected all the mathematical equations in the last section “Derivation of equations”, in which full details are provided to help readers for checking the validity of my mathematics.

• Relativistic dynamics
a) Velocity in local frame

Take an attracting body of mass M around which orbits a small body of mass m, see Figure 1. We work with a polar coordinate system of which the body M sits at the origin. The position of the body m with respect to M is specified by the radial position vector r, of which the magnitude is r and the polarangle is q.

Let the frame of reference “frame_m” be an inertial fame that instantaneously moves with m. Frame_m is the proper frame of m where the velocity of m is 0. So, Newton’s laws apply in this frame. Let am be the acceleration vector of m in frame_m and the inertial force of m is m·am, see equation (3). The gravitational force on m is given by equation (4). Equating (4) with (3), we get equation (5), the proper acceleration of m caused by gravitational force in frame_m.

Let “frame_l” be the local frame of reference in which M is stationary. In frame_l m is under the effect of gravity of M, the velocityvector of m is vl and the acceleration of m is a l. As frame_m moves with m, it moves at the velocity vl in frame_l.

The acceleration of m in frame_m and frame_l are respectivelyam and al. To transform al into am we use the transformation of acceleration between relatively moving frames which is the equation (18) in «Relativistic kinematics and gravity»[3][4], in which we replace a1 with al, a2 with am and u with vl. Then the transformation between am and a l is equation (6).

Equating (5) with (6) we get equation (7), both sides of which are then dotted by the vector vl·dtl, see equation (8). On the left hand side of (8), we find dr the variation of the radial distance r, see (9). On the right hand side, we find the variation vector of velocity dvl, see (10), which, dotted by the velocity vector vl, gives dvl2/2 in (11).

Plugging (9) and (11) into (8), we get (12), both sides of which are differential expressions, see (13) and (15). Then, plugging (13) and (15) into (12) gives (16) which is a differential equation. (16) is integrated to give (17), with K being the integration constant. Then, we rearrange (17) to express vl2/c2 in (18), which relates the local orbital velocity vl to the gravitational field of M.

The value of eK is determined at a known point 0 at which the velocity is v0 and the radial distance is r0, see (19).

General equation for Space-Time geodesics and orbit equation in relativistic gravity https://www.academia.edu/44540764/Analytical_orbit_equation_for_relativistic_gravity_without_using_Space_Time_geodesics

https://pengkuanonphysics.blogspot.com/2020/11/analytical-orbit-equation-for.html

# Explaining Oumuamua and Pioneer anomaly using Time relativity

I find that in theory the weird Speed Boost of the interstellar object ‘Oumuamua should be
0.217 mm/s above the prediction and that ‘Oumuamua should slow down less than prediction, in proportion of which the difference is 4.28×10-8 near the Sun. For Pioneer anomaly I have computed the gap between real and predicted acceleration and found the value 8.70×10-10 which is very close to the observation (8.74±1.33)×10−10 m/s2.

The mysterious interstellar object ‘Oumuamua confuses scientists because of its Speed Boost, which is an excess of velocity with respect to the expected one. In the past, the manmade Pioneer spacecrafts were also found to deviate from expected Newtonian trajectory.

One thinks the velocity of ‘Oumuamua is too high because it is faster than the expected velocity that the mass of the Sun allows. But if we have used a mass for the Sun slightly different from the real one, then the expected velocity would be not correct. So, let us see how the mass of the Sun is determined

# Relativistic kinematics and gravitation

Like in Newtonian kinematics, the relativistic change of reference frame must be a vector system of transformation laws for position, velocity and acceleration.

In special relativity, when changing the reference frame the coordinates of a moving point is transformed using Lorentz transformation. But the velocity-addition formula that transforms velocity is in a too different mathematical form than the Lorentz transformation. And for acceleration, there is not a transformation at all. The theory of Time relativity that I develop provides a coherent vector system of transformation laws for position, velocity and acceleration
https://pengkuanonphysics.blogspot.com/2020/05/relativistic-kinematics-and-gravitation.html

# ‘Oumuamua, Pioneer anomaly and solar mass with Time Relativity

The theory of Time relativity explains well the weird behavior of the interstellar object ‘Oumuamua. I find that the real solar mass is slightly higher than today’s value, which caused the mysterious Speed Boost of which the value should be 0.217 mm/s above the prediction at perihelion. Time relativity confirms that ‘Oumuamua should slow down less than prediction, in proportion of which the difference is 4.28 ×10-8 near the Sun. For Pioneer anomaly I have computed the gap between real and predicted acceleration and found the value 8.70 ×10-10 which is very close to the observation (8.74±1.33)×10−10 m/s2.

The mass of the Sun is not measured by weighting, but derived from the parameters of Earth’s orbit which is nearly circular. Let rM be the radius of the Earth’s orbit and uE its orbital velocity. By equating the orbital acceleration of the Earth (see equation (36)) with its Newtonian gravitational acceleration (see equation (34)), we obtain equation (48) which gives the today’s used value of the mass of the Sun, M0, in equation (49).

# Velocity, mass, momentum and energy of an accelerated object in relativity

Analytical derivation of relativistic velocity, mass, momentum and kinetic energy of an accelerated object. For Special relativity the momentum of an object of rest mass m0 and velocity u is expressed by equation (1) which is infinite when u equals c. Is it physically meaningful that the momentum of an object becomes infinite while its velocity stays finite? On the other hand, the principle of mass–energy equivalence proposed by Albert Einstein in his article “Does the Inertia of an object Depend Upon Its Energy Content?” has not been rigorously demonstrated, hence it is called a principle not a law. In the contrary, in the theory of Time relativity which is been developed here, momentum and kinetic energy are derived by direct integration and stays limited when u=c.…

The expression of velocity (equation (13)) is directly integrated and thus is mathematically exact. In the contrary, the velocity-addition formula in Special relativity cannot be analytically integrated and one had to make an approximation to compute the velocity of an object which is thus not exact (see section 5.3 of « Introduction to Special Relativity » by James H. Smith).

In Special relativity the expression of relativistic mass is derived with the help of a shock between 2 objects (see section 9.4 of « Introduction to Special Relativity » by James H. Smith). For Time relativity relativistic mass is the derivative of momentum with respect to velocity, which is exactly the definition of mass.

In Special relativity the expression of momentum was derived with the help of a shock between 2 objects (see section 9 of « Introduction to Special Relativity » by James H. Smith) and is infinite when the velocity equals c (see equation (1)). For Time relativity momentum is the integral of infinitesimal change of momentum (see equation (25)). When the velocity of the object equals c its momentum equals the constant π/2 m_0 c, which gives a negative answer to the question of the beginning: “Is it physically meaningful that the momentum of an object becomes infinite while its velocity stays finite? ”

For Time relativity the total kinetic energy of an object is the integral of the work done on it and thus, its expression is mathematically exact. Moreover, when the velocity of the object equals c, its expression equals m0c2 (see equation (45)), which is a proof for the the principle of mass–energy equivalence, while in Special relativity mass–energy equivalence does not has mathematical proof.

At the end, we have derived the momentum-kinetic energy relation for Time relativity, which reduces to the expression of kinetic energy for classical mechanics for small velocity, while the momentum- energy relation in Special relativity does not. In the contrary, this relation is infinite when the velocity equals c.

# Time relativity transformation of velocity

A discrepancy-free transformation of velocity is derived using the Time relativity transformation of coordinates because relativistic transformation of velocity creates a discrepancy. The relativistic transformation of velocity expresses the velocity u2 of an object q in frame 2 in terms of its velocity in frame 1. In frame 2 at time tq, the position of q is xq=u2*tq.

As q departs from the origin of frame 2, we may compute its position by using the time at the origin which is t2o. The so computed position is x2o=u2*t2o. Notice that x2o does not equal xq. So, the relativistic transformation of velocity gives 2 different positions to the object q, which we call the discrepancy of double position. The cause of this discrepancy is that the time at the origin and at the abscissa xq are not equal due to relativity of simultaneity.

Below, we derive a discrepancy-free transformation of velocity. Let us take 2 frames of reference with frame 2 moving at the velocity vo in frame 1. An object q moves at the velocity v2 in frame 2. We will derive the velocity of q in frame 1, which is v1.

The object q moves a distance during a time interval. At the start, q passes by the point x2a in frame 2 which is a fixed point of the x axis of frame 2. x2a coincides with the point x1a which is a fixed point of the x axis of frame 1, see Figure 1. At the end, q passes by the point x2b of frame 2. x2b coincides with the point x1b of frame 1, see Figure 2.

We emphasize that x2a, x2b, x1a and x1b are fixed points on their respective x axes because x2a and x2b move with frame 2 in frame 1, but x1a and x1b stay still in frame 1. x2a, x2b, x1a and x1b do not move with q. Because the point x2a moves with frame 2 at the velocity vo, it arrives at the point x’1a in frame 1 at the end of the time interval, see Figure 2.

# Time relativity transformation of coordinates

Without length contraction, time relativity transformation solves paradoxes and explains incongruent relativistic experiments, which allows us to build a transformation of coordinates without length contraction. For abscissa transformation, Figure 1 shows a spaceship in the frame of O1, its backend is at O1 and frontend at A1. At time zero the spaceship is stationary, from time zero to time t1, it is accelerated. The trajectories of its backend and frontend are the parallel curves from O1 to O2 and from A1 to A2. At time t1 the spaceship moves at the velocity v, its backend is at O2 and frontend at A2. So, the frame of O2 moves with the spaceship at the velocity v in the frame of O1.

Because the trajectories of the backend and frontend are parallel, the distance O2A2 constantly equals the distance O1A1 which is the proper length of the spaceship. O2A2 is the length of the moving spaceship in the stationary frame O1. So, the length of the moving spaceship constantly equals its proper length.

For Time transformation, The transformation of time from the stationary frame to the mobile frame is like that in special relativity. In Figure 3, at time zero a light signal is sent in the mobile frame from O2 to the mirror M which reflects it back to O2. The time the light signal takes to complete the journey is tO2=2L2/c, with c being the speed of light and L2 the distance from O2 to M. tO2 is also the time of O2. In the stationary frame the light signal is sent when O2 coincides with O1 and goes from O1 to the moving M which reflects it to O2. The time of this journey is tO1=2L1/c, with L1 being the distance from O1 to M. tO1 is also the time of O1. Because L2= L1√(1-v^2/c^2 ), tO1 is related to tO2 by equation (7).

# Length, distance and Michelson–Morley experiment

There are 2 types of length contraction, the physical meaning of each is explained below with the help of the example shown in Figure 1. The Earth and the star are stationary and the spaceship mobile. The 2 types of length contraction are:
1) Object contraction: In the frame of the Earth the length of the moving spaceship appears shorter than its proper length.
2) Distance contraction: In the frame of the spaceship the distance from the star to the Earth appears shorter than in the frame of the Earth.

a) Measurement of distance by travel
Let us focus on Distance contraction and see why distance is shorter in the moving frame. Suppose that the spaceship coincides with the Earth at time zero and coincides with the star at the end of its travel. The spaceship moves at the velocity v in the frame of the Earth and the star moves at the velocity –v in the frame of the spaceship.

In the frame of the Earth time is tE at the end of the travel and the spaceship has traveled the distance v•tE which is the distance from the Earth to the star measured by the travel of the spaceship. This distance is denoted by DE and expressed by equation (1).

In the frame of the spaceship time is tS at the end of the travel and the star has traveled the distance v•tS which is the distance from the Earth to the star measured by the travel of the spaceship. This distance is denoted by DS and expressed by equation (2)…
https://pengkuanonphysics.blogspot.com/2020/02/length-distance-and-michelsonmorley.html

# Analysis of Einstein’s derivation of the Lorentz Transformation

Einstein’s derivation of the Lorentz Transformation is purely theoretical. This study shows how it is related to the physical phenomenon of time dilation and length contraction. The Lorentz Transformation was first derived using the conditions of time dilation and length contraction. Later, Albert Einstein has given a different derivation of the Lorentz Transformation by using constancy of the speed of light only, making time dilation and length contraction subsequent to the Lorentz Transformation which then acquired the status of fundamental law.

Einstein exposed his derivation in the Appendix 1 of his 1920 book « Relativity: The Special and General Theory ». But the physical significance of this derivation is blurry. I will analyze Einstein’s theoretical derivation to find out how it is related to the physical conditions of time dilation and length contraction.

In his derivation Einstein only used the fact that the speed of light is c in K and K’. However, another claim was used without being properly declared, which is: (x,t) the co-ordinates of an event in K and (x’,t’) the co-ordinates of the same event in K’ satisfy equations (3) and (6) in general. As these equations are in direct proportional form, we give this claim the name “Proportionality assumption”. This claim is an assumption because it is not proven.

I analyze this assumption in the article below.
https://pengkuanonphysics.blogspot.com/2020/01/analysis-of-einsteins-derivation-of.html

# Synchronizing moving GPS clocks

1. Light pulse synchronization
Can we synchronize clocks of a moving frame? Let us see Figure 1 where we have stationary frame F1 and moving frame F’2. The 2 clocks in the frame F1 are synchronized with the master clock through a light pulse.

The 2 clocks of the frame F’2 are moving. At the time t0 in the frame F1, the 2 clocks in the moving frame F’2 coincide with that of the frame F1, but due to relativity of simultaneity the time of the moving clocks are different,
dt’=t’2-t’1=/=0. So, the light pulse does not synchronize the clocks in the moving frame.

2. Fixed clocks in orbit
GPS satellites are moving, so they suffer from relativity of simultaneity. Let us imagine a disk centered at the center of the Earth whose rim is the GPS orbit and on the rim are equally spaced clocks, see Figure 2. The disk does not rotate with the Earth and the frame of the disk is the frame F1. These clocks can be synchronized because they are fixed and they are synchronized with the master clock at the North Pole of the Earth through a light pulse. They all show the time t0 when each GPS satellite pass in front of each corresponding fixed clock. But the clocks of the GPS satellites are not synchronized with the master clock because they are moving.

3. Time in GPS satellites
The event the satellite 1 coincides with a fixed clock and the satellite 2 coincides with the next fixed clock are simultaneous in the frame F1. But these 2 events are not simultaneous in the frame F’2, which is the frame moving with the satellite 1 and containing the satellite 2, see Figure 3.

In the frame F’2, the clock of the satellite 1 reads t’1 and the clock of the satellite 2 reads t’2 because of non-simultaneity. So, we have the gap of time dt’=t’2-t’1=/=0.

In the same way, the event the satellite 2 coincides with a fixed clock and the satellite 3 coincides with the next fixed clock are simultaneous in the frame F1, but not in the frame F’3, which is the frame moving with the satellite 2 and containing the satellite 3, see Figure 3. Also for the same reason, the gap of time in the frame F’3 is dt’=t’3-t”2=/=0.

Note that here in the frame F’3, the time of the satellite 2 is denoted by t”2 with double prime to distinguish it from the time of the satellite 2 in the frame F’2.

4. Time of the satellite 2
We have 2 different notations of the time of the satellite 2: t’2 with single prime in the frame F’2 and t”2 with double prime in the frame F’3. The satellite 2 has only one time which is shown by its clock. For the event the satellite 2 coincides with the fixed clock in Figure 3, its value in the frame F’2 is t’2. Now we ask: does the reading of the clock of the satellite 2 change to a different value t”2 when seen in the frame F’3? Logically no, because t’2 is the time of the satellite 2 while t”2 is the time of the same satellite at the same event. So, we must have t’2 = t”2. In this case, we have
t’3 = t”2+dt’=t’2+dt’=t’1+dt’+dt’=t’1+2dt’

5. Time of the satellite n
For the same reason, the double-primed t”3 equals the single-primed t’3 and each double-primed t”i equals each corresponding single-primed t’i. Then, the n+1th satellite has the time: t'(n+1) =t’1+ndt’.

Because ndt’ is not zero, t'(n+1) =/=t’1 while the satellite n+1 being the satellite 1, hence contradiction.

6. If t'(n+1) =t’1?
We can assert that the satellite n+1 being the satellite 1 and having the same time, that is,
t'(n+1) =t’1. In this case, we first suppose that the single-primed t’2 does not equal the double-primed t”2, but instead their different cancels dt’, that is, t”2=t’2-dt’. So, t’3=t”2+dt’=t’2-dt’+dt’=t’1+dt’ .

In this case, we will have for the n+1th satellites t'(n+1) =t’1+dt’ .

However, we still do not have t'(n+1) =t’1. For achieving this equality, we have to suppose the difference of time between single-primed and double-primed time to make t”2 = t’2 -dt'(1+1/n). This way, we will have at the n+1th satellite: t'(n+1) =t’1+dt’-dt’*n/n=t’1

But this assumption is absurd and cannot be true. So, due to relativity of simultaneity, the satellite n+1 and the satellite 1 are the same satellite but cannot have the same time.

The only way for the satellite n+1 and the satellite 1 to have the same time is that relativity of simultaneity is zero, that is, dt’=0.