pengkuan

Yes. It is a specific solution for Schwarzschild. As stars and black holes give Schwarzschild metric, it is useful to have Schwarzschild solution. I had a notice from the moderator. For complying I have opened a new topic for my Schwarzschild solution of which the link is below. In the new topic, I have given a more concise version.

Orbit equation and orbital precession General Relativity explains gravity as Space-Time curvature and orbits of space objects as Space-Time geodesics. This concept is extremely hard to understand and geodesics hard to compute. However, if we can express Space-Time geodesics in analytical form, that is, as orbit equation like Newtonian orbit equation for planets, relativistic gravity will be much simpler and most people can understand. From gravitational and inertial accelerations, I have derived the orbit equation for relativistic gravity, see equation (57), of which I will explain the derivation below. I will refer this equation as the relativistic orbit equation. Albert Einstein had correctly predicted the orbital precession of planet Mercury which had definitively validated General Relativity. The orbital precession that this orbit equation gives is identical to the one Albert Einstein had given [1][2], see equation (65). Because the relativistic orbit equation gives the same result than Space-Time geodesics, it is an analytical expression of Space-Time geodesics which is so simple that personal computer can use to compute geodesics rather than big or super computer. The direct derivation from gravitational and inertial accelerations is a new insight into General Relativity. Relativistic dynamics Velocity in local frame Figure 1 shows a body of mass m orbiting around a body of mass M which sits at the origin of the polar coordinate system. The position of m with respect to M is specified by the radial position vector r, of which the magnitude is r and the polar angle is q. Let frame_m be an inertial frame of reference that instantaneously moves with m. Newton’s laws apply in frame_m because m’s velocity is 0, so the inertial acceleration vector of m am equals the gravitational acceleration ag, see (1). Let frame_l be the local frame of reference in which M is stationary, m move at the velocity vl and the acceleration of m is a l. As frame_m moves with m, its velocity is vl in frame_l. We transform al into am using the transformation of accelerations between relatively moving frames, see equation (18) in «Relativistic kinematics and gravity»[3][4]. This transformation is in (2) where we have replaced a1 with al, a2 with am and u with vl. Equating (1) with (2) we get (3), both sides of which are dotted with the vector vl·dtl in (4) to give on the left hand side, see (5), and on the right hand side, see (6). Plugging (5) and (6) into (4) gives (7), which we integrate to give (8), with K being the integration constant. The value of eK is given in (9). Rearranging (8) gives (10), which expresses the local orbital velocity vl due to the gravitational field of M. Velocity in the frame at infinitely far The orbit of a planet is the curve drawn by its position with respect to a frame in which the gravitational effect of M is 0. This frame is infinitely far from M and is labeled frame_¥. In this frame the time is t¥, the radial position vector is r and the velocity vector v¥ equals the derivative of r with respect to t¥, see (11). The velocity vector vl in frame_l equals the derivative of the same r with respect to the time in frame_l, tl. We transform vl into the product of v¥ and in (12). is the time dilation factor of the Schwarzschild metric around M, see (13). By plugging (13) into (12) we obtain the relation between vl and v¥ in (14). Substituting (14) for vl in (10) gives (15), which expresses v¥ with the gravitational field of M. In the case of planets, is very small and the term is simplified into the Taylor’s series in (16). Notice that the right hand side of (15) is the product of two parentheses, the first represents gravitational acceleration, see (10), the second represents the time dilation of Schwarzschild metric, see (13). Substituting (16) for in (15) gives (17), from which we drop out the third order term to get in (18), where u¥ is the velocity vector of m in frame_¥ correct to the 2nd order of . Differential equation for orbit The general expression of velocity in polar coordinates is given in (19) and the velocity squared u¥2 in (20). For expressing u¥2 with h the constant of Kepler’s second law given in (21) [5][6], we transform in (22), which is plugged with (21) into (20) to give (23), which expresses u¥2 in terms of instead of r. We divide (23) with c2 to express in terms of , see (24). For transforming (18), we plug (24) and into (18) to get (25) and (26), which is the differential equation for orbit in frame_¥. Solving (26) will give the relativistic orbit equation. … Figures and equations are in the pdf here:Analytical equation for Space-Time geodesics and relativistic orbit equation.pdf

Sorry. I will post in a new topic.

General equation for Space-Time geodesics and orbit equation in relativistic gravity 1. 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. 2. 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 polar angle 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 velocity vector 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. … Figures and equations are in the pdf below: General equation for Space-Time geodesics and orbit equation in relativistic gravity.pdf

We analyze the mathematical mechanism that slows the time of the traveler in the twin paradox and explain what distinguishes the traveler's frame from the Earth's frame Please read the article at https://www.scienceforums.net/applications/core/interface/file/attachment.php?id=18516 PDF: Twin paradox when Earth is the moving frame url removed or Word: https://www.academia.edu/39216040/Twin_paradox_when_Earth_is_the_moving_frame

Thanks Thanks I just want to know the name of this philosophy.

I agree with what you say. Is it a school of thought in mathematics? Are there many people who think like you ?

I know that this is the well accepted theory. But it would be great if the contrary could be proven. Yes, that is impossible. But one can always think otherwise.

I do not think that a line is a set of points just because points cannot be fill all holes. But this is another story. I agree that real numbers are countable.

I will change my theory to handle infinity.

No. Length is the number of member of a series. So, it is a natural number.

It seems that everyone thinks that natural numbers have finite values while the entire set in infinite. I'm OK with that. But in this case, the length of the set of all even numbers is finite, because it's a natural number. Actually, one cannot pass from a finite number, the length of a finite set, to infinite number, the length of a infinite set, which is the finite set when its length is stretched to infinity. Actually, if the price is 1111....., you can double it, 2*1111...=2222... Thanks for your help. I think within the set of natural number, the ordinal numbers are natural numbers.

I mean that because you say “i can be an arbitrarily large finite number.”, i and j cannot be infinitely big, for example, a number with infinitely many digits like 9517452…… or \(10^{\infty }\). In this case, can we write the set of natural numbers as {1,2,3,…,n-1,n}, with n being a finite number with arbitrarily large value? If i had value 9517452…… , then the ratio i/1 would not have corresponding natural number.

I was restating your explanation above with my words. So, in the counting of \(\mathbb{Q}\), the ratio i/j corresponds to the number n which is finite. For the ratio i/1, the number of count is: \[n=\frac{i(i+1)}{2}\] So, if n is a finite number, i is also finite. Does this mean that the number i is not allowed to go to infinity? In this case, \(\mathbb{Q}\) does not completely cover the plane \(\mathbb{N}\times \mathbb{N}\) . Can we say that \(\mathbb{Q}\) is countable only because i and j are not allowed to have infinite value?

Yes, you are right. So, the set of the rationals is countable because every i/j corresponds to a finite number n in the counting order. At nth step, we stop the count. Although the set itself is infinite, all counting numbers are finite. But for the power set of N, the set of all even numbers corresponds to the infinite binary sequence 1010101010...... To reach it we cannot stop counting because this sequence is not finite. Can we say that the power set of N is not countable because the counting numbers of infinite subsets are infinite? Yes. I agree that I have difficulty in seizing the exact sense of the discussion.