Jump to content

Twin paradox when Earth is the moving frame


pengkuan

Recommended Posts

 

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

Link to comment
Share on other sites

44 minutes ago, pengkuan said:

 

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

 

!

Moderator Note

Rule 2.7 says, in part, "We don't mind if you put a link to your noncommercial site (e.g. a blog) in your signature and/or profile, but don't go around making threads to advertise it."

Since there is no discussion in this thread, that's exactly what this is. Your blog's address has been removed.

You have been here long enough to know the rules. If you have something to discuss, you may post it — not links to it. If the result is inconsistent with mainstream science, it goes in speculations. If all you want to do is announce some mainstream result that's described elsewhere, though, that's not permitted.

 
Link to comment
Share on other sites

Looking at blogs.scienceforums.net/pengkuan/2019/05/ I hope I'm not violating the mod edit.

This does not agree with SR. If Betty is inertial and the Earth changes inertial frames, Betty will age more.

I haven't read the whole thing but I think you're making a mistake in section 2. You have the Earth and star S in an inertial frame, and they both change velocity by the composition of v and v, half way through the experiment. It looks like you're treating these two events as simultaneous in all frames, which you can't do. That will give you errors.

It looks like you have the Earth travel a length-contracted distance away, and then travel a length-contracted distance back, but measure that in Betty's frame. If you'd done it completely, it should work out to the same as if the Earth traveled away, stopped, and then came back, and the full separation between Earth and Betty when stopped would be the rest distance, ie. the distance between Earth and S in their frame. I suspect that you're essentially having Earth teleport twice between its length-contracted distance and rest distance, in a time that Betty counts as zero. If it was accounted, she'd age more then. This is a guess, I haven't been thorough.

Edited by md65536
Link to comment
Share on other sites

  • 1 year later...

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

Link to comment
Share on other sites

9 hours ago, pengkuan said:

I have derived the analytical orbit equation for relativistic gravity which is equation (1)

This seems to be a specific solution for a specific type of spacetime, presumably Schwarzschild (I haven’t gone through your derivation). That’s fine, assuming the derivation is algebraically correct, but only so long as your gravitational source actually does admit a Schwarzschild solution. If not, then you still need to fall back on solving the geodesic equation.

Link to comment
Share on other sites

On 11/24/2020 at 9:33 AM, Markus Hanke said:

This seems to be a specific solution for a specific type of spacetime, presumably Schwarzschild (I haven’t gone through your derivation). That’s fine, assuming the derivation is algebraically correct, but only so long as your gravitational source actually does admit a Schwarzschild solution. If not, then you still need to fall back on solving the geodesic equation.

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.

 

 

 

Link to comment
Share on other sites

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account

Sign up for a new account in our community. It's easy!

Register a new account

Sign in

Already have an account? Sign in here.

Sign In Now
×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.