Doppler Effect Of Gravitational Field

[TonyYuan]

We derive the Doppler effect of gravitational field from mathematics and physics, and apply it to the calculation of Mercury precession. The result is consistent with general relativity.

The formula F(v)= G*M*m/R^2 * (c - v)/c, which is not from guess or hypothesis, but from mathematical and physical derivation. And but we need courage to admit its correctness, because it challenges Newton and Einstein. I hope this article can be read by astronomers, maybe this will be an amazing discovery.

I uploaded the article to the sciforums, you can join us to promote Newton's gravitational correction. You can also raise objections, and hope that our efforts can promote the development of natural science.

 

[(Q)]

Can we conclude your paper never got published?

http://www.sciforums.com/threads/why-mercurys-precession-is-not-as-big-as-earth-5-75-11-45.163035/

 

[TonyYuan]

Can we conclude your paper never got published?

Yes, you can. I tried several top journals like nature, and they all refused.

 

[TonyYuan]

There is a very famous professor of astrophysics in China. After reading my paper, he gave this reply:
"Your research involves the modification and improvement of gravity in gravitational physics, which is beyond my knowledge. I recommend you two outstanding youth funds and young Yangtze River winners in gravitational physics. They definitely have a say over me."

Below are the responses from several journal editors I submitted:
"We do not normally publish papers on alternate theories of gravity. Those should be submitted to a general physics journal. However, I note that General Relativity explains a wide variety phenomena and has passed a number of precision tests that exclude most competing theories. Proposing to explain Mercury's precession, by itself, is not very interesting."

"I regret to inform you that your above submission is considered unsuitable for publication in MNRAS due to it being out of scope. This submission is too theoretical in nature, lacking a direct application of the result to a relevant astrophysical problem, and contains insufficient references to the published astronomy literature. A theoretical journal such as CMDA may be more suitable for this submission."

"Your paper `The Doppler Effect Of Gravitational Field' looks interesting for the theoretical physics community, however, our journal has changed its publication policy in the recent past as follows: `Papers in mathematical physics or in general relativity which do not establish clear astrophysical applications will no longer be considered.' Your paper studies the Doppler Effect and its possible effect on a Gravitational Field which is a very theoretical subject which does not fit into the journal. As such I regret to inform you that your paper will not be considered further for publication by Astrophysics and Space Science."

 

[mathman]

I have no way of judging your theory. I can recommend you show specifically a situation where your theory predicts something different from gen. rel. and an experiment where this difference can be tested.

 

[TonyYuan]

I have no way of judging your theory. I can recommend you show specifically a situation where your theory predicts something different from gen. rel. and an experiment where this difference can be tested.

The gravitational equation under the Doppler effect is derived from mathematics and basic physical laws, so it is suitable for general physics experiments.
At present, I have only verified Mercury precession, and the calculated results are consistent with GR. And this theory can also explain light bending and Morley's experiment.

This theory is not complicated, but rather simple. It is Newton's gravitational correction. If there are astronomers willing to apply this theory to do more verification, that is what I hope very much.

Several first-year college students have read this paper, and they can all understand the whole derivation and the calculation process of Mercury precession. This requires only a little knowledge of calculus and basic theoretical knowledge of physics.
In fact, many people can understand it, but this requires courage to believe that this is true.

If anyone can provide the precession deviation of other planets, I can also verify.

 

[paddoboy]

Why would the gravitational field have a "Doppler Effect"?
Gravity is spacetime: Spacetime does not have any Doppler/cosmological/gravitational effects, which apply to light travelling in geodesics in that gravitational field/spacetime.

 

[TonyYuan]

Why would the gravitational field have a "Doppler Effect"?
Gravity is spacetime: Spacetime does not have any Doppler/cosmological/gravitational effects, which apply to light travelling in geodesics in that gravitational field/spacetime.

You can read my article carefully, there is a detailed derivation process above, I believe you can understand. We must believe in mathematics and basic laws of physics.

The final result I derived is that the equation looks like the Doppler effect equation.
F(v) = G*M*m/R^2 * (c-v)/c

Newton's universal gravitation formula F = G * M * m / R ^ 2 is based on a special case where there is no relative velocity in the direction of the gravitational field.
On the contrary, we must consider the impact of this velocity.
Although the speed of physical movement is often much lower than the speed of light c, this leads to Newton's gravitational force can get almost close to the correct result.
However, in the calculation of the orbits of high-speed moving objects, deviations were indeed found.

 

[(Q)]

Yes, you can. I tried several top journals like nature, and they all refused.

Yes, and based on the responses you received, they're telling you what everyone else here was telling you. Of course, that won't stop you from posting more nonsense instead of learning something.

 

[TonyYuan]

Yes, and based on the responses you received, they're telling you what everyone else here was telling you. Of course, that won't stop you from posting more nonsense instead of learning something.

You can refute my theory. My derivation and argument are clearly written on it. If you think something is wrong, you can specify it.

I don't know if there will be supporters of this theory in this forum. There are at least some opponents, which is always better for keeping silent. This is one of the reasons why I have been thanking those who opposed it before. I can't write this paper without them.

The path established by the new theory must be tortuous, but it will make scientific exploration more enjoyable. Let's continue, come up with your true skills, use physics and mathematics to overthrow me, or support me. Thank you!

 

[(Q)]

You can refute my theory. My derivation and argument are clearly written on it. If you think something is wrong, you can specify it.

That's what several people have done here already, over and over and over, but you refuse to listen to anyone. Typical crank.

 

[Neddy Bate]

Tony,

Your derivation is confusing to me. It seems like you start off by saying the gravitational force is Newtonian when v=0:
F = G * M * m / (R)^2

Then you consider that R changes linearly with time due to a constant velocity v:
F = G * M * m / (R + (v * t))^2
Which is still Newtonian, just with R changing over time...

...And then you consider that maybe v=0, or maybe v=c, and then somehow you conclude:
F = G * M * m / (R)^2 * [(c - v) / c]

Your conclusion ends up contradicting your premises.

 

[TonyYuan]

Tony,

Your derivation is confusing to me. It seems like you start off by saying the gravitational force is Newtonian when v=0:
F = G * M * m / (R)^2

Then you consider that R changes linearly with time due to a constant velocity v:
F = G * M * m / (R + (v * t))^2
Which is still Newtonian, just with R changing over time...

...And then you consider that maybe v=0, or maybe v=c, and then somehow you conclude:
F = G * M * m / (R)^2 * [(c - v) / c]

Your conclusion ends up contradicting your premises.

For gravitational models such as M and m, in a very short time T, the motion of m can be regarded as a uniform linear motion. The definition of F (v) in the article is the average gravitational force of m in T time. So in a short time T, the distance between m and M is R, the velocity of m is v, and the average gravitational force it receives is
F (v) = G * M * m / R^ 2 * [ (c - v) / c ].

I hope my explanation will help you understand.

 

[Neddy Bate]

Except, according to your derivation, the premise is that the force is the following at any given moment in time (assuming constant velocity):
F = G * M * m / (R + (v * t))^2
Where R is the "first" separation distance, and (R + (v * t)) is the "second" separation distance.

And that is not equal to:
F = G * M * m / (R)^2 * [(c - v) / c]
Where R is the "second" separation distance.

Do you think that F(v) and F(t) are different somehow? Don't both represent the force between M and m at the moment in time when they are at the "second" separation distance?

 

[TonyYuan]

Except, according to your derivation, the premise is that the force is the following at any given moment in time:
F = G * M * m / (R + (v * t))^2
Which is not equal to
F = G * M * m / (R)^2 * [(c - v) / c]
And both equations already assume that the motion of m can be regarded as a uniform linear motion, for a short enough period of time.

Do you think that F(v) and F(t) are different somehow? For the force between M and m at any moment in time?

F (t) = G * M * m / (R + (v * t)) ^ 2, the gravitational force received at time t.
F (v) = G * M * m / (R) ^ 2 * [(c-v) / c], the average gravitational force received during the period T.

When the elapsed time T is very short,
F (v) shows the characteristic at time t = 0, and the velocity of m is v(t=0)=v.
F (t) shows the characteristic at time t, and the velocity v of m is v (t) which changes with time t.

 

[Neddy Bate]

F (t) = G * M * m / (R + (v * t)) ^ 2, the gravitational force received at time t.
F (v) = G * M * m / (R) ^ 2 * [(c-v) / c], the average gravitational force received during the period T.

When the elapsed time T is very short,
F (v) shows the characteristic at time t = 0, and the velocity of m is v(t=0)=v.
F (t) shows the characteristic at time t, and the velocity v of m is v (t) which changes with time t.

Assuming a very very short period of time, this equation...
F = G * M * m / (R + (v * t))^2 .........................................(1)
...would become this equation...
F = G * M * m / (R)^2 .........................................(2)

And you use both of those to conclude that, (somehow?), for a very very short period of time, the equation should be...
F = G * M * m / (R) ^ 2 * [(c-v) / c] .........................................(3)

Yet equation (2) is not equal to equation (3). So I would ask which equation you think is the correct equation for the force, assuming a very very short period of time?

 

[Neddy Bate]

Tony,

To make things perfectly clear, just use a time interval of zero. For m moving at great velocity toward M, what is the force between the two?

I assume you will say:
F = G * M * m / (R) ^ 2 * [(c-v) / c]

Yet, the equation you use as a premise to "derive" that equation, says it would be:
F = G * M * m / (R)^2
Which does not even have any velocity variables.

This is as nonsensical as starting with the premise 1=1, and then "deriving" (mistakenly) the false conclusion that 1=2.

 

[TonyYuan]

Assuming a very very short period of time, this equation...
F = G * M * m / (R + (v * t))^2 .........................................(1)
...would become this equation...
F = G * M * m / (R)^2 .........................................(2)

And you use both of those to conclude that, (somehow?), for a very very short period of time, the equation should be...
F = G * M * m / (R) ^ 2 * [(c-v) / c] .........................................(3)

Yet equation (2) is not equal to equation (3). So I would ask which equation you think is the correct equation for the force, assuming a very very short period of time?

F = G * M * m / (R + (v * t))^2 .........................................(1) if t=0, then F = G * M * m / (R)^2. Nothing wrong.
This is like an accelerating object. At t = 0 we cannot know his acceleration. We must know the acceleration of this object after a certain time.
The question you asked is a similar question.

F = G * M * m / (R) ^ 2 * [(c-v) / c] , This formula embodies an average gravitation, which reflects the characteristics within a certain time T.

 

[TonyYuan]

Yet, the equation you use as a premise to "derive" that equation, says it would be:
F = G * M * m / (R)^2

It is F = G * M * m / (R + (v * t))^2

You can understand it this way:
1. Newton's gravitation reflects a static characteristic of gravity.
2. Doppler gravitation reflects the dynamic characteristics of gravity.

Like the Doppler effect of sound. Listening to the whistle of a stationary car is different from listening to the whistle of a moving car.

 

[TonyYuan]

(somehow?)

Neddy , I described the derivation process.

The derivation process is to accumulate the impulse F (t) * t in the time T to obtain P, and then average it to the time T. When the time T is very small, m can be regarded as a uniform linear motion. Then we get P = F (v) * T and F (v) = P / T. Then we proved that [ F (v1) - F (v2) ] and (v1 - v2) are a linear relationship. Finally, we derive the final Doppler gravitation formula F(v) = G * M * m / (R) ^ 2 * [(c - v) / c] by the boundary condition F (v=0) = G * M * m / R ^ 2 and F (v=c) = 0.