Doppler Effect Of Gravitational Field

Discussion in 'Alternative Theories' started by TonyYuan, May 27, 2020.

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1. TonyYuanRegistered Senior Member

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Neddy, I know what you want to express now. Below is the explanation I gave.

In step1, in the integral calculation, the force used is based on the gravitational force when the instantaneous velocity v of the object m at any position is 0, and the average gravitational force reflects the equivalent gravitational force at different positions of the object. . The velocity of gravity relative to m is c-0.

The real situation is that the object has an instantaneous velocity v at any position. We introduced boundary conditions and finally obtained the relationship between gravity and distance and velocity according to the linear relationship between the equivalent gravity and the velocity v of m. .

Then you may ask, the premise of our deduced average gravity is that the instantaneous velocity of any position of the object is 0. But it is not. Will the change of this premise affect the correctness of the derivation?

In fact, we can continue to maintain the position change velocity v of the object, and let the instantaneous velocity of the object at any position continue to be 0. We only need to adjust the instantaneous velocity of the gravity at any position from the original c to the current c-v. Let's look at the Newtonian gravitational model. This model reflects the relationship between gravitation and distance. The velocity of gravitation relative to the object is not directly reflected, but is implicit in the gravitational constant G. So after the gravitational velocity changes now, if you continue to establish only the relationship between gravitation and distance, then G needs to be changed. We need to derive G after the gravitational velocity becomes c-v.

Because the gravitational velocity changes from c to c - v, it does not affect the linear relationship between the average gravitational force and the object velocity derived previously. So we get K = G*M*m/R^2/ (c*T) through the boundary conditions v=0, v=c, and then substitute it into F(v) = G*M*m/R^2 - K*v *T, get F(v) = G*M*m/R^2 * ((c - v)/c). If v is a constant, then we can write it like this F(v) = (G* (c - v)/c )*M*m/R^2, where G*(c - v)/c is the new gravitational constant.

Now can you understand why here F(0) / F(c) = (R + c*T) / R is right ?

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3. Neddy BateValued Senior Member

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Tony,

I have a couple of questions.

1. Why do you say that F(v) and v have a linear relationship, when it appears to me that F(v) and (R + v*T) have a linear relationship? (Where (R + v*T) is the final separation distance.)

2. Why do you say that F(c) = 0? Are you thinking that as mass approaches the speed of light it tends to gravitate less? Why not more, as one would think considering relativistic mass?

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5. TonyYuanRegistered Senior Member

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440
y = a*x + b, this is the most common linear equation, y and x is a linear relationship. If we draw a trajectory diagram of y, x, it is a straight line.
1. The premise of SR is that the speed of light is constant. But the assumption that the speed of light is constant does not hold. GR has made it clear. SR is a mathematical game.
2. I have done a lot of discussion about SR in the forum. In fact, the purpose of my coming here was to refute SR. Only later, as the discussion deepened, I proposed the Doppler effect of the gravitational field. After we finish this discussion, I would like to have a detailed analysis of SR with you again.

I hope that in the discussion of "Doppler effect of gravitational field", we do not introduce anything SR, we use classical physics and mathematics to analyze the discussion.

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7. Neddy BateValued Senior Member

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Are you saying this equation...
F(v) = (G*M*m) / (R^2 * (1 + ((v/R) * T)))
...is of the form...
F(v) = av + b
?
If so, what is a and what is b?

So, without getting into relativity theory, what is your motivation for setting F(c) = 0 instead of, for example F(c) = infinity? Where does this idea come from, and what is your reasoning?

8. TonyYuanRegistered Senior Member

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440
F(v1) - F(v2) = K*(v2-v1)*T. It can be seen that F(v1) - F(v2) and v2-v1 is proportional.

From this equation F = G*M*m/(R^2/(1+v/R*T)), we can't see the linear relationship between F and v (or 1+v/R*T).
But from F(v1) - F(v2) = K*(v2-v1)*T, we can see the linear relationship.

I have already explained on post#25 .
In addition, let me give another example: a bullet, its speed is v0 = 1000 m / s, to shoot a dog, then will this dog be hurt? I think everyone will say this depends on how fast the dog escapes. If the dog's escape speed is much less than v0, then it may be killed. But if the dog's escape speed is slightly less than v0, then although it will be hit, it will hardly be hurt.

If your speed is as fast as your voice, can you still hear the voice?
If you are as fast as light, can you still see the image?
If your speed is as fast as gravity, can gravity still affect you?

Last edited: Jun 5, 2020
9. Neddy BateValued Senior Member

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2,115
Okay, yes I see the linear relationship between F(v1)-F(v2) and (v2-v1). So, when you said, "This is just a mathematical proof that they ( F(v) and v ) are a linear relationship," you did not mean that F(v1) and v1 are in a linear relationship, or that F(v2) and v2 are in a linear relationship, correct?

Okay so you are assuming that gravity moves outward from mass M at the speed of light, c, and that if m is moving away from M at the speed v, then it experiences less gravity because the speed of that gravity, according to m, would be c-v, is that it? If so, shouldn't the gravity increase when m moves toward M instead of away from it? Does that happen with your equations when the sign of v is reversed?

By the way, to answer your last question, light moves at the speed of c and yet it is supposed to be affected by gravity. Do you account for that somehow?

10. TonyYuanRegistered Senior Member

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440
Yes, It does not matter whether F(v1) and v1 are linear.
Yes, the gravity increase when m moves toward M instead of away from it. It will happen. It can be seen from the formula F(v) = G*M*m/R^2*((c-v)/c). v is velocity, not speed.
If your speed is as fast as your voice, can you still hear the voice?
If you are as fast as light, can you still see the image?
If your speed is as fast as gravity, can gravity still affect you?

This requires looking at the component of the speed of light in the direction of the gravitational field. If the direction of the light is the same as the direction of the gravitational field, then the light will no longer be affected by gravity. Otherwise, it will continue to be affected by gravity.

Using the Doppler effect of gravitational field can easily explain the light bending and Morley experiment.
In particular, the Morley experiment does not need to introduce the assumption that the speed of light is constant. (When this topic is completed, I can do a detailed elaboration with you)

Last edited: Jun 5, 2020
11. Neddy BateValued Senior Member

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2,115
In that case, I think I have completed step 3, because this equation...
F(v1) / F(v2) = (R + v2*T) / (R + v1*T)

...logically leads to these equations...
F(v1) = K*(R + v2*T)
F(v2) = K*(R + v1*T)

...and these...
F(v1) - F(v2) = K*(R + v2*T) - K*(R + v1*T)
F(v1) - F(v2) = K*R + K*v2*T - K*R - K*v1*T
F(v1) - F(v2) = (K*v2*T) - (K*v1*T)
F(v1) - F(v2) = K*T*(v2 - v1)

Okay, I understand your idea a little better now.

And yet gravitational redshift occurs when light travels in the direction directly away from the center of the gravitational body. Do you have an explanation for that?

If the speed of light is not constant, then what number are we supposed to use for c in your equations?

12. TonyYuanRegistered Senior Member

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440
I haven't studied gravity redshift before. I need a little time to understand this aspect. Maybe you can give me some information about gravity redshift.

If a spaceship wants to fly away from the earth, and its speed direction is the same as the direction of the gravitational field, then under the condition that the engine power is unchanged, its acceleration will become larger and larger, and its speed will become faster and faster. When the speed reaches the speed of light, then it will no longer be affected by gravity, and the acceleration will reach the maximum value.

I think the photon's idea should be the same as the spaceship.

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, this is my interpretation of gravity redshift.

The speed of light in vacuum, without interference of gravitational field, is c=3*10^8 m/s.

Last edited: Jun 5, 2020
13. Neddy BateValued Senior Member

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2,115
I assume you mean 299792458 m/s but I would like to know how you think this speed was determined experimentally "without interference of gravitational field". If that number was measured on earth, in its own gravity, then does that mean we have to use a different number in your equations, when the gravity is different than the earth's?

14. TonyYuanRegistered Senior Member

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440
The velocity of the gravitational field is not measured by me, my data is also from some sci papers and google.
But it does not matter whether the velocity of the gravitational field is c = v0, c = v1, or c = vx. What is important is that it has a velocity c. As for how the velocity of the gravitational field is measured, I haven't studied it yet, but how much does not affect our formula.
From my verification of Mercury's precession deviation of 43", the gravitational field velocity c measured by the scientists is trustworthy.

For the measurement of the speed of light, I think that for most processes, the direction of light and the direction of the gravitational field are perpendicular, that is, the effect of the gravitational field on the speed of light is negligible. But if there is a certain angle, and the light travels far enough in this measurement system, then you must consider the light bending, which I have described in detail in other posts.

15. TonyYuanRegistered Senior Member

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440
Neddy, do you have any other questions? If my answer is unreasonable, please point out.

16. TonyYuanRegistered Senior Member

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440
So far, this paper has only received the affirmation and support of Dr. Heweiping of the Massachusetts Institute of Technology, and has not received direct support from other professors and scholars. But so far no professor or scholar can point out any errors.

The Doppler effect of the gravitational field improves Newton's gravitation. It comes from the derivation of mathematics and physics. All we have done is to perfect Newton's theory. This theory is simple and easy to understand, and it will be more conducive to the research of cosmic science.

Hope that more astronomers will join our discussion.

17. (Q)Encephaloid MartiniValued Senior Member

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If the paper wasn't rubbish and someone reading it thought the paper had merit and only required some corrections, you would hear about it. The only error is your expectation that someone who knows what they're talking would not consider it rubbish.

18. TonyYuanRegistered Senior Member

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440
I can feel that your theoretical GR has been challenged.

I hope you can be like Neddy, you can ask your questions, and I will be happy to explain to you.
We are all exploring science, why can't there be a new theory or an improvement of an old theory? Will this hurt your interests?

GR is good, is Newton's gravitation bad? The revision of Newton's theory was tried several hundred years ago, but the correct method has not been found. Now I have proposed the Doppler effect of the gravitational field, which can be a good correction for Newton's gravitational equation. Is not it good?

Countless funds have been invested in GR research every year. The complexity of this theory has led to endless research. But the physical model described by GR is actually the Doppler effect of gravitational field.
I think this will touch the cake of very many people. But this will promote human research on astrophysics and the universe.

Last edited: Jun 11, 2020
19. TonyYuanRegistered Senior Member

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440
Neddy, will we continue our discussion? I want to hear your final thoughts.

20. Write4UValued Senior Member

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15,761
Question; You see a Doppler effect, but relative to what? In a gravitational field all objects are subject to some form of Doppler effect, no? Does time "stretch" in a gravitational field? Gravitational Time dilation?
http://thescienceexplorer.com/unive...n-gravitational-time-dilation#:~:text=Gravity

Last edited: Jun 13, 2020
21. (Q)Encephaloid MartiniValued Senior Member

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20,834
You have challenged nothing other than everyone's patience.

That's a lie, you have been asked repeatedly to explain yourself and you never did.

That is a wonderful example of what someone with Dunning Kruger would say.

22. TonyYuanRegistered Senior Member

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440
Time expansion is something of SR. The basis of SR is the constant speed of light. Unfortunately, GR has denied this foundation.
SR is a theory derived from a wrong assumption.

Neddy,
Do we continue our discussion? You are one of the scholars who read my thesis seriously, and I hope more scholars can participate.

Unfortunately, there are always a few people who like to talk rubbish, I hope they have not affected your analysis and judgment.

23. TonyYuanRegistered Senior Member

Messages:
440
Newton believed that the speed of gravity is infinite, and gravity is an over-range effect. But in fact gravity has speed. Therefore, the gravitational equation proposed by Newton also needs to be revised.

You may say that F=GMm/R^2 does not use the velocity parameter, but the calculated result is already quite correct. That's because the constant G contains the relationship between the velocity of gravity and the velocity of the object. Everyone must know that G is measured in a state of rest, which reflects that the difference between the velocity of the gravitational field and the velocity of the object in the direction of gravitational field is equal to c, and G is measured on this premise.

If the object has a relative velocity in the direction of the gravitational field, then this G is no longer G, but G*[(c-v)/c].

My conclusion comes from the derivation of mathematics and physics. It is also common sense.

And this theory has been verified in the calculation of Mercury precession.
Perhaps emotionally, everyone cannot accept that Newton's gravitation is corrected, GR is challenged, and SR is denied. But this is a fact! We should not be emotional in our theoretical research.