Doppler Effect Of Gravitational Field

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

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  1. TonyYuan Registered Senior Member

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    This is obviously impossible. Although the experience time is very short, because of the existence of v, we must look at the size of v * t, so from the perspective of the average effect of force, it is the derivation process I did.
    First calculate the impulse P in T time, and then calculate the average F (v) in T time.
     
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  3. TonyYuan Registered Senior Member

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    I don't understand what you mean by a time interval of zero.

    Newton's gravitation is a force in a static state. If we consider the effect of speed, we need to calculate the total impulse, and then calculate the average force in T time. We can simply analyze from the boundary conditions of the gravitational field. If the speed of m leaving reaches the speed of light c, then m will no longer be affected by gravity.
    The speed direction I am talking about here is the same as the direction of the gravitational field.
     
    Last edited: May 29, 2020
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  5. TonyYuan Registered Senior Member

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  7. Neddy Bate Valued Senior Member

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    Just use t=0, like this:
    F = G * M * m / (R + (v * t))^2 .........................................(1)
    t = 0
    F = G * M * m / (R + (v * 0))^2
    F = G * M * m / (R + 0)^2
    F = G * M * m / (R)^2 .........................................(2)

    When t=0, the size of v*t is 0. You said that we must trust in mathematics. I don't understand how you can start with equations (1) and (2) as premises, and then "derive" equation (3):

    F = G * M * m / (R) ^ 2 * [(c-v) / c] .........................................(3)

    There must be some error in your derivation. Equation (3) is clearly different than equation (2), yet both of them are equations for the exact same thing, the gravitational force between M and m at a single instant of time.
     
  8. TonyYuan Registered Senior Member

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    1. First calculate the impulse P in time T
    2. Calculate the average value of gravity F (v) in time T
    3. Prove that F (v1)-F (v2) and v1-v2 are a linear relationship
    4. Use the boundary condition F (v = 0) = GMm / R ^ 2 and F (v = c) = 0
    5. Derive the final Doppler gravity equation F (v) = GMm / R ^ 2 * [(c-v) / c]

    I think that what troubles you should be at a certain moment, such as t = 0, why the following two equations are not equal.
    F (t = 0) = GMm / R ^ 2 ............................. (1)
    F (v) = GMm / R ^ 2 * [(c-v) / c] ............... (2)
    I said earlier that (1) reflects the gravity at time t, which is an instantaneous value. (2) What is reflected is the average value of equivalent gravitation in time T. Their physical meaning is different.
    Like there is a huge force F to strike someone, then this person will definitely hurt? If this force lasts for a time T, then the time effect of this force is F * T. But if the question now is how much damage this force has at time t, it can only be F * 0, because the duration of time t is 0, which is an instantaneous value.
    So the same force F, we have come up with different answers. This is why (1) and (2) are different.

    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.
    Let us come to a certain moment t, which is an instantaneous value. Can we tell if the dog will be killed? Obviously without their speed, we cannot know the damage effect of the bullet on the dog. So in order to describe this damage effect, we introduce an equivalent equation f () of damage. This equation definitely needs their relative speed as a parameter, so it becomes f (v). We then introduce an equation with time t as a parameter f (t), so the problem comes to the following scenario:
    f (t = 0) .................... (1)
    f (v) .......................... (2)
    Why is (1) and (2) not equal at t = 0?
     
  9. TonyYuan Registered Senior Member

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    Neddy,
    There are two cars, they have different initial speeds and different accelerations, then at a certain moment t, they may have the same speed, but during the time T, their average speed is different. So when we study the moving effect of cars, it is obviously more meaningful to look at their average speed.

    Similarly, the same is true for F (t) and F (v).
     
  10. TonyYuan Registered Senior Member

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    Newton's gravitational force reveals an instantaneous gravitational force between two objects M and m at a distance R. Just like speed, it is an instant value. We use the calculus method to calculate the total impulse during time T, and then to find the average gravity. Just like we integrate the velocity in the time domain to get the total displacement, and then calculate the average velocity during time T.

    From the relationship between F (v1)-F (v2) and v1-v2, they are a linear relationship, that is, F (v) and v are a linear relationship. When v = 0, the Newtonian equation of gravitation happens. So when v is equal to what, F (v) is exactly equal to 0? We know that the speed of gravity is c, so we can naturally think that if the speed of m is equal to c, m will no longer be affected by gravity. So we have a boundary condition F (v = c) = 0.
    https://photos.app.goo.gl/kQTLG4GTCgVHiqP1A

    I hope my explanation can help you understand. Thanks.
     
    Last edited: May 31, 2020
  11. TonyYuan Registered Senior Member

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    Newton's gravitation describes the equation between gravitation F and distance R, which is overrange. We write it as F (R).

    But in fact it is not the case, gravity has speed c. An object m moves at different speeds in the direction of the gravitational field, and its gravitational force F (v) is different. Take the most easily thinkable example, when v = c, m will no longer be affected by gravity, that is, F (v = c) = 0. Obviously, this result will be different from the result of Newton's gravitational calculation.

    Neddy,
    F = G * M * m / (R) ^ 2 * [(c-v) / c] , this formula reveals that gravitation is not only related to the distance R, but also to the moving speed v of the object. The characteristic it shows is the Doppler effect. If Newton could realize that the speed of gravity is limited hundreds of years ago, Newton should also propose gravitational waves and think of the existence of Doppler effect.
     
  12. exchemist Valued Senior Member

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    A gravitational field does not have speed.

    Nor does it have a wavelength.

    No wavelength, no Doppler effect.
     
    Last edited: May 31, 2020
  13. TonyYuan Registered Senior Member

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    Summarize in one sentence:
    Newton's gravitational force reveals the relationship between gravity F and distance R, which is a force under "static" conditions. Then these "static" forces are integrated through the time domain, and then averaged to time T, the average force obtained contains the influence of velocity.

    Finally, we can derive the relationship between gravity F and distance R and velocity v through the boundary conditions, F (v = 0) = GMm / R ^ 2 and F (v = c) = 0.
     
  14. exchemist Valued Senior Member

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    Gibberish.
     
  15. TonyYuan Registered Senior Member

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    Neddy,
    Can my explanation solve your doubt? Do you have any other questions?
     
  16. Neddy Bate Valued Senior Member

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    Okay, let's just say for the moment that the physical meanings are different for those equations, even though I do not necessarily agree.

    Later in your "derivation" you have these equations:
    F(v1) = p1/T = G * M * m / R^2/(1 + v1/R*T))
    F(v2) = p2/T = G * M * m / R^2/(1 + v2/R*T))

    Which means that now you essentially have this equation:
    F = G*M*m / R^2/(1 + v/R*T))

    So you are giving yet another equation for force, which is not algebraically the same as either of these:
    F = G*M*m / (R + (v * t))^2
    F = G*M*m / (R)^2 * [(c-v) / c]

    What is the difference, and why? What is the physical meaning in this case? Are t and T different also?
     
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  17. TonyYuan Registered Senior Member

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    Yes, they are different.
    t is a moment, a momentary value.
    T is the elapsed time.
    Within T seconds, there can be countless t.
    For example, T = 10 seconds, then t = 0s, t = 1.023s, t = 2.9s ...... these are all instant moments.
    For example, from 10:00 to 11:00, there is a total of 1 hour, which is T = 1 hour. But there can be countless t, t = 10:00, t = 10:05 , t = 10:39 ......


    The following F is the gravitational moment at time t.
    F = G * M * m / (R + (v * t)) ^ 2 , it's Newton's formula of gravitation. It reveals the relationship between gravity and distance.

    The following F physical meaning is the average gravity during time T.
    F = G * M * m / R ^ 2 / (1 + v / R * T)) ........................(1)
    The following F is derived from the above equation, which reveals the relationship between gravity and distance and velocity.
    F = G * M * m / (R) ^ 2 * [(c-v) / c] .............................(2)

    In the following process, you will see how to derive (2) from (1).
    1. Derive that F (v) and v are linear:
    First we derive from equation (a): F (v1) / F (v2) = (R + v2 * T) / (R + v1 * T),
    Then suppose F (v1) = K * (R + v2 * T), F (v2) = K * (R + v1 * T),
    Then by subtraction: F (v1)-F (v2) = K * (v2-v1) * T ................................(x).
    It can be seen that the difference between the equivalent average gravity and their speed difference is proportional.

    2. Further calculation by the boundary conditions of v = 0 and v = c.
    When v2 = 0 is substituted into (x), F (v1)-F (0) = -K * v1 * T,
    Because what v = 0 gets is Newton's universal gravitation F (0) = G * M * m / R,
    So F (v1) = G * M * m / R- K * v1 * T ...............................(a).
    When v2 = c, F (v1)-F (c) = K * (c-v1) * T,
    Because we know that the speed of the gravitational field is c, it is natural to think of F (c) = 0,
    So F (v1) = K * (c-v1) * T ..................................................(b).
    From equations (a), (b): K = G * M * m / R / (c * T) ......(c).
    Substituting (c) into (b) yields:
    F (v1) = K * (c-v1) * T = G * M * m / R / (c * T) * (c-v1) * T = G * M * m / R * [(c-v1 ) / c],
    Replace v1 with v, and get: F (v) = G * M * m / R * [(c-v) / c].
     
    Last edited: Jun 1, 2020
  18. Neddy Bate Valued Senior Member

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    Okay, so let me just test these two equations out. Let's say that R=10 length units and v=1 length unit per time unit. We will consider the following interval of time units:
    At time t=0 the distance between M and m = (R + (v * t)) = 10 + (1 * 0) = 10
    At time t=1 the distance between M and m = (R + (v * t)) = 10 + (1 * 1) = 11
    At time t=2 the distance between M and m = (R + (v * t)) = 10 + (1 * 2) = 12
    At time t=3 the distance between M and m = (R + (v * t)) = 10 + (1 * 3) = 13
    So the average distance is (10+11+12+13)/4 = 11.5 which when squared is 132.25.
    Or, if we square them first, and then average them, it is (100+121+144+169)/4 = 133.5.

    So let's try out your other equation:
    For a total time T=3 the squared distance between M and m = R^2 / (1 + v / R * T)) = 10^2 / (1 + 1 / 10 * 3)) is that correct?
    For a total time T=3 the squared distance between M and m = R^2 / (1 + v / R * T)) = 100 / (1 + 1 / 30)) is that correct?
    For a total time T=3 the squared distance between M and m = R^2 / (1 + v / R * T)) = 100 / (2 / 30)) is that correct?
    For a total time T=3 the squared distance between M and m = R^2 / (1 + v / R * T)) = 100 / 0.0666... is that correct?
    For a total time T=3 the squared distance between M and m = R^2 / (1 + v / R * T)) = 1500 which is nowhere near either 132.25 or 133.5.

    Or maybe you meant this?
    For a total time T=3 the squared distance between M and m = R^2 / (1 + v / R * T)) = 100 / (1 + 1 / 30)) is that correct?
    For a total time T=3 the squared distance between M and m = R^2 / (1 + v / R * T)) = 100 / (1 + 0.0333...)) is that correct?
    For a total time T=3 the squared distance between M and m = R^2 / (1 + v / R * T)) = 100 / 1.0333... is that correct?
    For a total time T=3 the squared distance between M and m = R^2 / (1 + v / R * T)) = 96.774... which is still nowhere near either 132.25 or 133.5.
     
    Last edited: Jun 1, 2020
  19. TonyYuan Registered Senior Member

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    You have a few mistakes here.
    mistake 1:(1 + 1/10 * 3) is not equal to (1 + 1/30); (1 + 1 / 30) is not equal to (2 / 30).

    mistake 2: "So the average distance is (10 + 11 + 12 + 13) / 4 = 11.5 which when squared is 132.25."
    You just take a few distance values discretely and then average. This calculation is inaccurate.

    According to your thinking, I make a modification
    For a total time T=3 the squared distance between M and m = R^2 * (1 + v / R * T)) = 10^2 * (1 + 1 / 10 * 3)) = 100*(1 + 0.3)= 130.
     
    Last edited: Jun 1, 2020
  20. Neddy Bate Valued Senior Member

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    Please tell me what should it be. There are not enough parenthesis to tell me.

    It should be close.

    Now you changed the division sign to multiplication? Your paper says R^2 is followed by a division sign, not a multiplication.

    Please Register or Log in to view the hidden image!

     
  21. TonyYuan Registered Senior Member

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    1 + 2 * 3 = 7, is this not the most basic mathematical operation? I was a little surprised.
    Yes
    F(v) = G*M*m/R^2/(1+v/R*T)) = G*M*m/[ R^2 * (1+v/R*T)].

    100/5/2 = 100/(5*2) = 10. Neddy, what's wrong with you? These are the most basic mathematical logic.
     
    Last edited: Jun 1, 2020
  22. Neddy Bate Valued Senior Member

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    Without parenthesis it can be interpreted different ways. In your paper you have one "(" followed by two "))" so it was not possible for me to understand it. Look at the screenshot from your paper.

    Good, we can agree on that.

    Again, without matching parenthesis, you make it ambiguous. But we are making progress, thank you.
     
  23. TonyYuan Registered Senior Member

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    This really surprised me. Since the first grade of elementary school in our country, we have taught students that multiplication and division have higher priority than addition and subtraction.
    I don’t know how to teach in your country, but now I know that mathematics is also regionally different.

    Can we make such a convention that multiplication and division have higher priority than addition and subtraction?
     
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