Doppler Effect Of Gravitational Field

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

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  1. Neddy Bate Valued Senior Member

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    Okay, but when there are parenthesis, they indicate that the operations inside of them have higher priority. So F = G*M*m/R^2/(1+v/R*T)) with the unmatched parenthesis could mean F = G*M*m/(R^2/(1+v/R*T)) or F = G*M*m/R^2/((1+v)/R*T) since I have to guess where the parenthesis might have been intended. And if you are prone to leaving the parentheses unmatched, then I might wonder if you mean something else. I did tell you that I was having trouble understanding your derivation.
     
    Last edited: Jun 2, 2020
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  3. TonyYuan Registered Senior Member

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    Ok, I see, this is indeed an editorial error in my article. Thanks very much for helping me discover this error. Let's continue the discussion.
     
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  5. TonyYuan Registered Senior Member

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    Neddy,
    It is very meaningful to discuss with you. The doubts you encounter are certainly also the doubts that other scholars will encounter. I hope to discuss with you to let everyone solve the doubts.

    Whether my previous answer can help you understand. Are there any other questions?

    If you have understood my article, please let me know, thank you.
     
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  7. Neddy Bate Valued Senior Member

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    Okay, so one more time, let me just try to 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.

    And I think this is your other equation:
    [For a total time T=3 the squared distance between M and m] = R^2 * (1 + ((v/R) * T))
    [For a total time T=3 the squared distance between M and m] = 10^2 * (1 + ((1/10) * 3))
    [For a total time T=3 the squared distance between M and m] = 10^2 * (1 + (0.1 * 3))
    [For a total time T=3 the squared distance between M and m] = 100 * (1 + 0.3)= 130.

    Okay, now I see. And for the case of only t=0 and T=0 we get:
    [For a total time T=0 the squared distance between M and m] = R^2 * (1 + ((v/R) * 0))
    [For a total time T=0 the squared distance between M and m] = 10^2 * (1 + ((1/10) * 0))
    [For a total time T=0 the squared distance between M and m] = 10^2 * (1 + (0.1 * 0))
    [For a total time T=0 the squared distance between M and m] = 100 * (1 + 0.0)= 100

    Okay, it makes sense now, but I am still thinking about the meaning...
     
  8. TonyYuan Registered Senior Member

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    Neddy, You only need to follow the steps below to understand:
    1. Calculate the total impulse P generated by the gravity during time T.
    2. P/T, calculate the average value of gravity F(v) in time T.
    3. Analyze whether F(v1)-F(v2) and v1-v2 are linear or nonlinear at different speeds v1 and v2.
    4. Use the boundary conditions v = 0 and v = c to derive the relationship between F(v) and R and v.
     
  9. Neddy Bate Valued Senior Member

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    2,110
    For R=10 length units and v=1 length unit per time unit, I have included the square, and also increased the number of data points in my interval of time units:
    [At time t=0.0 the squared distance between M and m] = (R + (v * t))^2 = (10 + (1 * 0.0))^2 = 10.0^2 = 100
    [At time t=0.5 the squared distance between M and m] = (R + (v * t))^2 = (10 + (1 * 0.5))^2 = 10.5^2 = 110.25
    [At time t=1.0 the squared distance between M and m] = (R + (v * t))^2 = (10 + (1 * 1.0))^2 = 11.0^2 = 121
    [At time t=1.5 the squared distance between M and m] = (R + (v * t))^2 = (10 + (1 * 1.5))^2 = 11.5^2 = 132.25
    [At time t=2.0 the squared distance between M and m] = (R + (v * t))^2 = (10 + (1 * 2.0))^2 = 12.0^2 = 144
    [At time t=2.5 the squared distance between M and m] = (R + (v * t))^2 = (10 + (1 * 2.5))^2 = 12.5^2 = 156.25
    [At time t=3.0 the squared distance between M and m] = (R + (v * t))^2 = (10 + (1 * 3.0))^2 = 13.0^2 = 169
    So the average is (100+110.25+121+132.25+144+156.25+169)/7 = 133.25.
    Before, with only 4 data points, the average was (100+121+144+169)/4 = 133.5.

    So I made a spreadsheet to see where the average converges with more and more data points:
    Average for 4 data points = 133.5
    Average for 7 data points = 133.25
    Average for 13 data points = 133.125
    Average for 25 data points = 133.0625
    Average for 49 data points = 133.03125
    .
    .
    .
    So the average is obviously converging on 133.000000000000

    And using your other equation:
    [For a total time T=3 the squared distance between M and m] = R^2 * (1 + ((v/R) * T))
    [For a total time T=3 the squared distance between M and m] = 10^2 * (1 + ((1/10) * 3))
    [For a total time T=3 the squared distance between M and m] = 10^2 * (1 + (0.1 * 3))
    [For a total time T=3 the squared distance between M and m] = 100 * (1 + 0.3)= 130.

    Why the discrepancy?
     
  10. TonyYuan Registered Senior Member

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    440
    F = G*M*m/(R^2/(1+v/R*T)), you should calculate 1/[R^2*(1+v/R*T)].

    (1/100 + 1/121 + 1/144+ 1/169)/4 = 0.00778151675441885232095022304813

    (1/100 + 1/110.25 + 1/121 + 1/132.25 + 1/144 + 1/156.25 + 1/169 ) = 0.00773682835360339600630770438548

    1/130 = 0.00769230769230769230769230769231

    When more samples, the closer to 1/130.
     
  11. Neddy Bate Valued Senior Member

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    2,110
    Interesting, thank you.

    Okay, so now I am at step 3.
     
  12. TonyYuan Registered Senior Member

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    440
    Neddy, I randomly calculated some data.
    samples num=193----------0.00769369--------1/130=0.00769231
    samples num=3073--------0.00769239-------- 1/130=0.00769231
    samples num=3145729----0.00769231---------1/130=0.00769231
     
  13. TonyYuan Registered Senior Member

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    440
    OK. Let's go.
     
  14. TonyYuan Registered Senior Member

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    440
    Neddy, do you have some new progress?
     
  15. Neddy Bate Valued Senior Member

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    2,110
    Thanks. Yes I understood after your post #47 that I should have been looking at the inverses instead.

    I have confirmed to my satisfaction that F(v1) / F(v2) = (R + v2*T) / (R + v1*T) as your paper says. However, I am not sure what that means. Does that let me answer the question in your step 3? I am not sure whether I even understand the question in step 3.
     
  16. TonyYuan Registered Senior Member

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    440
    This is just a mathematical proof that they ( F(v) and v ) are a linear relationship. Then use the step4 to draw the final conclusion.
    Have you started the verification of step 4 now?

    Neddy,
    You can raise your doubts and I will give you some explanations to help you understand.
     
    Last edited: Jun 3, 2020
  17. TonyYuan Registered Senior Member

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    440
    Some people I know, they told me, don’t waste time on the forum, there will be no professional physicists, no professional astronomers here, just a bunch of people who don’t know anything. Only if you publish this article in a journal can you prove that it is correct, otherwise no one will recognize your theory.

    I have tried some journals, but this theory challenges general relativity and is a gravitational revision of Newton. Its influence is so great that no journal dares to publish it.

    I went back to this forum and back to where I started. Here I feel the academic atmosphere, where I can freely discuss and freely refute. I hope that I believe that our efforts here can give the final answer. Is this theory right or wrong? If we finally confirm that it is correct, I hope that we can work together to get it recognized by professional journals and the physics community.

    Let's work hard together, come on.

    If you think something is wrong, please boldly point out.
    If you agree with my theory, please bravely support me.
     
    Last edited: Jun 4, 2020
  18. Neddy Bate Valued Senior Member

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    2,110
    Please correct me if I am wrong:
    F(v) is the average gravitational force over the time interval T.
    R is the initial distance between M and m.
    R + v*T is the final distance between M and m.

    So...
    F(v1) / F(v2) = (R + v2*T) / (R + v1*T)
    ...means that for two different velocities v1 and v2 with everything else held constant...
    ... the ratio of the two different average gravitational forces...
    ...is equal to the inverse of the ratio of the two different final distances between M and m.

    Correct?

    I am not on to step 4 yet, but getting closer, I think.
     
  19. TonyYuan Registered Senior Member

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    440
    Yes.
    Good, come on.
     
  20. Neddy Bate Valued Senior Member

    Messages:
    2,110
    F(v1) / F(v2) = (R + v2*T) / (R + v1*T)

    For the boundary conditions of v1=0 and v2=c, we have:

    F(0) / F(c) = (R + c*T) / (R + 0*T)
    F(0) / F(c) = (R + c*T) / (R)

    Correct?
     
  21. TonyYuan Registered Senior Member

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    440
    Yes.
     
  22. Neddy Bate Valued Senior Member

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    2,110
    so
    F(0) / F(c) = (R + c*T) / R
    now what?
     
  23. TonyYuan Registered Senior Member

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    440
    I don’t know what you want to do next? I have a detailed process for the step3 and the step4 in the article.

    F(v1) - F(v2) = K*(v2-v1)*T

    F(0) = GMm/R^2,
    F(v) - F(0) = F(v) - GMm/R^2 = - K*(v)*T.
    F(v) = GMm/R^2 - K*(v)*T

    F(c) = 0,
    F(v) - F(c) = K*(c-v)*T
    F(v) = K*(c-v)*T

    F(v) = GMm/R^2 - K*(v)*T = K*(c-v)*T
    K*c*T = GMm/R^2
    K = [GMm/R^2]/(c*T)

    F(v) = K*(c-v)*T = [GMm/R^2]/(c*T) * (c-v)*T = (GMm/R^2) * [(c-v)/c]

    The final equation:
    F(v) = (GMm/R^2) * [(c-v)/c]
     
    Last edited: Jun 4, 2020
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