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Transformation of acceleration between two reference frames

  1. May 15, 2014 #1
    hello

    i want to derive the Transformation of acceleration between two reference frames
    i searched in internet and found a book but i dont understand just one step

    i attach a picture so you can see what i found in the internet

    my problem is eq. (1.10) \begin{align}du=\frac{dU}{\gamma ^{2}(1-U\frac{v}{c^{2}})^{2}}\end{align}
    how did he get this factor
    can you please give me a tip


    thank you very much
     

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  3. May 15, 2014 #2

    Bill_K

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    There's a misprint. Calculus rule is d(x/y) = (y dx - x dy)/y2 where x = U - v and y = 1 - Uv/c2, so the first denominator in Eq.(1.10) should be (1 - U v/c2)2. The numerator is dU(1 - v2/c2), which reduces to dU(1/γ2).
     
  4. May 15, 2014 #3
    thank you for your answer
    where is this misprint? is it the first denominator? the square is missing right?
    my problem is this gamma in eq 1.10 how did he get it? i dont understand it
    also the next step with gamma
    can you show me how to get it?
    thank you
     
  5. May 15, 2014 #4

    xox

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    There are TWO errors, not one in the book:

    [tex]dU=\frac{dU(1-Uv/c^2)-(U-v)(-dUv/c^2)}{(1-Uv/c^2)^2}[/tex]

    The final result is correct, the derivation is riddled with errors.
     
    Last edited: May 15, 2014
  6. May 15, 2014 #5
    xox thank you for your answer but im sorry you are wrong
    the positive sign is wrong. it must be negative
    therefore the only one error is the square in the denominator

    but this is not my problem. i just want to know how to get the next step with gamma square.
    hope you can help me
     
  7. May 15, 2014 #6

    xox

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    There are TWO minus signs, they indeed cancel each other, so I was a little pedantic.


    [tex]du=\frac{dU(1-Uv/c^2)-(U-v)(-dUv/c^2)}{(1-Uv/c^2)^2}=\frac{dU(1-v^2/c^2)}{(1-Uv/c^2)^2}[/tex]

    But [tex]1-v^2/c^2=\frac{1}{\gamma^2}[/tex]
     
    Last edited: May 15, 2014
  8. May 15, 2014 #7
    yes you are right
    sorryyyy

    this gamma confused me xD
     
  9. May 15, 2014 #8

    xox

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    You are welcome!
     
  10. May 15, 2014 #9
    xox
    thank you very much. i get it
    so now i have to divide both sides by dt to get du/dt=a
    but again with this gamma. eq 1.11 i dont understand it
     
  11. May 15, 2014 #10

    xox

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    OK, let's help u a little more:

    We agree that:
    [tex]du=\frac{dU}{\gamma^2(1-Uv/c^2)^2}[/tex]

    Now,

    [tex]dt=\gamma(dT-dXv/c^2)[/tex]

    Therefore:

    [tex]a=\frac{du}{dt}=\frac{dU}{\gamma^3(1-Uv/c^2)^2(dT-dXv/c^2)}=\frac{dU/dT}{\gamma^3(1-Uv/c^2)^2(1-dX/dTv/c^2)}=\frac{dU/dT}{\gamma^3(1-Uv/c^2)^3}=\frac{A}{\gamma^3(1-Uv/c^2)^3}[/tex]

    If [itex]u=V[/itex] then [itex]a=\gamma^3A[/itex]
     
  12. May 15, 2014 #11
    thank you i got it finally
    thank you very much
     
  13. May 15, 2014 #12

    xox

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    Glad I could help.
     
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