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Special relativity-analogy of rotation

  1. May 28, 2017 #1
    1. The problem statement, all variables and given/known data
    Analogy to rotation:
    $$(x')^2+(y')^2+(z')^2-c^2(t')^2=x^2+y^2+z^2-c^2t^2$$
    It isn't

    2. Relevant equations
    Lorentz transformations:
    $$x'=\frac{x-ut}{\sqrt{1-u^2/c^2}}$$
    $$t'=\frac{t-ux/c^2}{\sqrt{1-u^2/c^2}}$$

    3. The attempt at a solution
    ##~(x')^2-c^2(t')^2~## must equal ##~x^2-c^2t^2## but it isn't so:
    $$\frac{(x-ut)^2}{1-u^2/c^2}-\frac{c^2(t-\frac{ux}{c^2})}{1-u^2/c^2}\neq x^2-c^2t^2$$
     
  2. jcsd
  3. May 28, 2017 #2

    BvU

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    Perhaps if you restore the square in the numerator of t'2 you'll fare better :rolleyes: $$
    \frac{(x-ut)^2}{1-u^2/c^2}-\frac{c^2(t-\frac{ux}{c^2})^2}{1-u^2/c^2}\ = x^2-c^2t^2 \ \ !$$
     
  4. May 29, 2017 #3
    $$\frac{(x-ut)^2}{1-u^2/c^2}-\frac{c^2(t-\frac{ux}{c^2})^2}{1-u^2/c^2}\ =\frac{x^2-2xut+u^2t^2-c^2t^2+2cxut-u^2x^2}{(c+u)(c-u)/c^2}$$
    $$=\frac{(1-u^2)x^2+(c-1)2xut+(u^2-c^2)t^2}{(c+u)(c-u)/c^2}$$
     
  5. May 29, 2017 #4

    BvU

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    Nonsense. ##c## only occurs as ##c^2## so you can't have a ##c^1## in there. Another check you should do: dimensions: they don't fit !
     
  6. May 29, 2017 #5
    $$\frac{(x-ut)^2}{1-u^2/c^2}-\frac{c^2(t-\frac{ux}{c^2})^2}{1-u^2/c^2}\ =...=\frac{x^2+u^2t^2-c^2t^2-u^2x^2}{(c+u)(c-u)/c^2}$$
    $$=\frac{(1-u^2)x^2+(u^2-c^2)t^2}{(c+u)(c-u)/c^2}$$
     
  7. May 29, 2017 #6

    BvU

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    Numerator dimensions aren't the same: can't have ##x^2## and ##u^2 x^2## side by side. :confused:
     
  8. May 29, 2017 #7
    $$\frac{(x-ut)^2}{1-u^2/c^2}-\frac{c^2(t-\frac{ux}{c^2})^2}{1-u^2/c^2}\ =...=\frac{x^2+u^2t^2-c^2t^2-\frac{u^2}{c^2}x^2}{(c+u)(c-u)/c^2}$$
    $$=\frac{(1-\frac{u^2}{c^2})x^2+(u^2-c^2)t^2}{1-u^2/c^2}=x^2-c^2t^2$$
     
  9. May 29, 2017 #8

    BvU

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    Bingo !
     
  10. May 29, 2017 #9
    You are great, BvU
     
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