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Performing a Taylor Series Expansion for Lorentz Factor

  1. Sep 15, 2016 #1
    1. The problem statement, all variables and given/known data
    Perform a Taylor Series expansion for γ in powers of β^2, keeping only the third terms (ie. powers up to β^4). We are assuming at β < 1.

    2. Relevant equations
    γ = (1-β^2)^(-1/2)

    3. The attempt at a solution
    I have no background in math so I do not know how to do Taylor expansion. Could you help me out? Thank you very much.
     
  2. jcsd
  3. Sep 15, 2016 #2

    fresh_42

    Staff: Mentor

    Have you looked up Taylor series in Wikipedia?
    Here is a link to the formula, which is clearer than on the English version. Just look at the formula.
    Which brings me to the next question. Do you know how to differentiate a / your function (with respect to ##\beta##)?
     
  4. Sep 15, 2016 #3
    Sorry, I am not sure. I have been away from all maths for a long time.
    But could you help me out? What is my x and a?
     
  5. Sep 15, 2016 #4

    fresh_42

    Staff: Mentor

    Your ##f## is ##\gamma## and your ##x## is ##\beta##. Unfortunately you didn't specify what your ##a## is, so I assume it is ##a=0##.
    If not mentioned otherwise it's almost always ##0##. But some functions aren't defined at ##0##. In these cases one cannot "automatically" take zero. However, your function is defined in ##0## and your ##x = \beta## is assumed smaller than one, so ##a=0## should be ok.

    Edit: don't forget the internal differentiation: ##\gamma' = \frac{d}{d \beta} (1- \beta^2)^{-\frac{1}{2}} = -\frac{1}{2}(1- \beta^2)^{-\frac{3}{2}} \, \cdot \, (-2 \beta)##
     
    Last edited: Sep 15, 2016
  6. Sep 15, 2016 #5
    Okay I am solving it, one moment please!
     
  7. Sep 15, 2016 #6
    https://wikimedia.org/api/rest_v1/media/math/render/svg/132e57acb643253e7810ee9702d9581f159a1c61 is γ = (1-β^2)^(-1/2)
    x = β
    and a = 0 so therefore my equation would look like this:

    T [PLAIN]https://wikimedia.org/api/rest_v1/media/math/render/svg/132e57acb643253e7810ee9702d9581f159a1c61(x[I][I][I];a) [Broken] = [PLAIN]https://wikimedia.org/api/rest_v1/media/math/render/svg/132e57acb643253e7810ee9702d9581f159a1c61(a) [Broken] + [PLAIN]https://wikimedia.org/api/rest_v1/media/math/render/svg/132e57acb643253e7810ee9702d9581f159a1c61'(a) [Broken] (x-a) + [[PLAIN]https://wikimedia.org/api/rest_v1/media/math/render/svg/132e57acb643253e7810ee9702d9581f159a1c61''(a) [Broken] / 2 ] (x-a)^2 + [[PLAIN]https://wikimedia.org/api/rest_v1/media/math/render/svg/132e57acb643253e7810ee9702d9581f159a1c61'''(a) [Broken] / 6] (x-a)^3 ...[/I][/I][/I]

    T [PLAIN]https://wikimedia.org/api/rest_v1/media/math/render/svg/132e57acb643253e7810ee9702d9581f159a1c61([I][I]β[I];0) [Broken] = [PLAIN]https://wikimedia.org/api/rest_v1/media/math/render/svg/132e57acb643253e7810ee9702d9581f159a1c61(0) [Broken] + [PLAIN]https://wikimedia.org/api/rest_v1/media/math/render/svg/132e57acb643253e7810ee9702d9581f159a1c61'(0) [Broken] (x-0) + [[PLAIN]https://wikimedia.org/api/rest_v1/media/math/render/svg/132e57acb643253e7810ee9702d9581f159a1c61''(0) [Broken] / 2 ] (β-0)^2 + [[PLAIN]https://wikimedia.org/api/rest_v1/media/math/render/svg/132e57acb643253e7810ee9702d9581f159a1c61'''(0) [Broken] / 6] (β-0)^3 ...
    [/I][/I][/I]

    I am stuck here. I can find each derivative of (1-β^2)^(-1/2) but there is nowhere to plug in "a" in this
    function. I apologize if I am making a very dumb mistake. Thanks for the help.
     
    Last edited by a moderator: May 8, 2017
  8. Sep 15, 2016 #7

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    Better: take ##x = \beta^2##. Expand ##(1-x)^{-1/2}## in powers of ##x##, then put ##x = \beta^2## later. That makes finding the needed derivatives a lot easier.
     
  9. Sep 15, 2016 #8
    I see, thanks. That's what the question meant!
    But I am still confused at how a = 0 fits into this equation.
    Looking at the equation, the functions do not have anywhere to plug in for a.
     
  10. Sep 15, 2016 #9

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    If your "##a##" refers to the formula
    $$f(x) = f(a) + f'(a) (x-a) + \frac{1}{2!} f''(a) (x-a)^2 + \cdots, $$
    then you are dealing with the case ##a = 0##
     
    Last edited: Sep 16, 2016
  11. Sep 15, 2016 #10

    fresh_42

    Staff: Mentor

    If you take ##\gamma(x) = (1-x)^{-\frac{1}{2}}## then ##\gamma'(x) = \frac{1}{2}(1-x)^{-\frac{3}{2}}## and ##\gamma'(0) = \frac{1}{2}(1-0)^{-\frac{3}{2}} = \frac{1}{2}##. Don't forget that there is a factor ##(x-a)^1= (x-0)^1=x##, too. (In the second term of the series.)
     
  12. Sep 15, 2016 #11
    Just a quick question...
    When do we use γ and when do we use its reciprocal α when solving for coordinates of S' frame through Lorentz transformations?
    My teacher used the reciprocal but my textbook uses γ in the same equation and my length contraction is actually going reverse and
    getting longer.
     
  13. Sep 15, 2016 #12
    Solving...
     
  14. Sep 15, 2016 #13

    fresh_42

    Staff: Mentor

    Yes. But you want to expand at ##a=0##, that is in the formula ##f^{(n)}(a)=f^{(n)}(0)##. So you have to substitute ##0## where ##x## was or ##\beta## is. And the exponents are still negative (although it doesn't matter in this case). You multiply with ##(x-a)^n=(x-0)^n=x^n=\beta^{2n}##. The differentials of ##f=\gamma## are evaluated at ##0##, not extra multiplied.

    So the first term of the sum is ##f(0)x^0 = 1## the second ##f'(0)x^1=\frac{1}{2}x## the third ##\frac{1}{2}f''(0)x^2=\frac{3}{8}x^2## and so on.

    Edit: Remember the formula is (with ##f=\gamma## and ##x=\beta^2##) ##\; Tf(x;0)= f(0)+f'(0)x+\frac{f''(0)}{2}x^2+\frac{f'''(0)}{6}x^3+ \dots##.
     
  15. Sep 15, 2016 #14
    Thanks so much...
     
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