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Velocity transformation using the chain rule

  1. Jul 15, 2017 #1
    1. The problem statement, all variables and given/known data
    How to obtain the famous formula of velocity transformation using a chain rule.
    I know that there is a straightforward way by dividing ##dx## as a function of ##dx`## and ##dt`## on ##dt## which is also a function of them. But I would rather try using the chain rule.
    2. Relevant equations
    ##x=\gamma(x`+vt`)##
    ##t=\gamma(t`+\frac{v}{c^2}x`)##

    3. The attempt at a solution
    I tried the following chain rule ##\frac{dx}{dt}=\frac{dx}{dx`}\frac{dx`}{dt`}\frac{dt`}{dt}## so ##u=\frac{dx}{dx`}\frac{dt`}{dt}u`##
    The first term ##\frac{dx}{dx`}=\gamma(1+\frac{v}{u`})##
    The second term ##\frac{dt`}{dt}## requires me to take a derivative of ##t`## with respect to ##t##. Here the equation, ##t=\gamma(t`+\frac{v}{c^2}x`)## will not help because I have to differentiate ##x`## with respect to ##t##.
     
  2. jcsd
  3. Jul 15, 2017 #2

    Orodruin

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    Your chain rules ignores the fact that it is a differentiation wrt many variables. You need the full chain rule with all partial derivatives. An important thing to keep in mind is when t and t' are used in their capacity as coordinates in the Lorentz transformation and when they are used to refer to acurve parameter.
     
  4. Jul 16, 2017 #3
    Here is another trial;
    ##\frac{dx}{dt}=\frac{\partial x}{\partial x`}\frac{\partial x`}{\partial t}+\frac{\partial x}{\partial t`}\frac{\partial t`}{\partial t}##
    But ##\frac{\partial x`}{\partial t}=\frac{\partial x`}{\partial t`}\frac{\partial t`}{\partial t}##
    So, ##\frac{dx}{dt}=\frac{\partial x}{\partial x`}\frac{\partial x`}{\partial t`}\frac{\partial t`}{\partial t}+\frac{\partial x}{\partial t`}\frac{\partial t`}{\partial t}##
    Substitute ##u## for ##\frac{dx}{dt}## and ##u`## for ##\frac{\partial x`}{\partial t`}##
    Also, from the two equation of LT ##\frac{\partial x}{\partial x`}=\gamma##, ##\frac{\partial t`}{\partial t}=\frac{1}{\gamma}##, ##\frac{\partial x}{\partial t`}=v\gamma## and ##\frac{\partial t`}{\partial t}=\frac{1}{\gamma}## (here, I did not use the inverse LT to substitute for ##\frac{\partial t`}{\partial t}##, instead I used the main transformation)
    yields; ##u=\gamma u`\frac{1}{\gamma}+v\gamma\frac{1}{\gamma}=u`+v## This is non-relativistic transformation.
     
    Last edited: Jul 16, 2017
  5. Jul 17, 2017 #4
    From some help, I found a way out.
    First the velocity should be represented by the total derivatives not partial derivatives.
    ##\frac{dx}{dt}=\frac{dx}{dx`}\frac{dx`}{dt`}\frac{d t`}{dt}##
    Now ##\frac{dx}{dx`}## and ##\frac{d t`}{dt}## are expressed in term of partial derivatives;
    ##\frac{dx}{dx`}=\frac{\partial x}{\partial x`}\frac{\partial x`}{\partial x`}+\frac{\partial x}{\partial t`}\frac{\partial t`}{\partial x`}=\frac{\partial x}{\partial x`}+\frac{\partial x}{\partial t`}\frac{\partial t`}{\partial x`}##
    ##\frac{dt`}{dt}=\frac{\partial t`}{\partial t}\frac{\partial t}{\partial t}+\frac{\partial t`}{\partial x}\frac{\partial x}{\partial t}=\frac{\partial t`}{\partial t}+\frac{\partial t`}{\partial x}\frac{\partial x}{\partial t}##
    Replacing; ##\frac{dx}{dt}## and ##\frac{dx`}{dt`}## by ##u## and ##u`##, and the other partial derivatives from Lorentz Transformation yields the proper result.
     
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