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Thread: Simultaneous equation from relativity

  1. #1
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    Simultaneous equation from relativity

    I need to solve the equations:

    $\displaystyle ct_1=l+\frac 12 \alpha t_1^2$

    $\displaystyle ct_2=l+\frac 12 \alpha t_1^2-\frac 12 \alpha (t_1+t_2)^2$


    First I try adding and subtracting the two equations to get:
    $\displaystyle c(t_1+t_2)=2l+\alpha t_1^2-\frac 12 \alpha (t_1+t_2)^2$

    $\displaystyle c(t_1-t_2)=\frac 12 \alpha (t_1+t_2)^2$


    But I can make no real progress from there.

    Now the context of the question means that:

    $\displaystyle \alpha << 1$, $\displaystyle l << 1$ and I am more than happy to make a few aproximations. I expect the binomial aproximation may be useful.

    The required solution is:

    $\displaystyle t_1+t_2 = \frac{2l}c (1+\frac {\alpha l}2) $

    I have posted the whole question on Physics help forum at:
    http://www.physicshelpforum.com/phys...lleration.html
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  2. #2
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    Finally solved it myself.

    eqn1 $\displaystyle ct_1=l+\frac 12 \alpha t_1^2$

    eqn2 $\displaystyle ct_2=l+\frac 12 \alpha t_1^2-\frac 12 \alpha (t_1+t_2)^2$

    eqn3 $\displaystyle c(t_1+t_2)=2l+\alpha t_1^2-\frac 12 \alpha (t_1+t_2)^2$

    Let
    $\displaystyle T=t_1+t_2$

    From eqn1:
    $\displaystyle t_1=\frac{c-\sqrt{c^2-2 \alpha l}}{\alpha}=\frac{c-c \sqrt{1-\frac {2 \alpha l}{c^2}}}{\alpha}$

    $\displaystyle t_1 \approx \frac {c-c+\frac {\alpha l}{c}+\frac{\alpha^2l^2}{2c^2}}{\alpha}=\frac lc+\frac{\alpha l^2}{2c^2}$
    So
    $\displaystyle t_1^2 \approx \frac {l^2}{c^2}+\frac{\alpha l^3}{c^3}$

    now substituting into eqn3

    $\displaystyle cT \approx 2l-\frac 12 \alpha T^2+ \alpha t_1^2=2l-\frac 12 \alpha T^2+\frac {\alpha l^2}{c^2}+\frac{\alpha^2 l^3}{c^3}$
    Rearanging
    $\displaystyle cT+\frac 12 \alpha T^2 \approx 2l+\frac {\alpha l^2}{c^2}+\frac{\alpha^2 l^3}{c^3}$

    Dumping the smallest terms:

    $\displaystyle cT \approx 2l+\frac {\alpha l^2}{c^2}$

    And finally

    $\displaystyle T \approx \frac {2l}{c}(1+\frac {\alpha l}{2c^2})$
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