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GR0877 Problem 94

PROBLEM STATEMENT: This problem is still being typed.

SOLUTION: (C) Lorentz transformation \displaystyle t' = \gamma (t - \frac{v}{c^2}x) suggests that for the observer O' moving at constant speed v parallel to x-axis the event (flash of light) that occurs at (t_0, x_0) = (0, 10) for the observer O is at time \displaystyle t' =  \gamma(0 -\frac{v}{c^2}x_0) (we assume O' is moving in the positive direction of x and at a zero time according to both observers the origins of their reference frames coincide, therefore t' is negative; this means that (0, 10) event (as observed by O') occurs |t'| seconds before the other). Let \displaystyle \alpha = \frac{ct'}{x_0} = \frac{3 \cdot 10^{8} \cdot 13 \cdot 10^{-9}}{10} = 0.39, then, using units in which c = 1, one has \displaystyle \alpha = \gamma v = \frac{v}{\sqrt{1 - v^2}} or \displaystyle \alpha^2 = \frac{v^2}{1-v^2} \implies \displaystyle \displaystyle v = \frac{\alpha}{\sqrt{\alpha^2 +1}} \approx \alpha = 0.39 (for such an approximation the calculation error will be \sim 7%). The closest choice is (C). If you calculate the square root in the above equation fairly, you will get v \approx 0.363c.

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  1. jim
    12.11.2011 at 01:06

    please define your variables! It makes it much more difficult to understand your entire set of solutions. there are more than one set of common variables to represent various physical properties and you generally use one of the less common ones.

    • 12.11.2011 at 16:43

      \alpha is the only ‘uncommon’ variable in this particular solution and I clearly defined it.

  2. Sam Fischer
    29.10.2012 at 09:07

    Somewhat confused about this solution. Why is alpha = (c*t)/x_0? It from your original t’ equation that it should be (c^2*t)/x_0 = -gamma*v. What happened to the c squared term, and the negative sign? Also, how come you set c = 1 for alpha in the last step but not in the one just previous to it.

  3. physics_dan
    01.04.2014 at 08:04

    Can you use space-time interval to solve this problem? Like find the contracted length and solve for v ?
    I somehow got 0.39c instead of 0.36.

  4. Alex Clain
    26.08.2015 at 18:23

    Approximating \alpha as 2/5 makes the calculation at the end a bit easier.

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