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Force on a relativistic particle

  1. Dec 24, 2016 #1
    1. The problem statement, all variables and given/known data

    A particle of rest mass ##m_0## is is caused to move along a line in such a way that its postion is $$x = \sqrt{b^2 + c^2 t^2} -b$$What force must be applied to the particle to produce this motion?


    Feynman_12_7.png

    2. Relevant equations


    The velocity of the particle as seen from the rest frame is ##v = \frac{dx}{dt}## and the acceleration is ##a = \frac{dv}{dt}##.

    The mass of the particle as seen in the rest frame is ##m = \gamma m_0##, where ##\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}##.

    Then the force should be ##F = m a##.

    3. The attempt at a solution

    Differentiating ##x##, I get ##v = \frac{c^2 t}{\sqrt{b^2 + c^2 t^2}}##.

    So, ##\gamma = \frac{\sqrt{b^2 + c^2 t^2}}{b}##.

    Differentiating ##v##, I get ##a = \frac{b^2 c^2}{(b^2 + c^2 t^2)^{\frac{3}{2}}}##.

    So, ##F = \frac{m_0 b c^2}{b^2 + c^2 t^2}##.

    The answer given in the book is ##F = \frac{m_0 c^2}{b}##. What did I do wrong?
     
  2. jcsd
  3. Dec 24, 2016 #2

    TSny

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    Force in special relativity is generally defined as rate of change of relativistic momentum: ##\vec{F} = \frac{d\vec{p}}{dt}##, where ##\vec{p} = \gamma m_0 \vec{v}##. This is not equivalent to ##\vec{F} = m \vec{a}##, where ##m = \gamma m_0## and ##\vec{a} = \frac{d\vec{v}}{dt}##.

    See
     
  4. Dec 24, 2016 #3
    OK, I get the book answer now. Thank you!
     
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