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4-vector law of motion in different inertial frames

  1. Apr 5, 2012 #1

    Wox

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    Newton's second law of motion is given in Minkowski space by
    [tex]\bar{F}=m(c\gamma\dot{\gamma}, \gamma\dot{\gamma}\tilde{v}+\gamma^{2}\tilde{a})[/tex]
    where [itex]\dot{\gamma}=\frac{d\gamma}{dt}=\frac{\gamma^{3}}{c^{2}}\tilde{v}\cdot\tilde{a}[/itex] and [itex]\tilde{v}(t)[/itex] and [itex]\tilde{a}(t)[/itex] the 3-velocity and 3-acceleration. How can I show now that this law has the same form in all inertial frames?
     
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  3. Apr 5, 2012 #2

    PAllen

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    Well, it depends on what you want to assume. If you start from 4-momentum being a vector, then the RHS is d/dĪ„ of 4-momentum (that should be easy for you to show). That is sufficient to to establish it has the same form in any inertial frame.

    If, instead, you are supposed to demonstrate by brute force, it will work that applying the Lorentz transform should produce the same form, but that will be a large amount of messy, error prone algebra.
     
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