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Minkowski force

  1. Dec 12, 2012 #1
    Our professor derived a Minkowski force like this:

    [itex]
    F^\mu = \left[ \gamma(e\vec{E} + e(\vec{v}\times \vec{B})) , \gamma \frac{e \vec{E} \vec{v}}{c} \right]
    [/itex]

    Does this mean that i can write 4-force like this?

    [itex]
    F^\mu =
    \begin{bmatrix}
    \gamma(e\vec{E} + e(\vec{v_x}\times \vec{B}))\\
    \gamma(e\vec{E} + e(\vec{v_y}\times \vec{B}))\\
    \gamma(e\vec{E} + e(\vec{v_z}\times \vec{B}))\\
    \gamma \frac{e \vec{E} \vec{v}}{c}
    \end{bmatrix}
    [/itex]

    Short anwser would be ok. How do i put this into a matrix form from which i can get Lorentz matrix ##\Lambda## for boost in $x$ direction?
     
  2. jcsd
  3. Dec 12, 2012 #2

    Mentz114

    User Avatar
    Gold Member

    It's easier if you start with the Faraday tensor (F) and write the force (f) as,
    [tex]
    f^\mu = F^{\mu\nu}J_\nu
    [/tex]
    then you can boost the field tensor F and J ( current) with [itex]\Lambda[/itex] to get the transformed force. The Wiki article might be relevant

    http://en.wikipedia.org/wiki/Electromagnetic_tensor
     
  4. Dec 12, 2012 #3
    I am new to this and have never encountered tensors before. This is why i have to use basic math. Some day when things are clear to me in simple math i can learn advanced math and swich to tensors.
     
  5. Dec 12, 2012 #4

    Mentz114

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    Gold Member

    OK.

    I think what your professor has written is the electrostatic force after the field has been boosted by v = √(vx2+vy2+vz2). A 'boost' is a Lorentz, transformation, in case you're not familiar with the term. In this case, the tensor expressions are like matrix and vector operations so that

    [itex]f'=F'\cdot J'[/itex], [itex]F' = \Lambda(v)\cdot F\cdot\Lambda(v)[/itex] and [itex]J' = \Lambda(-v)\cdot J[/itex].
     
    Last edited: Dec 12, 2012
  6. Dec 12, 2012 #5

    Meir Achuz

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    Science Advisor
    Homework Helper
    Gold Member

    You have some misprints.
    It should be
    [itex]
    F^\mu =
    \begin{bmatrix}
    \gamma(e\vec{E} + e(\vec{v}\times \vec{B}))_x\\
    \gamma(e\vec{E} + e(\vec{v}\times \vec{B}))_y\\
    \gamma(e\vec{E} + e(\vec{v}\times \vec{B}))_z\\
    \gamma \frac{e \vec{v}\cdot\vec{E} }{c}
    \end{bmatrix}
    [/itex].
    Then F^\mu transforms like any other four-vector.
     
    Last edited: Dec 12, 2012
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