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Minkowsky force and physical interpretation.

  1. May 26, 2010 #1
    1. The problem statement, all variables and given/known data

    Compute the [itex] \mu=0 [/itex] component of the Minkowski force law [itex]K^\mu=q\eta_\nu F^{\mu\nu}.[/itex] (Einstien summation convention applies.)

    2. Relevant equations

    [tex]\eta_\nu=\frac{1}{\sqrt{1-u^2/c^2}}(-c,u_x,u_y,u_z) [/tex]
    [itex] F^{\mu\nu} [/itex] is the field tensor where
    [tex]F^{00}=0,F^{01}=\frac{E_x}{c},F^{02}=\frac{E_y}{c},F^{03}=\frac{E_x}{c}.[/tex]

    3. The attempt at a solution

    [tex] K^0=q(\eta_0 F^{00} +\eta_1 F^{01} +\eta_2 F^{02} +\eta_3 F^{03}) = \frac {q \gamma}{c}(u_x E_x + u_y E_y +u_z E_z) = \frac {q \gamma}{c}(\bf{u}.\bf{E})[/tex]


    This all seems ok to me, but I have no idea what it actually means. What does [itex] K^0 [/itex] physically represent and what does [itex] \bf{u}.\bf{E} [/itex] mean.

    Thank you for your help.
     
  2. jcsd
  3. May 27, 2010 #2
  4. May 27, 2010 #3
    I think Chris's answer is okay, but not nearly sufficient.

    The dot product is the magnitude of the projection of one vector onto the other. In this case, we are projecting the electric field vector onto the velocity vector (or vice versa). So the time component of the Minkowski 4-force is the magnitude of the electric field vector in direction of the velocity.
     
  5. May 27, 2010 #4

    vela

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    Newton's second law

    [tex]F^\mu = \frac{dp^\mu}{d\tau}[/tex]

    still holds in special relativity. Think about what the time component of the four-momentum is. Also, it might help to rewrite K0 slightly to get

    [tex]K^0 = \frac{\gamma}{c}[\textbf{u}\cdot (q\textbf{E})][/tex]

    to see what it physically means.
     
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