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Relativity - Lorentz Force Law

  1. May 20, 2007 #1
    Assuming the Lorentz force law and also that in the rest frame of the particle the 3 acceleration is zero, we need to explain why the following equations hold:

    E.v = 0 and E + v.B = 0

    where v is the velocity.

    I think this is because g(A,A) = -a squared is invariant. Therefore if a=0, I think this means that A must equal zero in every frame. Is this true, or can A be non zero and we get g(A,A) = 0 (i.e A is null).
     
  2. jcsd
  3. May 21, 2007 #2

    Meir Achuz

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    I am not sure what you mean by A and a, but acceleration is not a 4-vector, and has complicated LT properties.
    In the rest frame for a=0, E=0, and B is unknown.
    Since E=0, E' in a system moving with velocity v is given by
    [tex]{\vec E}=\gamma{\vec v}\times{\vec B}[/tex], so
    [tex]{\vec v}\cdot{\vec E'}=0.[/tex]
     
  4. May 21, 2007 #3

    Meir Achuz

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    [tex]E=|{\vec E'}|=\gamma v B[/tex],
    but [tex]{\vec v}\cdot{\vec B'}=v B[/tex].

    Therefore [tex]E'+{\vec v}\cdot{\vec B'}[/tex]
    does not equal zero.
     
    Last edited: May 21, 2007
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