Why Do the Equations E.v = 0 and E + v.B = 0 Hold in the Rest Frame?

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SUMMARY

The discussion focuses on the validity of the equations E.v = 0 and E + v.B = 0 in the rest frame of a particle, assuming the Lorentz force law. It is established that in the rest frame, where acceleration (a) is zero, the electric field (E) must also be zero. The transformation of the electric field in a moving system is described by the equation E' = γ(v × B), leading to the conclusion that E' remains zero when E is zero. The relationship between the electric field and magnetic field in different frames is clarified, emphasizing the invariance of certain quantities.

PREREQUISITES
  • Understanding of the Lorentz force law
  • Familiarity with four-vector notation and transformations
  • Knowledge of electromagnetic field transformations in special relativity
  • Basic concepts of acceleration in relativistic physics
NEXT STEPS
  • Study the implications of the Lorentz force law in different inertial frames
  • Learn about four-vector transformations and their applications in special relativity
  • Explore the relationship between electric and magnetic fields in moving reference frames
  • Investigate the concept of invariant quantities in relativistic physics
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This discussion is beneficial for physicists, students of theoretical physics, and anyone interested in the principles of electromagnetism and special relativity.

deadringer
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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).
 
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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]
 
[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:

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