From: PointedEars@web.de   
      
   [Supersedes because of too many typos]   
      
   Thomas 'PointedEars' Lahn wrote:   
   > What I have not shown is how to calculate the energy density itself because   
   > it is a rather lengthy calculation.   
      
   By that I meant how to _derive the formula_ for the energy density of the EM   
   field in Gaussian units,   
      
     u(X, t) = 1/(8π) [E²(X, t) + B²(X, t)].   
      
   Once you have that formula, its value is rather trivial to calculate, of   
   course.   
      
   > But it can be shown by integrating the Lorentz force   
   >   
   >   F = q (E + V × B)   
   >   
   > over the path of a charged particle,   
      
   Hint: The absolute value of the work done by the electromagnetic field on a   
   charged particle,   
      
     W = ∫_P dS ⋅ F = ∫_P dS ⋅ q (E + V × B)   
      
   is equal to the difference between the energy that was stored in the field   
   before it did that work and after that.  Notice that if dS is an   
   infinitesimal line element of the particle's spatial path P, then dS || V   
   which means that the magnetic field does not do work on the particle:   
   dS ⋅ q (V × B) = 0.  (But that does not mean that no energy is stored in   
   it.)   
      
   > which in turn can be derived from the special-relativistic (Lo   
   entz-covariant)   
   > Lagrangian for a charged particle coupled to the electromagnetic field (given   
   > by the Maxwell tensor with components F_ab = ∂_a A_b − ∂_b A_a, where   
   > A_a = [−ϕ/c, A] is the four-potential, where ϕ is the electric potential   
   in   
   > E = −∇ϕ, and A is the magnetic vector potential in B = ∇ × A).   
      
   Hint: The total action is the sum of the action for a free particle,   
      
     S_0[x] = -m c^2 ∫_W dτ,   
      
   where W is a section of the (timelike) worldline of the particle, τ is   
   proper time, and the action for the interaction with the EM field,   
      
     S_I[x; A] = q ∫_W dτ A_a dx^a/dτ.   
      
   As usual, one can find the Euler--Lagrange equations by calculating the   
   variation of the total action, here   
      
     S[x; A] = S_0[x] + S_I[x; A].   
      
   Because the variation is a *linear* differential operation, and the   
   Euler--Lagrange equations are *linear* differential equations, knowing the   
   free relativistic Lagrangian, it suffices to vary S_I[x; A] to obtain the   
   equations of motion for the interaction and add the non-interacting ones to   
   obtain the Lorentz force equations.   
      
   --   
   PointedEars   
      
   Twitter: @PointedEars2   
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