Guess solution to fields by minimizing the action.by Spinnor Tags: action, fields, guess, minimizing, solution 

#1
Jan313, 06:51 PM

P: 1,362

Suppose I want to try and guess the fields of a short laser pulse. We know that fields that satisfy Maxwell's equations minimize the action E^2  B^2 (say far from charge and current)?
For a plane wave E^2  B^2 = 0? Will a general solution of maxwell's equations satisfy E^2  B^2 = 0 If I have a set of fields (say the approximate solution given by Jackson above) can I be guaranteed that the action for the actual solution will be less then the action for the approximate solution? Can I consider E^2 the "kinetic" part of the energy and consider B^2 the "potential" part of the energy? Thanks for any help or suggestions! 



#2
Jan413, 04:09 PM

Sci Advisor
HW Helper
PF Gold
P: 2,606

[tex] \mathbf{E} = \hat{\mathbf{i}} E_0 \sin[k (zct)] [/tex] satisfies [itex] c\mathbf{B} = \hat{\mathbf{k}}\times \mathbf{E}[/itex], so that [itex] \mathbf{E}^2  c^2\mathbf{B}^2 = 0[/itex] [tex] \mathbf{E}_1+\mathbf{E}_2^2  c^2\mathbf{B}_1+\mathbf{B}_2^2 = (1 \hat{\mathbf{e}}_1\cdot\hat{\mathbf{e}}_2)\mathbf{E}_1\cdot \mathbf{E}_2.[/tex] The approximation where the plane waves have the same directions, [itex]\hat{\mathbf{e}}_1\cdot\hat{\mathbf{e}}_2=1[/itex], has vanishing action, whereas the realistic case where they do not has a finite action (which could be positive). In this case, we would conclude that the approximation of the sourcefree, vacuum field equations is incorrect. Instead, we must more accurately describe the source of the pulse, as well as the nature of the cavity. As an idealized mathematical problem, an approximate solution (taken for instance in the sense of perturbation theory) will always have an action that is larger than the exact solution. Some more or lessrelevant references can be found at http://www.scholarpedia.org/article/...f_least_action 



#3
Jan413, 05:20 PM

P: 1,362

Thank you fzero for quite a lesson! I just need to learn faster then I forget.



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