Deriving Lorentz Force Formula: Explanation of Elementary Electromagnetics

[karlzr]

In elementary electromagnetics, we learned the formula of Lorentz force. While I am always confused about the origin of this formula. Because when studying electrodynamics, I always thought Maxwell equations incorporate all information we need to know about electromagnetic interaction, but I fail to derive the Lorentz force formula from Maxwell equations or from the action [tex]S=-\frac{1}{4}\int d^4x F_{\mu\nu}F^{\mu\nu}[/tex].
Hoping anyone give me an explanation on the derivation of Lorentz force formula
[tex]\vec{F}=q(\vec{E}+\vec{v}\times\vec{B})[/tex]

 

[vanhees71]

If you want to derive the force of the electromagnetic field acting on a point charge from fundamental principles of Poincare and gauge invariance of electromagnetics (which latter itself is derivable from relativistic quantum field theory by investigating the unitary representations of the Poincare group a la Wigner), you have to put the particle into the action. The complete action (in Heaviside-Lorentz units with c=1) reads

[tex]S[A,\vec{\xi}]=-\frac{1}{4} \int d^4 x F_{\mu \nu} F^{\mu \nu}-m \int d t \sqrt{1-\dot{\vec{\xi}}^2} - q \int d t A_{\mu}(\xi) \frac{d \xi^{\mu}}{d t}.[/tex]

Of course, it's understood that

[tex]F_{\mu \nu}=\partial_{\mu} A_{\nu} - \partial_{\nu} A_{\mu}[/tex]

and

[tex]\xi^{\mu}=\begin{pmatrix} t \\ \vec{\xi} \end{pmatrix}[/tex]

As you see, this action is Poincare and gauge invariant as it should be.

Taking the variations with respect to the trajectory of the particle, you get the correct (relativistic) equation of motion for a particle in the em. field with the Lorentz force. Taking the variations with respect to the field you get the Maxwell equations with the four-current of the point particle given by

[tex]j^{\mu}(x)=\int d t \frac{d\xi^{\mu}}{d t} \delta^{(4)}[x-\xi(t)].[/tex]

You should be aware that a fully self-consistent solution of this set of equations is very much complicated by the issue of the interaction of the particle with its own radiation field (see, e.g., Jackson, Classical Electrodynamics, 3rd edition or The Feynman Lectures, Vol II for a thorough discussion of this point).

 

[karlzr]

Thank you so much!
By the way, the third term in the action denotes the interaction of em field with the source, what about the second term? what does [tex]m\sqrt{1-\dot{\vec{\xi}}\ ^2}[/tex] mean? Here "m" represents the rest mass, right ?

 

[vanhees71]

This is just the relativistic kinetic term for a point particle of invariant mass, m. A free particle's Lagrangian is given by

[tex]L_0=-m \int \mathrm{d} t \sqrt{1-\dot{\vec{\xi}}^2}[/tex].

From this you get via Noether's theorem the (canonical) momenta and the energy of the free particle,

[tex]\vec{p}=m\frac{\dot{\vec{\xi}}}{\sqrt{1-\dot{\vec{\xi}}^2}}[/tex]

and

[tex]E=\dot{\vec{\xi}} \cdot \vec{p}-L=\frac{m}{\sqrt{1-\dot{\vec{\xi}}^2}}.[/tex]

 

[edevault]

Hello, either of you may be able to answer this. I am not sure if its legal to ask a related question off of someone elses question or if I have to start my own post, so let me know if I am breaking etiquette. I am trying to determine the force on an Air Core Reactor (sometimes called a shunt reactor) during a short circiut. Obviosuly this will depend on variables such as # of cable turns, the current, the diameter of the Air Core etc. I was lead to Lorentz Force. However most of the examples for Lorentz Force involved 2 parrellel wires. An air core reactor is a loop of one wire around and around the core. I found a version of Lorentz force for current moving through a magnetic through a curved wire. It is: F = I (current) integral of dl (infentesimal wire segment) X (cross product) B (magnetic field). Sorry the copy and paste is not working. Am I on the right track? Would the limits of the integral be 0 and 2pie? Any help is greatly appreciated.