What is the lagrangian of a free relativistic particle?

AI Thread Summary
The Lagrangian for a free relativistic particle in an electromagnetic field is expressed as L = mc^2*sqrt(1-(v/c)^2) + q*(A*v - φ). The discussion highlights the confusion surrounding the concept of a "free" particle in the presence of electromagnetic fields, as all particles exhibit electromagnetic properties. The kinetic energy is defined as T = m*[1 - {1/(1-(v²/c²))}]*c², emphasizing the role of rest mass. The invariant nature of the Lagrangian under Lorentz transformations is noted, with proper time parameterization leading to the formulation L = m∫dτ√((dt/dτ)² - (dx/dτ)²). The conversation concludes with a reaffirmation of the Lagrangian's applicability to free particles.
Lior Fa
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Homework Statement


What is the lagrangian of a free reletavistic particle in a electro-magnetic field?
And what are the v(t) equations that come from the Euler-Lagrange equations (given A(x) = B0/2 crosProduct x)
(B/2 is at z direction)

Homework Equations

The Attempt at a Solution


I've got to: L = mc^2*sqrt(1-(v/c)^2)+q*(A*v-φ)

But don't get to the velocity equations
 
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The problem is quite confusing a particle cannot be free if it is in an electromagnetic field because every conceivable particle has electromagnetic attributes.
Kinetic energy, T = m*[1 - {1/(1-(v²/c²))}]*c², where m is rest mass.
 
For a free relativistic massive particle ,lagrangian would be ##m\int ds## where ds is the proper time and m is the rest mass..So it is invariant under lorentz transformation. ##ds=\sqrt{\eta_{\mu \nu}dx^{\mu}dx^{\nu}}## So if I parametrize the whole thing with parameter ##\tau## then we have ##ds=\sqrt{\eta_{\mu \nu}\frac{dx^{\mu}}{d\tau}\frac{dx^{\nu}}{d\tau}}d\tau## Then we can write $$L=m\int d\tau \sqrt{(\frac{dt}{d\tau})^{2}-(\frac{dx}{d\tau})^{2}}$$ where I have taken mostly negative sign convention...
I think This should be the case with free particle...
 
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