What is the lagrangian of a free relativistic particle?

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SUMMARY

The Lagrangian of a free relativistic particle in an electromagnetic field is expressed as L = mc²√(1 - (v/c)²) + q(Av - φ), where A(x) = B₀/2 × x and B₀ is directed along the z-axis. The discussion highlights the confusion surrounding the concept of a "free" particle in the presence of electromagnetic attributes. The kinetic energy is defined as T = m[1 - (1/(1 - (v²/c²)))]c², and the invariant Lagrangian for a free relativistic massive particle is given by L = m∫ds, where ds represents proper time. The parametrization of the Lagrangian leads to L = m∫dτ√((dt/dτ)² - (dx/dτ)²) with a negative sign convention.

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This discussion is beneficial for physicists, particularly those specializing in theoretical physics, as well as students studying advanced mechanics and electromagnetism. It is also relevant for anyone interested in the mathematical foundations of relativistic particle dynamics.

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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