What is the lagrangian of a free relativistic particle?

In summary, the Lagrangian of a free relativistic particle in an electromagnetic field is given by L = mc^2*sqrt(1-(v/c)^2)+q*(A*v-φ), where m is the rest mass, c is the speed of light, q is the charge of the particle, A is the magnetic vector potential, and φ is the scalar potential. The Euler-Lagrange equations for this system yield the equations of motion for the velocity, which can be expressed as d^2x/dt^2 = q/m * (E + v x B), where E is the electric field and B is the magnetic field.
  • #1
Lior Fa
4
0

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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  • #2
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.
 
  • #3
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...
 

Related to What is the lagrangian of a free relativistic particle?

1. What is the Lagrangian of a free relativistic particle?

The Lagrangian of a free relativistic particle is a mathematical function that describes the dynamics of the particle in a relativistic framework. It is represented by L = -mc²√(1-v²/c²), where m is the mass of the particle, c is the speed of light, and v is the velocity of the particle.

2. How is the Lagrangian derived for a free relativistic particle?

The Lagrangian for a free relativistic particle is derived using the principle of least action, which states that the actual path taken by a particle is the one for which the action (a quantity that represents the total energy of the particle) is minimized. This leads to the equation L = -mc²√(1-v²/c²).

3. What is the significance of the Lagrangian in the study of free relativistic particles?

The Lagrangian is significant because it allows us to describe the motion and behavior of a free relativistic particle in a consistent and elegant way. It also serves as the basis for the development of other important concepts, such as the Hamiltonian and the equations of motion.

4. Can the Lagrangian be used to describe the behavior of particles with massless or infinite mass?

Yes, the Lagrangian can be used to describe the behavior of both massless and infinitely massive particles. For massless particles, the expression reduces to L = E/c², where E is the energy of the particle. For infinitely massive particles, the expression becomes L = -∞.

5. How does the Lagrangian approach differ from the Newtonian approach in describing the motion of a free relativistic particle?

The Lagrangian approach differs from the Newtonian approach in that it takes into account the effects of relativity, such as time dilation and length contraction, in describing the motion of a free relativistic particle. It also allows for a more elegant and general description of the particle's behavior, as it is not limited by specific reference frames or coordinate systems.

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