Relativistically correct exression for Lorentz force

In summary, the correct expression for the Lorentz force on a relativistic particle in an EM field in cgs units, in terms of vector potential A and scalar potential phi, can be derived from a given Lagrangian. This can be done by writing out the Euler-Lagrange equations and identifying which partial derivatives of the four-potential correspond to the E and B fields. The Lorentz force law is already relativistic and can be obtained from the Lagrangian by adding the charge times the four-potential to the four-momentum.
  • #1
mlh3789
1
0
I need to know the correct expression for the Lorentz force on a relativistic particle in an EM field, in cgs units, in terms of vector potential A and scalar potential phi, to prove that a given Lagrangian produces the correct equations of motion. I don't need it in covariant form with tensors and all of that because we haven't learned that yet. Thanks everyone.
 
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  • #2
mlh3789 said:
I need to know the correct expression for the Lorentz force on a relativistic particle in an EM field, in cgs units, in terms of vector potential A and scalar potential phi, to prove that a given Lagrangian produces the correct equations of motion. I don't need it in covariant form with tensors and all of that because we haven't learned that yet. Thanks everyone.


The the correct expression for the Lorentz force on a relativistic particle in an EM field is no different than the same expression for a non-relativistic particle, i.e. in MKS units

f = dp/dt = q(E + vxB)

This is not a 4-tensor equation though. If you're speaking about the tensor formulation then its taks on a different form. The tensor form is dP/d[itex]\tau[/itex] = qF*U where P = 4-momentum of particle, F = Faraday tensor and U = particle's 4-velocity.

Pete
 
  • #3
In cgs gaussian units, just divide v by c, unless you have set c=1.
 
  • #4
Because of my poor editing, I cannot write it down here. In fact, I am not quite sure what you are asking. But I believe you can find your answer by looking at the differential expression of conservation of momentum of EM field which is in the form of divergence equation on the internet or other sources. The momentum density is something like 1/(4times pi times c) of cross product of E and B, the Lorentz force density is something like rho times E plus j/c cross B etc. If you have that equation in SI unit, you can transform it in Gaussian unit by replacing epsilon with 1/4pi, mu with 4pi / c^2, and B with B/c. If you find this is not helpful, I am sorry but I already did the best I can.
 
  • #5
mlh3789 said:
I need to know the correct expression for the Lorentz force on a relativistic particle in an EM field, in cgs units, in terms of vector potential A and scalar potential phi, to prove that a given Lagrangian produces the correct equations of motion. I don't need it in covariant form with tensors and all of that because we haven't learned that yet. Thanks everyone.

I'm interested to see that someone is being taught about this way of getting the Lorentz force law from a Lagrangian. It makes it possible to see how adding the charge times the four-potential to the four-momentum (as in the Pauli "minimal electromagnetic coupling" assumption) gives a Lagrangian which results in exactly the same law of motion for charged particles as the Lorentz force law. That puzzled me for a long time until I worked it out for myself, which isn't difficult but I didn't see it mentioned in the textbooks which I used.

As mentioned in a previous response, the Lorentz force law is already relativistic. That means all you have to do is write out the Euler-Lagrange equations, recognize which partial derivatives of the four-potential correspond to the E and B fields and reorganize the result to show that it matches the Lorentz force law. I'm leaving the details as an exercise for the student.
 

1. What is the formula for calculating the Lorentz force?

The formula for calculating the Lorentz force, which describes the force experienced by a charged particle moving through an electromagnetic field, is F = q(E + v x B), where q is the charge of the particle, E is the electric field, v is the velocity of the particle, and B is the magnetic field.

2. How does the Lorentz force change at relativistic speeds?

At relativistic speeds, the Lorentz force formula must be modified to account for the effects of special relativity. The relativistically correct expression for the Lorentz force is F = q(E + v x B) + (q/c)(v x E), where c is the speed of light. This additional term accounts for the fact that electric fields can produce a force on a moving charged particle, as observed in the particle's rest frame.

3. What is the significance of the relativistically correct expression for Lorentz force?

The relativistically correct expression for Lorentz force allows for more accurate calculations of the forces experienced by charged particles at high speeds. It is essential for understanding and predicting the behavior of particles in particle accelerators, as well as in other areas of physics such as astrophysics and plasma physics.

4. Can the relativistically correct expression for Lorentz force be used for both positive and negative charges?

Yes, the relativistically correct expression for Lorentz force can be used for both positive and negative charges. The charge of the particle is accounted for in the formula, so it does not matter if it is positive or negative.

5. Is the relativistically correct expression for Lorentz force consistent with Maxwell's equations?

Yes, the relativistically correct expression for Lorentz force is consistent with Maxwell's equations, which describe the fundamental laws of electromagnetism. In fact, the relativistically correct expression was derived from Maxwell's equations, taking into account the effects of special relativity.

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