Potential Energy of Relativistic Particles in Coulomb Field

In summary: Spin_ parity_ violation/aef5d6e6_39cb_4c11_b7a0_2b1dc2c4ddd1/In summary, the potential energy of a particle in a static field is not renormalized.
  • #1
reterty
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Let us consider relativistic particle (electron) which moves with relativistic speed ##v## in the Coulomb field (in the field of a fixed heavy nucleus). The main question is what is the potential energy of a particle in such a static field? Landau and Lifshitz in their book "Field Theory" believe that the potential energy is not renormalized in any way and is equal to ##\frac{qQ}{r}##. At the same time, a number of authors of original articles on this topic introduce a reduced distance ##r\sqrt{1-v^2/c^2}## into the denominator of this fraction due to the relativistic effect of the reduction in linear dimensions. Which of them is right?
 
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  • #2
I guess you mean the relativistic motion of a charged particle in the coulomb field of a very much heavier particle, neglecting the radiation reaction. The relativistic equation of motion in the non-covariant formalism is derived from the Lagrangian
$$L=-mc^2 \sqrt{1-\dot{\vec{x}}^2} + \frac{q Q}{4 \pi \epsilon_0 |\vec{x}|}.$$
It's of some historical interest since it was Sommerfeld's derivation of the fine structure of the hydrogen-atom spectrum within old quantum theory. It's kind of surprising that he got the correct result although the model is, of course, entirely wrong, i.e., it doesn't take into account the spin 1/2 of the electron and the gyrofactor 2 (both of which weren't known in 1916). That's why you find the solution in Wikipedia here:

https://en.wikipedia.org/wiki/Bohr–Sommerfeld_model#Relativistic_orbit
 
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  • #3
reterty said:
$r\sqrt{1-v^2/c^2}$
Please note that on this website you need to use a double-$ instead of a single-$ for LaTeX to work.
$$r\sqrt{1-v^2/c^2}$$
 
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  • #4
DrGreg said:
on this website you need to use a double-$ instead of a single-$ for LaTeX to work
Or a double # for inline LaTeX (the double $ means an equation in its own paragraph).
 
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  • #5
For example: https://www.researchgate.net/publication/305345527_A_New_Relativistic_Extension_of_the_Harmonic_Oscillator_Satisfying_an_Isochronicity_Principle
 

FAQ: Potential Energy of Relativistic Particles in Coulomb Field

What is the definition of potential energy in a Coulomb field?

Potential energy in a Coulomb field is the energy that a charged particle possesses by virtue of its position in an electric field created by a stationary charged particle.

How is the potential energy of a relativistic particle in a Coulomb field calculated?

The potential energy of a relativistic particle in a Coulomb field is calculated using the formula U = q1q2/4πε₀r, where q1 and q2 are the charges of the two particles, ε₀ is the permittivity of free space, and r is the distance between the particles.

What is the relationship between potential energy and kinetic energy in a Coulomb field?

In a Coulomb field, potential energy and kinetic energy are interconvertible. As the potential energy of a charged particle increases, its kinetic energy decreases and vice versa.

How does the potential energy of a relativistic particle change as it approaches a stationary charged particle in a Coulomb field?

The potential energy of a relativistic particle increases as it approaches a stationary charged particle in a Coulomb field. This is because the distance between the particles decreases, resulting in a stronger electric field and a higher potential energy.

Can the potential energy of a relativistic particle in a Coulomb field be negative?

Yes, the potential energy of a relativistic particle in a Coulomb field can be negative. This occurs when the two particles have opposite charges and are attracted to each other, resulting in a negative potential energy value.

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