Where is potential energy in relativistic formula of energy?

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SUMMARY

The discussion centers on the role of potential energy in the relativistic energy formula, specifically E=m²c²/√(1-v²/c²). It highlights that potential energy in relativistic contexts varies based on the type of force involved, such as gravitational or electromagnetic. For gravitational interactions, the potential energy is generalized, while for electromagnetic interactions, it is represented as a 4-vector (φ, A). The Hamiltonian for electromagnetic potential is expressed as H = √(m²c⁴ + (p - eA)²c²) + eφ.

PREREQUISITES
  • Understanding of relativistic energy equations, specifically E=m²c²/√(1-v²/c²).
  • Familiarity with Hamiltonian mechanics and its application in physics.
  • Knowledge of electromagnetic theory, particularly scalar and vector potentials.
  • Basic principles of general relativity, especially in relation to gravitational potential energy.
NEXT STEPS
  • Study the derivation of the relativistic energy-momentum relation.
  • Learn about the Hamiltonian formulation in classical mechanics and its implications in relativistic contexts.
  • Explore the concept of 4-vectors in electromagnetism, focusing on scalar and vector potentials.
  • Investigate the principles of general relativity, particularly how potential energy is treated in curved spacetime.
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Physicists, students of theoretical physics, and anyone interested in the interplay between kinetic and potential energy in relativistic frameworks.

ndung200790
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Please teach me this:
The relativistic energy E=m.square(c)/squareroot(1-square(v)/square(c)) is determined by momentum p(because of ''square''(4-p)=square(m)).Then what is the role of potential in relativistic energy?When we consider the interaction between particles,how can we express the kinetic energy plus potential energy in the same formula?
Thank you very much in advance.
 
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ndung200790 said:
Please teach me this:
The relativistic energy E=m.square(c)/squareroot(1-square(v)/square(c)) is determined by momentum p(because of ''square''(4-p)=square(m)).Then what is the role of potential in relativistic energy?When we consider the interaction between particles,how can we express the kinetic energy plus potential energy in the same formula?
Thank you very much in advance.

The question has to be more specific: what kind of force? Because if it is gravity, you need to do some GR and it's not exactly a potential anymore. If it's EM, the potential is a 4-vector: (\phi, \mathbf{A}), where \phi is the scalar potential and \mathbf{A} the vector potential. The Hamiltonian in this case is:
H = \sqrt{m^2 c^4 + \left(\mathbf{p} - e\mathbf{A}\right)^2 c^2} + e\phi
 
If it's gravity, then both potential and kinetic energy are generalised as (letting c=1):
-m\frac{d\tau}{dt}​
where
d\tau^2=g_{\mu\nu}dx^{\mu}dx^{\nu}​
We can see this clearly when we assume spherical symmetry and consider the Newtonian limit:
-m\frac{d\tau}{dt}=-m\sqrt{\frac{g_{tt}dt^2+g_{mm}(dx^m)^2}{dt^2}}​
=-m\sqrt{g_{tt}-\dot{x}^2}​
=-m\sqrt{1-\frac{2GM}{r}-\dot{x}^2}​
\approx-m+\frac{GMm}{r}+\frac{1}{2}m\dot{x}^2​
And given that the action in a gravity well is
-m\int{d\tau}=0​
then we recover the non-relativistic
E=\Delta{U}+KE​
 
Last edited:

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