Calculating Momentum of a Particle: Q&A

  • Thread starter Thread starter Stickybees
  • Start date Start date
  • Tags Tags
    Momentum Particle
Stickybees
Messages
36
Reaction score
0
Say I have the rest energy(mc^2) and the total energy of a particle (E), would getting the momentum energy of the particle be as simple as doing (E^2-(mc^2)^2)^(1/2) = pc?

And when accelerating a electron through a potential difference how would I work out its momentum, given I have its rest energy and the value for potential difference?

Thanks.
 
Physics news on Phys.org
That looks like the correct equation to use.

When an electron is accelerated through a potential difference, what happens to its total energy?
 
c=1 you have
E2= p2+m2

so you want to know the momentum you do what you said

p=√ [E2-m2]

by the time you say "total energy E" it means that the kinetical energy is within your already known parameter, so is there a potential, is there not- you know E, you know mass, so you know its mommentum.
 
I am assuming that for a p.d. of for example 1 MV, this will create an energy difference of 1 MeV for an electron and with the rest energy of around 0.5 MeV for an electron, it would simply equal 1.5 MeV/c for the momentum, but I think this is wrong, but I don't understand why.
 
Assuming all the energy was kinetic, your formula works. If there is also potential energy involved, then you have to take that into account as well.
 
Not an expert in QM. AFAIK, Schrödinger's equation is quite different from the classical wave equation. The former is an equation for the dynamics of the state of a (quantum?) system, the latter is an equation for the dynamics of a (classical) degree of freedom. As a matter of fact, Schrödinger's equation is first order in time derivatives, while the classical wave equation is second order. But, AFAIK, Schrödinger's equation is a wave equation; only its interpretation makes it non-classical...
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. Towards the end of the first lecture for the Qiskit Global Summer School 2025, Foundations of Quantum Mechanics, Olivia Lanes (Global Lead, Content and Education IBM) stated... Source: https://www.physicsforums.com/insights/quantum-entanglement-is-a-kinematic-fact-not-a-dynamical-effect/ by @RUTA
Is it possible, and fruitful, to use certain conceptual and technical tools from effective field theory (coarse-graining/integrating-out, power-counting, matching, RG) to think about the relationship between the fundamental (quantum) and the emergent (classical), both to account for the quasi-autonomy of the classical level and to quantify residual quantum corrections? By “emergent,” I mean the following: after integrating out fast/irrelevant quantum degrees of freedom (high-energy modes...
Back
Top