What is the initial kinetic energy of a point charge traveling at velocity v?

  • Context: Graduate 
  • Thread starter Thread starter Ed Quanta
  • Start date Start date
  • Tags Tags
    Formula
Click For Summary
SUMMARY

The initial kinetic energy of a point charge traveling at velocity v can be calculated using relativistic mechanics. For velocities approaching the speed of light, the kinetic energy is determined by the difference between total energy (mc²) and rest energy (m₀c²), where m is the relativistic mass given by m = m₀/sqrt(1 - (v/c)²). The discussion highlights that traditional kinetic energy formulas, such as E = 1/2 mv², are not applicable at relativistic speeds. Instead, the focus should be on the relationship between charge, velocity, and energy, particularly for particles like electrons.

PREREQUISITES
  • Understanding of Larmor's formula for radiated power
  • Familiarity with relativistic mass and energy concepts
  • Knowledge of kinetic energy equations for both classical and relativistic contexts
  • Basic principles of electromagnetism related to point charges
NEXT STEPS
  • Study the derivation and implications of Larmor's formula in electromagnetic theory
  • Learn about relativistic energy-momentum relations in particle physics
  • Explore the concept of relativistic mass and its impact on kinetic energy calculations
  • Investigate the role of charge in electric forces and energy calculations for charged particles
USEFUL FOR

Physicists, electrical engineers, and students studying electromagnetism and relativistic mechanics will benefit from this discussion, particularly those focusing on the energy dynamics of charged particles at high velocities.

Ed Quanta
Messages
296
Reaction score
0
I understand that when a point charge accelerates that it radiates energy due to the acceleration field. Larmor's formula shows the power which is radiated, and is written: P=uoq^2a^2/6(pi)c. My question is this. How can you calculate the initial kinetic energy of a point charge which is traveling at a velocity v? I understand that if the point charge is decelerating until it reaches v=0 then its kinetic energy will be less because it radiated energy away while accelerating. But I am just not sure how to numerically represent the total kinetic energy of the particle prior to decelerating. Is it just E= 1/2qv^2?
 
Physics news on Phys.org
Kinetic energy for particles going at ordinary speeds is m0v2/2.

However, when getting close to the speed of light, the kinetic energy is the difference between the total energy and the rest energy. The total energy is mc2 while the rest energy is m0c2. m=m0/sqrt(1-(v/c)2).
 
Isn't 1/2mv^2=gravitational kinetic energy? If talking about the kinetic energy of an electron, wouldn't it be more accurate to talk of kinetic energy in terms of charge and velocity since charge is what causes the electric force?
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 21 ·
Replies
21
Views
4K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 41 ·
2
Replies
41
Views
5K
  • · Replies 68 ·
3
Replies
68
Views
5K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 13 ·
Replies
13
Views
3K