Showing particle travels at constant speed (geometry)

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SUMMARY

The discussion focuses on the motion of a charged particle in a uniform magnetic field, described by the differential equation γ'' = B × γ'(t), where B is a constant 3-vector. It is established that the particle travels at constant speed, and the component of its velocity in the direction of the magnetic field remains constant. The key insight is that demonstrating the constancy of the squared velocity is a more straightforward approach than finding the full trajectory.

PREREQUISITES
  • Understanding of differential equations, specifically second-order equations.
  • Familiarity with vector calculus and the vector product.
  • Knowledge of the physics of charged particles in magnetic fields.
  • Basic concepts of velocity and acceleration in classical mechanics.
NEXT STEPS
  • Study the properties of vector products in three-dimensional space.
  • Learn about the conservation of energy and momentum in magnetic fields.
  • Explore the derivation of the Lorentz force law and its implications for particle motion.
  • Investigate the mathematical techniques for analyzing the constancy of velocity in differential equations.
USEFUL FOR

This discussion is beneficial for physics students, particularly those studying electromagnetism, as well as educators and anyone interested in the dynamics of charged particles in magnetic fields.

jack1990
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Homework Statement


the trajectory γ: ℝ→ℝ3 of a charged particle moving in a uniform magnetic field satisfies the differential equation γ''= B x γ'(t) . where B = (B1, B2, B3) is a constant 3-vector describing the magnetic field, and × denotes the vector product.

(a) Show that the particle travels at constant speed.
(b) Show that the component of the particle’s velocity in the direction of the magnetic field B is also constant.

Homework Equations

The Attempt at a Solution


Apart from knowing γ'' should =0 I have no idea how to show that, I just need a bump in the right direction on how to start

Thanks for any help
 
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jack1990 said:
Apart from knowing γ'' should =0
That is not true. In general, acceleration won't be zero.

You can find the full trajectory, but you can also show that the magnitude of the velocity is constant without the trajectory. For mathematical reasons, it is easier to show that the squared velocity is constant, which is equivalent to the previous statement.
 

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