Phy 340 class notes for Fall, 1999

Aug 25	Sep 8	Oct 4	Nov 1	Phy 340 page
Aug 30	Sep 13	Oct 11	Nov 8
	Sep 20	Oct 18	Nov 15
	Sep 27	Oct 25	Nov 22
			Nov 29

Phy 340, Week 9

Complete the solution of the point charge in a uniform magnetic field. Quickly re-derive the differential equations and assume an exponential form of solution. Substitute and get the corresponding algebraic equations.

x( t ) = C₁ e^at, y( t ) = C₂ e^at

When you plug these into the above differential equation, you get

a² C₁ - a w C₂ = 0, a² C₂ + a w C₁ = 0,

where w = qB/m is called the "cyclotron frequency." These are two simultaneous, linear, algebraicequations for the unknowns C₁ and C₂.

The condition needed so that there are solutions other than the trivial zero solution is that the two equations are not really independent; that happens when the determinant of the coefficients is zero.

This determines (a set of) a's. For each a, solve the resulting equations for C₁ and C₂. In each case, there are fewer than two simultaneous equaitons! That is, one or zero equations.

a = i w gives C₂ = i C₁

a = - i w gives C₂ = - i C₁

a = 0 gives C₂ = anything, and the same for C₁

Write the total solution for x and y in terms of the four arbitrary constants. Pick some initial conditions to determine all these arbitrary constants, and see what the solution looks like. I chose to start the charge at the origin, moving toward the +x-axis with speed v₀.

20 Oct Discuss the interpretations of some of the problems and how to attack them.
Start on doing mechanics in a non-inertial frame. Take the simplest case first, in which the coordinate system has constant acceleration. In this case, the change of variables is simple:

z' = z - a₀t²/2 x' = x

where z and x are the coordinates in the original, inertial system and z' and x' are in the accelerated system. With these variables, you have

The equations of motion are now

F_x = m d²x/ dt², F_z - ma₀ = d²z/ dt²,

where I have put the m a₀ term on the other side of the equation in order to make it look more like the usual F = ma.

This is the patched-up version of the equations of motion in this non-inertial frame. The added term has the same form as that of a perfectly uniform gravitational field.

The next problem is to see what happens in a rotating coordinate system. As viewed from the inertial system, the motion of a particle in time dt is the combination of two parts. First, from the motion as viewed by the rotating observer, and second as caused by the rotation of the other's coordinate system.

In time dt, the rotating observer says that a mass moves by an amount v' dt, where v' is the mass's velocity in the rotating frame. In the same time, a point that the rotating observer says is fixed will move by an amount w x r dt acording to the stationary observer.

Combine these to get the velocity transformation