Undulator Homework

Undulator Homework

Mexican Particle Accelerator School 2011

2Undulators Mexican Particle Accelerator School Sept. 26 - Oct 7, 2011

Undulator Homework problem 1a

Assume a sinusoidal magnetic field in the vertical (y) direction

where z is the direction of travel of the electron beam and lu is the period length of the undulator.Calculate the effect that this magnetic field has on the passing electrons. Remember that the electrons are relativistic, so use F=gma, where g is the relativistic g = 1 / sqrt (1 – b2). The velocity of the electron v divided by the speed of light c is b.

Show that the transverse (x) velocity of the electron satisfies

e is the charge of the electron, c is the speed of light, and m is the electron mass.


Undulator Homework problem 1bAs the electron travels through the undulator, it will emit a photon pulse as it goes through a wiggle. One period later, it will emit a second photon pulse. What is the distance between these successive photon pulses as seen by a downstream observer (in the lab frame of reference)?

This distance is the wavelength of the first harmonic of the light that the undulator produces. The goal of this problem is to derive the Undulator Equation,

Some hints to help:

What is the time it takes an electron to travel one period lu , in terms of the average z-component of the electron’s velocity, vz avg? How far does the photon pulse travel during this time?

Show that the distance between successive photon pulses is (lu c / vz avg) – lu .


Undulator Homework problem 1b, cont.Now relate vz avg to the vx found in part a.

The electron speed |v| doesn’t change in the undulator since the (magnetic) force is always perpendicular to the motion, but the direction of the velocity changes. (The energy lost in the undulator radiation is negligible.) Use v2 = vz

2 + vx2 to relate these, and use g to

eliminate v .

Show that

Solve for vz.

Rewrite the expression for vz so it has a constant part and an oscillating cosine term that averages to zero. The identity cos 2q = 2 cos2 q – 1 may prove useful. Also

Drop the oscillating term in vz to get vz avg , plug it into the photon pulse separation above, and show that the result is the Undulator Equation.


Undulator Homework problem 2

A user who knows all the details of his/her experimental program but nothing about insertion devices comes to ask for an insertion device for a new proposed beamline.What questions do you ask, what information do you give, and what discussions do you have with the user in order to determine what device is the best match to the user’s needs?


Undulator Homework 1a solution

Assume a sinusoidal magnetic field in the vertical (y) direction

where z is the direction of travel of the electron beam and lu is the period length of the undulator.Calculate the effect that this magnetic field has on the passing electrons. Remember that the electrons are relativistic, so use gm for the mass, where g is the relativistic g = 1 / sqrt (1 – b2). The velocity of the electron divided by the speed of light c is b.

Solve for vx :


Undulator Homework 1a solution, cont

Since the electron is highly relativistic, and vz is effectively (here, anyway) equal to c, vx/c is essentially the angle of the electron’s trajectory from the z axis.Thus, K/g is the maximum angle between the trajectory and the forward direction. This is also taken to be the maximum half-width of the undulator beam.

K is an important parameter of an undulator. It is characteristic of the undulator and does not change if the undulator is installed on a different storage ring. It is typically around 1. If the K of a device gets to be above about 5, the photon radiation characteristics are more wiggler-like.


Undulator Homework 1b, solutionAs the electron travels through the undulator, it will emit a photon pulse as it goes through a wiggle. One period later, it will emit a second photon pulse. What is the distance between these successive photon pulses as seen by a downstream observer (in the lab frame of reference)?

The time it takes an electron to travel one period is lu divided by the average z-component of the electron’s velocity, vz avg. During this time, the first photon pulse is traveling in the same direction as the electron’s average z velocity, traveling the distance lu c / vz avg. The distance between successive photon pulses is then the difference (lu c / vz avg) – lu .

Note that vz varies with z because vx varies with z. The electron speed v doesn’t change in the undulator since the force is always perpendicular to the motion, but the direction of the velocity changes. (The energy lost in the undulator radiation is negligible.) We will use v2 = vz

2 + vx2 to relate these, and use g to eliminate v .


Undulator Homework 1b, solution p. 2

Solving for vz and substituting the expression for vx found in part a) gives:

Taking the square root,

g is typically very large (for the 7 GeV Advanced Photon Source, it is 13699), so we can approximate

Use the identity



The last term is the oscillating part that averages to zero. The other terms are vz avg. Thus, (lu c / vz avg) – lu becomes (using 1/(1-e) = 1+e, and remembering that g is large):



Finally, this leads to the “Undulator Equation”. That characteristic distance between photon pulses is the wavelength of the first-harmonic radiation produced by the undulator.

Undulator Homework

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