Emittance Test

Design

New deisign: We changed to use just one magnet.

File:Emittance Test Optim File.txt

HRRL had steer magnets on itself. We don't need external steer magnets. But Dr. Chouffani said that we might need them after quads, if the quads are not on axis. He will think about it.

Beam Line design on 27th, Jul. 2010:

Optim File on 27th, Jul. 2010:

File:Emittance Test Optim File 27th Jul 2010.txt

File:HRRL Emittance.pdf

Faraday Cup

Old Beam Line

Beam Line Parts

Constructed Beam Line for Emittance Test

Vacuum pressure on the beam line is less than [math] 10^{-8} [/math] torr (on 2010-July-7th).

Leveling Survey

Downstream of the beam up by 2.6 mm.

How to do Leveling Survey and Alignment with Theodolite on e- Beam Line

A. Leveling Beam Pipe

1. Put the theodolite as high as the center of the beam line at upstream. Turn theodolite up and down. If the up angle is equal to down angle from the center of the pipe, theodolite is exactly as high as beam pipe. If not, adjust theodolite, until up and down angles are equal.

2. Once first step finished, then rotate theodolite horizontally to down stream. Turn it up and down, if up and down angles are not equal, adjust the stand till they are equal.

B. How to Make Sure the Quad is Perpendicular to the Beam Line

If the distances between quads and end of cavity at any given points are equal, then quad and end of cavity are parallel. Pipe is perpendicular to the end of the cavity, thus, magnet would also be perpendicular to the pipe.

I will find at lest three random points quads and three corresponding points at the end of the cavity, make sure they are equal, then I can say, magnet is perpendicular to the beam line.

I am planning to stick a metal ruler two, so two faces are parallel.

C. Centering the Quad Magnet with Theodolite

Vertical Centering

1. Finish the step 1 in A.

2. Then turn up the theodolite from the upper edge of the pipe to the magnet hole, call this angle upper angle. Do the same at lower part, call it lower angle. If the this two angles are not equal adjust the stand till they are equal.

Horizantal Centering

We know the diameter of the pipe [math] d_p [/math]. So, we can calculate from the pipe to the edge of the quad as:

We need to make sure this distance is same when measure it horizontally, so that our quad is centered on the pipe. I am going to use a ruler to ...(thinking)

July 2010 Run

Initial Accelerator Settings

E = 10 MeV [math]\Rightarrow[/math] Control Voltage= 4.09 V

Pulse Width <= 50 ns

Average Beam Current = least amount needed to light up YaG crystal, start at 1 mA and Rep rate of 100 Hz.

At the same time we want to do radiation test on HRRL. This is for predicting the radiation after shifting cavity to new position. Radiation test run plan is at: [1]

7/26/10

Alignment check

1.) Survey the Quad to be sure it is centered on the beam pipe

2.) survey the flag to determine if target is centered on the beam pipe

Transmission tuning

1.) Using the quads and accelerator coils, focus the beam on the YaG crystal to as fine a spot as possible. Monitor the FC during the focusing.

changed control Voltage from

Control Voltage= 4.9 V [math]\Rightarrow[/math] E = 16 MeV

to

4.09 V [math]\Rightarrow[/math] E = 10 MeV

Tune the 4 accelerator coils for max transmission.

We want to maximize the current reach the Farady Cup. This can be determined by the ratio on Channels of FC and Gun Current.

Gun current is the current on the pick up coil, which is input current for cavity. Current on Farady cup is the current reach the end of the beam line (FC is at the end of the beam line). We want to maximize the the ratio of Farady cup current to Gun current (FCC/GC), so there are more electrons will make it to the target. Thus, we enlarge our positron yield.

Basic principle: Fix three coils, change one till it maximized.

Coil 1 upstream left/right (mA)	Coil 2 upstream UP/down (mA)	Coil 3 downstream up/down	Coil 4 downstream left right	Scope Picture	Gun Current	FC current	Transmission [math]\equiv[/math] Gun / FC
	10				15 +/- 0.65 nV s	-5.4[math]\pm[/math]1.8 nV s	36 [math]\pm[/math] 12 %

The HRRL went down after 2 hours of operation.

7/27/10

No HRRL beam time today. Chad tested the Thermatron module and believes it is not the problem. Chad will continue debugging tomorrow.

Filament

convert files to pnm.

pnm: portable network map

JPG format:

BMP format:

Wiki won't allow me to upload with bmp extension. So, I used png extension to upload. After download files, you will find they are still with bmp extension.

Tiff

300 px

7/28/10

Maximize Transmission

Maximized setup according to the accelerator operator.

Beam Energy: 16 MeV

Rep Rate: 95 Hz

Pulse Width: 100 ns

Coil 1 upstream left/right (A)	Coil 2 upstream UP/down (A)	Coil 3 downstream up/down (A)	Coil 4 downstream left right (A)	Scope Picture	Gun Current	FC current	Ratio	Spot Size
0.1	0.5	3.1	0.7

Reversed the polarities on coil # 3 and coil # 4. Rep rate: 3Hz

Coil 1 upstream left/right (A)	Coil 2 upstream UP/down (A)	Coil 3 downstream up/down (A)	Coil 4 downstream left right (A)	Scope Picture	Gun Current	FC current	Ratio	Spot Size
1.2	1.2	0.0	3

It appears the the image from electrons hitting the YaG crystal is offset (to the right) compare to the image from the filament which appears near the center of the cross hairs.

Electron beam deflection due to Earth's B-field.

Is the earths Magnetic field causing the electrons to move this far beam right?

[math]B_{Earth}[/math] = 0.00005 Tesla ( [math]\frac{ \mbox{kg}}{\mbox{C}\cdot \mbox{s}}[/math]) = 0.5 Gauss

E_{electron} = 10 MeV = Pc [math]\Rightarrow[/math] P = 10 MeV/c

Estimation using Classical Lorentz Force (forget about angle of B and V)

[math]\vec{F} = q \vec{v} \times \vec{B}[/math]

assuming [math]v[/math] and [math]B[/math] are orthogonal then we get the cyclotron radius

[math]F= qvB = m\frac{v^2}{r} \Rightarrow r = \frac{qB}{mv}[/math]

[math]r = \frac{(1.6 \times 10^{-19} \mbox{C}) \times ( 5 \times 10^{-5}\frac{\mbox{kg}}{\mbox{C} \cdot \mbox{s}})}{mv}[/math]

[math]mv = P = 10 \frac{ \mbox{MeV}}{\mbox{c}} = 5.6 \times 10^{-21} \frac{\mbox{kg} \cdot \mbox{m} }{\mbox{s}} [/math]

[math]\Rightarrow r = 700 \mbox{m}[/math]

Now some Geometry.

Assuming the orbit is a circle with a 700 m radius, then the question becomes how far does the electron move to maintain the orbit if it travels 1 m in a strait line.

let

[math]r[/math]= radius of circle

[math]l[/math]= distance along the beam line

[math]d[/math] = deflection due to gravity

Note

[math]l[/math] is a tangent to the orbit

one can construct a right angle triangle with side of length d, the [math]l[/math] as the base and [math]r+d[/math] as the hypotenuse.

Then Pythagorean's theorem gives

[math]r^2 + l^2 = (r+d)^2[/math]

Giving

[math]d^2+ (2r)d +l^2 =0[/math]

using quadratic formula you get

[math]2d = -(2r)/2 \pm \sqrt{(2r)^2 + 4\frac{l^2}{r^2}}/2[/math]

since l/r is very small (1/700) we can approximate this as

[math]d = -r \pm r \sqrt{1+ \frac{l^2}{r^2}} \approx \frac{l^2}{r} = \frac{l^2}\frac{P}{qB}= \frac{qBl^2}{P}[/math]

for 10 MeV and a 1 m long beam pipe

[math]d = (l\mbox {m})^2 \frac{(1.6 \times 10^{-19} \mbox{C}) \times ( 5 \times 10^{-5}\frac{\mbox{kg}}{\mbox{C} \cdot \mbox{s}})}{ 5.6 \times 10^{-21} \frac{\mbox{kg} \cdot \mbox{m} }{\mbox{s}} } = l^2 \times 1.42 \mbox{mm}[/math] where [math]l[/math] is in units of m

Spent the remaining beam time doing radiation footprint measurements.

HRRL_radiation_tests#Run_Plan

7/29/10

Today we first tune up the beam for max transmission.

If the above calculation is correct we can expect the electron beam to move at least 2 mm away from the center of the FC. We could test it the Earth's B-field is the cause of the deflection by watch the deflection as a function of the beam energy. If we had enough [math]\mu[/math]-metal we could wrap the beam pipe in it and see if the deflection goes away.

Maximize Transmission

Change the linac coils to determine a setting which delivers the maximum current to the FC and has a reasonable spot size on the YaG viewer.

Coil 1 upstream left/right (mA)	Coil 2 upstream UP/down (mA)	Coil 3 downstream up/down	Coil 4 downstream left right	Scope Picture	Gun Current	FC current	Ratio	YaG Image
	10			100px				100px

a.) Does a small beam spot size on the crystal give maximum FC signal?

b.) Is there a point where the spot size changes and you can't resolve a change in the FC output?

c.) Remove the YaG target and determine if above observations are unchanged.

2.) After tuning the beam to deliver the Maximum beam current to the FC, measure the FC and green pickup coil output as a function of current by integrating the pulse with an oscilloscope.

FC output	Green Pickup Coil output
(mA)	(mA)
	10
	20
	30
	40
	50
	60
	70

a.) does the FC output change with rep rate? It should not but maybe there is loss at high rates?

Quad scan

First attempt to measure the spot size change as a function of Quad current

Quad Current	Beam Spot RMS (0 degrees= Left/right)	Beam Spot RMS (90 degrees= Up/Down)
(Amps)	(mm)	(mm)

Shall we do both polarities?

Imaging

Background subtraction

The filament produces at least a 2 mm diameter spot on the YaG crystal. Beam dominates the image.

7/30/10

Open day to finish leftover tasks

HRRL_radiation_tests#Run_Plan

Consider emittance measurements at different energies.

Imaging scratch area

After image manipulation in GIMP to remove spot from filament. Obtained reasonable projections for horizontal and vertical beam sizes. The quad was off, low rep rate to avoid saturation, 10MeV. This was the spot for the initial setup on Wednesday. Go back Positrons