# Bivariate gaussians for ion beam optics

1. Apr 6, 2013

### LinguisticM

Hello,

here is something that's been bothering me for a while:
say I have a charged particle beam that can be described by one bivariate Gaussian in X dimension and another bivariate Gaussian in Y dimension. The parameters of the bivariate gaussian are the position (X) of an individual particle and the projection of the tangent of the angle onto the X axis (Xnu). The angle and position can be correlated, hence the covariance between X and Xnu may be not zero. The same for the Y axis: Y and Ynu. Each distribution is characterized by its own variance, so there are var(X), var(Xnu), var(Y), and var(Ynu) as well as cov(x,Xnu), cov(Y,Ynu). Covariance between X and Y is not accounted for but also not important in this problem.

Now, I would like to experimentally measure the beam spot size and I have two available techniques:
(1) a 2D scintillation screen that gives me X,Y map of beam intensities in the plane perpendicular to XY
(2) a pinhole detector that moves along X or Y to collect the beam profile. Neither of the two detectors gives the particle direction information.

In (1) I get the map of the intensities and fit 2D gaussian in X and Y to give me var(X) and var(Y) (actually I'm after beam spot sizes, so sqrt(var(X)) and sqrt(var(Y))).
In (2) I get a profile that typically runs across the center of the spot along X and Y directions, so for Y=0 and X=0, respectively. These are the 'conditional' profiles and they should also give me the same values for var(X) and var(Y).

Should or should not?

I looked into a statistics textbook and found the proof that the variance of a projected bivariate gaussian (which is what I measure by fitting the 2D gaussian in (1)) and the conditional gaussian (which is what I get from my experiment in (2)) should be equal.
I also found that in the beam optics physics literature there is a distinction between the 'conditional' (aka plane) and the 'projected' (aka space) beam spot standard deviations. The article that I've found says that the conditional beam spot size is sqrt(2) times bigger than the projected beam spot size (variance(conditional) = 2* variance (projected)).

I'm confused. Are the variances equal or are they not? Your input is appreciated.

LinguisticM

2. Apr 7, 2013

### Staff: Mentor

If X an Y are independent, both measurements should give the same result.
That could refer to the expectation value of |r|, which is sqrt(2) the standard deviation in your individual coordinates. If you scan along an axis, you do not measure the expectation value of |r|, so this is not relevant here.

3. Apr 7, 2013

### Stephen Tashi

What angle are you talking about? What do you mean by "the projection" of an angle onto an axis?

4. Apr 7, 2013

### LinguisticM

Does |r| denote the radial distance?

I understand that the measurements should give the same results. But still, in the literature I can find "plane" and "space" sigmas (standard deviations) for characterization of beam spot sizes. Read this e.g. this quote:

<quote begin>
Random variables x and thetaX are projected, or 'plane' variables. For example, x is the x coordinate of the random variable r. The variance of x, denoted a, is however different from what would be measured in a typical beam pro file measurement. In a such a measurement, assuming measurement in x direction along the y = 0 line, one obtains a conditional probability distribution of x coordinate with the condition y = 0. This is not the same as the probability distribution for the x coordinate; they are both Gaussians, but the variance of the conditional is larger by a factor of 2. Sometimes such conditional variables are called 'space' variables
<quote end>

I can't see how the conditional can have variance larger by the factor of 2 than the space.

thetaX which is mentioned in the quote (strictly speaking this is the tan(thetaX), but since the angles are very small, this does not matter) comes from the direction of motion of the particle, so this is the angle between the projection of the direction vector onto the y=0 plane and the z direction. If theta and phi are the polar and azimuthal angles in the spherical coordinate system then after conversion to the cartesian coordinate system thetaX is:

thetaX = tan(theta)cos(phi)

As it says in the quote, X and thetaX describe the motion of the particle in the X plane.

5. Apr 7, 2013

### Stephen Tashi

It would help if you precisely described the data. By "motion", do you mean velocity? Is the "X plane" the YZ-plane? Does one vector in the data give both the position and velocity of a particle? Can you give a link to a diagram that illustrates the situation?

6. Apr 7, 2013

### Staff: Mentor

Right.
That makes perfectly sense in terms of the standard deviation of individual coordinates, and the standard deviation of the radial distance.

7. Apr 7, 2013

### LinguisticM

Not precise language again, I apologize.
- yes, by motion I mean the momentum vector
- yes, X plane is the YZ-plane
- the measured data do not give the momentum vector, only position. Using the 2D scintillation screen I got the 2D fluence map, which is a square array. Using the pinhole detector I got a vector showing fluence along one of the axes (Y=0 or X=0).
- this picture illustrates the use of the X,thetaX beam description I quoted earlier: http://www.desy.de/fel-beam/s2e/data/astra_simu_flash/astra_transv_phase_space_1nC.jpg

does it help?

8. Apr 7, 2013

### LinguisticM

This occurred to me as well but this would mean that the author in this text messed up definitions. The space variance is the variance of the radial distribution, and the plane is the variance of the cartesian projection of this distribution.