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Gaussian distributed falling drops on a non-orthogonal plane

  1. Aug 29, 2011 #1
    Hi all,

    I'm considering a problem of a bunch of particles (say electrons) arriving on a plate. If the plate is orthogonal to the direction of motion of the particles their distribution (on the plate) is a gaussian distribution centered in a point (x,y) of the plate with standard deviation "sigma".
    Using cylindrical coordinates (r,phi) and taking as (0,0) the center of the distribution:

    [itex]\frac{1}{\sqrt{2\pi}\sigma}\cdot exp(\frac{-(r)^{2}}{2\sigma^{2}} )[/itex]

    Now my question is, what happen to the distribution if the plate is tilted of an angle theta?
    I expect that the cylindrical symmetry is broken, but cannot explain it in "numbers"...

    Does anybody have useful ideas?

    Thank you
  2. jcsd
  3. Aug 31, 2011 #2
    You should probably check your original equation first - the radius of a multivariate gaussian r.v. isn't gaussian.
  4. Aug 31, 2011 #3
    But I have Density(x,y) = G(x)*g(y), where G and g are two normal distributions with equal sigma and centered in (0,0)...
    then the correlation matrix is diagonal so why I can't say that I have a normal distribution on the radius?

    Then I've tried to answer by myself and found that


    because the density has to be the same as before (having tilted the plate at an angle [itex]\alpha[/itex] with respect to the x-axis)

    [itex]\rho(r)=\frac{dx}{dr}\cdot G(x)[/itex]
    [itex]=sin(\alpha)G(r)\cdot cos(\alpha))[/itex]

    Is that correct?

    I think the Jacobian is necessary!
  5. Aug 31, 2011 #4


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    Hey condorino and welcome to the forums.

    With the assumption you have presented, you can't really extrapolate the probability information for when the surface is not orthogonal: there is not really enough information.

    My guess is if it depended on the dot (or inner) product of the light vector and the surface normal vector, then there would be some kind of "weird" interpolation going on, and this would be factored in to calculating the distribution, but this would most likely be very complicated.

    If I were to have a "guess" as what to do, it would be that the mean and the standard deviation would depend on -cos(a,b) = -a . b / |a||b| where a is the surface normal and b is the light vector. Anything negative would result in nonsense.

    Also the distribution would not be normal, but would be skewed according to the angle and orientation of the surface.

    Again I emphasize the above is a "guess" based on geometry, but formulating the actual distribution would be an interesting and complicated exercise.
  6. Aug 31, 2011 #5

    Maybe I come to a solution...

    The distribution is skewed according to the tilt angle, and I have found between x->x', y=y' (where ' means "tilted world")... Than with this geometrical relation I calculate G(x') supposing dxG(x)=dx'G(x').

    Here I understand that there's a big difference in result depending on wheter the particles are arriving from an infinite distance or not.

    Once I have G(x')*g(y') I can move to a different coordinate system (cylindrical) with a rotation of (x,y)... but this is another story...

    My big question is if the assumption dxG(x)= dx' G(x') is correct.
    I have plotted the solution with Matlab and it seems to have sense (skewed gaussian with maximum moved in the zone "near to the source").
  7. Aug 31, 2011 #6


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    That does sound intuitive from a geometric standpoint, but in terms of getting an actual pdf in closed form that takes into account your angle, I can't see how you would do that with what you have presented.

    Also the fact the distribution would be skewed means that it transforms from Gaussian to something that looks like a chi-square distribution. Generalizing the distribution should give distributions that look like a normal or a chi-square (or similar) for specific values of the angle.

    If you end up getting a closed form expression, I'd be very interested to see what you come up with.
  8. Aug 31, 2011 #7
    Well, that's what I ended up with... It's demanding translating it in TeX form, and, by the way, does somebody know how to translate a matlab equation in a LaTeX one?

    I started with G(x,y) (lower case xy) and assume a gaussian distribution in both direction. Particles are generated by a source placed d=5cm far from the plate and I suppose (for simplicity) sigma=d and center=(0,0). Knowing this I have tilted the plate by 30° and calculated G(X,Y) as g(X)*g(Y), with dXg(X)=dxg(x) and g(Y)=g(y). Now I calculate G(X,Y)=g(X)*g(Y) obtaining this formula:

    G(X,Y)=(574936417714633*cos(pi/6 + atan((3^(1/2)*X)/(2*(X/2 - 5))))*((3*X^2)/(4*(X/2 - 5)^2) + 1)^(1/2))/(90310801308676176*exp(Y^2/50)*exp((3*X^2)/(8*(X/2 – 5)^2)))

    I know it's difficult to read, so any help in translating it is welcome...

    Thank you for your feedback to the discussion!
  9. Aug 31, 2011 #8
    [itex]\frac{A\, \cos\!\left(\frac{\pi}{6} + \arctan\!\left(\frac{\sqrt{3}\, X}{X - 10}\right)\right)\, \sqrt{\frac{3\, X^2}{4\, {\left(\frac{X}{2} - 5\right)}^2} + 1}}{B\, \mathrm{e}^{\frac{Y^2}{50}}\, \mathrm{e}^{\frac{3\, X^2}{8\, {\left(\frac{X}{2} - 5\right)}^2}}}[/itex]
  10. Aug 31, 2011 #9

    Stephen Tashi

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    That describes a Rayleigh distribution. See the section "Related Distributions - Rayleigh([itex]\sigma[/itex]) in http://en.wikipedia.org/wiki/Rayleigh_distribution,

    Your reasoning about the correlation matrix might tell you something about the random variable [itex] X + Y [/itex], but what is involved is [itex] \sqrt{X^2 + Y^2} [/itex].

    I agree that the probablity density on the tilted plate can be computed from the probability density on the plate that is orthogonal to the trajectories of the particles. I'll have to think more about that to be of any help. It looks like a change-of-variables type of problem. If the PDF f(x,y) on the orthogonal plate is integrated over the any area A on the orthogonal plate, we want a function h(p,q) defined at points on the tilted plate such at when h is integrated over the projection of the area A on the tilted plate, it gives the same value as the f(x,y) integrated over A.
  11. Aug 31, 2011 #10
    Great! That Helps a Lot!

    Yes you are right... the problem is divided in two parts:

    1- changing the (x,y) -orthogonal plate- problem in the (X,Y) -tilted one. That seems to have a result in my previous post.

    2- describing the problem in a cylindrical coordinate system instead of cartesian. And here your answer convince me!

    Now, what I was doing was Attempting point 1 and after point 2. But, in this way the (X,Y) distributions are not normal anymore (well, Y is still normal) and in this case I cannot use Raileigh in point 2. But I will try to change the order 2-1.

    So G(x,y) ---> G(X,Y) ---> G(R,THETA)
    or G(x,y) ---> (Rayleigh) G(r,theta) ---> G(R,THETA)

    What about the last arrow? Now the geometry is less intuitive for me, maybe not for everyone....


    Anyway, The discussion has interested me more than the problem itself, I like this forum and the way of sharing Ideas... Thank you all, I really appreciate your help!

    PS: the last thing you said is correct, that's the rason of assuming g(x)dx=g(X)dX... Not the same for y because dy/dY=1.
    I think this ensures the continuity of "flux" between the tilted and nottilted worlds
  12. Aug 31, 2011 #11

    Stephen Tashi

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    Do you mean a "polar" coordinate system [itex](r,\theta)[/itex] ?
    or an actual "cylindrical" coordinate system [itex](r,\theta,h)[/itex] ?
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