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Determining radius at which 50% of energy is in the profile.

  1. Nov 7, 2014 #1
    1. The problem statement, all variables and given/known data
    I have to determine the radius at which 50% of energy is in a Gaussian profile.


    2. Relevant equations
    The intensity is given by I=Ioe^(-r/2c)^2. This is just a gaussian function ofcourse.

    3. The attempt at a solution
    I know c is the standard deviation. I searched through charts that 50% of the data set is contained within about 0.67 standard deviations. So can i set I/i0 to 0.5 and set c=0.67? I feel like this is not the correct method though. What I need is the area under the gaussian curve that is 50%. So I need to integrate between values of -r and r that will give me an answer of 50 % but I dont know what values those are.
     
  2. jcsd
  3. Nov 7, 2014 #2

    vela

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    ##c## sets the length scale of the problem. Your value for ##r## will be expressed as some multiple of ##c##.

    You need to be a little more careful here. Your Gaussian is a two-dimensional Gaussian, not the one-dimensional Gaussian you've assumed. That is, you actually have
    $$e^{-\frac{r^2}{2\sigma^2}} = e^{-\frac{x^2+y^2}{2\sigma^2}}.$$ You have to integrate with respect to polar coordinates ##r## and ##\theta##, not just ##r## alone.
     
  4. Nov 7, 2014 #3
    right so I can just multiply by 2pi and integrate with respect to r. But I don't know how to integrate this. My value of r will be a multiple of c...so I can rewrite r as n*c? and try integrating that?
     
  5. Nov 7, 2014 #4

    vela

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    Not exactly. What's the area element for polar coordinates?
     
  6. Nov 8, 2014 #5
    rdrdtheta but what are the limits?
     
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