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

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Homework Help Overview

The discussion revolves around determining the radius at which 50% of the energy is contained within a Gaussian profile, specifically a two-dimensional Gaussian function. The original poster attempts to find this radius using the intensity equation and seeks to understand the integration required to find the area under the curve that corresponds to 50% of the total energy.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the need to integrate the Gaussian function in polar coordinates and question the correct limits for integration. There is also a consideration of expressing the radius as a multiple of the standard deviation.

Discussion Status

The discussion is ongoing, with participants providing guidance on the integration process and clarifying the dimensionality of the Gaussian function. There is no explicit consensus yet, as participants are still exploring various aspects of the problem.

Contextual Notes

Participants note the importance of correctly identifying the area element in polar coordinates and the need for careful consideration of the integration limits, which remain unspecified in the discussion.

nabeel17
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Homework Statement


I have to determine the radius at which 50% of energy is in a Gaussian profile.

Homework Equations


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

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 don't know what values those are.
 
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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.
 
vela said:
##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.
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?
 
nabeel17 said:
right so I can just multiply by 2pi and integrate with respect to r.
Not exactly. What's the area element for polar coordinates?
 
vela said:
Not exactly. What's the area element for polar coordinates?
rdrdtheta but what are the limits?
 

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