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

In summary, the conversation is discussing the process of determining the radius at which 50% of energy is in a Gaussian profile. This involves using the Gaussian function I=Ioe^(-r/2c)^2, where c is the standard deviation. The speaker suggests setting I/i0 to 0.5 and solving for c, but the expert points out that this is a two-dimensional Gaussian and requires integrating with respect to polar coordinates r and theta. The expert suggests using the area element for polar coordinates, rdrdtheta, to find the correct limits for the integration.
  • #1
nabeel17
57
1

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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  • #2
##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.
 
  • #3
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?
 
  • #4
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?
 
  • #5
vela said:
Not exactly. What's the area element for polar coordinates?
rdrdtheta but what are the limits?
 

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

What is the meaning of "determining radius at which 50% of energy is in the profile"?

This phrase refers to finding the distance from the center of a profile at which half of the total energy is contained. This is often used in scientific studies to analyze the distribution of energy or matter within a system.

Why is it important to determine this radius?

Determining the radius at which 50% of energy is in the profile can provide valuable information about the structure and properties of a system. It can also help to identify patterns or trends in the distribution of energy, which can be useful in understanding various phenomena.

How is this radius typically determined?

This radius is usually determined through mathematical calculations based on the data collected from a system. It may involve analyzing graphs, equations, or other types of data to determine the distance at which 50% of energy is contained.

What factors can affect the radius at which 50% of energy is in the profile?

The radius at which 50% of energy is in the profile can be affected by various factors, such as the shape and size of the system, the type of energy being studied, and any external influences or forces acting on the system.

Are there any limitations to using this method of determining energy distribution?

Like any scientific method, there may be limitations to using this method to determine energy distribution. It is important to consider the accuracy and precision of the data, as well as any potential sources of error in the calculations. Additionally, this method may not be applicable to all systems or types of energy.

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