Integral Calculation for Yukawa Potential Differential Cross Section

In summary, the conversation is about a calculation involving the differential cross section of a Yukawa potential. The highlighted part needs to be -1, and the value at r=0 needs to be subtracted. In order for the value at r=∞ to disappear, | e(iq-1/a) | must be less than 1. This means that if q and a are real, a must be greater than 0. The conversation also discusses a correction to a previous post, where the value of | iq-1/a | is shown to be equal to 1/| e(1/a) |.
  • #1
BREAD
50
0

Homework Statement


upload_2017-12-20_21-19-23.png


Homework Equations


This is a calculation about differential cross section of Yukawa potential.

The Attempt at a Solution


I can't understand how that highlighted part can be -1 ,
we don't know if the parenthesis term (iq-1/a) is negative or positive tho.
 

Attachments

  • upload_2017-12-20_21-19-23.png
    upload_2017-12-20_21-19-23.png
    19.1 KB · Views: 1,162
Physics news on Phys.org
  • #2
Subtracting the value at r=0 would give you -1. For the value at r=∞ to disappear, we must have | e(iq - 1/a) | < 1. So if q and a are real, that means | e1/a | > 1, so a>0.
 
  • #3
FactChecker said:
Subtracting the value at r=0 would give you -1. For the value at r=∞ to disappear, we must have | e(iq - 1/a) | < 1.
==========================================
I don't know why | iq - 1/a | should be smaller than 1 ?
 
  • #4
BREAD said:
==========================================
I don't know why | iq - 1/a | should be smaller than 1 ?
Sorry. You may have looked at my post as I was correcting it. | e(iq-1/a)r | = | (e(iq-1/a))r | = | (e(iq-1/a)) |r = | (e(-1/a)) |r= 1/| (e(1/a)) |r
 
  • #5
FactChecker said:
Sorry. You may have looked at my post as I was correcting it. | e(iq-1/a)r | = | (e(iq-1/a))r | = | (e(iq-1/a)) |r = | (e(-1/a)) |r= 1/| (e(1/a)) |r

===================

I appreciate for your quick reply
 

Related to Integral Calculation for Yukawa Potential Differential Cross Section

1. What is a simple integral calculation?

A simple integral calculation is a mathematical process used to find the area under a curve on a graph. It involves breaking down the curve into smaller, simpler shapes and adding up their areas to find the total area.

2. Why is simple integral calculation important?

Simple integral calculation is important because it allows us to solve real-world problems involving rates of change, such as finding the distance traveled by an object or the amount of water in a tank. It is also a fundamental concept in calculus, which is used in many fields of science and engineering.

3. How do you perform a simple integral calculation?

To perform a simple integral calculation, you first need to determine the limits of integration, which are the points where the curve begins and ends. Then, you use integration rules to find a general formula for the area under the curve. Finally, you plug in the limits of integration to calculate the specific area.

4. What are some common integration rules used in simple integral calculation?

Some common integration rules used in simple integral calculation include the power rule, the constant multiple rule, and the sum and difference rules. These rules allow you to find the integral of basic functions, such as polynomials, exponentials, and trigonometric functions.

5. Can you use technology to perform simple integral calculations?

Yes, technology such as graphing calculators and computer software can be used to perform simple integral calculations. They are especially useful for more complex functions where manual calculations may be difficult or time-consuming.

Similar threads

  • Introductory Physics Homework Help
Replies
1
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
1K
  • Introductory Physics Homework Help
Replies
2
Views
686
  • Introductory Physics Homework Help
Replies
4
Views
158
  • Introductory Physics Homework Help
Replies
6
Views
205
  • Introductory Physics Homework Help
Replies
6
Views
3K
  • Advanced Physics Homework Help
Replies
4
Views
1K
  • Introductory Physics Homework Help
Replies
19
Views
835
  • Introductory Physics Homework Help
Replies
4
Views
1K
  • Introductory Physics Homework Help
Replies
2
Views
3K
Back
Top