Stuck on a change of variables

In summary, the step in the equation above asks to integrate the following equation: \phi(\vec{x}) = \int \frac{d^3k}{(2\pi)^3}\frac{e^{i\vec{k}\cdot\vec{x}}}{\vec{x}^2 + m^2}changing to polar coordinates, and writing \vec{k}\cdot\vec{x}=kr \cos\theta, we have:\phi(\vec{x}) = \frac{1}{(2\pi)^2} \int_0^\infty dk \frac{k^2}{k^2 + m
  • #1
VantagePoint72
821
34
I'm reading some course notes for a physics class that contain the following step in a derivation:

[itex]\phi(\vec{x}) = \int \frac{d^3k}{(2\pi)^3}\frac{e^{i\vec{k}\cdot\vec{x}}}{\vec{x}^2 + m^2}[/itex]
Changing to polar coordinates, and writing [itex]\vec{k}\cdot\vec{x}=kr \cos\theta[/itex], we have:
[itex]\phi(\vec{x}) = \frac{1}{(2\pi)^2} \int_0^\infty dk \frac{k^2}{k^2 + m^2}\frac{2\sin kr}{kr}[/itex]

I'm having a bit of difficulty seeing this step. Could someone please show some of the intermediate steps between these two and explain what's happening? I understand that k2 in the numerator comes from converting to spherical coordinates, but that's about all I follow.
 
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  • #2
LastOneStanding said:
[itex]\phi(\vec{x}) = \int \frac{d^3k}{(2\pi)^3}\frac{e^{i\vec{k}\cdot\vec{x}}}{\vec{x}^2 + m^2}[/itex]
This seems strange. How can it be meaningful to integrate w.r.t ##d^3k## ?
 
  • #3
Oh dear, you're right of course. I made a transcription error, it should be:
[itex]\phi(\vec{x}) = \int \frac{d^3k}{(2\pi)^3}\frac{e^{i\vec{k}\cdot\vec{x} }}{\vec{k}^2 + m^2}[/itex]

It's a Fourier transform. Now that it's fixed, can you see how to proceed?
 
  • #4
LastOneStanding said:
Oh dear, you're right of course. I made a transcription error, it should be:
[itex]\phi(\vec{x}) = \int \frac{d^3k}{(2\pi)^3}\frac{e^{i\vec{k}\cdot\vec{x} }}{\vec{k}^2 + m^2}[/itex]

It's a Fourier transform. Now that it's fixed, can you see how to proceed?
I still don't understand what ##d^3k## is. Shouldn't it be ##dx## or something like that?
 
  • #5
Erland said:
I still don't understand what ##d^3k## is. Shouldn't it be ##dx## or something like that?

It is a short hand notation to indicate that the integral is 3 dimensional, dx would be 1 dimensional.
 
  • #6
Hi LastOneStanding! :smile:
LastOneStanding said:
… I understand that k2 in the numerator comes from converting to spherical coordinates …

no, only one k comes from the conversion, the other k cancels with the k in the denominator :wink:

(d3k = ksinθ dkdθdφ, and so you have to ∫ ksinθ eikrcosθ dθ)
 
  • #7
mathman said:
It is a short hand notation to indicate that the integral is 3 dimensional, dx would be 1 dimensional.
Ok, I did never see this notation before. Now, I got it:

First, ##\phi(\vec{x})## depends only upon the radial coordinate ##r## of ##\vec{x}##. (Why?) We can therefore assume that ##\vec{x}## is ##(0,0,r)##. Then ## \theta## is the colatitude of ##\vec{k}## and in spherical coordinates ##d^3 k## becomes ##k^2 \sin \theta \,dk\,d\theta\,d\phi##. Now, it is easy to integrate ##\sin\theta\,e^{ikrcos\theta}## wrt ##\theta##. This will give the desired the result.
 
  • #8
tiny-tim said:
Hi LastOneStanding! :smile:


no, only one k comes from the conversion, the other k cancels with the k in the denominator :wink:

(d3k = ksinθ dkdθdφ, and so you have to ∫ ksinθ eikrcosθ dθ)

I don't follow...the spherical volume element is [itex]d^3x = r^2 \sin\theta dr d\theta d\phi[/itex]. For an integral in k-space, this means a factor of k2. What do you mean "the other k cancels with the k in the denominator"?
 
  • #9
Erland said:
Ok, I did never see this notation before. Now, I got it:

First, ##\phi(\vec{x})## depends only upon the radial coordinate ##r## of ##\vec{x}##. (Why?) We can therefore assume that ##\vec{x}## is ##(0,0,r)##. Then ## \theta## is the colatitude of ##\vec{k}## and in spherical coordinates ##d^3 k## becomes ##k^2 \sin \theta \,dk\,d\theta\,d\phi##. Now, it is easy to integrate ##\sin\theta\,e^{ikrcos\theta}## wrt ##\theta##. This will give the desired the result.

As I said in the original question, I understand the part of the step that comes from converting to spherical coordinates. So, unfortunately it is precisely the part of the calculation you suppressed by saying "it is easy to integrate..." which is what I'm stuck on.
 
  • #10
Oh, bother. Undone by a simple u-subsitution, sigh. Thank you, Erland, I see how the rest of the integration goes.
 

1. What is a change of variables?

A change of variables is a mathematical concept that involves substituting one set of variables with another set in a given equation or problem. It is used to simplify or solve complex equations and problems.

2. Why is a change of variables important in science?

A change of variables is important in science because it allows us to transform complex problems into simpler forms that are easier to understand and solve. It also helps us to establish relationships between different sets of variables and make predictions.

3. How do I know when to use a change of variables?

You can use a change of variables when you encounter a problem or equation that involves multiple variables, and it seems difficult to solve using traditional methods. In such cases, a change of variables can help simplify the problem and make it more manageable.

4. What are the common types of changes of variables used in science?

The most common types of changes of variables used in science are linear transformations, logarithmic transformations, and trigonometric transformations. Each type has its own specific use and can be applied in different situations.

5. Are there any limitations to using a change of variables?

Yes, there are some limitations to using a change of variables. For example, it may not always be possible to find a suitable change of variables for a given problem. Also, in some cases, the transformed equation may be more difficult to solve than the original one. It is important to carefully consider the benefits and limitations before using a change of variables.

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