Imaginary sinusoidal exponential

  • #1
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Main Question or Discussion Point

I need to convince myself that

##e^{ikr\text{cos}\ \theta}=\frac{\text{sin}\ \theta}{kr}##.

Could you please help me with it?
 

Answers and Replies

  • #2
blue_leaf77
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Cannot be true, the left side is complex whereas the right side is real.
 
  • #3
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Sorry, there was a typo.

How about this one:

##\text{exp}({ikr\text{cos}\ \theta}) = \frac{\text{exp}(ikr)-\text{exp}(-ikr)}{2ikr}##?
 
  • #4
blue_leaf77
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What do you know about the Euler formula for a complex exponential ##e^{i\phi}##?
 
  • #5
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The Euler formula is ##e^{i\phi}= \text{cos}\ \phi + \text{sin}\ \phi##.

I don't see how ##\text{exp}(ikr\text{cos}\ \theta) = \frac{\text{sin}(kr)}{kr}##.
 
  • #6
blue_leaf77
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The Euler formula is ##e^{i\phi}= \text{cos}\ \phi + \text{sin}\ \phi##.
It's ##e^{i\phi}= \text{cos}\ \phi + i\text{sin}\ \phi##.
I don't see how ##\text{exp}(ikr\text{cos}\ \theta) = \frac{\text{sin}(kr)}{kr}##.
Indeed, I too, do not see how that can be true.
 
  • #7
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The following is taken from page 39 of Matthew Schwartz's 'Introduction to Quantum Field Theory.' My above posts refer to the third and fourth lines of the following calculation:

Capture.jpg
a

Would you please have a look and let me know what I'm missing in the previous posts?
 
  • #8
blue_leaf77
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$$\int_{-1}^1 d(\cos \theta) e^{ikr\cos \theta} = \frac{1}{ikr}e^{ikr\cos \theta}\Big|_{-1}^1 = \frac{e^{ikr}-e^{-ikr}}{ikr}$$
Then in the third line you have ##\frac{\sin kr}{k}## as the integrand to be integrated from 0 to infiinity. The function ##\frac{\sin kr}{k}## is an even function, thus
$$
\int_0^\infty \frac{\sin kr}{k} \, dk = \frac{1}{2} \int_{-\infty}^\infty \frac{\sin kr}{k} \, dk
$$
 
  • #9
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Thanks! I feel like a total idiot now! :frown:
 

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