So we can solve for ##c_1## and find ##x_1(\omega)##.

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In summary, the Kramers-Kronig relation allows us to find the real part, x1(ω), of a complex function, x(ω), by using an integral involving the imaginary part, x2(ω). This can be applied when we know the relationship between x2(ω) and another function, alpha(ω), defined on a specific interval. To solve the integral, we can use the Cauchy principal value and the residue theorem.
  • #1
Another
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Homework Statement
##x(\omega) = x_1(\omega)+ix_2(\omega)##

##x_1(\omega ) = \frac{2}{\pi} P \int_{0}^{∞} \frac{\omega 'x_2(\omega ') }{ \omega '^2 - \omega ^2} d \omega '##

Where ##P## are Cauchy principal value.
##P \int_a^b g(x)dx = lim_{\epsilon → 0} \int_a^{k - \epsilon} g(x) dx + \int_{k + \epsilon}^b g(x) dx ##
Relevant Equations
##x(\omega) = x_1(\omega)+ix_2(\omega)##
##P \int_a^b g(x)dx = lim_{\epsilon → 0} \int_a^{k - \epsilon} g(x) dx + \int_{k + \epsilon}^b g(x) dx ##
In the Kramers-Kroning relation
Let ##x(\omega) = x_1(\omega)+ix_2(\omega)## be a complex function of the complex variable ##\omega## , Where ## x_1(\omega) ## and ## x_2(\omega) ## are real
We can find ##x_1(\omega) ## from this integral
##x_1(\omega ) = \frac{2}{\pi} P \int_{0}^{∞} \frac{\omega 'x_2(\omega ') }{ \omega '^2 - \omega ^2} d \omega '##
Where ##P## are Cauchy principal value.

If we know ##\alpha (\omega)## In relation to ## x_2(\omega) =\frac{k}{2 \omega} \alpha(\omega) ## this function defined interval ##a ≤ \omega ≤ b## and ##k## are real constant.

So I need find ##x_1(\omega)##
Let
##x_1(\omega ) = \frac{k}{\pi} P \int_{0}^{∞} \frac{\omega 'x_2(\omega ') }{ \omega '^2 - \omega ^2} d \omega '##
##x_1(\omega ) = \frac{k}{\pi} P \int_{a}^{b} \frac{\alpha (\omega ') }{ \omega '^2 - \omega ^2} d \omega '##

Let ##f(\omega') = \frac{\alpha (\omega ') }{ \omega '^2 - \omega ^2}##

if ##a ≤ \omega ≤ b## same ##a ≤ \omega '≤ b## Lead to ##f(\omega ')## go to infinity

But ##P \int_a^b g(x)dx = lim_{\epsilon → 0} \int_a^{c - \epsilon} g(x) dx + \int_{c + \epsilon}^b g(x) dx ##

How can i application the Cauchy principal value. to this problem.
I THINK

##x_1(\omega ) = \frac{k}{\pi} P \int_{a}^{b} \frac{\alpha (\omega ') }{ \omega '^2 - \omega ^2} d \omega '##
##x_1(\omega ) = \frac{k}{\pi} lim_{\epsilon → 0}( \int_{a}^{c - \epsilon} \frac{\alpha (\omega ') }{ \omega '^2 - \omega ^2} d \omega ' + \int_{c + \epsilon}^{b} \frac{\alpha (\omega ') }{ \omega '^2 - \omega ^2} d \omega ' )##

if ##\alpha (\omega') ## maybe a ##c_1## constant we get
##x_1(\omega ) = \frac{k c_1}{\pi} lim_{\epsilon → 0}( \int_{a}^{c - \epsilon} \frac{1}{ \omega '^2 - \omega ^2} d \omega ' + \int_{c + \epsilon}^{b} \frac{ 1 }{ \omega '^2 - \omega ^2} d \omega ' )##

I don't know how can i set ##c## value
 
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  • #2
to solve this problem##x_1(\omega ) = \frac{k c_1}{\pi} lim_{\epsilon → 0}( \int_{a}^{c - \epsilon} \frac{1}{ \omega '^2 - \omega ^2} d \omega ' + \int_{c + \epsilon}^{b} \frac{ 1 }{ \omega '^2 - \omega ^2} d \omega ' )##I think we can use the residue theorem to solve this problem. Let ##f(\omega) = \frac{1}{ \omega '^2 - \omega ^2}## we can find the residue of this function at ##\omega## .Res(f,ω) = lim (ω-→ω) (ω-ω)f(ω) =1/2ωThen##x_1(\omega ) = \frac{k c_1}{2\pi} \left(Res(f,a) + Res(f,b)\right) = \frac{k c_1}{\pi} \left(\frac{1}{2a} + \frac{1}{2b}\right)##
 

Related to So we can solve for ##c_1## and find ##x_1(\omega)##.

1. What is the Kramers-Kronig relation?

The Kramers-Kronig relation is a fundamental mathematical relationship between the real and imaginary parts of a complex function. It states that the real part of a function is related to its imaginary part by a Hilbert transform.

2. Who discovered the Kramers-Kronig relation?

The Kramers-Kronig relation was first proposed by Dutch physicist Hendrik Kramers in 1927 and was later independently derived by Swedish physicist Oskar Klein and German physicist Ralph Kronig in 1928.

3. What is the significance of the Kramers-Kronig relation?

The Kramers-Kronig relation has important applications in various fields of physics, such as optics, electromagnetism, and statistical mechanics. It can be used to relate the real and imaginary parts of a complex refractive index, which is crucial in understanding the behavior of light in materials.

4. How is the Kramers-Kronig relation used in experimental data analysis?

The Kramers-Kronig relation is often used to analyze experimental data, especially in spectroscopy and other optical measurements. By measuring either the real or imaginary part of a complex function, the other part can be calculated using the Kramers-Kronig relation, providing a more complete understanding of the system being studied.

5. Are there any limitations to the Kramers-Kronig relation?

While the Kramers-Kronig relation is a powerful tool, it does have some limitations. It assumes that the function being analyzed is analytic, meaning it has no singularities or branch cuts. In addition, it is only valid for linear systems and may not hold for highly nonlinear systems.

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