Understanding Kramers-Kronig relationsThe Kramers-Kronig relationsProving Kramers-Kronig relationships from Fourier sine / cosine transformShow that the imaginary part of $fracz^2z-z_p$ is harmonicContinuity of principal complex arccosWhy do I get different results using different formulae of Kramer-Kronig relation on the same function?Can you explain us a simple example, for a mathematical function, of Kramers–Kronig relations?Calculate $int_- infty^infty fracsin omega xx dx$ for $omega$ real and positiveA weird value obtained by using Cauchy Principal Value on $int_-infty^inftyfrac1x^2dx$Help with contour integral popping up in Kramers-Kronig problemEvaluating Cauchy principal value
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Understanding Kramers-Kronig relations
The Kramers-Kronig relationsProving Kramers-Kronig relationships from Fourier sine / cosine transformShow that the imaginary part of $fracz^2z-z_p$ is harmonicContinuity of principal complex arccosWhy do I get different results using different formulae of Kramer-Kronig relation on the same function?Can you explain us a simple example, for a mathematical function, of Kramers–Kronig relations?Calculate $int_- infty^infty fracsin omega xx dx$ for $omega$ real and positiveA weird value obtained by using Cauchy Principal Value on $int_-infty^inftyfrac1x^2dx$Help with contour integral popping up in Kramers-Kronig problemEvaluating Cauchy principal value
$begingroup$
I'm a bit confused on how to calculate the Kramer-Kronig relations using the principal values, so I was hoping if someone could explain me how to proceed.
As you know, for a function $chi(omega)=chi_1(omega)+ichi_2(omega)$, the Kramer-Kronig relations are given by:
$$chi_1(omega)=frac1piPint_-infty^inftydOmegafracchi_2(Omega)Omega-omega$$
$$chi_2(omega)=-frac1piPint_-infty^inftydOmegafracchi_1(Omega)Omega-omega$$
where the principal value for a singularity at $eta$ can be calculated as,
$$Pint_-infty^inftydx f(x)=lim_crightarrow 0^+left[ int_-infty^eta-c/2dx f(x)+int_eta+c/2^inftydx f(x)right ]$$
However, just following from this definition I'm not able to make any calculation for the cases I've tested so far, so I would like to understand how is this done (all the books I've reviewed so far only include theory).
For example, take the most simple case for real part of a susceptibility function $chi_1(omega)=omega_p^2/omega^2$. If I insert it into the second equation and I try calculating the corresponding real part,
$$chi_2(omega)=fracomega_p^2piPint_-infty^inftydOmegafrac1Omega^2(Omega-omega)$$
Which gives me $chi_2(omega)=-infty (sgn(omega))$ upon calculation of the integrals for both singularities at $Omega=0$ and $Omega=omega$. This is strange, as it doesn't make sense that an imaginary part of a real quantity is infinite.
Therefore, tried using an alternative definition found in the French version of Wikipedia,
$$chi_2(omega)=-frac2omegapiint_0^inftydOmegafracchi_2(Omega)Omega^2-omega^2$$
which doesn't have the principal value for some reason. If I assume symmetry and I take the limit from zero to infinity, I can use the residu theorem to calculate that integral, I actually get $chi_2(omega)=-2iomega_p^2/omega^2$. It also doesn't make too much sense, since I'm now getting an imaginary number which will cancel with the $i$ of the imaginary part ($ichi_2(omega)$).
Therefore, I know I'm doing something wrong or lacking some mathematical tool to do the correct computations.
complex-analysis
edited 20 mins ago
J.G.
32.1k23250
asked 2 hours ago
CharlieCharlie
1496
$endgroup$
add a comment |
$begingroup$
I'm a bit confused on how to calculate the Kramer-Kronig relations using the principal values, so I was hoping if someone could explain me how to proceed.
As you know, for a function $chi(omega)=chi_1(omega)+ichi_2(omega)$, the Kramer-Kronig relations are given by:
$$chi_1(omega)=frac1piPint_-infty^inftydOmegafracchi_2(Omega)Omega-omega$$
$$chi_2(omega)=-frac1piPint_-infty^inftydOmegafracchi_1(Omega)Omega-omega$$
where the principal value for a singularity at $eta$ can be calculated as,
$$Pint_-infty^inftydx f(x)=lim_crightarrow 0^+left[ int_-infty^eta-c/2dx f(x)+int_eta+c/2^inftydx f(x)right ]$$
However, just following from this definition I'm not able to make any calculation for the cases I've tested so far, so I would like to understand how is this done (all the books I've reviewed so far only include theory).
For example, take the most simple case for real part of a susceptibility function $chi_1(omega)=omega_p^2/omega^2$. If I insert it into the second equation and I try calculating the corresponding real part,
$$chi_2(omega)=fracomega_p^2piPint_-infty^inftydOmegafrac1Omega^2(Omega-omega)$$
Which gives me $chi_2(omega)=-infty (sgn(omega))$ upon calculation of the integrals for both singularities at $Omega=0$ and $Omega=omega$. This is strange, as it doesn't make sense that an imaginary part of a real quantity is infinite.
Therefore, tried using an alternative definition found in the French version of Wikipedia,
$$chi_2(omega)=-frac2omegapiint_0^inftydOmegafracchi_2(Omega)Omega^2-omega^2$$
which doesn't have the principal value for some reason. If I assume symmetry and I take the limit from zero to infinity, I can use the residu theorem to calculate that integral, I actually get $chi_2(omega)=-2iomega_p^2/omega^2$. It also doesn't make too much sense, since I'm now getting an imaginary number which will cancel with the $i$ of the imaginary part ($ichi_2(omega)$).
Therefore, I know I'm doing something wrong or lacking some mathematical tool to do the correct computations.
complex-analysis
edited 20 mins ago
J.G.
32.1k23250
asked 2 hours ago
CharlieCharlie
1496
$endgroup$
add a comment |
$begingroup$
I'm a bit confused on how to calculate the Kramer-Kronig relations using the principal values, so I was hoping if someone could explain me how to proceed.
As you know, for a function $chi(omega)=chi_1(omega)+ichi_2(omega)$, the Kramer-Kronig relations are given by:
$$chi_1(omega)=frac1piPint_-infty^inftydOmegafracchi_2(Omega)Omega-omega$$
$$chi_2(omega)=-frac1piPint_-infty^inftydOmegafracchi_1(Omega)Omega-omega$$
where the principal value for a singularity at $eta$ can be calculated as,
$$Pint_-infty^inftydx f(x)=lim_crightarrow 0^+left[ int_-infty^eta-c/2dx f(x)+int_eta+c/2^inftydx f(x)right ]$$
However, just following from this definition I'm not able to make any calculation for the cases I've tested so far, so I would like to understand how is this done (all the books I've reviewed so far only include theory).
For example, take the most simple case for real part of a susceptibility function $chi_1(omega)=omega_p^2/omega^2$. If I insert it into the second equation and I try calculating the corresponding real part,
$$chi_2(omega)=fracomega_p^2piPint_-infty^inftydOmegafrac1Omega^2(Omega-omega)$$
Which gives me $chi_2(omega)=-infty (sgn(omega))$ upon calculation of the integrals for both singularities at $Omega=0$ and $Omega=omega$. This is strange, as it doesn't make sense that an imaginary part of a real quantity is infinite.
Therefore, tried using an alternative definition found in the French version of Wikipedia,
$$chi_2(omega)=-frac2omegapiint_0^inftydOmegafracchi_2(Omega)Omega^2-omega^2$$
which doesn't have the principal value for some reason. If I assume symmetry and I take the limit from zero to infinity, I can use the residu theorem to calculate that integral, I actually get $chi_2(omega)=-2iomega_p^2/omega^2$. It also doesn't make too much sense, since I'm now getting an imaginary number which will cancel with the $i$ of the imaginary part ($ichi_2(omega)$).
Therefore, I know I'm doing something wrong or lacking some mathematical tool to do the correct computations.
complex-analysis
edited 20 mins ago
J.G.
32.1k23250
asked 2 hours ago
CharlieCharlie
1496
$endgroup$
I'm a bit confused on how to calculate the Kramer-Kronig relations using the principal values, so I was hoping if someone could explain me how to proceed.
As you know, for a function $chi(omega)=chi_1(omega)+ichi_2(omega)$, the Kramer-Kronig relations are given by:
$$chi_1(omega)=frac1piPint_-infty^inftydOmegafracchi_2(Omega)Omega-omega$$
$$chi_2(omega)=-frac1piPint_-infty^inftydOmegafracchi_1(Omega)Omega-omega$$
where the principal value for a singularity at $eta$ can be calculated as,
$$Pint_-infty^inftydx f(x)=lim_crightarrow 0^+left[ int_-infty^eta-c/2dx f(x)+int_eta+c/2^inftydx f(x)right ]$$
However, just following from this definition I'm not able to make any calculation for the cases I've tested so far, so I would like to understand how is this done (all the books I've reviewed so far only include theory).
For example, take the most simple case for real part of a susceptibility function $chi_1(omega)=omega_p^2/omega^2$. If I insert it into the second equation and I try calculating the corresponding real part,
$$chi_2(omega)=fracomega_p^2piPint_-infty^inftydOmegafrac1Omega^2(Omega-omega)$$
Which gives me $chi_2(omega)=-infty (sgn(omega))$ upon calculation of the integrals for both singularities at $Omega=0$ and $Omega=omega$. This is strange, as it doesn't make sense that an imaginary part of a real quantity is infinite.
Therefore, tried using an alternative definition found in the French version of Wikipedia,
$$chi_2(omega)=-frac2omegapiint_0^inftydOmegafracchi_2(Omega)Omega^2-omega^2$$
which doesn't have the principal value for some reason. If I assume symmetry and I take the limit from zero to infinity, I can use the residu theorem to calculate that integral, I actually get $chi_2(omega)=-2iomega_p^2/omega^2$. It also doesn't make too much sense, since I'm now getting an imaginary number which will cancel with the $i$ of the imaginary part ($ichi_2(omega)$).
Therefore, I know I'm doing something wrong or lacking some mathematical tool to do the correct computations.
complex-analysis
complex-analysis
edited 20 mins ago
J.G.
32.1k23250
asked 2 hours ago
CharlieCharlie
1496
edited 20 mins ago
J.G.
32.1k23250
asked 2 hours ago
CharlieCharlie
1496
edited 20 mins ago
J.G.
32.1k23250
edited 20 mins ago
J.G.
32.1k23250
edited 20 mins ago
J.G.
32.1k23250
32.1k23250
asked 2 hours ago
CharlieCharlie
1496
asked 2 hours ago
CharlieCharlie
1496
asked 2 hours ago
CharlieCharlie
1496
1496
add a comment |
add a comment |
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