Mixing
How is this integral computed??
$begingroup$
I'm reading this book about electrical properties of materials where the electron is introduced as a wave. Using the equation of a wave, they bring about the "envelope" of a wave. So here is how the derivation goes:
1)Consider the equation of a wave traveling in one dimension:
$ u = ae^{-i(omega t - kz)} $ where $omega = 2 pi f$ is the frequency, $k$ is the wave number, and $a$ is the amplitude. In addition to this information, the book refers to the phase velocity defined as: ${partial z}/{partial t} = {omega}/{k}$
2)Instead of 1 single wave, let there be multiple waves that are superimposed such that $u = sum a_n e^{-i(omega_n t - k_n z)}$
3)Consider the continuous case : $$ u(z) = int_{-infty}^infty a(k)e^{-i(omega t - kz)}dk $$
4)Make two more assumptions. First assumption: the waves are zero everywhere except at a small interval $Delta k $ and here the amplitude is unity such at $a(k)=1$. This brings the equation to $$ u(z) = int_{k_0 - Delta k /2}^{k_0 + Delta k /2} e^{-i(omega t - kz)}dk $$
5)Suppose that $t=0$, thus yielding the endpoint: $$ u(z) = int_{k_0 - Delta k /2}^{k_0 + Delta k /2} e^{i kz}dk $$
When the integration is carried out, the result is $$ u(z) = Delta k e^{ik_0 z} frac {sin 1/2(Delta kz)}{1/2(Delta k z)}$$.
My question is how is the integration done here? I do notice that u is a function of z and that the integrand integrates over the wave number k. Is there some change of variables going on here? By the way I only posses knowledge of ODEs and only know how to do basic Separation of variables for PDEs.
integration definite-integrals wave-equation change-of-variable envelope
edited Jan 16 at 21:40
clathratus
4,615337
asked Jan 16 at 20:04
Jonathan AguileraJonathan Aguilera
466
$endgroup$
add a comment |
$begingroup$
I'm reading this book about electrical properties of materials where the electron is introduced as a wave. Using the equation of a wave, they bring about the "envelope" of a wave. So here is how the derivation goes:
1)Consider the equation of a wave traveling in one dimension:
$ u = ae^{-i(omega t - kz)} $ where $omega = 2 pi f$ is the frequency, $k$ is the wave number, and $a$ is the amplitude. In addition to this information, the book refers to the phase velocity defined as: ${partial z}/{partial t} = {omega}/{k}$
2)Instead of 1 single wave, let there be multiple waves that are superimposed such that $u = sum a_n e^{-i(omega_n t - k_n z)}$
3)Consider the continuous case : $$ u(z) = int_{-infty}^infty a(k)e^{-i(omega t - kz)}dk $$
4)Make two more assumptions. First assumption: the waves are zero everywhere except at a small interval $Delta k $ and here the amplitude is unity such at $a(k)=1$. This brings the equation to $$ u(z) = int_{k_0 - Delta k /2}^{k_0 + Delta k /2} e^{-i(omega t - kz)}dk $$
5)Suppose that $t=0$, thus yielding the endpoint: $$ u(z) = int_{k_0 - Delta k /2}^{k_0 + Delta k /2} e^{i kz}dk $$
When the integration is carried out, the result is $$ u(z) = Delta k e^{ik_0 z} frac {sin 1/2(Delta kz)}{1/2(Delta k z)}$$.
My question is how is the integration done here? I do notice that u is a function of z and that the integrand integrates over the wave number k. Is there some change of variables going on here? By the way I only posses knowledge of ODEs and only know how to do basic Separation of variables for PDEs.
integration definite-integrals wave-equation change-of-variable envelope
edited Jan 16 at 21:40
clathratus
4,615337
asked Jan 16 at 20:04
Jonathan AguileraJonathan Aguilera
466
$endgroup$
add a comment |
$begingroup$
I'm reading this book about electrical properties of materials where the electron is introduced as a wave. Using the equation of a wave, they bring about the "envelope" of a wave. So here is how the derivation goes:
1)Consider the equation of a wave traveling in one dimension:
$ u = ae^{-i(omega t - kz)} $ where $omega = 2 pi f$ is the frequency, $k$ is the wave number, and $a$ is the amplitude. In addition to this information, the book refers to the phase velocity defined as: ${partial z}/{partial t} = {omega}/{k}$
2)Instead of 1 single wave, let there be multiple waves that are superimposed such that $u = sum a_n e^{-i(omega_n t - k_n z)}$
3)Consider the continuous case : $$ u(z) = int_{-infty}^infty a(k)e^{-i(omega t - kz)}dk $$
4)Make two more assumptions. First assumption: the waves are zero everywhere except at a small interval $Delta k $ and here the amplitude is unity such at $a(k)=1$. This brings the equation to $$ u(z) = int_{k_0 - Delta k /2}^{k_0 + Delta k /2} e^{-i(omega t - kz)}dk $$
5)Suppose that $t=0$, thus yielding the endpoint: $$ u(z) = int_{k_0 - Delta k /2}^{k_0 + Delta k /2} e^{i kz}dk $$
When the integration is carried out, the result is $$ u(z) = Delta k e^{ik_0 z} frac {sin 1/2(Delta kz)}{1/2(Delta k z)}$$.
My question is how is the integration done here? I do notice that u is a function of z and that the integrand integrates over the wave number k. Is there some change of variables going on here? By the way I only posses knowledge of ODEs and only know how to do basic Separation of variables for PDEs.
integration definite-integrals wave-equation change-of-variable envelope
edited Jan 16 at 21:40
clathratus
4,615337
asked Jan 16 at 20:04
Jonathan AguileraJonathan Aguilera
466
$endgroup$
I'm reading this book about electrical properties of materials where the electron is introduced as a wave. Using the equation of a wave, they bring about the "envelope" of a wave. So here is how the derivation goes:
1)Consider the equation of a wave traveling in one dimension:
$ u = ae^{-i(omega t - kz)} $ where $omega = 2 pi f$ is the frequency, $k$ is the wave number, and $a$ is the amplitude. In addition to this information, the book refers to the phase velocity defined as: ${partial z}/{partial t} = {omega}/{k}$
2)Instead of 1 single wave, let there be multiple waves that are superimposed such that $u = sum a_n e^{-i(omega_n t - k_n z)}$
3)Consider the continuous case : $$ u(z) = int_{-infty}^infty a(k)e^{-i(omega t - kz)}dk $$
4)Make two more assumptions. First assumption: the waves are zero everywhere except at a small interval $Delta k $ and here the amplitude is unity such at $a(k)=1$. This brings the equation to $$ u(z) = int_{k_0 - Delta k /2}^{k_0 + Delta k /2} e^{-i(omega t - kz)}dk $$
5)Suppose that $t=0$, thus yielding the endpoint: $$ u(z) = int_{k_0 - Delta k /2}^{k_0 + Delta k /2} e^{i kz}dk $$
When the integration is carried out, the result is $$ u(z) = Delta k e^{ik_0 z} frac {sin 1/2(Delta kz)}{1/2(Delta k z)}$$.
My question is how is the integration done here? I do notice that u is a function of z and that the integrand integrates over the wave number k. Is there some change of variables going on here? By the way I only posses knowledge of ODEs and only know how to do basic Separation of variables for PDEs.
integration definite-integrals wave-equation change-of-variable envelope
integration definite-integrals wave-equation change-of-variable envelope
edited Jan 16 at 21:40
clathratus
4,615337
asked Jan 16 at 20:04
Jonathan AguileraJonathan Aguilera
466
edited Jan 16 at 21:40
clathratus
4,615337
asked Jan 16 at 20:04
Jonathan AguileraJonathan Aguilera
466
edited Jan 16 at 21:40
clathratus
4,615337
edited Jan 16 at 21:40
clathratus
4,615337
edited Jan 16 at 21:40
clathratus
4,615337
4,615337
asked Jan 16 at 20:04
Jonathan AguileraJonathan Aguilera
466
asked Jan 16 at 20:04
Jonathan AguileraJonathan Aguilera
466
asked Jan 16 at 20:04
Jonathan AguileraJonathan Aguilera
466
466
add a comment |
add a comment |
1 Answer
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$begingroup$
The antiderivative of $e^{ikz}$ with respect to $k$ is given by $frac{e^{ikz}}{iz}$ for $z neq 0$, so if we impose this condition on $z$, we have
begin{align}int_{k_0 - Delta k /2}^{k_0 + Delta k /2} e^{i kz}dk &= frac{1}{iz} left[e^{i k z} right]_{k_0 - frac{Delta k}{2}}^{k_0 + frac{Delta k}{2}} \
&= frac{1}{iz}e^{i k_0 z}left(e^{i frac{Delta k}{2}} - e^{- i frac{Delta k}{2}} right).
end{align}
Can you continue from here, perhaps using the fact that for any $theta in mathbb{R}$, $e^{i theta} = cos{theta} + i sin{theta}$?
answered Jan 16 at 21:03
E-muE-mu
787417
$endgroup$
1
$begingroup$
Thanks a lot it worked out well!
$endgroup$
– Jonathan Aguilera
Jan 18 at 0:18
add a comment |
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1 Answer
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$begingroup$
The antiderivative of $e^{ikz}$ with respect to $k$ is given by $frac{e^{ikz}}{iz}$ for $z neq 0$, so if we impose this condition on $z$, we have
begin{align}int_{k_0 - Delta k /2}^{k_0 + Delta k /2} e^{i kz}dk &= frac{1}{iz} left[e^{i k z} right]_{k_0 - frac{Delta k}{2}}^{k_0 + frac{Delta k}{2}} \
&= frac{1}{iz}e^{i k_0 z}left(e^{i frac{Delta k}{2}} - e^{- i frac{Delta k}{2}} right).
end{align}
Can you continue from here, perhaps using the fact that for any $theta in mathbb{R}$, $e^{i theta} = cos{theta} + i sin{theta}$?
answered Jan 16 at 21:03
E-muE-mu
787417
$endgroup$
1
$begingroup$
Thanks a lot it worked out well!
$endgroup$
– Jonathan Aguilera
Jan 18 at 0:18
add a comment |
$begingroup$
The antiderivative of $e^{ikz}$ with respect to $k$ is given by $frac{e^{ikz}}{iz}$ for $z neq 0$, so if we impose this condition on $z$, we have
begin{align}int_{k_0 - Delta k /2}^{k_0 + Delta k /2} e^{i kz}dk &= frac{1}{iz} left[e^{i k z} right]_{k_0 - frac{Delta k}{2}}^{k_0 + frac{Delta k}{2}} \
&= frac{1}{iz}e^{i k_0 z}left(e^{i frac{Delta k}{2}} - e^{- i frac{Delta k}{2}} right).
end{align}
Can you continue from here, perhaps using the fact that for any $theta in mathbb{R}$, $e^{i theta} = cos{theta} + i sin{theta}$?
answered Jan 16 at 21:03
E-muE-mu
787417
$endgroup$
1
$begingroup$
Thanks a lot it worked out well!
$endgroup$
– Jonathan Aguilera
Jan 18 at 0:18
add a comment |
$begingroup$
The antiderivative of $e^{ikz}$ with respect to $k$ is given by $frac{e^{ikz}}{iz}$ for $z neq 0$, so if we impose this condition on $z$, we have
begin{align}int_{k_0 - Delta k /2}^{k_0 + Delta k /2} e^{i kz}dk &= frac{1}{iz} left[e^{i k z} right]_{k_0 - frac{Delta k}{2}}^{k_0 + frac{Delta k}{2}} \
&= frac{1}{iz}e^{i k_0 z}left(e^{i frac{Delta k}{2}} - e^{- i frac{Delta k}{2}} right).
end{align}
Can you continue from here, perhaps using the fact that for any $theta in mathbb{R}$, $e^{i theta} = cos{theta} + i sin{theta}$?
answered Jan 16 at 21:03
E-muE-mu
787417
$endgroup$
The antiderivative of $e^{ikz}$ with respect to $k$ is given by $frac{e^{ikz}}{iz}$ for $z neq 0$, so if we impose this condition on $z$, we have
begin{align}int_{k_0 - Delta k /2}^{k_0 + Delta k /2} e^{i kz}dk &= frac{1}{iz} left[e^{i k z} right]_{k_0 - frac{Delta k}{2}}^{k_0 + frac{Delta k}{2}} \
&= frac{1}{iz}e^{i k_0 z}left(e^{i frac{Delta k}{2}} - e^{- i frac{Delta k}{2}} right).
end{align}
Can you continue from here, perhaps using the fact that for any $theta in mathbb{R}$, $e^{i theta} = cos{theta} + i sin{theta}$?
answered Jan 16 at 21:03
E-muE-mu
787417
edited Jan 30 at 14:21
answered Jan 16 at 21:03
E-muE-mu
787417
answered Jan 16 at 21:03
E-muE-mu
787417
answered Jan 16 at 21:03
E-muE-mu
787417
787417
1
$begingroup$
Thanks a lot it worked out well!
$endgroup$
– Jonathan Aguilera
Jan 18 at 0:18
add a comment |
1
$begingroup$
Thanks a lot it worked out well!
$endgroup$
– Jonathan Aguilera
Jan 18 at 0:18
1
1
$begingroup$
Thanks a lot it worked out well!
$endgroup$
– Jonathan Aguilera
Jan 18 at 0:18
$begingroup$
Thanks a lot it worked out well!
$endgroup$
– Jonathan Aguilera
Jan 18 at 0:18
add a comment |
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