Stationary phase for retarded potentials in electromagnetism
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I want to apply something like a stationary phase approximation to the following expression
$int_V d^3x' frac{B(x')}{|x-x'|}e^{ik|x-x'|}$
with $xin mathbb{R}^3$, $krightarrow infty$ and $B$ is slowly varying function over the Volume $V$.
For better clearness one can assume the volume $V$ to be a finite box $V=[0,L_x]times [0,L_y]times [0,L_z]$ and one can for example assume $B=sin(pi x/L_x)^2sin(pi y/L_y)^2sin(pi z/L_z)^2$, where $L_{x,y,z}gg frac{2pi}{k}$.
My problem is that I cannot apply the standard method for stationary phases, e.g. https://en.wikipedia.org/wiki/Stationary_phase_approximation because my function $g=frac{B(x')}{|x-x'|}$ is not smooth at $x=x'$ and the phase function $f=|x-x'|$ is not differentiable at $x=x'$...
The answer to my question at lowest order should be
$int_V d^3x' frac{B(x')}{|x-x'|}e^{ik|x-x'|}=-frac{4pi}{k^2}B(x)$, but I don't know how to show that rigorously...
Many thanks in advance
greens-function stationary-point oscillatory-integral
asked Dec 14 '18 at 11:04
Jan SEJan SE
114
$endgroup$
add a comment |
$begingroup$
I want to apply something like a stationary phase approximation to the following expression
$int_V d^3x' frac{B(x')}{|x-x'|}e^{ik|x-x'|}$
with $xin mathbb{R}^3$, $krightarrow infty$ and $B$ is slowly varying function over the Volume $V$.
For better clearness one can assume the volume $V$ to be a finite box $V=[0,L_x]times [0,L_y]times [0,L_z]$ and one can for example assume $B=sin(pi x/L_x)^2sin(pi y/L_y)^2sin(pi z/L_z)^2$, where $L_{x,y,z}gg frac{2pi}{k}$.
My problem is that I cannot apply the standard method for stationary phases, e.g. https://en.wikipedia.org/wiki/Stationary_phase_approximation because my function $g=frac{B(x')}{|x-x'|}$ is not smooth at $x=x'$ and the phase function $f=|x-x'|$ is not differentiable at $x=x'$...
The answer to my question at lowest order should be
$int_V d^3x' frac{B(x')}{|x-x'|}e^{ik|x-x'|}=-frac{4pi}{k^2}B(x)$, but I don't know how to show that rigorously...
Many thanks in advance
greens-function stationary-point oscillatory-integral
asked Dec 14 '18 at 11:04
Jan SEJan SE
114
$endgroup$
$begingroup$
Are you also looking for a formula for $x$ outside of the volume? If not, and you are looking for $x$ inside the volume, then you will get reasonable formulas of the type you describe only if your integrand falls off with $1/|x-x'|^3$.
$endgroup$
– Void
Dec 14 '18 at 12:26
$begingroup$
Hi, inside the Volume $xin V$. I think your assumption is fulfilled, because sin falls off with $1/x^2$ to the boundray of my volume $V$. Can you explain how you came to this statement?
$endgroup$
– Jan SE
Dec 14 '18 at 12:30
$begingroup$
Ok, if you transform to spherical coordinates such that $r = |x-x'|$ you will get an extra $r^2$ from the volume element and you will have $sim B r e^{ikr}$ in your integrand. However, if you have start with $B e^{ik|x-x'|}/|x-x'|^3$ in your integrand and transform to spherical coordinates you get $sim B e^{ikr}/r$, the real part of which will integrate exactly to $sim B(x)$ as $k to infty$.
$endgroup$
– Void
Dec 14 '18 at 12:36
$begingroup$
Unfortunately it is not written like that. One just has $int_V d^3x' frac{B(x')}{|x-x'|}e^{ik|x-x'|}$. But the result has to be $sim B(x)$ in the limit of large $k$. I checked this by evaluating the integral numerically for many points.
$endgroup$
– Jan SE
Dec 14 '18 at 13:43
add a comment |
$begingroup$
I want to apply something like a stationary phase approximation to the following expression
$int_V d^3x' frac{B(x')}{|x-x'|}e^{ik|x-x'|}$
with $xin mathbb{R}^3$, $krightarrow infty$ and $B$ is slowly varying function over the Volume $V$.
For better clearness one can assume the volume $V$ to be a finite box $V=[0,L_x]times [0,L_y]times [0,L_z]$ and one can for example assume $B=sin(pi x/L_x)^2sin(pi y/L_y)^2sin(pi z/L_z)^2$, where $L_{x,y,z}gg frac{2pi}{k}$.
My problem is that I cannot apply the standard method for stationary phases, e.g. https://en.wikipedia.org/wiki/Stationary_phase_approximation because my function $g=frac{B(x')}{|x-x'|}$ is not smooth at $x=x'$ and the phase function $f=|x-x'|$ is not differentiable at $x=x'$...
The answer to my question at lowest order should be
$int_V d^3x' frac{B(x')}{|x-x'|}e^{ik|x-x'|}=-frac{4pi}{k^2}B(x)$, but I don't know how to show that rigorously...
Many thanks in advance
greens-function stationary-point oscillatory-integral
asked Dec 14 '18 at 11:04
Jan SEJan SE
114
$endgroup$
I want to apply something like a stationary phase approximation to the following expression
$int_V d^3x' frac{B(x')}{|x-x'|}e^{ik|x-x'|}$
with $xin mathbb{R}^3$, $krightarrow infty$ and $B$ is slowly varying function over the Volume $V$.
For better clearness one can assume the volume $V$ to be a finite box $V=[0,L_x]times [0,L_y]times [0,L_z]$ and one can for example assume $B=sin(pi x/L_x)^2sin(pi y/L_y)^2sin(pi z/L_z)^2$, where $L_{x,y,z}gg frac{2pi}{k}$.
My problem is that I cannot apply the standard method for stationary phases, e.g. https://en.wikipedia.org/wiki/Stationary_phase_approximation because my function $g=frac{B(x')}{|x-x'|}$ is not smooth at $x=x'$ and the phase function $f=|x-x'|$ is not differentiable at $x=x'$...
The answer to my question at lowest order should be
$int_V d^3x' frac{B(x')}{|x-x'|}e^{ik|x-x'|}=-frac{4pi}{k^2}B(x)$, but I don't know how to show that rigorously...
Many thanks in advance
greens-function stationary-point oscillatory-integral
greens-function stationary-point oscillatory-integral
asked Dec 14 '18 at 11:04
Jan SEJan SE
114
asked Dec 14 '18 at 11:04
Jan SEJan SE
114
edited Dec 14 '18 at 11:21
Jan SE
asked Dec 14 '18 at 11:04
Jan SEJan SE
114
asked Dec 14 '18 at 11:04
Jan SEJan SE
114
asked Dec 14 '18 at 11:04
Jan SEJan SE
114
114
$begingroup$
Are you also looking for a formula for $x$ outside of the volume? If not, and you are looking for $x$ inside the volume, then you will get reasonable formulas of the type you describe only if your integrand falls off with $1/|x-x'|^3$.
$endgroup$
– Void
Dec 14 '18 at 12:26
$begingroup$
Hi, inside the Volume $xin V$. I think your assumption is fulfilled, because sin falls off with $1/x^2$ to the boundray of my volume $V$. Can you explain how you came to this statement?
$endgroup$
– Jan SE
Dec 14 '18 at 12:30
$begingroup$
Ok, if you transform to spherical coordinates such that $r = |x-x'|$ you will get an extra $r^2$ from the volume element and you will have $sim B r e^{ikr}$ in your integrand. However, if you have start with $B e^{ik|x-x'|}/|x-x'|^3$ in your integrand and transform to spherical coordinates you get $sim B e^{ikr}/r$, the real part of which will integrate exactly to $sim B(x)$ as $k to infty$.
$endgroup$
– Void
Dec 14 '18 at 12:36
$begingroup$
Unfortunately it is not written like that. One just has $int_V d^3x' frac{B(x')}{|x-x'|}e^{ik|x-x'|}$. But the result has to be $sim B(x)$ in the limit of large $k$. I checked this by evaluating the integral numerically for many points.
$endgroup$
– Jan SE
Dec 14 '18 at 13:43
add a comment |
$begingroup$
Are you also looking for a formula for $x$ outside of the volume? If not, and you are looking for $x$ inside the volume, then you will get reasonable formulas of the type you describe only if your integrand falls off with $1/|x-x'|^3$.
$endgroup$
– Void
Dec 14 '18 at 12:26
$begingroup$
Hi, inside the Volume $xin V$. I think your assumption is fulfilled, because sin falls off with $1/x^2$ to the boundray of my volume $V$. Can you explain how you came to this statement?
$endgroup$
– Jan SE
Dec 14 '18 at 12:30
$begingroup$
Ok, if you transform to spherical coordinates such that $r = |x-x'|$ you will get an extra $r^2$ from the volume element and you will have $sim B r e^{ikr}$ in your integrand. However, if you have start with $B e^{ik|x-x'|}/|x-x'|^3$ in your integrand and transform to spherical coordinates you get $sim B e^{ikr}/r$, the real part of which will integrate exactly to $sim B(x)$ as $k to infty$.
$endgroup$
– Void
Dec 14 '18 at 12:36
$begingroup$
Unfortunately it is not written like that. One just has $int_V d^3x' frac{B(x')}{|x-x'|}e^{ik|x-x'|}$. But the result has to be $sim B(x)$ in the limit of large $k$. I checked this by evaluating the integral numerically for many points.
$endgroup$
– Jan SE
Dec 14 '18 at 13:43
$begingroup$
Are you also looking for a formula for $x$ outside of the volume? If not, and you are looking for $x$ inside the volume, then you will get reasonable formulas of the type you describe only if your integrand falls off with $1/|x-x'|^3$.
$endgroup$
– Void
Dec 14 '18 at 12:26
$begingroup$
Are you also looking for a formula for $x$ outside of the volume? If not, and you are looking for $x$ inside the volume, then you will get reasonable formulas of the type you describe only if your integrand falls off with $1/|x-x'|^3$.
$endgroup$
– Void
Dec 14 '18 at 12:26
$begingroup$
Hi, inside the Volume $xin V$. I think your assumption is fulfilled, because sin falls off with $1/x^2$ to the boundray of my volume $V$. Can you explain how you came to this statement?
$endgroup$
– Jan SE
Dec 14 '18 at 12:30
$begingroup$
Hi, inside the Volume $xin V$. I think your assumption is fulfilled, because sin falls off with $1/x^2$ to the boundray of my volume $V$. Can you explain how you came to this statement?
$endgroup$
– Jan SE
Dec 14 '18 at 12:30
$begingroup$
Ok, if you transform to spherical coordinates such that $r = |x-x'|$ you will get an extra $r^2$ from the volume element and you will have $sim B r e^{ikr}$ in your integrand. However, if you have start with $B e^{ik|x-x'|}/|x-x'|^3$ in your integrand and transform to spherical coordinates you get $sim B e^{ikr}/r$, the real part of which will integrate exactly to $sim B(x)$ as $k to infty$.
$endgroup$
– Void
Dec 14 '18 at 12:36
$begingroup$
Ok, if you transform to spherical coordinates such that $r = |x-x'|$ you will get an extra $r^2$ from the volume element and you will have $sim B r e^{ikr}$ in your integrand. However, if you have start with $B e^{ik|x-x'|}/|x-x'|^3$ in your integrand and transform to spherical coordinates you get $sim B e^{ikr}/r$, the real part of which will integrate exactly to $sim B(x)$ as $k to infty$.
$endgroup$
– Void
Dec 14 '18 at 12:36
$begingroup$
Unfortunately it is not written like that. One just has $int_V d^3x' frac{B(x')}{|x-x'|}e^{ik|x-x'|}$. But the result has to be $sim B(x)$ in the limit of large $k$. I checked this by evaluating the integral numerically for many points.
$endgroup$
– Jan SE
Dec 14 '18 at 13:43
$begingroup$
Unfortunately it is not written like that. One just has $int_V d^3x' frac{B(x')}{|x-x'|}e^{ik|x-x'|}$. But the result has to be $sim B(x)$ in the limit of large $k$. I checked this by evaluating the integral numerically for many points.
$endgroup$
– Jan SE
Dec 14 '18 at 13:43
add a comment |
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$begingroup$
Are you also looking for a formula for $x$ outside of the volume? If not, and you are looking for $x$ inside the volume, then you will get reasonable formulas of the type you describe only if your integrand falls off with $1/|x-x'|^3$.
$endgroup$
– Void
Dec 14 '18 at 12:26
$begingroup$
Hi, inside the Volume $xin V$. I think your assumption is fulfilled, because sin falls off with $1/x^2$ to the boundray of my volume $V$. Can you explain how you came to this statement?
$endgroup$
– Jan SE
Dec 14 '18 at 12:30
$begingroup$
Ok, if you transform to spherical coordinates such that $r = |x-x'|$ you will get an extra $r^2$ from the volume element and you will have $sim B r e^{ikr}$ in your integrand. However, if you have start with $B e^{ik|x-x'|}/|x-x'|^3$ in your integrand and transform to spherical coordinates you get $sim B e^{ikr}/r$, the real part of which will integrate exactly to $sim B(x)$ as $k to infty$.
$endgroup$
– Void
Dec 14 '18 at 12:36
$begingroup$
Unfortunately it is not written like that. One just has $int_V d^3x' frac{B(x')}{|x-x'|}e^{ik|x-x'|}$. But the result has to be $sim B(x)$ in the limit of large $k$. I checked this by evaluating the integral numerically for many points.
$endgroup$
– Jan SE
Dec 14 '18 at 13:43