Mixing
Surface integral of a partially constant Dirac delta
$begingroup$
I am trying to integrate the product of a function and a partially constant delta function over a sphere of constant radius $r$.
The integral is of the form
$$int^{2pi}_0 int^{pi}_0 f(mathbf{r})delta(x-x_0)delta(y-y_0) sintheta,mathrm{d}theta ,mathrm{d}phi. $$
I know that this integral is very simple when the delta function is fully three-dimensional, such as $$delta^3(mathbf{r}-mathbf{r}_0)=delta(x-x_0)delta(y-y_0)delta(z-z_0),$$ but what I have here is something a bit different. Does anyone have anyone have any suggestions?
coordinate-systems distribution-theory dirac-delta spherical-coordinates surface-integrals
edited Jan 23 at 8:58
Qmechanic
5,13211858
asked Nov 10 '15 at 8:21
aquarkydudeaquarkydude
184
$endgroup$
add a comment |
$begingroup$
I am trying to integrate the product of a function and a partially constant delta function over a sphere of constant radius $r$.
The integral is of the form
$$int^{2pi}_0 int^{pi}_0 f(mathbf{r})delta(x-x_0)delta(y-y_0) sintheta,mathrm{d}theta ,mathrm{d}phi. $$
I know that this integral is very simple when the delta function is fully three-dimensional, such as $$delta^3(mathbf{r}-mathbf{r}_0)=delta(x-x_0)delta(y-y_0)delta(z-z_0),$$ but what I have here is something a bit different. Does anyone have anyone have any suggestions?
coordinate-systems distribution-theory dirac-delta spherical-coordinates surface-integrals
edited Jan 23 at 8:58
Qmechanic
5,13211858
asked Nov 10 '15 at 8:21
aquarkydudeaquarkydude
184
$endgroup$
1
$begingroup$
should you have also integral in $r$ for spherical coordinates?
$endgroup$
– Michael Medvinsky
Nov 10 '15 at 8:41
$begingroup$
no, it's a surface integral over the sphere of constant radius $r$, so there is no integration on $,mathrm{d}r$.
$endgroup$
– aquarkydude
Nov 10 '15 at 8:42
1
$begingroup$
if $r$ is constant $f(r)$ is constant should it be $f(r_0)$?
$endgroup$
– Michael Medvinsky
Nov 10 '15 at 8:44
$begingroup$
it's not $f(r)$ but rather $f(mathbf{r})$. That is, the function $f$ takes as an argument the vector $mathbf{r}$, not just the distance from the origin $r$
$endgroup$
– aquarkydude
Nov 10 '15 at 8:45
$begingroup$
Moreover, $r_0$ is not defined in the context of the problem, but only $x_0$ and $y_0$.
$endgroup$
– aquarkydude
Nov 10 '15 at 8:47
add a comment |
$begingroup$
I am trying to integrate the product of a function and a partially constant delta function over a sphere of constant radius $r$.
The integral is of the form
$$int^{2pi}_0 int^{pi}_0 f(mathbf{r})delta(x-x_0)delta(y-y_0) sintheta,mathrm{d}theta ,mathrm{d}phi. $$
I know that this integral is very simple when the delta function is fully three-dimensional, such as $$delta^3(mathbf{r}-mathbf{r}_0)=delta(x-x_0)delta(y-y_0)delta(z-z_0),$$ but what I have here is something a bit different. Does anyone have anyone have any suggestions?
coordinate-systems distribution-theory dirac-delta spherical-coordinates surface-integrals
edited Jan 23 at 8:58
Qmechanic
5,13211858
asked Nov 10 '15 at 8:21
aquarkydudeaquarkydude
184
$endgroup$
I am trying to integrate the product of a function and a partially constant delta function over a sphere of constant radius $r$.
The integral is of the form
$$int^{2pi}_0 int^{pi}_0 f(mathbf{r})delta(x-x_0)delta(y-y_0) sintheta,mathrm{d}theta ,mathrm{d}phi. $$
I know that this integral is very simple when the delta function is fully three-dimensional, such as $$delta^3(mathbf{r}-mathbf{r}_0)=delta(x-x_0)delta(y-y_0)delta(z-z_0),$$ but what I have here is something a bit different. Does anyone have anyone have any suggestions?
coordinate-systems distribution-theory dirac-delta spherical-coordinates surface-integrals
coordinate-systems distribution-theory dirac-delta spherical-coordinates surface-integrals
edited Jan 23 at 8:58
Qmechanic
5,13211858
asked Nov 10 '15 at 8:21
aquarkydudeaquarkydude
184
edited Jan 23 at 8:58
Qmechanic
5,13211858
asked Nov 10 '15 at 8:21
aquarkydudeaquarkydude
184
edited Jan 23 at 8:58
Qmechanic
5,13211858
edited Jan 23 at 8:58
Qmechanic
5,13211858
edited Jan 23 at 8:58
Qmechanic
5,13211858
5,13211858
asked Nov 10 '15 at 8:21
aquarkydudeaquarkydude
184
asked Nov 10 '15 at 8:21
aquarkydudeaquarkydude
184
asked Nov 10 '15 at 8:21
aquarkydudeaquarkydude
184
184
1
$begingroup$
should you have also integral in $r$ for spherical coordinates?
$endgroup$
– Michael Medvinsky
Nov 10 '15 at 8:41
$begingroup$
no, it's a surface integral over the sphere of constant radius $r$, so there is no integration on $,mathrm{d}r$.
$endgroup$
– aquarkydude
Nov 10 '15 at 8:42
1
$begingroup$
if $r$ is constant $f(r)$ is constant should it be $f(r_0)$?
$endgroup$
– Michael Medvinsky
Nov 10 '15 at 8:44
$begingroup$
it's not $f(r)$ but rather $f(mathbf{r})$. That is, the function $f$ takes as an argument the vector $mathbf{r}$, not just the distance from the origin $r$
$endgroup$
– aquarkydude
Nov 10 '15 at 8:45
$begingroup$
Moreover, $r_0$ is not defined in the context of the problem, but only $x_0$ and $y_0$.
$endgroup$
– aquarkydude
Nov 10 '15 at 8:47
add a comment |
1
$begingroup$
should you have also integral in $r$ for spherical coordinates?
$endgroup$
– Michael Medvinsky
Nov 10 '15 at 8:41
$begingroup$
no, it's a surface integral over the sphere of constant radius $r$, so there is no integration on $,mathrm{d}r$.
$endgroup$
– aquarkydude
Nov 10 '15 at 8:42
1
$begingroup$
if $r$ is constant $f(r)$ is constant should it be $f(r_0)$?
$endgroup$
– Michael Medvinsky
Nov 10 '15 at 8:44
$begingroup$
it's not $f(r)$ but rather $f(mathbf{r})$. That is, the function $f$ takes as an argument the vector $mathbf{r}$, not just the distance from the origin $r$
$endgroup$
– aquarkydude
Nov 10 '15 at 8:45
$begingroup$
Moreover, $r_0$ is not defined in the context of the problem, but only $x_0$ and $y_0$.
$endgroup$
– aquarkydude
Nov 10 '15 at 8:47
1
1
$begingroup$
should you have also integral in $r$ for spherical coordinates?
$endgroup$
– Michael Medvinsky
Nov 10 '15 at 8:41
$begingroup$
should you have also integral in $r$ for spherical coordinates?
$endgroup$
– Michael Medvinsky
Nov 10 '15 at 8:41
$begingroup$
no, it's a surface integral over the sphere of constant radius $r$, so there is no integration on $,mathrm{d}r$.
$endgroup$
– aquarkydude
Nov 10 '15 at 8:42
$begingroup$
no, it's a surface integral over the sphere of constant radius $r$, so there is no integration on $,mathrm{d}r$.
$endgroup$
– aquarkydude
Nov 10 '15 at 8:42
1
1
$begingroup$
if $r$ is constant $f(r)$ is constant should it be $f(r_0)$?
$endgroup$
– Michael Medvinsky
Nov 10 '15 at 8:44
$begingroup$
if $r$ is constant $f(r)$ is constant should it be $f(r_0)$?
$endgroup$
– Michael Medvinsky
Nov 10 '15 at 8:44
$begingroup$
it's not $f(r)$ but rather $f(mathbf{r})$. That is, the function $f$ takes as an argument the vector $mathbf{r}$, not just the distance from the origin $r$
$endgroup$
– aquarkydude
Nov 10 '15 at 8:45
$begingroup$
it's not $f(r)$ but rather $f(mathbf{r})$. That is, the function $f$ takes as an argument the vector $mathbf{r}$, not just the distance from the origin $r$
$endgroup$
– aquarkydude
Nov 10 '15 at 8:45
$begingroup$
Moreover, $r_0$ is not defined in the context of the problem, but only $x_0$ and $y_0$.
$endgroup$
– aquarkydude
Nov 10 '15 at 8:47
$begingroup$
Moreover, $r_0$ is not defined in the context of the problem, but only $x_0$ and $y_0$.
$endgroup$
– aquarkydude
Nov 10 '15 at 8:47
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Here we give a rough sketch, and leave it to the reader to iron out various technical issues.
Define the spherical coordinates $(r,theta,phi)$ as
$$tag{1} x~=~rxi, qquad xi~=~sinthetacosphi, qquad y~=~reta, qquad eta~=~sinthetasinphi, qquad z~=~rcostheta. $$Define reflection $$tag{2}sigma: (x,y,z)mapsto (x,y,-z)$$ in the $(x,y)$-plane.
Define function $$tag{3}g({bf r})~:=~f({bf r})+fcircsigma({bf r}), qquad {bf r}~=~(x,y,z).$$
Define function $$tag{4}h({bf r})~:=~frac{g({bf r})}{rz}.$$
Define unit disk
$$ tag{5} D~:=~{(xi,eta)inmathbb{R}^2 mid xi^2+eta^2~<~1}. $$OP's integral then reads
$$ I:~=~ int_0^{2pi}!dphi~int_0^{pi}!dtheta~sintheta ~f({bf r}) delta(x-x_0)delta(y-y_0)$$
$$ ~=~int_0^{2pi}!dphi~int_0^{frac{pi}{2}}!dtheta~sinthetacostheta ~h({bf r})deltaleft(xi-frac{x_0}{r}right)deltaleft(eta-frac{x_0}{r}right)$$
$$~=~iint_D !dxi~deta ~h({bf r})deltaleft(xi-frac{x_0}{r}right)deltaleft(eta-frac{x_0}{r}right)$$
$$tag{6}~=~left{begin{array}{ccc} hleft(x_0, y_0, sqrt{r^2-x_0^2-y_0^2}right) &{rm for}& x_0^2+y_0^2<r^2, cr
frac{1}{2}h(x_0, y_0, 0) &{rm for}& x_0^2+y_0^2=r^2, cr
0 &{rm for}& x_0^2+y_0^2>r^2.
end{array}right. $$
answered Nov 10 '15 at 15:39
QmechanicQmechanic
5,13211858
$endgroup$
$begingroup$
Wow. Thank you. Tremendously.
$endgroup$
– aquarkydude
Nov 11 '15 at 17:01
$begingroup$
Sorry, I'm not great with proper mathematics notations. The open circle in equation 5: does this indicate applying the reflection operation to the argument of the function $f(mathbf{r})$?
$endgroup$
– aquarkydude
Nov 11 '15 at 17:08
$begingroup$
Also, it appears you've defined $eta$ and $xi$ identically. I think $eta = sin{theta} sin{phi}$?
$endgroup$
– aquarkydude
Nov 11 '15 at 17:15
$begingroup$
Ups. Yes. Corrected. And yes: $fcircsigma({bf r})=f(sigma({bf r}))$.
$endgroup$
– Qmechanic
Nov 11 '15 at 17:20
1
$begingroup$
1. Well, that was one of the reasons to put the disclaimer in the first line of the answer :) Note however that the fraction $h$ could be extended to a well-defined regular function if the numerator $f$ has a matching zero at $z=0$. 2. Finally, your last intuitive comment about a delta function seems to make no mathematical sense.
$endgroup$
– Qmechanic
Nov 11 '15 at 20:42
|
show 2 more comments
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f1522166%2fsurface-integral-of-a-partially-constant-dirac-delta%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Here we give a rough sketch, and leave it to the reader to iron out various technical issues.
Define the spherical coordinates $(r,theta,phi)$ as
$$tag{1} x~=~rxi, qquad xi~=~sinthetacosphi, qquad y~=~reta, qquad eta~=~sinthetasinphi, qquad z~=~rcostheta. $$Define reflection $$tag{2}sigma: (x,y,z)mapsto (x,y,-z)$$ in the $(x,y)$-plane.
Define function $$tag{3}g({bf r})~:=~f({bf r})+fcircsigma({bf r}), qquad {bf r}~=~(x,y,z).$$
Define function $$tag{4}h({bf r})~:=~frac{g({bf r})}{rz}.$$
Define unit disk
$$ tag{5} D~:=~{(xi,eta)inmathbb{R}^2 mid xi^2+eta^2~<~1}. $$OP's integral then reads
$$ I:~=~ int_0^{2pi}!dphi~int_0^{pi}!dtheta~sintheta ~f({bf r}) delta(x-x_0)delta(y-y_0)$$
$$ ~=~int_0^{2pi}!dphi~int_0^{frac{pi}{2}}!dtheta~sinthetacostheta ~h({bf r})deltaleft(xi-frac{x_0}{r}right)deltaleft(eta-frac{x_0}{r}right)$$
$$~=~iint_D !dxi~deta ~h({bf r})deltaleft(xi-frac{x_0}{r}right)deltaleft(eta-frac{x_0}{r}right)$$
$$tag{6}~=~left{begin{array}{ccc} hleft(x_0, y_0, sqrt{r^2-x_0^2-y_0^2}right) &{rm for}& x_0^2+y_0^2<r^2, cr
frac{1}{2}h(x_0, y_0, 0) &{rm for}& x_0^2+y_0^2=r^2, cr
0 &{rm for}& x_0^2+y_0^2>r^2.
end{array}right. $$
answered Nov 10 '15 at 15:39
QmechanicQmechanic
5,13211858
$endgroup$
$begingroup$
Wow. Thank you. Tremendously.
$endgroup$
– aquarkydude
Nov 11 '15 at 17:01
$begingroup$
Sorry, I'm not great with proper mathematics notations. The open circle in equation 5: does this indicate applying the reflection operation to the argument of the function $f(mathbf{r})$?
$endgroup$
– aquarkydude
Nov 11 '15 at 17:08
$begingroup$
Also, it appears you've defined $eta$ and $xi$ identically. I think $eta = sin{theta} sin{phi}$?
$endgroup$
– aquarkydude
Nov 11 '15 at 17:15
$begingroup$
Ups. Yes. Corrected. And yes: $fcircsigma({bf r})=f(sigma({bf r}))$.
$endgroup$
– Qmechanic
Nov 11 '15 at 17:20
1
$begingroup$
1. Well, that was one of the reasons to put the disclaimer in the first line of the answer :) Note however that the fraction $h$ could be extended to a well-defined regular function if the numerator $f$ has a matching zero at $z=0$. 2. Finally, your last intuitive comment about a delta function seems to make no mathematical sense.
$endgroup$
– Qmechanic
Nov 11 '15 at 20:42
|
show 2 more comments
$begingroup$
Here we give a rough sketch, and leave it to the reader to iron out various technical issues.
Define the spherical coordinates $(r,theta,phi)$ as
$$tag{1} x~=~rxi, qquad xi~=~sinthetacosphi, qquad y~=~reta, qquad eta~=~sinthetasinphi, qquad z~=~rcostheta. $$Define reflection $$tag{2}sigma: (x,y,z)mapsto (x,y,-z)$$ in the $(x,y)$-plane.
Define function $$tag{3}g({bf r})~:=~f({bf r})+fcircsigma({bf r}), qquad {bf r}~=~(x,y,z).$$
Define function $$tag{4}h({bf r})~:=~frac{g({bf r})}{rz}.$$
Define unit disk
$$ tag{5} D~:=~{(xi,eta)inmathbb{R}^2 mid xi^2+eta^2~<~1}. $$OP's integral then reads
$$ I:~=~ int_0^{2pi}!dphi~int_0^{pi}!dtheta~sintheta ~f({bf r}) delta(x-x_0)delta(y-y_0)$$
$$ ~=~int_0^{2pi}!dphi~int_0^{frac{pi}{2}}!dtheta~sinthetacostheta ~h({bf r})deltaleft(xi-frac{x_0}{r}right)deltaleft(eta-frac{x_0}{r}right)$$
$$~=~iint_D !dxi~deta ~h({bf r})deltaleft(xi-frac{x_0}{r}right)deltaleft(eta-frac{x_0}{r}right)$$
$$tag{6}~=~left{begin{array}{ccc} hleft(x_0, y_0, sqrt{r^2-x_0^2-y_0^2}right) &{rm for}& x_0^2+y_0^2<r^2, cr
frac{1}{2}h(x_0, y_0, 0) &{rm for}& x_0^2+y_0^2=r^2, cr
0 &{rm for}& x_0^2+y_0^2>r^2.
end{array}right. $$
answered Nov 10 '15 at 15:39
QmechanicQmechanic
5,13211858
$endgroup$
$begingroup$
Wow. Thank you. Tremendously.
$endgroup$
– aquarkydude
Nov 11 '15 at 17:01
$begingroup$
Sorry, I'm not great with proper mathematics notations. The open circle in equation 5: does this indicate applying the reflection operation to the argument of the function $f(mathbf{r})$?
$endgroup$
– aquarkydude
Nov 11 '15 at 17:08
$begingroup$
Also, it appears you've defined $eta$ and $xi$ identically. I think $eta = sin{theta} sin{phi}$?
$endgroup$
– aquarkydude
Nov 11 '15 at 17:15
$begingroup$
Ups. Yes. Corrected. And yes: $fcircsigma({bf r})=f(sigma({bf r}))$.
$endgroup$
– Qmechanic
Nov 11 '15 at 17:20
1
$begingroup$
1. Well, that was one of the reasons to put the disclaimer in the first line of the answer :) Note however that the fraction $h$ could be extended to a well-defined regular function if the numerator $f$ has a matching zero at $z=0$. 2. Finally, your last intuitive comment about a delta function seems to make no mathematical sense.
$endgroup$
– Qmechanic
Nov 11 '15 at 20:42
|
show 2 more comments
$begingroup$
Here we give a rough sketch, and leave it to the reader to iron out various technical issues.
Define the spherical coordinates $(r,theta,phi)$ as
$$tag{1} x~=~rxi, qquad xi~=~sinthetacosphi, qquad y~=~reta, qquad eta~=~sinthetasinphi, qquad z~=~rcostheta. $$Define reflection $$tag{2}sigma: (x,y,z)mapsto (x,y,-z)$$ in the $(x,y)$-plane.
Define function $$tag{3}g({bf r})~:=~f({bf r})+fcircsigma({bf r}), qquad {bf r}~=~(x,y,z).$$
Define function $$tag{4}h({bf r})~:=~frac{g({bf r})}{rz}.$$
Define unit disk
$$ tag{5} D~:=~{(xi,eta)inmathbb{R}^2 mid xi^2+eta^2~<~1}. $$OP's integral then reads
$$ I:~=~ int_0^{2pi}!dphi~int_0^{pi}!dtheta~sintheta ~f({bf r}) delta(x-x_0)delta(y-y_0)$$
$$ ~=~int_0^{2pi}!dphi~int_0^{frac{pi}{2}}!dtheta~sinthetacostheta ~h({bf r})deltaleft(xi-frac{x_0}{r}right)deltaleft(eta-frac{x_0}{r}right)$$
$$~=~iint_D !dxi~deta ~h({bf r})deltaleft(xi-frac{x_0}{r}right)deltaleft(eta-frac{x_0}{r}right)$$
$$tag{6}~=~left{begin{array}{ccc} hleft(x_0, y_0, sqrt{r^2-x_0^2-y_0^2}right) &{rm for}& x_0^2+y_0^2<r^2, cr
frac{1}{2}h(x_0, y_0, 0) &{rm for}& x_0^2+y_0^2=r^2, cr
0 &{rm for}& x_0^2+y_0^2>r^2.
end{array}right. $$
answered Nov 10 '15 at 15:39
QmechanicQmechanic
5,13211858
$endgroup$
Here we give a rough sketch, and leave it to the reader to iron out various technical issues.
Define the spherical coordinates $(r,theta,phi)$ as
$$tag{1} x~=~rxi, qquad xi~=~sinthetacosphi, qquad y~=~reta, qquad eta~=~sinthetasinphi, qquad z~=~rcostheta. $$Define reflection $$tag{2}sigma: (x,y,z)mapsto (x,y,-z)$$ in the $(x,y)$-plane.
Define function $$tag{3}g({bf r})~:=~f({bf r})+fcircsigma({bf r}), qquad {bf r}~=~(x,y,z).$$
Define function $$tag{4}h({bf r})~:=~frac{g({bf r})}{rz}.$$
Define unit disk
$$ tag{5} D~:=~{(xi,eta)inmathbb{R}^2 mid xi^2+eta^2~<~1}. $$OP's integral then reads
$$ I:~=~ int_0^{2pi}!dphi~int_0^{pi}!dtheta~sintheta ~f({bf r}) delta(x-x_0)delta(y-y_0)$$
$$ ~=~int_0^{2pi}!dphi~int_0^{frac{pi}{2}}!dtheta~sinthetacostheta ~h({bf r})deltaleft(xi-frac{x_0}{r}right)deltaleft(eta-frac{x_0}{r}right)$$
$$~=~iint_D !dxi~deta ~h({bf r})deltaleft(xi-frac{x_0}{r}right)deltaleft(eta-frac{x_0}{r}right)$$
$$tag{6}~=~left{begin{array}{ccc} hleft(x_0, y_0, sqrt{r^2-x_0^2-y_0^2}right) &{rm for}& x_0^2+y_0^2<r^2, cr
frac{1}{2}h(x_0, y_0, 0) &{rm for}& x_0^2+y_0^2=r^2, cr
0 &{rm for}& x_0^2+y_0^2>r^2.
end{array}right. $$
answered Nov 10 '15 at 15:39
QmechanicQmechanic
5,13211858
edited Nov 11 '15 at 17:20
answered Nov 10 '15 at 15:39
QmechanicQmechanic
5,13211858
answered Nov 10 '15 at 15:39
QmechanicQmechanic
5,13211858
answered Nov 10 '15 at 15:39
QmechanicQmechanic
5,13211858
5,13211858
$begingroup$
Wow. Thank you. Tremendously.
$endgroup$
– aquarkydude
Nov 11 '15 at 17:01
$begingroup$
Sorry, I'm not great with proper mathematics notations. The open circle in equation 5: does this indicate applying the reflection operation to the argument of the function $f(mathbf{r})$?
$endgroup$
– aquarkydude
Nov 11 '15 at 17:08
$begingroup$
Also, it appears you've defined $eta$ and $xi$ identically. I think $eta = sin{theta} sin{phi}$?
$endgroup$
– aquarkydude
Nov 11 '15 at 17:15
$begingroup$
Ups. Yes. Corrected. And yes: $fcircsigma({bf r})=f(sigma({bf r}))$.
$endgroup$
– Qmechanic
Nov 11 '15 at 17:20
1
$begingroup$
1. Well, that was one of the reasons to put the disclaimer in the first line of the answer :) Note however that the fraction $h$ could be extended to a well-defined regular function if the numerator $f$ has a matching zero at $z=0$. 2. Finally, your last intuitive comment about a delta function seems to make no mathematical sense.
$endgroup$
– Qmechanic
Nov 11 '15 at 20:42
|
show 2 more comments
$begingroup$
Wow. Thank you. Tremendously.
$endgroup$
– aquarkydude
Nov 11 '15 at 17:01
$begingroup$
Sorry, I'm not great with proper mathematics notations. The open circle in equation 5: does this indicate applying the reflection operation to the argument of the function $f(mathbf{r})$?
$endgroup$
– aquarkydude
Nov 11 '15 at 17:08
$begingroup$
Also, it appears you've defined $eta$ and $xi$ identically. I think $eta = sin{theta} sin{phi}$?
$endgroup$
– aquarkydude
Nov 11 '15 at 17:15
$begingroup$
Ups. Yes. Corrected. And yes: $fcircsigma({bf r})=f(sigma({bf r}))$.
$endgroup$
– Qmechanic
Nov 11 '15 at 17:20
1
$begingroup$
1. Well, that was one of the reasons to put the disclaimer in the first line of the answer :) Note however that the fraction $h$ could be extended to a well-defined regular function if the numerator $f$ has a matching zero at $z=0$. 2. Finally, your last intuitive comment about a delta function seems to make no mathematical sense.
$endgroup$
– Qmechanic
Nov 11 '15 at 20:42
$begingroup$
Wow. Thank you. Tremendously.
$endgroup$
– aquarkydude
Nov 11 '15 at 17:01
$begingroup$
Wow. Thank you. Tremendously.
$endgroup$
– aquarkydude
Nov 11 '15 at 17:01
$begingroup$
Sorry, I'm not great with proper mathematics notations. The open circle in equation 5: does this indicate applying the reflection operation to the argument of the function $f(mathbf{r})$?
$endgroup$
– aquarkydude
Nov 11 '15 at 17:08
$begingroup$
Sorry, I'm not great with proper mathematics notations. The open circle in equation 5: does this indicate applying the reflection operation to the argument of the function $f(mathbf{r})$?
$endgroup$
– aquarkydude
Nov 11 '15 at 17:08
$begingroup$
Also, it appears you've defined $eta$ and $xi$ identically. I think $eta = sin{theta} sin{phi}$?
$endgroup$
– aquarkydude
Nov 11 '15 at 17:15
$begingroup$
Also, it appears you've defined $eta$ and $xi$ identically. I think $eta = sin{theta} sin{phi}$?
$endgroup$
– aquarkydude
Nov 11 '15 at 17:15
$begingroup$
Ups. Yes. Corrected. And yes: $fcircsigma({bf r})=f(sigma({bf r}))$.
$endgroup$
– Qmechanic
Nov 11 '15 at 17:20
$begingroup$
Ups. Yes. Corrected. And yes: $fcircsigma({bf r})=f(sigma({bf r}))$.
$endgroup$
– Qmechanic
Nov 11 '15 at 17:20
1
1
$begingroup$
1. Well, that was one of the reasons to put the disclaimer in the first line of the answer :) Note however that the fraction $h$ could be extended to a well-defined regular function if the numerator $f$ has a matching zero at $z=0$. 2. Finally, your last intuitive comment about a delta function seems to make no mathematical sense.
$endgroup$
– Qmechanic
Nov 11 '15 at 20:42
$begingroup$
1. Well, that was one of the reasons to put the disclaimer in the first line of the answer :) Note however that the fraction $h$ could be extended to a well-defined regular function if the numerator $f$ has a matching zero at $z=0$. 2. Finally, your last intuitive comment about a delta function seems to make no mathematical sense.
$endgroup$
– Qmechanic
Nov 11 '15 at 20:42
|
show 2 more comments
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f1522166%2fsurface-integral-of-a-partially-constant-dirac-delta%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
1
$begingroup$
should you have also integral in $r$ for spherical coordinates?
$endgroup$
– Michael Medvinsky
Nov 10 '15 at 8:41
$begingroup$
no, it's a surface integral over the sphere of constant radius $r$, so there is no integration on $,mathrm{d}r$.
$endgroup$
– aquarkydude
Nov 10 '15 at 8:42
1
$begingroup$
if $r$ is constant $f(r)$ is constant should it be $f(r_0)$?
$endgroup$
– Michael Medvinsky
Nov 10 '15 at 8:44
$begingroup$
it's not $f(r)$ but rather $f(mathbf{r})$. That is, the function $f$ takes as an argument the vector $mathbf{r}$, not just the distance from the origin $r$
$endgroup$
– aquarkydude
Nov 10 '15 at 8:45
$begingroup$
Moreover, $r_0$ is not defined in the context of the problem, but only $x_0$ and $y_0$.
$endgroup$
– aquarkydude
Nov 10 '15 at 8:47