Mixing
Uniform PDF of offset sphere
$begingroup$
Note that throughout this I use the spherical mapping convention: $$(x,y,z) = (rcosphisintheta,rcostheta,rsinphisintheta)$$
I have derived that the uniform pdf for a sphere $S_1$ with radius $rho$ and center $(0,0,0)$ is $p_A(r,phi,theta) = frac{delta(r-rho)}{4pirho^2}$. I want to translate this sphere by $(0,rho,0)$ and find the corresponding pdf.
What I have tried is to rewrite the pdf in cartesian coordinates, that is:
$$p_B(x,y,z) = frac{p_A(r, phi, theta)}{|r^2sintheta|}$$
And translate it: $p_C(x,y,z) = p_B(x,y-rho,z)$.
Then I believe that the pdf I am looking for is: $p_D(r',phi',theta') = p_C(x,y,z)|r'^2sintheta'|$.
I have $r^2 = x^2 + (y-rho)^2 + z^2$, $costheta = frac{y-rho}{r}$, $r'^2 = x^2+y^2 + z^2$, $costheta' = frac{y}{r'}$. Assuming that this is correct I get the relationship:
$$r^2 = r'^2 - 2rho r'costheta' + rho^2$$
$$costheta = costheta' - frac{rho}{r}$$
Is everything correct? Is this the correct expression for the pdf I am looking for?
$$p_D(r',phi,theta') = p_A(r(r',theta'),phi,theta(r',theta'))frac{|r'^2sintheta'|}{|r^2(r',theta')sin(theta(r',theta'))|}$$
Edit:
Just for completeness I provide the derivation of the uniform pdf on the sphere centered at $(0,0,0)$ with radius $rho$. Since I only want to have a constant probability over the surface of the sphere I can write $p_A(r,phi,theta) = Cdelta(r-rho)$. In order to derive the pdf normalization constant $C$ I integrate the pdf:
$$int_{0}^{2pi}{int_{0}^{pi}{int_{0}^{infty}{p_A(r,phi,theta)r^2sintheta dr}dtheta}dphi} = $$
$$2pi Cint_{0}^{pi}{rho^2sintheta dtheta} =$$
$$4Cpirho^2 = 1$$
$$C = frac{1}{4pirho^2}$$
probability-distributions
asked Jan 20 at 17:14
lightxbulblightxbulb
945311
$endgroup$
add a comment |
$begingroup$
Note that throughout this I use the spherical mapping convention: $$(x,y,z) = (rcosphisintheta,rcostheta,rsinphisintheta)$$
I have derived that the uniform pdf for a sphere $S_1$ with radius $rho$ and center $(0,0,0)$ is $p_A(r,phi,theta) = frac{delta(r-rho)}{4pirho^2}$. I want to translate this sphere by $(0,rho,0)$ and find the corresponding pdf.
What I have tried is to rewrite the pdf in cartesian coordinates, that is:
$$p_B(x,y,z) = frac{p_A(r, phi, theta)}{|r^2sintheta|}$$
And translate it: $p_C(x,y,z) = p_B(x,y-rho,z)$.
Then I believe that the pdf I am looking for is: $p_D(r',phi',theta') = p_C(x,y,z)|r'^2sintheta'|$.
I have $r^2 = x^2 + (y-rho)^2 + z^2$, $costheta = frac{y-rho}{r}$, $r'^2 = x^2+y^2 + z^2$, $costheta' = frac{y}{r'}$. Assuming that this is correct I get the relationship:
$$r^2 = r'^2 - 2rho r'costheta' + rho^2$$
$$costheta = costheta' - frac{rho}{r}$$
Is everything correct? Is this the correct expression for the pdf I am looking for?
$$p_D(r',phi,theta') = p_A(r(r',theta'),phi,theta(r',theta'))frac{|r'^2sintheta'|}{|r^2(r',theta')sin(theta(r',theta'))|}$$
Edit:
Just for completeness I provide the derivation of the uniform pdf on the sphere centered at $(0,0,0)$ with radius $rho$. Since I only want to have a constant probability over the surface of the sphere I can write $p_A(r,phi,theta) = Cdelta(r-rho)$. In order to derive the pdf normalization constant $C$ I integrate the pdf:
$$int_{0}^{2pi}{int_{0}^{pi}{int_{0}^{infty}{p_A(r,phi,theta)r^2sintheta dr}dtheta}dphi} = $$
$$2pi Cint_{0}^{pi}{rho^2sintheta dtheta} =$$
$$4Cpirho^2 = 1$$
$$C = frac{1}{4pirho^2}$$
probability-distributions
asked Jan 20 at 17:14
lightxbulblightxbulb
945311
$endgroup$
add a comment |
$begingroup$
Note that throughout this I use the spherical mapping convention: $$(x,y,z) = (rcosphisintheta,rcostheta,rsinphisintheta)$$
I have derived that the uniform pdf for a sphere $S_1$ with radius $rho$ and center $(0,0,0)$ is $p_A(r,phi,theta) = frac{delta(r-rho)}{4pirho^2}$. I want to translate this sphere by $(0,rho,0)$ and find the corresponding pdf.
What I have tried is to rewrite the pdf in cartesian coordinates, that is:
$$p_B(x,y,z) = frac{p_A(r, phi, theta)}{|r^2sintheta|}$$
And translate it: $p_C(x,y,z) = p_B(x,y-rho,z)$.
Then I believe that the pdf I am looking for is: $p_D(r',phi',theta') = p_C(x,y,z)|r'^2sintheta'|$.
I have $r^2 = x^2 + (y-rho)^2 + z^2$, $costheta = frac{y-rho}{r}$, $r'^2 = x^2+y^2 + z^2$, $costheta' = frac{y}{r'}$. Assuming that this is correct I get the relationship:
$$r^2 = r'^2 - 2rho r'costheta' + rho^2$$
$$costheta = costheta' - frac{rho}{r}$$
Is everything correct? Is this the correct expression for the pdf I am looking for?
$$p_D(r',phi,theta') = p_A(r(r',theta'),phi,theta(r',theta'))frac{|r'^2sintheta'|}{|r^2(r',theta')sin(theta(r',theta'))|}$$
Edit:
Just for completeness I provide the derivation of the uniform pdf on the sphere centered at $(0,0,0)$ with radius $rho$. Since I only want to have a constant probability over the surface of the sphere I can write $p_A(r,phi,theta) = Cdelta(r-rho)$. In order to derive the pdf normalization constant $C$ I integrate the pdf:
$$int_{0}^{2pi}{int_{0}^{pi}{int_{0}^{infty}{p_A(r,phi,theta)r^2sintheta dr}dtheta}dphi} = $$
$$2pi Cint_{0}^{pi}{rho^2sintheta dtheta} =$$
$$4Cpirho^2 = 1$$
$$C = frac{1}{4pirho^2}$$
probability-distributions
asked Jan 20 at 17:14
lightxbulblightxbulb
945311
$endgroup$
Note that throughout this I use the spherical mapping convention: $$(x,y,z) = (rcosphisintheta,rcostheta,rsinphisintheta)$$
I have derived that the uniform pdf for a sphere $S_1$ with radius $rho$ and center $(0,0,0)$ is $p_A(r,phi,theta) = frac{delta(r-rho)}{4pirho^2}$. I want to translate this sphere by $(0,rho,0)$ and find the corresponding pdf.
What I have tried is to rewrite the pdf in cartesian coordinates, that is:
$$p_B(x,y,z) = frac{p_A(r, phi, theta)}{|r^2sintheta|}$$
And translate it: $p_C(x,y,z) = p_B(x,y-rho,z)$.
Then I believe that the pdf I am looking for is: $p_D(r',phi',theta') = p_C(x,y,z)|r'^2sintheta'|$.
I have $r^2 = x^2 + (y-rho)^2 + z^2$, $costheta = frac{y-rho}{r}$, $r'^2 = x^2+y^2 + z^2$, $costheta' = frac{y}{r'}$. Assuming that this is correct I get the relationship:
$$r^2 = r'^2 - 2rho r'costheta' + rho^2$$
$$costheta = costheta' - frac{rho}{r}$$
Is everything correct? Is this the correct expression for the pdf I am looking for?
$$p_D(r',phi,theta') = p_A(r(r',theta'),phi,theta(r',theta'))frac{|r'^2sintheta'|}{|r^2(r',theta')sin(theta(r',theta'))|}$$
Edit:
Just for completeness I provide the derivation of the uniform pdf on the sphere centered at $(0,0,0)$ with radius $rho$. Since I only want to have a constant probability over the surface of the sphere I can write $p_A(r,phi,theta) = Cdelta(r-rho)$. In order to derive the pdf normalization constant $C$ I integrate the pdf:
$$int_{0}^{2pi}{int_{0}^{pi}{int_{0}^{infty}{p_A(r,phi,theta)r^2sintheta dr}dtheta}dphi} = $$
$$2pi Cint_{0}^{pi}{rho^2sintheta dtheta} =$$
$$4Cpirho^2 = 1$$
$$C = frac{1}{4pirho^2}$$
probability-distributions
probability-distributions
asked Jan 20 at 17:14
lightxbulblightxbulb
945311
asked Jan 20 at 17:14
lightxbulblightxbulb
945311
edited Jan 21 at 2:38
lightxbulb
asked Jan 20 at 17:14
lightxbulblightxbulb
945311
asked Jan 20 at 17:14
lightxbulblightxbulb
945311
asked Jan 20 at 17:14
lightxbulblightxbulb
945311
945311
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The denominator seems to be redundant, since it actually cancels out with the term $|r^2sintheta|$ in the numerator, which I forgot to add in the pdf it seems: $p_A(r,phi,theta) = delta(r-rho)frac{|r^2sintheta|}{4pirho^2}$.
answered Feb 8 at 18:34
lightxbulblightxbulb
945311
$endgroup$
add a comment |
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%2f3080839%2funiform-pdf-of-offset-sphere%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
$begingroup$
The denominator seems to be redundant, since it actually cancels out with the term $|r^2sintheta|$ in the numerator, which I forgot to add in the pdf it seems: $p_A(r,phi,theta) = delta(r-rho)frac{|r^2sintheta|}{4pirho^2}$.
answered Feb 8 at 18:34
lightxbulblightxbulb
945311
$endgroup$
add a comment |
$begingroup$
The denominator seems to be redundant, since it actually cancels out with the term $|r^2sintheta|$ in the numerator, which I forgot to add in the pdf it seems: $p_A(r,phi,theta) = delta(r-rho)frac{|r^2sintheta|}{4pirho^2}$.
answered Feb 8 at 18:34
lightxbulblightxbulb
945311
$endgroup$
add a comment |
$begingroup$
The denominator seems to be redundant, since it actually cancels out with the term $|r^2sintheta|$ in the numerator, which I forgot to add in the pdf it seems: $p_A(r,phi,theta) = delta(r-rho)frac{|r^2sintheta|}{4pirho^2}$.
answered Feb 8 at 18:34
lightxbulblightxbulb
945311
$endgroup$
The denominator seems to be redundant, since it actually cancels out with the term $|r^2sintheta|$ in the numerator, which I forgot to add in the pdf it seems: $p_A(r,phi,theta) = delta(r-rho)frac{|r^2sintheta|}{4pirho^2}$.
answered Feb 8 at 18:34
lightxbulblightxbulb
945311
answered Feb 8 at 18:34
lightxbulblightxbulb
945311
answered Feb 8 at 18:34
lightxbulblightxbulb
945311
answered Feb 8 at 18:34
lightxbulblightxbulb
945311
945311
add a comment |
add a comment |
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%2f3080839%2funiform-pdf-of-offset-sphere%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
