Why are we not using Dirac delta and ignoring the contribution to the surface integral from the point $r=0$? The 2019 Stack Overflow Developer Survey Results Are In Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Calculate surface integral of point charge located outside the surfaceHow do I convert a vector field in Cartesian coordinates to spherical coordinates?Directional derivative conceptual questionPotential theory solution for Variable coefficient Poisson with Dirichlet Boundary conditionsDivergence change of variables (to polar)What is wrong with this line integral? (Line integral change of variables)How can I find the curl of velocity in spherical coordinates?A question about gradient fieldIf $vecnabla cdot vecV neq 0$ at only one point, will this prevent us from saying that $vecV=vecnabla times vecU$?Showing volume and surface integration is unaffected by the singularity at $mathbfr'=mathbfr$
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Why are we not using Dirac delta and ignoring the contribution to the surface integral from the point $r=0$?
The 2019 Stack Overflow Developer Survey Results Are In
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Calculate surface integral of point charge located outside the surfaceHow do I convert a vector field in Cartesian coordinates to spherical coordinates?Directional derivative conceptual questionPotential theory solution for Variable coefficient Poisson with Dirichlet Boundary conditionsDivergence change of variables (to polar)What is wrong with this line integral? (Line integral change of variables)How can I find the curl of velocity in spherical coordinates?A question about gradient fieldIf $vecnabla cdot vecV neq 0$ at only one point, will this prevent us from saying that $vecV=vecnabla times vecU$?Showing volume and surface integration is unaffected by the singularity at $mathbfr'=mathbfr$
$begingroup$
Let $V'$ be the volume of dipole distribution and $S'$ be the boundary.
The potential of a dipole distribution at a point $P$ is:
$$psi=-k int_V'
dfracvecnabla'.vecM'rdV'
+k oint_S'dfracvecM'.hatnrdS'$$
If $Pin V'$ and $Pin S'$, the integrand is discontinuous (infinite) at the point $r=0$. If we use spherical and polar coordinates, the integrand is continuous everywhere:
beginalign
psi &=bbox[orange,5px]-k int_V' dfracvecnabla'.vecM'r r^2 sin theta dtheta dphi dr\
&bbox[pink,5px]+koint_S'_1 dfracvecM'.hatnr sqrtf_x^2+f_y^2+1 R dR dtheta'
bbox[yellow,5px]+ k oint_S'_2dfracvecM'.hatnrdS'\
&=bbox[orange,5px]-k int_V' vecnabla'.vecM' r sin theta dtheta dphi dr\
&bbox[pink,5px] + k oint_S'_1 vecM'.hatn sqrtf_x^2+f_y^2+1 dfracRsqrtR^2+f^2 dR dtheta'
bbox[yellow,5px] + k oint_S'_2dfracvecM'.hatnrdS'
endalign
The field of a dipole distribution at a point $P$ is:
$$nablapsi=-k int_V'
(vecnabla'.vecM') nabla left( dfrac1r right) dV'
+k oint_S' (vecM'.hatn) nabla left( dfrac1r right) dS'$$
If $Pin V'$ and $Pin S'$, the integrand is discontinuous (infinite) at the point $r=0$. If we use spherical and polar coordinates:
beginalign
nablapsi&=bbox[orange,5px]-k int_V'
(vecnabla'.vecM') left( dfrachatrr^2 right) r^2 sin theta dtheta dphi dr\
&bbox[pink,5px]+ k oint_S'_1 (vecM'.hatn) left( dfrachatrr right) sqrtf_x^2+f_y^2+1 dfracRsqrtR^2+f^2 dR dtheta'
bbox[yellow,5px] + k oint_S'_2dfracvecM'.hatnrdS'
endalign
The first term has the integrand continuous everywhere. The second term has the integrand discontinuous (infinite) at the point $r=0$.
Question:
Is it necessary to remove a small circle around $R=0$ in order to remove the singularity in the second term so that we can compute the integral? Then how shall we show that the second term of $nabla psi$ is convergent after removing a small circle around $R=0$ and then taking the limit as the radius of the small circle tends to zero?
multivariable-calculus polar-coordinates spherical-coordinates singularity potential-theory
asked Mar 31 at 14:31
JoeJoe
245214
$endgroup$
add a comment |
$begingroup$
Let $V'$ be the volume of dipole distribution and $S'$ be the boundary.
The potential of a dipole distribution at a point $P$ is:
$$psi=-k int_V'
dfracvecnabla'.vecM'rdV'
+k oint_S'dfracvecM'.hatnrdS'$$
If $Pin V'$ and $Pin S'$, the integrand is discontinuous (infinite) at the point $r=0$. If we use spherical and polar coordinates, the integrand is continuous everywhere:
beginalign
psi &=bbox[orange,5px]-k int_V' dfracvecnabla'.vecM'r r^2 sin theta dtheta dphi dr\
&bbox[pink,5px]+koint_S'_1 dfracvecM'.hatnr sqrtf_x^2+f_y^2+1 R dR dtheta'
bbox[yellow,5px]+ k oint_S'_2dfracvecM'.hatnrdS'\
&=bbox[orange,5px]-k int_V' vecnabla'.vecM' r sin theta dtheta dphi dr\
&bbox[pink,5px] + k oint_S'_1 vecM'.hatn sqrtf_x^2+f_y^2+1 dfracRsqrtR^2+f^2 dR dtheta'
bbox[yellow,5px] + k oint_S'_2dfracvecM'.hatnrdS'
endalign
The field of a dipole distribution at a point $P$ is:
$$nablapsi=-k int_V'
(vecnabla'.vecM') nabla left( dfrac1r right) dV'
+k oint_S' (vecM'.hatn) nabla left( dfrac1r right) dS'$$
If $Pin V'$ and $Pin S'$, the integrand is discontinuous (infinite) at the point $r=0$. If we use spherical and polar coordinates:
beginalign
nablapsi&=bbox[orange,5px]-k int_V'
(vecnabla'.vecM') left( dfrachatrr^2 right) r^2 sin theta dtheta dphi dr\
&bbox[pink,5px]+ k oint_S'_1 (vecM'.hatn) left( dfrachatrr right) sqrtf_x^2+f_y^2+1 dfracRsqrtR^2+f^2 dR dtheta'
bbox[yellow,5px] + k oint_S'_2dfracvecM'.hatnrdS'
endalign
The first term has the integrand continuous everywhere. The second term has the integrand discontinuous (infinite) at the point $r=0$.
Question:
Is it necessary to remove a small circle around $R=0$ in order to remove the singularity in the second term so that we can compute the integral? Then how shall we show that the second term of $nabla psi$ is convergent after removing a small circle around $R=0$ and then taking the limit as the radius of the small circle tends to zero?
multivariable-calculus polar-coordinates spherical-coordinates singularity potential-theory
asked Mar 31 at 14:31
JoeJoe
245214
$endgroup$
$begingroup$
Crossposted to physics.stackexchange.com/q/470123/2451
$endgroup$
– Qmechanic
Apr 2 at 21:40
$begingroup$
Can anybody answer???
$endgroup$
– Joe
Apr 4 at 7:16
$begingroup$
Please somebody answer....
$endgroup$
– Joe
Apr 9 at 11:43
add a comment |
$begingroup$
Let $V'$ be the volume of dipole distribution and $S'$ be the boundary.
The potential of a dipole distribution at a point $P$ is:
$$psi=-k int_V'
dfracvecnabla'.vecM'rdV'
+k oint_S'dfracvecM'.hatnrdS'$$
If $Pin V'$ and $Pin S'$, the integrand is discontinuous (infinite) at the point $r=0$. If we use spherical and polar coordinates, the integrand is continuous everywhere:
beginalign
psi &=bbox[orange,5px]-k int_V' dfracvecnabla'.vecM'r r^2 sin theta dtheta dphi dr\
&bbox[pink,5px]+koint_S'_1 dfracvecM'.hatnr sqrtf_x^2+f_y^2+1 R dR dtheta'
bbox[yellow,5px]+ k oint_S'_2dfracvecM'.hatnrdS'\
&=bbox[orange,5px]-k int_V' vecnabla'.vecM' r sin theta dtheta dphi dr\
&bbox[pink,5px] + k oint_S'_1 vecM'.hatn sqrtf_x^2+f_y^2+1 dfracRsqrtR^2+f^2 dR dtheta'
bbox[yellow,5px] + k oint_S'_2dfracvecM'.hatnrdS'
endalign
The field of a dipole distribution at a point $P$ is:
$$nablapsi=-k int_V'
(vecnabla'.vecM') nabla left( dfrac1r right) dV'
+k oint_S' (vecM'.hatn) nabla left( dfrac1r right) dS'$$
If $Pin V'$ and $Pin S'$, the integrand is discontinuous (infinite) at the point $r=0$. If we use spherical and polar coordinates:
beginalign
nablapsi&=bbox[orange,5px]-k int_V'
(vecnabla'.vecM') left( dfrachatrr^2 right) r^2 sin theta dtheta dphi dr\
&bbox[pink,5px]+ k oint_S'_1 (vecM'.hatn) left( dfrachatrr right) sqrtf_x^2+f_y^2+1 dfracRsqrtR^2+f^2 dR dtheta'
bbox[yellow,5px] + k oint_S'_2dfracvecM'.hatnrdS'
endalign
The first term has the integrand continuous everywhere. The second term has the integrand discontinuous (infinite) at the point $r=0$.
Question:
Is it necessary to remove a small circle around $R=0$ in order to remove the singularity in the second term so that we can compute the integral? Then how shall we show that the second term of $nabla psi$ is convergent after removing a small circle around $R=0$ and then taking the limit as the radius of the small circle tends to zero?
multivariable-calculus polar-coordinates spherical-coordinates singularity potential-theory
asked Mar 31 at 14:31
JoeJoe
245214
$endgroup$
Let $V'$ be the volume of dipole distribution and $S'$ be the boundary.
The potential of a dipole distribution at a point $P$ is:
$$psi=-k int_V'
dfracvecnabla'.vecM'rdV'
+k oint_S'dfracvecM'.hatnrdS'$$
If $Pin V'$ and $Pin S'$, the integrand is discontinuous (infinite) at the point $r=0$. If we use spherical and polar coordinates, the integrand is continuous everywhere:
beginalign
psi &=bbox[orange,5px]-k int_V' dfracvecnabla'.vecM'r r^2 sin theta dtheta dphi dr\
&bbox[pink,5px]+koint_S'_1 dfracvecM'.hatnr sqrtf_x^2+f_y^2+1 R dR dtheta'
bbox[yellow,5px]+ k oint_S'_2dfracvecM'.hatnrdS'\
&=bbox[orange,5px]-k int_V' vecnabla'.vecM' r sin theta dtheta dphi dr\
&bbox[pink,5px] + k oint_S'_1 vecM'.hatn sqrtf_x^2+f_y^2+1 dfracRsqrtR^2+f^2 dR dtheta'
bbox[yellow,5px] + k oint_S'_2dfracvecM'.hatnrdS'
endalign
The field of a dipole distribution at a point $P$ is:
$$nablapsi=-k int_V'
(vecnabla'.vecM') nabla left( dfrac1r right) dV'
+k oint_S' (vecM'.hatn) nabla left( dfrac1r right) dS'$$
If $Pin V'$ and $Pin S'$, the integrand is discontinuous (infinite) at the point $r=0$. If we use spherical and polar coordinates:
beginalign
nablapsi&=bbox[orange,5px]-k int_V'
(vecnabla'.vecM') left( dfrachatrr^2 right) r^2 sin theta dtheta dphi dr\
&bbox[pink,5px]+ k oint_S'_1 (vecM'.hatn) left( dfrachatrr right) sqrtf_x^2+f_y^2+1 dfracRsqrtR^2+f^2 dR dtheta'
bbox[yellow,5px] + k oint_S'_2dfracvecM'.hatnrdS'
endalign
The first term has the integrand continuous everywhere. The second term has the integrand discontinuous (infinite) at the point $r=0$.
Question:
Is it necessary to remove a small circle around $R=0$ in order to remove the singularity in the second term so that we can compute the integral? Then how shall we show that the second term of $nabla psi$ is convergent after removing a small circle around $R=0$ and then taking the limit as the radius of the small circle tends to zero?
multivariable-calculus polar-coordinates spherical-coordinates singularity potential-theory
multivariable-calculus polar-coordinates spherical-coordinates singularity potential-theory
asked Mar 31 at 14:31
JoeJoe
245214
asked Mar 31 at 14:31
JoeJoe
245214
edited Apr 8 at 16:26
Joe
asked Mar 31 at 14:31
JoeJoe
245214
asked Mar 31 at 14:31
JoeJoe
245214
asked Mar 31 at 14:31
JoeJoe
245214
245214
$begingroup$
Crossposted to physics.stackexchange.com/q/470123/2451
$endgroup$
– Qmechanic
Apr 2 at 21:40
$begingroup$
Can anybody answer???
$endgroup$
– Joe
Apr 4 at 7:16
$begingroup$
Please somebody answer....
$endgroup$
– Joe
Apr 9 at 11:43
add a comment |
$begingroup$
Crossposted to physics.stackexchange.com/q/470123/2451
$endgroup$
– Qmechanic
Apr 2 at 21:40
$begingroup$
Can anybody answer???
$endgroup$
– Joe
Apr 4 at 7:16
$begingroup$
Please somebody answer....
$endgroup$
– Joe
Apr 9 at 11:43
$begingroup$
Crossposted to physics.stackexchange.com/q/470123/2451
$endgroup$
– Qmechanic
Apr 2 at 21:40
$begingroup$
Crossposted to physics.stackexchange.com/q/470123/2451
$endgroup$
– Qmechanic
Apr 2 at 21:40
$begingroup$
Can anybody answer???
$endgroup$
– Joe
Apr 4 at 7:16
$begingroup$
Can anybody answer???
$endgroup$
– Joe
Apr 4 at 7:16
$begingroup$
Please somebody answer....
$endgroup$
– Joe
Apr 9 at 11:43
$begingroup$
Please somebody answer....
$endgroup$
– Joe
Apr 9 at 11:43
add a comment |
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$begingroup$
Crossposted to physics.stackexchange.com/q/470123/2451
$endgroup$
– Qmechanic
Apr 2 at 21:40
$begingroup$
Can anybody answer???
$endgroup$
– Joe
Apr 4 at 7:16
$begingroup$
Please somebody answer....
$endgroup$
– Joe
Apr 9 at 11:43