Delta distribution and divergence theorem

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Let's say we have some vector field $vec C$ such that $$operatornamedivvec C=-mu_0vec j=-mu_0 I,delta(x),delta(y),vec e_z.$$ where $mu_0$ and $I$ are constants.



I am interested in the value of $int_V d^3roperatornamedivvec C$ where $V$ is a cylindrical volume centered at the $z$-axis. On the one hand, I can simply plug it in and obtain $$int_V d^3roperatornamedivvec C=-mu_0 I,intdelta(x),dxintdelta(y),dyint dz=-mu_0 I z$$ but I should get the same result using polar coordinates. For the infinitesimal Volume I obtain $d^3 r=rho , drho , dvarphi , dz$. And (this is probably where my error is) for the current density we get $vec j=I,delta(rho),vec e_z$. Using this though yields the integral $$int_V d^3roperatornamedivvec C=-mu_0 Iint_0^rhoint_0^2piint _-z/2^z/2rho',delta(rho') ,drho',dvarphi ,dz$$ in the $z$-component. But if I'm not mistaken that would be $0$ since $int_0^rhorho',delta(rho'),drho'=0$. What do I have to do to get the same result in polar coordinates?




differential-geometry mathematical-physics electromagnetism




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  • The volume element is ok. The problem is the transformation for the delta. Here is the solution
    – Rafa Budría
    2 days ago











  • Yes, as I thought. How do I fix it?
    – Buh
    2 days ago






  • 1




    $delta(x)delta(y) = dfrac1rhodelta(rho)delta(varphi)$ as cylindrical coordinates restricted to $x$ and $y$ are the same as polar.
    – Rafa Budría
    2 days ago







  • 1




    Your first equation doesn't make sense; the left-hand side is a scalar and the middle and right-hand side are vectors.
    – joriki
    2 days ago










  • Hence I wrote "in the $z$-component" in the end.
    – Buh
    2 days ago














up vote
0
down vote

favorite












Let's say we have some vector field $vec C$ such that $$operatornamedivvec C=-mu_0vec j=-mu_0 I,delta(x),delta(y),vec e_z.$$ where $mu_0$ and $I$ are constants.



I am interested in the value of $int_V d^3roperatornamedivvec C$ where $V$ is a cylindrical volume centered at the $z$-axis. On the one hand, I can simply plug it in and obtain $$int_V d^3roperatornamedivvec C=-mu_0 I,intdelta(x),dxintdelta(y),dyint dz=-mu_0 I z$$ but I should get the same result using polar coordinates. For the infinitesimal Volume I obtain $d^3 r=rho , drho , dvarphi , dz$. And (this is probably where my error is) for the current density we get $vec j=I,delta(rho),vec e_z$. Using this though yields the integral $$int_V d^3roperatornamedivvec C=-mu_0 Iint_0^rhoint_0^2piint _-z/2^z/2rho',delta(rho') ,drho',dvarphi ,dz$$ in the $z$-component. But if I'm not mistaken that would be $0$ since $int_0^rhorho',delta(rho'),drho'=0$. What do I have to do to get the same result in polar coordinates?




differential-geometry mathematical-physics electromagnetism




share|cite|improve this question





















  • The volume element is ok. The problem is the transformation for the delta. Here is the solution
    – Rafa Budría
    2 days ago











  • Yes, as I thought. How do I fix it?
    – Buh
    2 days ago






  • 1




    $delta(x)delta(y) = dfrac1rhodelta(rho)delta(varphi)$ as cylindrical coordinates restricted to $x$ and $y$ are the same as polar.
    – Rafa Budría
    2 days ago







  • 1




    Your first equation doesn't make sense; the left-hand side is a scalar and the middle and right-hand side are vectors.
    – joriki
    2 days ago










  • Hence I wrote "in the $z$-component" in the end.
    – Buh
    2 days ago












up vote
0
down vote

favorite









up vote
0
down vote

favorite











Let's say we have some vector field $vec C$ such that $$operatornamedivvec C=-mu_0vec j=-mu_0 I,delta(x),delta(y),vec e_z.$$ where $mu_0$ and $I$ are constants.



I am interested in the value of $int_V d^3roperatornamedivvec C$ where $V$ is a cylindrical volume centered at the $z$-axis. On the one hand, I can simply plug it in and obtain $$int_V d^3roperatornamedivvec C=-mu_0 I,intdelta(x),dxintdelta(y),dyint dz=-mu_0 I z$$ but I should get the same result using polar coordinates. For the infinitesimal Volume I obtain $d^3 r=rho , drho , dvarphi , dz$. And (this is probably where my error is) for the current density we get $vec j=I,delta(rho),vec e_z$. Using this though yields the integral $$int_V d^3roperatornamedivvec C=-mu_0 Iint_0^rhoint_0^2piint _-z/2^z/2rho',delta(rho') ,drho',dvarphi ,dz$$ in the $z$-component. But if I'm not mistaken that would be $0$ since $int_0^rhorho',delta(rho'),drho'=0$. What do I have to do to get the same result in polar coordinates?




differential-geometry mathematical-physics electromagnetism




share|cite|improve this question













Let's say we have some vector field $vec C$ such that $$operatornamedivvec C=-mu_0vec j=-mu_0 I,delta(x),delta(y),vec e_z.$$ where $mu_0$ and $I$ are constants.



I am interested in the value of $int_V d^3roperatornamedivvec C$ where $V$ is a cylindrical volume centered at the $z$-axis. On the one hand, I can simply plug it in and obtain $$int_V d^3roperatornamedivvec C=-mu_0 I,intdelta(x),dxintdelta(y),dyint dz=-mu_0 I z$$ but I should get the same result using polar coordinates. For the infinitesimal Volume I obtain $d^3 r=rho , drho , dvarphi , dz$. And (this is probably where my error is) for the current density we get $vec j=I,delta(rho),vec e_z$. Using this though yields the integral $$int_V d^3roperatornamedivvec C=-mu_0 Iint_0^rhoint_0^2piint _-z/2^z/2rho',delta(rho') ,drho',dvarphi ,dz$$ in the $z$-component. But if I'm not mistaken that would be $0$ since $int_0^rhorho',delta(rho'),drho'=0$. What do I have to do to get the same result in polar coordinates?





differential-geometry mathematical-physics electromagnetism





share|cite|improve this question












share|cite|improve this question




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edited 2 days ago









md2perpe

5,5791821




5,5791821









asked 2 days ago









Buh

59427




59427











  • The volume element is ok. The problem is the transformation for the delta. Here is the solution
    – Rafa Budría
    2 days ago











  • Yes, as I thought. How do I fix it?
    – Buh
    2 days ago






  • 1




    $delta(x)delta(y) = dfrac1rhodelta(rho)delta(varphi)$ as cylindrical coordinates restricted to $x$ and $y$ are the same as polar.
    – Rafa Budría
    2 days ago







  • 1




    Your first equation doesn't make sense; the left-hand side is a scalar and the middle and right-hand side are vectors.
    – joriki
    2 days ago










  • Hence I wrote "in the $z$-component" in the end.
    – Buh
    2 days ago
















  • The volume element is ok. The problem is the transformation for the delta. Here is the solution
    – Rafa Budría
    2 days ago











  • Yes, as I thought. How do I fix it?
    – Buh
    2 days ago






  • 1




    $delta(x)delta(y) = dfrac1rhodelta(rho)delta(varphi)$ as cylindrical coordinates restricted to $x$ and $y$ are the same as polar.
    – Rafa Budría
    2 days ago







  • 1




    Your first equation doesn't make sense; the left-hand side is a scalar and the middle and right-hand side are vectors.
    – joriki
    2 days ago










  • Hence I wrote "in the $z$-component" in the end.
    – Buh
    2 days ago















The volume element is ok. The problem is the transformation for the delta. Here is the solution
– Rafa Budría
2 days ago





The volume element is ok. The problem is the transformation for the delta. Here is the solution
– Rafa Budría
2 days ago













Yes, as I thought. How do I fix it?
– Buh
2 days ago




Yes, as I thought. How do I fix it?
– Buh
2 days ago




1




1




$delta(x)delta(y) = dfrac1rhodelta(rho)delta(varphi)$ as cylindrical coordinates restricted to $x$ and $y$ are the same as polar.
– Rafa Budría
2 days ago





$delta(x)delta(y) = dfrac1rhodelta(rho)delta(varphi)$ as cylindrical coordinates restricted to $x$ and $y$ are the same as polar.
– Rafa Budría
2 days ago





1




1




Your first equation doesn't make sense; the left-hand side is a scalar and the middle and right-hand side are vectors.
– joriki
2 days ago




Your first equation doesn't make sense; the left-hand side is a scalar and the middle and right-hand side are vectors.
– joriki
2 days ago












Hence I wrote "in the $z$-component" in the end.
– Buh
2 days ago




Hence I wrote "in the $z$-component" in the end.
– Buh
2 days ago















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