How can we show that $I_{yy}=I_{zz}=I_{xx}$ and $I_{xy}=0$ using symmetry arguments?












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Consider two integrals of the form $$I_{xx}=int x^2 f(r)d^3r,~~I^prime=int xy ~f(r)d^3r$$ where $f(r)$ is a function of $r=|vec{r}|$ only and has no dependence on $theta,phi$ in pherical polar coordinates. I have two questions. Can we show that $I_{yy}=I_{zz}=I_{xx}$ and $I_{xy}=I_{yz}=I_{zx}=0$ using symmetry arguments?






definite-integrals coordinate-systems spherical-coordinates symmetry multiple-integral







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  • $begingroup$
    Are you thrice integrating over $r$? Your question would be better answered if you clarify what exactly $I_{yy},I_{zz},I_{xy},I_{yz},I_{zx}$ are......
    $endgroup$
    – Mostafa Ayaz
    Jan 6 at 12:33


















0












$begingroup$


Consider two integrals of the form $$I_{xx}=int x^2 f(r)d^3r,~~I^prime=int xy ~f(r)d^3r$$ where $f(r)$ is a function of $r=|vec{r}|$ only and has no dependence on $theta,phi$ in pherical polar coordinates. I have two questions. Can we show that $I_{yy}=I_{zz}=I_{xx}$ and $I_{xy}=I_{yz}=I_{zx}=0$ using symmetry arguments?






definite-integrals coordinate-systems spherical-coordinates symmetry multiple-integral







share|cite|improve this question









$endgroup$












  • $begingroup$
    Are you thrice integrating over $r$? Your question would be better answered if you clarify what exactly $I_{yy},I_{zz},I_{xy},I_{yz},I_{zx}$ are......
    $endgroup$
    – Mostafa Ayaz
    Jan 6 at 12:33
















0












0








0





$begingroup$


Consider two integrals of the form $$I_{xx}=int x^2 f(r)d^3r,~~I^prime=int xy ~f(r)d^3r$$ where $f(r)$ is a function of $r=|vec{r}|$ only and has no dependence on $theta,phi$ in pherical polar coordinates. I have two questions. Can we show that $I_{yy}=I_{zz}=I_{xx}$ and $I_{xy}=I_{yz}=I_{zx}=0$ using symmetry arguments?






definite-integrals coordinate-systems spherical-coordinates symmetry multiple-integral







share|cite|improve this question









$endgroup$




Consider two integrals of the form $$I_{xx}=int x^2 f(r)d^3r,~~I^prime=int xy ~f(r)d^3r$$ where $f(r)$ is a function of $r=|vec{r}|$ only and has no dependence on $theta,phi$ in pherical polar coordinates. I have two questions. Can we show that $I_{yy}=I_{zz}=I_{xx}$ and $I_{xy}=I_{yz}=I_{zx}=0$ using symmetry arguments?






definite-integrals coordinate-systems spherical-coordinates symmetry multiple-integral




definite-integrals coordinate-systems spherical-coordinates symmetry multiple-integral






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asked Dec 15 '18 at 13:48









mithusengupta123mithusengupta123

1069




1069












  • $begingroup$
    Are you thrice integrating over $r$? Your question would be better answered if you clarify what exactly $I_{yy},I_{zz},I_{xy},I_{yz},I_{zx}$ are......
    $endgroup$
    – Mostafa Ayaz
    Jan 6 at 12:33




















  • $begingroup$
    Are you thrice integrating over $r$? Your question would be better answered if you clarify what exactly $I_{yy},I_{zz},I_{xy},I_{yz},I_{zx}$ are......
    $endgroup$
    – Mostafa Ayaz
    Jan 6 at 12:33


















$begingroup$
Are you thrice integrating over $r$? Your question would be better answered if you clarify what exactly $I_{yy},I_{zz},I_{xy},I_{yz},I_{zx}$ are......
$endgroup$
– Mostafa Ayaz
Jan 6 at 12:33






$begingroup$
Are you thrice integrating over $r$? Your question would be better answered if you clarify what exactly $I_{yy},I_{zz},I_{xy},I_{yz},I_{zx}$ are......
$endgroup$
– Mostafa Ayaz
Jan 6 at 12:33












2 Answers
2






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$begingroup$

Since $f = f(|{bf r}|)$ it does not depend on the angles $(theta, phi)$, so at a given radius $f$ is a constant over the sphere defined by such radius. Now consider the term $x$, over the same sphere, for each $x$ there will be a $-x$, so for each $x f$ there will be a a $-x f$, the result after adding them up all together is zero. You can extend the argument to show that



$$
I_{ij} = int x_i x_j f(r) {rm d}^3{bf r} = 0 ~~~mbox{for}~~inot= j
$$



And from this is also easy to see why $I_{ii} not = 0$






share|cite|improve this answer









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    $begingroup$

    $f(r)$ has two relevant symmetries: rotational and reflection. From the rotational symmetry we can conclude that $int x^n f(r) d^3r=int y^n f(r) d^3r=int z^n f(r) d^3r$. explicitly $f(sqrt{x^2+y^2+z^2})=f(sqrt{y^2+x^2+z^2})$ etc (in fact we didn't use the rotational symmetry but a permutation symmetry).



    From the reflection symmetry we conclude that $int xy f(r) d^3r=int (-x)y f(r) d^3r=-int xy f(r) d^3r$ so that $I_{xy}=-I_{xy}=0$, since $f(sqrt{x^2+y^2+z^2})=f(sqrt{(-x)^2+y^2+z^2}) $ etc.






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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      1












      $begingroup$

      Since $f = f(|{bf r}|)$ it does not depend on the angles $(theta, phi)$, so at a given radius $f$ is a constant over the sphere defined by such radius. Now consider the term $x$, over the same sphere, for each $x$ there will be a $-x$, so for each $x f$ there will be a a $-x f$, the result after adding them up all together is zero. You can extend the argument to show that



      $$
      I_{ij} = int x_i x_j f(r) {rm d}^3{bf r} = 0 ~~~mbox{for}~~inot= j
      $$



      And from this is also easy to see why $I_{ii} not = 0$






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$

        Since $f = f(|{bf r}|)$ it does not depend on the angles $(theta, phi)$, so at a given radius $f$ is a constant over the sphere defined by such radius. Now consider the term $x$, over the same sphere, for each $x$ there will be a $-x$, so for each $x f$ there will be a a $-x f$, the result after adding them up all together is zero. You can extend the argument to show that



        $$
        I_{ij} = int x_i x_j f(r) {rm d}^3{bf r} = 0 ~~~mbox{for}~~inot= j
        $$



        And from this is also easy to see why $I_{ii} not = 0$






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          Since $f = f(|{bf r}|)$ it does not depend on the angles $(theta, phi)$, so at a given radius $f$ is a constant over the sphere defined by such radius. Now consider the term $x$, over the same sphere, for each $x$ there will be a $-x$, so for each $x f$ there will be a a $-x f$, the result after adding them up all together is zero. You can extend the argument to show that



          $$
          I_{ij} = int x_i x_j f(r) {rm d}^3{bf r} = 0 ~~~mbox{for}~~inot= j
          $$



          And from this is also easy to see why $I_{ii} not = 0$






          share|cite|improve this answer









          $endgroup$



          Since $f = f(|{bf r}|)$ it does not depend on the angles $(theta, phi)$, so at a given radius $f$ is a constant over the sphere defined by such radius. Now consider the term $x$, over the same sphere, for each $x$ there will be a $-x$, so for each $x f$ there will be a a $-x f$, the result after adding them up all together is zero. You can extend the argument to show that



          $$
          I_{ij} = int x_i x_j f(r) {rm d}^3{bf r} = 0 ~~~mbox{for}~~inot= j
          $$



          And from this is also easy to see why $I_{ii} not = 0$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 15 '18 at 14:06









          caveraccaverac

          14.6k31130




          14.6k31130























              1












              $begingroup$

              $f(r)$ has two relevant symmetries: rotational and reflection. From the rotational symmetry we can conclude that $int x^n f(r) d^3r=int y^n f(r) d^3r=int z^n f(r) d^3r$. explicitly $f(sqrt{x^2+y^2+z^2})=f(sqrt{y^2+x^2+z^2})$ etc (in fact we didn't use the rotational symmetry but a permutation symmetry).



              From the reflection symmetry we conclude that $int xy f(r) d^3r=int (-x)y f(r) d^3r=-int xy f(r) d^3r$ so that $I_{xy}=-I_{xy}=0$, since $f(sqrt{x^2+y^2+z^2})=f(sqrt{(-x)^2+y^2+z^2}) $ etc.






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                $f(r)$ has two relevant symmetries: rotational and reflection. From the rotational symmetry we can conclude that $int x^n f(r) d^3r=int y^n f(r) d^3r=int z^n f(r) d^3r$. explicitly $f(sqrt{x^2+y^2+z^2})=f(sqrt{y^2+x^2+z^2})$ etc (in fact we didn't use the rotational symmetry but a permutation symmetry).



                From the reflection symmetry we conclude that $int xy f(r) d^3r=int (-x)y f(r) d^3r=-int xy f(r) d^3r$ so that $I_{xy}=-I_{xy}=0$, since $f(sqrt{x^2+y^2+z^2})=f(sqrt{(-x)^2+y^2+z^2}) $ etc.






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  $f(r)$ has two relevant symmetries: rotational and reflection. From the rotational symmetry we can conclude that $int x^n f(r) d^3r=int y^n f(r) d^3r=int z^n f(r) d^3r$. explicitly $f(sqrt{x^2+y^2+z^2})=f(sqrt{y^2+x^2+z^2})$ etc (in fact we didn't use the rotational symmetry but a permutation symmetry).



                  From the reflection symmetry we conclude that $int xy f(r) d^3r=int (-x)y f(r) d^3r=-int xy f(r) d^3r$ so that $I_{xy}=-I_{xy}=0$, since $f(sqrt{x^2+y^2+z^2})=f(sqrt{(-x)^2+y^2+z^2}) $ etc.






                  share|cite|improve this answer









                  $endgroup$



                  $f(r)$ has two relevant symmetries: rotational and reflection. From the rotational symmetry we can conclude that $int x^n f(r) d^3r=int y^n f(r) d^3r=int z^n f(r) d^3r$. explicitly $f(sqrt{x^2+y^2+z^2})=f(sqrt{y^2+x^2+z^2})$ etc (in fact we didn't use the rotational symmetry but a permutation symmetry).



                  From the reflection symmetry we conclude that $int xy f(r) d^3r=int (-x)y f(r) d^3r=-int xy f(r) d^3r$ so that $I_{xy}=-I_{xy}=0$, since $f(sqrt{x^2+y^2+z^2})=f(sqrt{(-x)^2+y^2+z^2}) $ etc.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 31 '18 at 6:27









                  user617446user617446

                  4443




                  4443






























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