Grad in polar coordinates












0












$begingroup$


As part of my lectures, it is noted that $nabla sin theta propto frac{1}{r^2}$ and $nabla phi propto frac{1}{r^2}$ where we are working in spherical polar coordinates with $theta$ as the polar angle and $phi$ as the azimuthal (so the physics convention)...



Now I can’t seem to see why this is true. I’ve tried $$nabla sin theta = frac{partial}{partial r} (sin theta) + frac{partial}{partial theta} (sin theta) + frac{partial}{partial phi} (sin theta)$$ but I can’t see how a $frac{1}{r^2}$ is going to come out of this.



I’ve also tried to work with grad in spherical polars but I still can’t seem to get the $frac{1}{r^2}$, likewise for $nabla phi$



Help would be appreciated....






vector-analysis polar-coordinates spherical-coordinates







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$endgroup$












  • $begingroup$
    Are you sure that your lecturer was talking about the gradient, and not the divergence, curl (with some included unit vector) or laplacian? For the gradient the magnitudes are proportional to $1/r$, for the Laplacian $1/r^2$, and for the others it depends on the unit vector, but should be $1/r$ also.
    $endgroup$
    – Eddy
    Jan 5 at 20:04
















0












$begingroup$


As part of my lectures, it is noted that $nabla sin theta propto frac{1}{r^2}$ and $nabla phi propto frac{1}{r^2}$ where we are working in spherical polar coordinates with $theta$ as the polar angle and $phi$ as the azimuthal (so the physics convention)...



Now I can’t seem to see why this is true. I’ve tried $$nabla sin theta = frac{partial}{partial r} (sin theta) + frac{partial}{partial theta} (sin theta) + frac{partial}{partial phi} (sin theta)$$ but I can’t see how a $frac{1}{r^2}$ is going to come out of this.



I’ve also tried to work with grad in spherical polars but I still can’t seem to get the $frac{1}{r^2}$, likewise for $nabla phi$



Help would be appreciated....






vector-analysis polar-coordinates spherical-coordinates







share|cite|improve this question









$endgroup$












  • $begingroup$
    Are you sure that your lecturer was talking about the gradient, and not the divergence, curl (with some included unit vector) or laplacian? For the gradient the magnitudes are proportional to $1/r$, for the Laplacian $1/r^2$, and for the others it depends on the unit vector, but should be $1/r$ also.
    $endgroup$
    – Eddy
    Jan 5 at 20:04














0












0








0





$begingroup$


As part of my lectures, it is noted that $nabla sin theta propto frac{1}{r^2}$ and $nabla phi propto frac{1}{r^2}$ where we are working in spherical polar coordinates with $theta$ as the polar angle and $phi$ as the azimuthal (so the physics convention)...



Now I can’t seem to see why this is true. I’ve tried $$nabla sin theta = frac{partial}{partial r} (sin theta) + frac{partial}{partial theta} (sin theta) + frac{partial}{partial phi} (sin theta)$$ but I can’t see how a $frac{1}{r^2}$ is going to come out of this.



I’ve also tried to work with grad in spherical polars but I still can’t seem to get the $frac{1}{r^2}$, likewise for $nabla phi$



Help would be appreciated....






vector-analysis polar-coordinates spherical-coordinates







share|cite|improve this question









$endgroup$




As part of my lectures, it is noted that $nabla sin theta propto frac{1}{r^2}$ and $nabla phi propto frac{1}{r^2}$ where we are working in spherical polar coordinates with $theta$ as the polar angle and $phi$ as the azimuthal (so the physics convention)...



Now I can’t seem to see why this is true. I’ve tried $$nabla sin theta = frac{partial}{partial r} (sin theta) + frac{partial}{partial theta} (sin theta) + frac{partial}{partial phi} (sin theta)$$ but I can’t see how a $frac{1}{r^2}$ is going to come out of this.



I’ve also tried to work with grad in spherical polars but I still can’t seem to get the $frac{1}{r^2}$, likewise for $nabla phi$



Help would be appreciated....






vector-analysis polar-coordinates spherical-coordinates




vector-analysis polar-coordinates spherical-coordinates






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 5 at 11:38









PhysicsMathsLovePhysicsMathsLove

1,208414




1,208414












  • $begingroup$
    Are you sure that your lecturer was talking about the gradient, and not the divergence, curl (with some included unit vector) or laplacian? For the gradient the magnitudes are proportional to $1/r$, for the Laplacian $1/r^2$, and for the others it depends on the unit vector, but should be $1/r$ also.
    $endgroup$
    – Eddy
    Jan 5 at 20:04


















  • $begingroup$
    Are you sure that your lecturer was talking about the gradient, and not the divergence, curl (with some included unit vector) or laplacian? For the gradient the magnitudes are proportional to $1/r$, for the Laplacian $1/r^2$, and for the others it depends on the unit vector, but should be $1/r$ also.
    $endgroup$
    – Eddy
    Jan 5 at 20:04
















$begingroup$
Are you sure that your lecturer was talking about the gradient, and not the divergence, curl (with some included unit vector) or laplacian? For the gradient the magnitudes are proportional to $1/r$, for the Laplacian $1/r^2$, and for the others it depends on the unit vector, but should be $1/r$ also.
$endgroup$
– Eddy
Jan 5 at 20:04




$begingroup$
Are you sure that your lecturer was talking about the gradient, and not the divergence, curl (with some included unit vector) or laplacian? For the gradient the magnitudes are proportional to $1/r$, for the Laplacian $1/r^2$, and for the others it depends on the unit vector, but should be $1/r$ also.
$endgroup$
– Eddy
Jan 5 at 20:04










2 Answers
2






active

oldest

votes


















0












$begingroup$

The $mathrm{grad}$ operator is of the form $partial_xmathbf{i}+partial_ymathbf{j}+partial_zmathbf{k}$ in Catesian coordinates. In other coordinate systems, the operator has different forms. See this wikipedia article for a list of them.



You can easily see the reason for this when you write out the conversion formula from Cartesian coordinates to polar/spherical coordinates, and differentiate it. You get extra factors on the polar coordinates side, because the relationship is no longer linear. To systematically address this problem, we can introduce connections. Furthermore, this can be generalized to curved spaces.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    The thing is that



    $$
    nabla f = frac{partial f}{partial u_1}hat{bf e}_1 + frac{partial f}{partial u_2}hat{bf e}_2 + frac{partial f}{partial u_3}hat{bf e}_3
    $$



    only works when $u$ are cartesian coordinates. In spherical coordinates is not as trivial,



    $$
    nabla f = frac{partial f}{partial r}hat{bf e}_r + frac{1}{r}frac{partial f}{partial theta} hat{bf e}_theta + frac{1}{rsintheta}frac{partial f}{partial phi}hat{bf e}_phi
    $$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Ok but I still don’t see how I’ll get $frac{1}{r^2}$? All the derivatives apart from the $theta$ derivative will evaluate to 0, and I’ll just have $cos theta hat{e_{theta}}$
      $endgroup$
      – PhysicsMathsLove
      Jan 5 at 11:57












    • $begingroup$
      @PhysicsMathsLove That may be a problem with your notes. You see that dimensionally it does not make sense $nabla$ has units of $L^{-1}$ and $sintheta$ is dimensionless, so $nabla sintheta$ must have dimensions $L^{-1}$, so it should be something like $1/r$
      $endgroup$
      – caverac
      Jan 5 at 12:21











    Your Answer





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    2 Answers
    2






    active

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    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    0












    $begingroup$

    The $mathrm{grad}$ operator is of the form $partial_xmathbf{i}+partial_ymathbf{j}+partial_zmathbf{k}$ in Catesian coordinates. In other coordinate systems, the operator has different forms. See this wikipedia article for a list of them.



    You can easily see the reason for this when you write out the conversion formula from Cartesian coordinates to polar/spherical coordinates, and differentiate it. You get extra factors on the polar coordinates side, because the relationship is no longer linear. To systematically address this problem, we can introduce connections. Furthermore, this can be generalized to curved spaces.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      The $mathrm{grad}$ operator is of the form $partial_xmathbf{i}+partial_ymathbf{j}+partial_zmathbf{k}$ in Catesian coordinates. In other coordinate systems, the operator has different forms. See this wikipedia article for a list of them.



      You can easily see the reason for this when you write out the conversion formula from Cartesian coordinates to polar/spherical coordinates, and differentiate it. You get extra factors on the polar coordinates side, because the relationship is no longer linear. To systematically address this problem, we can introduce connections. Furthermore, this can be generalized to curved spaces.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        The $mathrm{grad}$ operator is of the form $partial_xmathbf{i}+partial_ymathbf{j}+partial_zmathbf{k}$ in Catesian coordinates. In other coordinate systems, the operator has different forms. See this wikipedia article for a list of them.



        You can easily see the reason for this when you write out the conversion formula from Cartesian coordinates to polar/spherical coordinates, and differentiate it. You get extra factors on the polar coordinates side, because the relationship is no longer linear. To systematically address this problem, we can introduce connections. Furthermore, this can be generalized to curved spaces.






        share|cite|improve this answer









        $endgroup$



        The $mathrm{grad}$ operator is of the form $partial_xmathbf{i}+partial_ymathbf{j}+partial_zmathbf{k}$ in Catesian coordinates. In other coordinate systems, the operator has different forms. See this wikipedia article for a list of them.



        You can easily see the reason for this when you write out the conversion formula from Cartesian coordinates to polar/spherical coordinates, and differentiate it. You get extra factors on the polar coordinates side, because the relationship is no longer linear. To systematically address this problem, we can introduce connections. Furthermore, this can be generalized to curved spaces.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 5 at 11:49









        TreborTrebor

        85415




        85415























            0












            $begingroup$

            The thing is that



            $$
            nabla f = frac{partial f}{partial u_1}hat{bf e}_1 + frac{partial f}{partial u_2}hat{bf e}_2 + frac{partial f}{partial u_3}hat{bf e}_3
            $$



            only works when $u$ are cartesian coordinates. In spherical coordinates is not as trivial,



            $$
            nabla f = frac{partial f}{partial r}hat{bf e}_r + frac{1}{r}frac{partial f}{partial theta} hat{bf e}_theta + frac{1}{rsintheta}frac{partial f}{partial phi}hat{bf e}_phi
            $$






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Ok but I still don’t see how I’ll get $frac{1}{r^2}$? All the derivatives apart from the $theta$ derivative will evaluate to 0, and I’ll just have $cos theta hat{e_{theta}}$
              $endgroup$
              – PhysicsMathsLove
              Jan 5 at 11:57












            • $begingroup$
              @PhysicsMathsLove That may be a problem with your notes. You see that dimensionally it does not make sense $nabla$ has units of $L^{-1}$ and $sintheta$ is dimensionless, so $nabla sintheta$ must have dimensions $L^{-1}$, so it should be something like $1/r$
              $endgroup$
              – caverac
              Jan 5 at 12:21
















            0












            $begingroup$

            The thing is that



            $$
            nabla f = frac{partial f}{partial u_1}hat{bf e}_1 + frac{partial f}{partial u_2}hat{bf e}_2 + frac{partial f}{partial u_3}hat{bf e}_3
            $$



            only works when $u$ are cartesian coordinates. In spherical coordinates is not as trivial,



            $$
            nabla f = frac{partial f}{partial r}hat{bf e}_r + frac{1}{r}frac{partial f}{partial theta} hat{bf e}_theta + frac{1}{rsintheta}frac{partial f}{partial phi}hat{bf e}_phi
            $$






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Ok but I still don’t see how I’ll get $frac{1}{r^2}$? All the derivatives apart from the $theta$ derivative will evaluate to 0, and I’ll just have $cos theta hat{e_{theta}}$
              $endgroup$
              – PhysicsMathsLove
              Jan 5 at 11:57












            • $begingroup$
              @PhysicsMathsLove That may be a problem with your notes. You see that dimensionally it does not make sense $nabla$ has units of $L^{-1}$ and $sintheta$ is dimensionless, so $nabla sintheta$ must have dimensions $L^{-1}$, so it should be something like $1/r$
              $endgroup$
              – caverac
              Jan 5 at 12:21














            0












            0








            0





            $begingroup$

            The thing is that



            $$
            nabla f = frac{partial f}{partial u_1}hat{bf e}_1 + frac{partial f}{partial u_2}hat{bf e}_2 + frac{partial f}{partial u_3}hat{bf e}_3
            $$



            only works when $u$ are cartesian coordinates. In spherical coordinates is not as trivial,



            $$
            nabla f = frac{partial f}{partial r}hat{bf e}_r + frac{1}{r}frac{partial f}{partial theta} hat{bf e}_theta + frac{1}{rsintheta}frac{partial f}{partial phi}hat{bf e}_phi
            $$






            share|cite|improve this answer











            $endgroup$



            The thing is that



            $$
            nabla f = frac{partial f}{partial u_1}hat{bf e}_1 + frac{partial f}{partial u_2}hat{bf e}_2 + frac{partial f}{partial u_3}hat{bf e}_3
            $$



            only works when $u$ are cartesian coordinates. In spherical coordinates is not as trivial,



            $$
            nabla f = frac{partial f}{partial r}hat{bf e}_r + frac{1}{r}frac{partial f}{partial theta} hat{bf e}_theta + frac{1}{rsintheta}frac{partial f}{partial phi}hat{bf e}_phi
            $$







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Jan 5 at 12:26

























            answered Jan 5 at 11:49









            caveraccaverac

            14.8k31130




            14.8k31130












            • $begingroup$
              Ok but I still don’t see how I’ll get $frac{1}{r^2}$? All the derivatives apart from the $theta$ derivative will evaluate to 0, and I’ll just have $cos theta hat{e_{theta}}$
              $endgroup$
              – PhysicsMathsLove
              Jan 5 at 11:57












            • $begingroup$
              @PhysicsMathsLove That may be a problem with your notes. You see that dimensionally it does not make sense $nabla$ has units of $L^{-1}$ and $sintheta$ is dimensionless, so $nabla sintheta$ must have dimensions $L^{-1}$, so it should be something like $1/r$
              $endgroup$
              – caverac
              Jan 5 at 12:21


















            • $begingroup$
              Ok but I still don’t see how I’ll get $frac{1}{r^2}$? All the derivatives apart from the $theta$ derivative will evaluate to 0, and I’ll just have $cos theta hat{e_{theta}}$
              $endgroup$
              – PhysicsMathsLove
              Jan 5 at 11:57












            • $begingroup$
              @PhysicsMathsLove That may be a problem with your notes. You see that dimensionally it does not make sense $nabla$ has units of $L^{-1}$ and $sintheta$ is dimensionless, so $nabla sintheta$ must have dimensions $L^{-1}$, so it should be something like $1/r$
              $endgroup$
              – caverac
              Jan 5 at 12:21
















            $begingroup$
            Ok but I still don’t see how I’ll get $frac{1}{r^2}$? All the derivatives apart from the $theta$ derivative will evaluate to 0, and I’ll just have $cos theta hat{e_{theta}}$
            $endgroup$
            – PhysicsMathsLove
            Jan 5 at 11:57






            $begingroup$
            Ok but I still don’t see how I’ll get $frac{1}{r^2}$? All the derivatives apart from the $theta$ derivative will evaluate to 0, and I’ll just have $cos theta hat{e_{theta}}$
            $endgroup$
            – PhysicsMathsLove
            Jan 5 at 11:57














            $begingroup$
            @PhysicsMathsLove That may be a problem with your notes. You see that dimensionally it does not make sense $nabla$ has units of $L^{-1}$ and $sintheta$ is dimensionless, so $nabla sintheta$ must have dimensions $L^{-1}$, so it should be something like $1/r$
            $endgroup$
            – caverac
            Jan 5 at 12:21




            $begingroup$
            @PhysicsMathsLove That may be a problem with your notes. You see that dimensionally it does not make sense $nabla$ has units of $L^{-1}$ and $sintheta$ is dimensionless, so $nabla sintheta$ must have dimensions $L^{-1}$, so it should be something like $1/r$
            $endgroup$
            – caverac
            Jan 5 at 12:21


















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