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Error in TransformedField


Using spherical derivatives with NDSolveIs there an easy way to transform unit vectors from spherical to Cartesian coordinates?Why does a transform from polar coordinates using `TransformedField` not agree with by-hand calculation?ListDensityPlot in Polar coordinates with a higher efficiencyChanging from Cartesian to polar coordinates in partial differential equationA little confused about coordinate conversionInverse substitution polar-cartesianIntegration using polar coordinatesPlotting a function $psi(rho,theta,phi)$ in spherical coordinatesTransform a field with derivatives













6












$begingroup$


I am using TransformedField to convert a system of ODEs from Cartesian to polar coordinates:



TransformedField[
"Cartesian" -> "Polar",
{μ x1 - x2 - σ x1 (x1^2 + x2^2), x1 + μ x2 - σ x2 (x1^2 + x2^2)},
{x1, x2} -> {r, θ}
] // Simplify


and I get the result



{r μ - r^3 σ, r}


but I am pretty sure that the right answer should be



{r μ - r^3 σ, 1}


Where is the error?










share|improve this question











$endgroup$

















    6












    $begingroup$


    I am using TransformedField to convert a system of ODEs from Cartesian to polar coordinates:



    TransformedField[
    "Cartesian" -> "Polar",
    {μ x1 - x2 - σ x1 (x1^2 + x2^2), x1 + μ x2 - σ x2 (x1^2 + x2^2)},
    {x1, x2} -> {r, θ}
    ] // Simplify


    and I get the result



    {r μ - r^3 σ, r}


    but I am pretty sure that the right answer should be



    {r μ - r^3 σ, 1}


    Where is the error?










    share|improve this question











    $endgroup$















      6












      6








      6


      1



      $begingroup$


      I am using TransformedField to convert a system of ODEs from Cartesian to polar coordinates:



      TransformedField[
      "Cartesian" -> "Polar",
      {μ x1 - x2 - σ x1 (x1^2 + x2^2), x1 + μ x2 - σ x2 (x1^2 + x2^2)},
      {x1, x2} -> {r, θ}
      ] // Simplify


      and I get the result



      {r μ - r^3 σ, r}


      but I am pretty sure that the right answer should be



      {r μ - r^3 σ, 1}


      Where is the error?










      share|improve this question











      $endgroup$




      I am using TransformedField to convert a system of ODEs from Cartesian to polar coordinates:



      TransformedField[
      "Cartesian" -> "Polar",
      {μ x1 - x2 - σ x1 (x1^2 + x2^2), x1 + μ x2 - σ x2 (x1^2 + x2^2)},
      {x1, x2} -> {r, θ}
      ] // Simplify


      and I get the result



      {r μ - r^3 σ, r}


      but I am pretty sure that the right answer should be



      {r μ - r^3 σ, 1}


      Where is the error?







      coordinate-transformation






      share|improve this question















      share|improve this question













      share|improve this question




      share|improve this question








      edited 6 hours ago









      MarcoB

      36.9k556113




      36.9k556113










      asked 6 hours ago









      rparpa

      856




      856






















          2 Answers
          2






          active

          oldest

          votes


















          5












          $begingroup$

          We can test the built-in TransformedField by defining our own functions.




          From $x',y'$ to $r',theta'$, we derive:
          $$
          r' = left(sqrt{x^2 +y^2} right)'
          = frac{(x^2 +y^2)'}{2
          sqrt{x^2 +y^2}}=frac{xx' +yy'}{r}
          $$

          and
          $$
          theta' = left(arctan frac{y}{x} right)'
          = frac{(y/x)'}{1+(y/x)^2} = frac{y' x -x' y}{r^2}.
          $$




          First, we define



          rdot[x1_, x2_] := (x1 (μ x1 - x2 - σ x1 (x1^2 + x2^2)) + x2 (x1 + μ x2 - σ x2 (x1^2 + x2^2)))/r


          We now make the substitution and simplify



          rdot[r Cos[t], r Sin[t]] // FullSimplify


          This yields (matches Mathematica)



          $$r' = mu r-r^3 sigma$$



          We now do the same for the other



          thetadot[x1_,x2_]:=(x1 (x1+μ x2-σ x2 (x1^2+x2^2)) - x2(μ x1-x2-σ x1 (x1^2+x2^2)))/r^2


          We now make the substitution and simplify



          thetadot[r Cos[t], r Sin[t]] // FullSimplify


          This yields (does not match Mathematica - it is as though they are forgetting to divide by $r$)



          $$theta'= 1$$



          I have asked this question before on this site in two different ways and have never gotten an answer that resolves the matter, but that could just be my denseness!






          share|improve this answer











          $endgroup$









          • 1




            $begingroup$
            Have you reported it to the Wolfram tech support?
            $endgroup$
            – Alexey Popkov
            2 hours ago










          • $begingroup$
            @AlexeyPopkov: I have not. I have had many issues with it when transforming between different methods. These days, I don't trust it and just create my own transformation rules to do it.
            $endgroup$
            – Moo
            2 hours ago










          • $begingroup$
            It is worth to write them about it in order to get it finally fixed. You can even write a short letter to support@wolfram.com with a link to this post.
            $endgroup$
            – Alexey Popkov
            2 hours ago












          • $begingroup$
            @AlexeyPopkov: I sent them an email per your suggestion.
            $endgroup$
            – Moo
            2 hours ago



















          0












          $begingroup$

          Mathematica's answer is correct and consistent with your expectations, but you are not accounting for the basis of the vector field.



          TransformedField transforms a vector field between two coordinate systems and bases. In this case, it is converting from $f(x,y)hat x+g(x,y)hat y$ to the same geometrical vector field expressed as $u(r,theta)hat r + v(r,theta) hat theta$. Mathematica's answer can therefore be interpreted as saying
          $$left(μ x_1 - x_2 - σ x_1 (x_1^2 + x_2^2)right)hat x + left( x_1 + μ x_2 - σ x_2 (x_1^2 + x_2^2)right)hat y = left(r μ - r^3 σright)hat r + r hattheta$$



          Notice that the expressions $r'$ and $theta'$ don't appear anyhwere. Those are dynamical quantities, not geometrical ones (unless working in the jet bundle, but let's not go there). Also notice the hats! As stated in the documentation, TransformedField assumes inputs are in an orthonormal basis, and returns outputs in the same basis. That will be important for later on.



          Now, you are dealing with a differential equation, and based on your expected answer I'll assume what you have is a first-order system and you are transforming the associated vector field (AKA the "right-hand side"). Finding solutions means find the integral curves of the vector field. This gives as a nice relationship between the geometric variables the dynamical ones, except that this relationship is of necessity expressed in the so called coordinate basis, written $(r',theta') = a frac{partial}{partial r} + b frac{partial}{partial theta}$. So to get the answer expressed in your desired basis, we need the relationship between the coordinate and orthonormal basis vectors. As is covered in books on vector calculus (and elsewhere), the relationsip is $hat r = frac{partial}{partial r}$ and $hat theta = frac{1}{r}frac{partial}{partial theta}$. Substituting this into the answer Mathematica gave above, we get
          $$left(r μ - r^3 σright)hat r + r hattheta = left(r μ - r^3 σright) frac{partial}{partial r} + (1) frac{partial}{partial theta},$$
          which is the answer you expected.






          share|improve this answer









          $endgroup$













            Your Answer





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            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            5












            $begingroup$

            We can test the built-in TransformedField by defining our own functions.




            From $x',y'$ to $r',theta'$, we derive:
            $$
            r' = left(sqrt{x^2 +y^2} right)'
            = frac{(x^2 +y^2)'}{2
            sqrt{x^2 +y^2}}=frac{xx' +yy'}{r}
            $$

            and
            $$
            theta' = left(arctan frac{y}{x} right)'
            = frac{(y/x)'}{1+(y/x)^2} = frac{y' x -x' y}{r^2}.
            $$




            First, we define



            rdot[x1_, x2_] := (x1 (μ x1 - x2 - σ x1 (x1^2 + x2^2)) + x2 (x1 + μ x2 - σ x2 (x1^2 + x2^2)))/r


            We now make the substitution and simplify



            rdot[r Cos[t], r Sin[t]] // FullSimplify


            This yields (matches Mathematica)



            $$r' = mu r-r^3 sigma$$



            We now do the same for the other



            thetadot[x1_,x2_]:=(x1 (x1+μ x2-σ x2 (x1^2+x2^2)) - x2(μ x1-x2-σ x1 (x1^2+x2^2)))/r^2


            We now make the substitution and simplify



            thetadot[r Cos[t], r Sin[t]] // FullSimplify


            This yields (does not match Mathematica - it is as though they are forgetting to divide by $r$)



            $$theta'= 1$$



            I have asked this question before on this site in two different ways and have never gotten an answer that resolves the matter, but that could just be my denseness!






            share|improve this answer











            $endgroup$









            • 1




              $begingroup$
              Have you reported it to the Wolfram tech support?
              $endgroup$
              – Alexey Popkov
              2 hours ago










            • $begingroup$
              @AlexeyPopkov: I have not. I have had many issues with it when transforming between different methods. These days, I don't trust it and just create my own transformation rules to do it.
              $endgroup$
              – Moo
              2 hours ago










            • $begingroup$
              It is worth to write them about it in order to get it finally fixed. You can even write a short letter to support@wolfram.com with a link to this post.
              $endgroup$
              – Alexey Popkov
              2 hours ago












            • $begingroup$
              @AlexeyPopkov: I sent them an email per your suggestion.
              $endgroup$
              – Moo
              2 hours ago
















            5












            $begingroup$

            We can test the built-in TransformedField by defining our own functions.




            From $x',y'$ to $r',theta'$, we derive:
            $$
            r' = left(sqrt{x^2 +y^2} right)'
            = frac{(x^2 +y^2)'}{2
            sqrt{x^2 +y^2}}=frac{xx' +yy'}{r}
            $$

            and
            $$
            theta' = left(arctan frac{y}{x} right)'
            = frac{(y/x)'}{1+(y/x)^2} = frac{y' x -x' y}{r^2}.
            $$




            First, we define



            rdot[x1_, x2_] := (x1 (μ x1 - x2 - σ x1 (x1^2 + x2^2)) + x2 (x1 + μ x2 - σ x2 (x1^2 + x2^2)))/r


            We now make the substitution and simplify



            rdot[r Cos[t], r Sin[t]] // FullSimplify


            This yields (matches Mathematica)



            $$r' = mu r-r^3 sigma$$



            We now do the same for the other



            thetadot[x1_,x2_]:=(x1 (x1+μ x2-σ x2 (x1^2+x2^2)) - x2(μ x1-x2-σ x1 (x1^2+x2^2)))/r^2


            We now make the substitution and simplify



            thetadot[r Cos[t], r Sin[t]] // FullSimplify


            This yields (does not match Mathematica - it is as though they are forgetting to divide by $r$)



            $$theta'= 1$$



            I have asked this question before on this site in two different ways and have never gotten an answer that resolves the matter, but that could just be my denseness!






            share|improve this answer











            $endgroup$









            • 1




              $begingroup$
              Have you reported it to the Wolfram tech support?
              $endgroup$
              – Alexey Popkov
              2 hours ago










            • $begingroup$
              @AlexeyPopkov: I have not. I have had many issues with it when transforming between different methods. These days, I don't trust it and just create my own transformation rules to do it.
              $endgroup$
              – Moo
              2 hours ago










            • $begingroup$
              It is worth to write them about it in order to get it finally fixed. You can even write a short letter to support@wolfram.com with a link to this post.
              $endgroup$
              – Alexey Popkov
              2 hours ago












            • $begingroup$
              @AlexeyPopkov: I sent them an email per your suggestion.
              $endgroup$
              – Moo
              2 hours ago














            5












            5








            5





            $begingroup$

            We can test the built-in TransformedField by defining our own functions.




            From $x',y'$ to $r',theta'$, we derive:
            $$
            r' = left(sqrt{x^2 +y^2} right)'
            = frac{(x^2 +y^2)'}{2
            sqrt{x^2 +y^2}}=frac{xx' +yy'}{r}
            $$

            and
            $$
            theta' = left(arctan frac{y}{x} right)'
            = frac{(y/x)'}{1+(y/x)^2} = frac{y' x -x' y}{r^2}.
            $$




            First, we define



            rdot[x1_, x2_] := (x1 (μ x1 - x2 - σ x1 (x1^2 + x2^2)) + x2 (x1 + μ x2 - σ x2 (x1^2 + x2^2)))/r


            We now make the substitution and simplify



            rdot[r Cos[t], r Sin[t]] // FullSimplify


            This yields (matches Mathematica)



            $$r' = mu r-r^3 sigma$$



            We now do the same for the other



            thetadot[x1_,x2_]:=(x1 (x1+μ x2-σ x2 (x1^2+x2^2)) - x2(μ x1-x2-σ x1 (x1^2+x2^2)))/r^2


            We now make the substitution and simplify



            thetadot[r Cos[t], r Sin[t]] // FullSimplify


            This yields (does not match Mathematica - it is as though they are forgetting to divide by $r$)



            $$theta'= 1$$



            I have asked this question before on this site in two different ways and have never gotten an answer that resolves the matter, but that could just be my denseness!






            share|improve this answer











            $endgroup$



            We can test the built-in TransformedField by defining our own functions.




            From $x',y'$ to $r',theta'$, we derive:
            $$
            r' = left(sqrt{x^2 +y^2} right)'
            = frac{(x^2 +y^2)'}{2
            sqrt{x^2 +y^2}}=frac{xx' +yy'}{r}
            $$

            and
            $$
            theta' = left(arctan frac{y}{x} right)'
            = frac{(y/x)'}{1+(y/x)^2} = frac{y' x -x' y}{r^2}.
            $$




            First, we define



            rdot[x1_, x2_] := (x1 (μ x1 - x2 - σ x1 (x1^2 + x2^2)) + x2 (x1 + μ x2 - σ x2 (x1^2 + x2^2)))/r


            We now make the substitution and simplify



            rdot[r Cos[t], r Sin[t]] // FullSimplify


            This yields (matches Mathematica)



            $$r' = mu r-r^3 sigma$$



            We now do the same for the other



            thetadot[x1_,x2_]:=(x1 (x1+μ x2-σ x2 (x1^2+x2^2)) - x2(μ x1-x2-σ x1 (x1^2+x2^2)))/r^2


            We now make the substitution and simplify



            thetadot[r Cos[t], r Sin[t]] // FullSimplify


            This yields (does not match Mathematica - it is as though they are forgetting to divide by $r$)



            $$theta'= 1$$



            I have asked this question before on this site in two different ways and have never gotten an answer that resolves the matter, but that could just be my denseness!







            share|improve this answer














            share|improve this answer



            share|improve this answer








            edited 4 hours ago

























            answered 4 hours ago









            MooMoo

            7511515




            7511515








            • 1




              $begingroup$
              Have you reported it to the Wolfram tech support?
              $endgroup$
              – Alexey Popkov
              2 hours ago










            • $begingroup$
              @AlexeyPopkov: I have not. I have had many issues with it when transforming between different methods. These days, I don't trust it and just create my own transformation rules to do it.
              $endgroup$
              – Moo
              2 hours ago










            • $begingroup$
              It is worth to write them about it in order to get it finally fixed. You can even write a short letter to support@wolfram.com with a link to this post.
              $endgroup$
              – Alexey Popkov
              2 hours ago












            • $begingroup$
              @AlexeyPopkov: I sent them an email per your suggestion.
              $endgroup$
              – Moo
              2 hours ago














            • 1




              $begingroup$
              Have you reported it to the Wolfram tech support?
              $endgroup$
              – Alexey Popkov
              2 hours ago










            • $begingroup$
              @AlexeyPopkov: I have not. I have had many issues with it when transforming between different methods. These days, I don't trust it and just create my own transformation rules to do it.
              $endgroup$
              – Moo
              2 hours ago










            • $begingroup$
              It is worth to write them about it in order to get it finally fixed. You can even write a short letter to support@wolfram.com with a link to this post.
              $endgroup$
              – Alexey Popkov
              2 hours ago












            • $begingroup$
              @AlexeyPopkov: I sent them an email per your suggestion.
              $endgroup$
              – Moo
              2 hours ago








            1




            1




            $begingroup$
            Have you reported it to the Wolfram tech support?
            $endgroup$
            – Alexey Popkov
            2 hours ago




            $begingroup$
            Have you reported it to the Wolfram tech support?
            $endgroup$
            – Alexey Popkov
            2 hours ago












            $begingroup$
            @AlexeyPopkov: I have not. I have had many issues with it when transforming between different methods. These days, I don't trust it and just create my own transformation rules to do it.
            $endgroup$
            – Moo
            2 hours ago




            $begingroup$
            @AlexeyPopkov: I have not. I have had many issues with it when transforming between different methods. These days, I don't trust it and just create my own transformation rules to do it.
            $endgroup$
            – Moo
            2 hours ago












            $begingroup$
            It is worth to write them about it in order to get it finally fixed. You can even write a short letter to support@wolfram.com with a link to this post.
            $endgroup$
            – Alexey Popkov
            2 hours ago






            $begingroup$
            It is worth to write them about it in order to get it finally fixed. You can even write a short letter to support@wolfram.com with a link to this post.
            $endgroup$
            – Alexey Popkov
            2 hours ago














            $begingroup$
            @AlexeyPopkov: I sent them an email per your suggestion.
            $endgroup$
            – Moo
            2 hours ago




            $begingroup$
            @AlexeyPopkov: I sent them an email per your suggestion.
            $endgroup$
            – Moo
            2 hours ago











            0












            $begingroup$

            Mathematica's answer is correct and consistent with your expectations, but you are not accounting for the basis of the vector field.



            TransformedField transforms a vector field between two coordinate systems and bases. In this case, it is converting from $f(x,y)hat x+g(x,y)hat y$ to the same geometrical vector field expressed as $u(r,theta)hat r + v(r,theta) hat theta$. Mathematica's answer can therefore be interpreted as saying
            $$left(μ x_1 - x_2 - σ x_1 (x_1^2 + x_2^2)right)hat x + left( x_1 + μ x_2 - σ x_2 (x_1^2 + x_2^2)right)hat y = left(r μ - r^3 σright)hat r + r hattheta$$



            Notice that the expressions $r'$ and $theta'$ don't appear anyhwere. Those are dynamical quantities, not geometrical ones (unless working in the jet bundle, but let's not go there). Also notice the hats! As stated in the documentation, TransformedField assumes inputs are in an orthonormal basis, and returns outputs in the same basis. That will be important for later on.



            Now, you are dealing with a differential equation, and based on your expected answer I'll assume what you have is a first-order system and you are transforming the associated vector field (AKA the "right-hand side"). Finding solutions means find the integral curves of the vector field. This gives as a nice relationship between the geometric variables the dynamical ones, except that this relationship is of necessity expressed in the so called coordinate basis, written $(r',theta') = a frac{partial}{partial r} + b frac{partial}{partial theta}$. So to get the answer expressed in your desired basis, we need the relationship between the coordinate and orthonormal basis vectors. As is covered in books on vector calculus (and elsewhere), the relationsip is $hat r = frac{partial}{partial r}$ and $hat theta = frac{1}{r}frac{partial}{partial theta}$. Substituting this into the answer Mathematica gave above, we get
            $$left(r μ - r^3 σright)hat r + r hattheta = left(r μ - r^3 σright) frac{partial}{partial r} + (1) frac{partial}{partial theta},$$
            which is the answer you expected.






            share|improve this answer









            $endgroup$


















              0












              $begingroup$

              Mathematica's answer is correct and consistent with your expectations, but you are not accounting for the basis of the vector field.



              TransformedField transforms a vector field between two coordinate systems and bases. In this case, it is converting from $f(x,y)hat x+g(x,y)hat y$ to the same geometrical vector field expressed as $u(r,theta)hat r + v(r,theta) hat theta$. Mathematica's answer can therefore be interpreted as saying
              $$left(μ x_1 - x_2 - σ x_1 (x_1^2 + x_2^2)right)hat x + left( x_1 + μ x_2 - σ x_2 (x_1^2 + x_2^2)right)hat y = left(r μ - r^3 σright)hat r + r hattheta$$



              Notice that the expressions $r'$ and $theta'$ don't appear anyhwere. Those are dynamical quantities, not geometrical ones (unless working in the jet bundle, but let's not go there). Also notice the hats! As stated in the documentation, TransformedField assumes inputs are in an orthonormal basis, and returns outputs in the same basis. That will be important for later on.



              Now, you are dealing with a differential equation, and based on your expected answer I'll assume what you have is a first-order system and you are transforming the associated vector field (AKA the "right-hand side"). Finding solutions means find the integral curves of the vector field. This gives as a nice relationship between the geometric variables the dynamical ones, except that this relationship is of necessity expressed in the so called coordinate basis, written $(r',theta') = a frac{partial}{partial r} + b frac{partial}{partial theta}$. So to get the answer expressed in your desired basis, we need the relationship between the coordinate and orthonormal basis vectors. As is covered in books on vector calculus (and elsewhere), the relationsip is $hat r = frac{partial}{partial r}$ and $hat theta = frac{1}{r}frac{partial}{partial theta}$. Substituting this into the answer Mathematica gave above, we get
              $$left(r μ - r^3 σright)hat r + r hattheta = left(r μ - r^3 σright) frac{partial}{partial r} + (1) frac{partial}{partial theta},$$
              which is the answer you expected.






              share|improve this answer









              $endgroup$
















                0












                0








                0





                $begingroup$

                Mathematica's answer is correct and consistent with your expectations, but you are not accounting for the basis of the vector field.



                TransformedField transforms a vector field between two coordinate systems and bases. In this case, it is converting from $f(x,y)hat x+g(x,y)hat y$ to the same geometrical vector field expressed as $u(r,theta)hat r + v(r,theta) hat theta$. Mathematica's answer can therefore be interpreted as saying
                $$left(μ x_1 - x_2 - σ x_1 (x_1^2 + x_2^2)right)hat x + left( x_1 + μ x_2 - σ x_2 (x_1^2 + x_2^2)right)hat y = left(r μ - r^3 σright)hat r + r hattheta$$



                Notice that the expressions $r'$ and $theta'$ don't appear anyhwere. Those are dynamical quantities, not geometrical ones (unless working in the jet bundle, but let's not go there). Also notice the hats! As stated in the documentation, TransformedField assumes inputs are in an orthonormal basis, and returns outputs in the same basis. That will be important for later on.



                Now, you are dealing with a differential equation, and based on your expected answer I'll assume what you have is a first-order system and you are transforming the associated vector field (AKA the "right-hand side"). Finding solutions means find the integral curves of the vector field. This gives as a nice relationship between the geometric variables the dynamical ones, except that this relationship is of necessity expressed in the so called coordinate basis, written $(r',theta') = a frac{partial}{partial r} + b frac{partial}{partial theta}$. So to get the answer expressed in your desired basis, we need the relationship between the coordinate and orthonormal basis vectors. As is covered in books on vector calculus (and elsewhere), the relationsip is $hat r = frac{partial}{partial r}$ and $hat theta = frac{1}{r}frac{partial}{partial theta}$. Substituting this into the answer Mathematica gave above, we get
                $$left(r μ - r^3 σright)hat r + r hattheta = left(r μ - r^3 σright) frac{partial}{partial r} + (1) frac{partial}{partial theta},$$
                which is the answer you expected.






                share|improve this answer









                $endgroup$



                Mathematica's answer is correct and consistent with your expectations, but you are not accounting for the basis of the vector field.



                TransformedField transforms a vector field between two coordinate systems and bases. In this case, it is converting from $f(x,y)hat x+g(x,y)hat y$ to the same geometrical vector field expressed as $u(r,theta)hat r + v(r,theta) hat theta$. Mathematica's answer can therefore be interpreted as saying
                $$left(μ x_1 - x_2 - σ x_1 (x_1^2 + x_2^2)right)hat x + left( x_1 + μ x_2 - σ x_2 (x_1^2 + x_2^2)right)hat y = left(r μ - r^3 σright)hat r + r hattheta$$



                Notice that the expressions $r'$ and $theta'$ don't appear anyhwere. Those are dynamical quantities, not geometrical ones (unless working in the jet bundle, but let's not go there). Also notice the hats! As stated in the documentation, TransformedField assumes inputs are in an orthonormal basis, and returns outputs in the same basis. That will be important for later on.



                Now, you are dealing with a differential equation, and based on your expected answer I'll assume what you have is a first-order system and you are transforming the associated vector field (AKA the "right-hand side"). Finding solutions means find the integral curves of the vector field. This gives as a nice relationship between the geometric variables the dynamical ones, except that this relationship is of necessity expressed in the so called coordinate basis, written $(r',theta') = a frac{partial}{partial r} + b frac{partial}{partial theta}$. So to get the answer expressed in your desired basis, we need the relationship between the coordinate and orthonormal basis vectors. As is covered in books on vector calculus (and elsewhere), the relationsip is $hat r = frac{partial}{partial r}$ and $hat theta = frac{1}{r}frac{partial}{partial theta}$. Substituting this into the answer Mathematica gave above, we get
                $$left(r μ - r^3 σright)hat r + r hattheta = left(r μ - r^3 σright) frac{partial}{partial r} + (1) frac{partial}{partial theta},$$
                which is the answer you expected.







                share|improve this answer












                share|improve this answer



                share|improve this answer










                answered 36 mins ago









                Itai SeggevItai Seggev

                9,6883964




                9,6883964






























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