Basis vectors as a differentiation of position vector











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Basis vectors are defined as $vec {E_i}=vec{E_i}(x_1,x_2,x_3)=frac{partialvec{r}}{partial x^i}$ i=1,2,3.
In spherical coordinate system $x^1=r, x^2=theta, x^3=phi$



position vector $vec {R}=r hat{r}$.



That means $hat {r}=frac{partialvec{R}}{partial r}=hat {r}$



But $hat {theta}=frac{partialvec{R}}{partial theta}=rhat {theta}$



Similarly $hat {phi}=frac{partialvec{R}}{partial phi}=rhat {phi}$



Where am I wrong? Why is r appearing?










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  • $hat{r} = sin (theta) cos (phi) hat{x} + sin (theta) sin(phi) hat{y} + cos (theta) hat{z}$
    – user42298
    Nov 26 at 9:26












  • Differentiating this gives what we need. But we are differentiating position vector and it doesn't give the answer.
    – Asit Srivastava
    Nov 26 at 9:29















up vote
0
down vote

favorite












Basis vectors are defined as $vec {E_i}=vec{E_i}(x_1,x_2,x_3)=frac{partialvec{r}}{partial x^i}$ i=1,2,3.
In spherical coordinate system $x^1=r, x^2=theta, x^3=phi$



position vector $vec {R}=r hat{r}$.



That means $hat {r}=frac{partialvec{R}}{partial r}=hat {r}$



But $hat {theta}=frac{partialvec{R}}{partial theta}=rhat {theta}$



Similarly $hat {phi}=frac{partialvec{R}}{partial phi}=rhat {phi}$



Where am I wrong? Why is r appearing?










share|cite|improve this question






















  • $hat{r} = sin (theta) cos (phi) hat{x} + sin (theta) sin(phi) hat{y} + cos (theta) hat{z}$
    – user42298
    Nov 26 at 9:26












  • Differentiating this gives what we need. But we are differentiating position vector and it doesn't give the answer.
    – Asit Srivastava
    Nov 26 at 9:29













up vote
0
down vote

favorite









up vote
0
down vote

favorite











Basis vectors are defined as $vec {E_i}=vec{E_i}(x_1,x_2,x_3)=frac{partialvec{r}}{partial x^i}$ i=1,2,3.
In spherical coordinate system $x^1=r, x^2=theta, x^3=phi$



position vector $vec {R}=r hat{r}$.



That means $hat {r}=frac{partialvec{R}}{partial r}=hat {r}$



But $hat {theta}=frac{partialvec{R}}{partial theta}=rhat {theta}$



Similarly $hat {phi}=frac{partialvec{R}}{partial phi}=rhat {phi}$



Where am I wrong? Why is r appearing?










share|cite|improve this question













Basis vectors are defined as $vec {E_i}=vec{E_i}(x_1,x_2,x_3)=frac{partialvec{r}}{partial x^i}$ i=1,2,3.
In spherical coordinate system $x^1=r, x^2=theta, x^3=phi$



position vector $vec {R}=r hat{r}$.



That means $hat {r}=frac{partialvec{R}}{partial r}=hat {r}$



But $hat {theta}=frac{partialvec{R}}{partial theta}=rhat {theta}$



Similarly $hat {phi}=frac{partialvec{R}}{partial phi}=rhat {phi}$



Where am I wrong? Why is r appearing?







vector-spaces tensors






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share|cite|improve this question




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asked Nov 26 at 9:14









Asit Srivastava

256




256












  • $hat{r} = sin (theta) cos (phi) hat{x} + sin (theta) sin(phi) hat{y} + cos (theta) hat{z}$
    – user42298
    Nov 26 at 9:26












  • Differentiating this gives what we need. But we are differentiating position vector and it doesn't give the answer.
    – Asit Srivastava
    Nov 26 at 9:29


















  • $hat{r} = sin (theta) cos (phi) hat{x} + sin (theta) sin(phi) hat{y} + cos (theta) hat{z}$
    – user42298
    Nov 26 at 9:26












  • Differentiating this gives what we need. But we are differentiating position vector and it doesn't give the answer.
    – Asit Srivastava
    Nov 26 at 9:29
















$hat{r} = sin (theta) cos (phi) hat{x} + sin (theta) sin(phi) hat{y} + cos (theta) hat{z}$
– user42298
Nov 26 at 9:26






$hat{r} = sin (theta) cos (phi) hat{x} + sin (theta) sin(phi) hat{y} + cos (theta) hat{z}$
– user42298
Nov 26 at 9:26














Differentiating this gives what we need. But we are differentiating position vector and it doesn't give the answer.
– Asit Srivastava
Nov 26 at 9:29




Differentiating this gives what we need. But we are differentiating position vector and it doesn't give the answer.
– Asit Srivastava
Nov 26 at 9:29










2 Answers
2






active

oldest

votes

















up vote
0
down vote



accepted










You're missing the scale factors



$$
hat{e}_alpha = frac{1}{h^alpha}frac{partial {bf x}}{partial u^alpha}
$$



As an example



$$
{bf x} = pmatrix{rsinthetacosphi \ rsinthetasinphi \ r costheta}
$$



So



$$
frac{partial {bf x}}{partial theta} = frac{partial}{partial theta}pmatrix{rsinthetacosphi \ rsinthetasinphi \ r costheta} = pmatrix{rcosthetasinphi \ rcosthetacosphi \ -rsintheta}
$$



The scale factor is just



$$
h^2 = [r^2cos^2theta(sin^2phi + cos^2phi) + r^2sin^2theta]^{1/2} = r
$$



So the unit vector is



$$
hat{e}_2 = pmatrix{costhetasinphi \ costhetacosphi \ -sintheta}
$$






share|cite|improve this answer





















  • But what about the way of defining basis vectors? As mentioned above it doesn't mention scale factors.
    – Asit Srivastava
    Nov 26 at 11:07










  • @AsitSrivastava The vectors are defined through the derivatives as you said, but they are not unitary. You need to scale them first, hence the scale factors
    – caverac
    Nov 26 at 11:18


















up vote
0
down vote













In Cartesian co-ordinates the unit vectors parallel to the co-ordinate axes $hat{x}, hat{y}, hat{z}$ do not change with position. So if we have $vec{R}=xhat{x}+yhat{y}+zhat{z}$ then we are used to seeing



$frac{partialvec{R}}{partial x}=hat{x}$ etc.



However, in other co-ordinate systems this is not true. In particular, in spherical co-ordinates the direction of the $hat{r}$ vector changes depending on $theta$ and $phi$ - so we should more properly write $hat{r}$ as $hat{r}(theta, phi)$. So



$frac{partialvec{R}}{partial theta}=rfrac{partialvec{r}}{partial theta}=r hat{theta}$



$Rightarrow hat{theta} =frac{1}{r}frac{partialvec{R}}{partial theta}$



and similarly for $phi$:



$frac{partialvec{R}}{partial phi}=rfrac{partialvec{r}}{partial phi}=rsin theta hat{phi}$



$Rightarrow hat{phi} =frac{1}{r sin theta}frac{partialvec{R}}{partial phi}$






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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    0
    down vote



    accepted










    You're missing the scale factors



    $$
    hat{e}_alpha = frac{1}{h^alpha}frac{partial {bf x}}{partial u^alpha}
    $$



    As an example



    $$
    {bf x} = pmatrix{rsinthetacosphi \ rsinthetasinphi \ r costheta}
    $$



    So



    $$
    frac{partial {bf x}}{partial theta} = frac{partial}{partial theta}pmatrix{rsinthetacosphi \ rsinthetasinphi \ r costheta} = pmatrix{rcosthetasinphi \ rcosthetacosphi \ -rsintheta}
    $$



    The scale factor is just



    $$
    h^2 = [r^2cos^2theta(sin^2phi + cos^2phi) + r^2sin^2theta]^{1/2} = r
    $$



    So the unit vector is



    $$
    hat{e}_2 = pmatrix{costhetasinphi \ costhetacosphi \ -sintheta}
    $$






    share|cite|improve this answer





















    • But what about the way of defining basis vectors? As mentioned above it doesn't mention scale factors.
      – Asit Srivastava
      Nov 26 at 11:07










    • @AsitSrivastava The vectors are defined through the derivatives as you said, but they are not unitary. You need to scale them first, hence the scale factors
      – caverac
      Nov 26 at 11:18















    up vote
    0
    down vote



    accepted










    You're missing the scale factors



    $$
    hat{e}_alpha = frac{1}{h^alpha}frac{partial {bf x}}{partial u^alpha}
    $$



    As an example



    $$
    {bf x} = pmatrix{rsinthetacosphi \ rsinthetasinphi \ r costheta}
    $$



    So



    $$
    frac{partial {bf x}}{partial theta} = frac{partial}{partial theta}pmatrix{rsinthetacosphi \ rsinthetasinphi \ r costheta} = pmatrix{rcosthetasinphi \ rcosthetacosphi \ -rsintheta}
    $$



    The scale factor is just



    $$
    h^2 = [r^2cos^2theta(sin^2phi + cos^2phi) + r^2sin^2theta]^{1/2} = r
    $$



    So the unit vector is



    $$
    hat{e}_2 = pmatrix{costhetasinphi \ costhetacosphi \ -sintheta}
    $$






    share|cite|improve this answer





















    • But what about the way of defining basis vectors? As mentioned above it doesn't mention scale factors.
      – Asit Srivastava
      Nov 26 at 11:07










    • @AsitSrivastava The vectors are defined through the derivatives as you said, but they are not unitary. You need to scale them first, hence the scale factors
      – caverac
      Nov 26 at 11:18













    up vote
    0
    down vote



    accepted







    up vote
    0
    down vote



    accepted






    You're missing the scale factors



    $$
    hat{e}_alpha = frac{1}{h^alpha}frac{partial {bf x}}{partial u^alpha}
    $$



    As an example



    $$
    {bf x} = pmatrix{rsinthetacosphi \ rsinthetasinphi \ r costheta}
    $$



    So



    $$
    frac{partial {bf x}}{partial theta} = frac{partial}{partial theta}pmatrix{rsinthetacosphi \ rsinthetasinphi \ r costheta} = pmatrix{rcosthetasinphi \ rcosthetacosphi \ -rsintheta}
    $$



    The scale factor is just



    $$
    h^2 = [r^2cos^2theta(sin^2phi + cos^2phi) + r^2sin^2theta]^{1/2} = r
    $$



    So the unit vector is



    $$
    hat{e}_2 = pmatrix{costhetasinphi \ costhetacosphi \ -sintheta}
    $$






    share|cite|improve this answer












    You're missing the scale factors



    $$
    hat{e}_alpha = frac{1}{h^alpha}frac{partial {bf x}}{partial u^alpha}
    $$



    As an example



    $$
    {bf x} = pmatrix{rsinthetacosphi \ rsinthetasinphi \ r costheta}
    $$



    So



    $$
    frac{partial {bf x}}{partial theta} = frac{partial}{partial theta}pmatrix{rsinthetacosphi \ rsinthetasinphi \ r costheta} = pmatrix{rcosthetasinphi \ rcosthetacosphi \ -rsintheta}
    $$



    The scale factor is just



    $$
    h^2 = [r^2cos^2theta(sin^2phi + cos^2phi) + r^2sin^2theta]^{1/2} = r
    $$



    So the unit vector is



    $$
    hat{e}_2 = pmatrix{costhetasinphi \ costhetacosphi \ -sintheta}
    $$







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Nov 26 at 9:29









    caverac

    12.4k21027




    12.4k21027












    • But what about the way of defining basis vectors? As mentioned above it doesn't mention scale factors.
      – Asit Srivastava
      Nov 26 at 11:07










    • @AsitSrivastava The vectors are defined through the derivatives as you said, but they are not unitary. You need to scale them first, hence the scale factors
      – caverac
      Nov 26 at 11:18


















    • But what about the way of defining basis vectors? As mentioned above it doesn't mention scale factors.
      – Asit Srivastava
      Nov 26 at 11:07










    • @AsitSrivastava The vectors are defined through the derivatives as you said, but they are not unitary. You need to scale them first, hence the scale factors
      – caverac
      Nov 26 at 11:18
















    But what about the way of defining basis vectors? As mentioned above it doesn't mention scale factors.
    – Asit Srivastava
    Nov 26 at 11:07




    But what about the way of defining basis vectors? As mentioned above it doesn't mention scale factors.
    – Asit Srivastava
    Nov 26 at 11:07












    @AsitSrivastava The vectors are defined through the derivatives as you said, but they are not unitary. You need to scale them first, hence the scale factors
    – caverac
    Nov 26 at 11:18




    @AsitSrivastava The vectors are defined through the derivatives as you said, but they are not unitary. You need to scale them first, hence the scale factors
    – caverac
    Nov 26 at 11:18










    up vote
    0
    down vote













    In Cartesian co-ordinates the unit vectors parallel to the co-ordinate axes $hat{x}, hat{y}, hat{z}$ do not change with position. So if we have $vec{R}=xhat{x}+yhat{y}+zhat{z}$ then we are used to seeing



    $frac{partialvec{R}}{partial x}=hat{x}$ etc.



    However, in other co-ordinate systems this is not true. In particular, in spherical co-ordinates the direction of the $hat{r}$ vector changes depending on $theta$ and $phi$ - so we should more properly write $hat{r}$ as $hat{r}(theta, phi)$. So



    $frac{partialvec{R}}{partial theta}=rfrac{partialvec{r}}{partial theta}=r hat{theta}$



    $Rightarrow hat{theta} =frac{1}{r}frac{partialvec{R}}{partial theta}$



    and similarly for $phi$:



    $frac{partialvec{R}}{partial phi}=rfrac{partialvec{r}}{partial phi}=rsin theta hat{phi}$



    $Rightarrow hat{phi} =frac{1}{r sin theta}frac{partialvec{R}}{partial phi}$






    share|cite|improve this answer

























      up vote
      0
      down vote













      In Cartesian co-ordinates the unit vectors parallel to the co-ordinate axes $hat{x}, hat{y}, hat{z}$ do not change with position. So if we have $vec{R}=xhat{x}+yhat{y}+zhat{z}$ then we are used to seeing



      $frac{partialvec{R}}{partial x}=hat{x}$ etc.



      However, in other co-ordinate systems this is not true. In particular, in spherical co-ordinates the direction of the $hat{r}$ vector changes depending on $theta$ and $phi$ - so we should more properly write $hat{r}$ as $hat{r}(theta, phi)$. So



      $frac{partialvec{R}}{partial theta}=rfrac{partialvec{r}}{partial theta}=r hat{theta}$



      $Rightarrow hat{theta} =frac{1}{r}frac{partialvec{R}}{partial theta}$



      and similarly for $phi$:



      $frac{partialvec{R}}{partial phi}=rfrac{partialvec{r}}{partial phi}=rsin theta hat{phi}$



      $Rightarrow hat{phi} =frac{1}{r sin theta}frac{partialvec{R}}{partial phi}$






      share|cite|improve this answer























        up vote
        0
        down vote










        up vote
        0
        down vote









        In Cartesian co-ordinates the unit vectors parallel to the co-ordinate axes $hat{x}, hat{y}, hat{z}$ do not change with position. So if we have $vec{R}=xhat{x}+yhat{y}+zhat{z}$ then we are used to seeing



        $frac{partialvec{R}}{partial x}=hat{x}$ etc.



        However, in other co-ordinate systems this is not true. In particular, in spherical co-ordinates the direction of the $hat{r}$ vector changes depending on $theta$ and $phi$ - so we should more properly write $hat{r}$ as $hat{r}(theta, phi)$. So



        $frac{partialvec{R}}{partial theta}=rfrac{partialvec{r}}{partial theta}=r hat{theta}$



        $Rightarrow hat{theta} =frac{1}{r}frac{partialvec{R}}{partial theta}$



        and similarly for $phi$:



        $frac{partialvec{R}}{partial phi}=rfrac{partialvec{r}}{partial phi}=rsin theta hat{phi}$



        $Rightarrow hat{phi} =frac{1}{r sin theta}frac{partialvec{R}}{partial phi}$






        share|cite|improve this answer












        In Cartesian co-ordinates the unit vectors parallel to the co-ordinate axes $hat{x}, hat{y}, hat{z}$ do not change with position. So if we have $vec{R}=xhat{x}+yhat{y}+zhat{z}$ then we are used to seeing



        $frac{partialvec{R}}{partial x}=hat{x}$ etc.



        However, in other co-ordinate systems this is not true. In particular, in spherical co-ordinates the direction of the $hat{r}$ vector changes depending on $theta$ and $phi$ - so we should more properly write $hat{r}$ as $hat{r}(theta, phi)$. So



        $frac{partialvec{R}}{partial theta}=rfrac{partialvec{r}}{partial theta}=r hat{theta}$



        $Rightarrow hat{theta} =frac{1}{r}frac{partialvec{R}}{partial theta}$



        and similarly for $phi$:



        $frac{partialvec{R}}{partial phi}=rfrac{partialvec{r}}{partial phi}=rsin theta hat{phi}$



        $Rightarrow hat{phi} =frac{1}{r sin theta}frac{partialvec{R}}{partial phi}$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 26 at 10:16









        gandalf61

        7,365623




        7,365623






























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