Special curves in geometry and kinematics












0












$begingroup$


Let us consider a particle $vec{w} = (w_1,w_2,w_3)$ in three dimensional space, whose trajectory is parametrized by the spherical coordinate system $(rho,theta,varphi)$:



begin{equation*}
w_1 = rho sin theta cosvarphi,quad w_2 = rho sin theta sinvarphi,quad w_3 = rho cos theta
end{equation*}



Now the acceleration vector of this curve, in the spherical coordinates, is given by



begin{equation*}
begin{split}
vec{a} &= (rho '' - rho theta'^2 - rho varphi'^2 sin^2theta) vec{e}_r \
&quad + (2rho'theta' + rhotheta'' - rho varphi'^2 costhetasintheta) vec{e}_theta \
&quad + ((2rho'varphi' + rhovarphi'')sintheta + 2 rhotheta'varphi'costheta)vec{e}_varphi,
end{split}
end{equation*}
where ${ vec{e}_r, vec{e}_theta,vec{e}_varphi}$ is the standard basis in the spherical coordinate system; for more details, please see https://en.wikipedia.org/wiki/Spherical_coordinate_system.



My question is, is there a name for the curves whose $(theta,varphi)$ components of the acceleration vector are constantly zero, that is, $vec{a}cdot vec{e}_theta = vec{a} cdot vec{e}_varphi = 0$ for all time $t$?



Or is there any result concerning the property or the classification of such a curve? It would be really appreciated if one can provide any reference in this direction.



Many thanks in advance.






geometry kinematics







share|cite|improve this question









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    0












    $begingroup$


    Let us consider a particle $vec{w} = (w_1,w_2,w_3)$ in three dimensional space, whose trajectory is parametrized by the spherical coordinate system $(rho,theta,varphi)$:



    begin{equation*}
    w_1 = rho sin theta cosvarphi,quad w_2 = rho sin theta sinvarphi,quad w_3 = rho cos theta
    end{equation*}



    Now the acceleration vector of this curve, in the spherical coordinates, is given by



    begin{equation*}
    begin{split}
    vec{a} &= (rho '' - rho theta'^2 - rho varphi'^2 sin^2theta) vec{e}_r \
    &quad + (2rho'theta' + rhotheta'' - rho varphi'^2 costhetasintheta) vec{e}_theta \
    &quad + ((2rho'varphi' + rhovarphi'')sintheta + 2 rhotheta'varphi'costheta)vec{e}_varphi,
    end{split}
    end{equation*}
    where ${ vec{e}_r, vec{e}_theta,vec{e}_varphi}$ is the standard basis in the spherical coordinate system; for more details, please see https://en.wikipedia.org/wiki/Spherical_coordinate_system.



    My question is, is there a name for the curves whose $(theta,varphi)$ components of the acceleration vector are constantly zero, that is, $vec{a}cdot vec{e}_theta = vec{a} cdot vec{e}_varphi = 0$ for all time $t$?



    Or is there any result concerning the property or the classification of such a curve? It would be really appreciated if one can provide any reference in this direction.



    Many thanks in advance.






    geometry kinematics







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      Let us consider a particle $vec{w} = (w_1,w_2,w_3)$ in three dimensional space, whose trajectory is parametrized by the spherical coordinate system $(rho,theta,varphi)$:



      begin{equation*}
      w_1 = rho sin theta cosvarphi,quad w_2 = rho sin theta sinvarphi,quad w_3 = rho cos theta
      end{equation*}



      Now the acceleration vector of this curve, in the spherical coordinates, is given by



      begin{equation*}
      begin{split}
      vec{a} &= (rho '' - rho theta'^2 - rho varphi'^2 sin^2theta) vec{e}_r \
      &quad + (2rho'theta' + rhotheta'' - rho varphi'^2 costhetasintheta) vec{e}_theta \
      &quad + ((2rho'varphi' + rhovarphi'')sintheta + 2 rhotheta'varphi'costheta)vec{e}_varphi,
      end{split}
      end{equation*}
      where ${ vec{e}_r, vec{e}_theta,vec{e}_varphi}$ is the standard basis in the spherical coordinate system; for more details, please see https://en.wikipedia.org/wiki/Spherical_coordinate_system.



      My question is, is there a name for the curves whose $(theta,varphi)$ components of the acceleration vector are constantly zero, that is, $vec{a}cdot vec{e}_theta = vec{a} cdot vec{e}_varphi = 0$ for all time $t$?



      Or is there any result concerning the property or the classification of such a curve? It would be really appreciated if one can provide any reference in this direction.



      Many thanks in advance.






      geometry kinematics







      share|cite|improve this question









      $endgroup$




      Let us consider a particle $vec{w} = (w_1,w_2,w_3)$ in three dimensional space, whose trajectory is parametrized by the spherical coordinate system $(rho,theta,varphi)$:



      begin{equation*}
      w_1 = rho sin theta cosvarphi,quad w_2 = rho sin theta sinvarphi,quad w_3 = rho cos theta
      end{equation*}



      Now the acceleration vector of this curve, in the spherical coordinates, is given by



      begin{equation*}
      begin{split}
      vec{a} &= (rho '' - rho theta'^2 - rho varphi'^2 sin^2theta) vec{e}_r \
      &quad + (2rho'theta' + rhotheta'' - rho varphi'^2 costhetasintheta) vec{e}_theta \
      &quad + ((2rho'varphi' + rhovarphi'')sintheta + 2 rhotheta'varphi'costheta)vec{e}_varphi,
      end{split}
      end{equation*}
      where ${ vec{e}_r, vec{e}_theta,vec{e}_varphi}$ is the standard basis in the spherical coordinate system; for more details, please see https://en.wikipedia.org/wiki/Spherical_coordinate_system.



      My question is, is there a name for the curves whose $(theta,varphi)$ components of the acceleration vector are constantly zero, that is, $vec{a}cdot vec{e}_theta = vec{a} cdot vec{e}_varphi = 0$ for all time $t$?



      Or is there any result concerning the property or the classification of such a curve? It would be really appreciated if one can provide any reference in this direction.



      Many thanks in advance.






      geometry kinematics




      geometry kinematics






      share|cite|improve this question













      share|cite|improve this question











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      asked Jan 24 at 3:56









      Sunghan KimSunghan Kim

      588




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          1 Answer
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          $begingroup$

          You have radial acceleration only in the case the force acting on the particle is radial. The curves you are looking for are called spherical harmonics and are of great importance for quantum mechanics (think hydrogen orbitals). For classical mechanics, see https://en.wikipedia.org/wiki/Classical_central-force_problem. For $1/r^2$ type of force you get for example the Keplerian trajectories.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thanks a lot for the comment. Could you please elaborate more on why the curves are spherical harmonics. For instance, do you mean that the curve is a nodal set of a spherical harmonic?
            $endgroup$
            – Sunghan Kim
            Jan 24 at 6:25










          • $begingroup$
            The angular momentum is conserved. So the trajectory will be an expansion of spherical harmonics (weighted by some radial functions) with the same indices. Obviously, in classical mechanics you can just use simple planar motion. The plane of the orbit is given by the initial velocity direction and the initial position. Since the acceleration is in the plane with the initial velocity, the particle cannot leave the plane.
            $endgroup$
            – Andrei
            Jan 24 at 7:10











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          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          2












          $begingroup$

          You have radial acceleration only in the case the force acting on the particle is radial. The curves you are looking for are called spherical harmonics and are of great importance for quantum mechanics (think hydrogen orbitals). For classical mechanics, see https://en.wikipedia.org/wiki/Classical_central-force_problem. For $1/r^2$ type of force you get for example the Keplerian trajectories.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thanks a lot for the comment. Could you please elaborate more on why the curves are spherical harmonics. For instance, do you mean that the curve is a nodal set of a spherical harmonic?
            $endgroup$
            – Sunghan Kim
            Jan 24 at 6:25










          • $begingroup$
            The angular momentum is conserved. So the trajectory will be an expansion of spherical harmonics (weighted by some radial functions) with the same indices. Obviously, in classical mechanics you can just use simple planar motion. The plane of the orbit is given by the initial velocity direction and the initial position. Since the acceleration is in the plane with the initial velocity, the particle cannot leave the plane.
            $endgroup$
            – Andrei
            Jan 24 at 7:10
















          2












          $begingroup$

          You have radial acceleration only in the case the force acting on the particle is radial. The curves you are looking for are called spherical harmonics and are of great importance for quantum mechanics (think hydrogen orbitals). For classical mechanics, see https://en.wikipedia.org/wiki/Classical_central-force_problem. For $1/r^2$ type of force you get for example the Keplerian trajectories.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thanks a lot for the comment. Could you please elaborate more on why the curves are spherical harmonics. For instance, do you mean that the curve is a nodal set of a spherical harmonic?
            $endgroup$
            – Sunghan Kim
            Jan 24 at 6:25










          • $begingroup$
            The angular momentum is conserved. So the trajectory will be an expansion of spherical harmonics (weighted by some radial functions) with the same indices. Obviously, in classical mechanics you can just use simple planar motion. The plane of the orbit is given by the initial velocity direction and the initial position. Since the acceleration is in the plane with the initial velocity, the particle cannot leave the plane.
            $endgroup$
            – Andrei
            Jan 24 at 7:10














          2












          2








          2





          $begingroup$

          You have radial acceleration only in the case the force acting on the particle is radial. The curves you are looking for are called spherical harmonics and are of great importance for quantum mechanics (think hydrogen orbitals). For classical mechanics, see https://en.wikipedia.org/wiki/Classical_central-force_problem. For $1/r^2$ type of force you get for example the Keplerian trajectories.






          share|cite|improve this answer











          $endgroup$



          You have radial acceleration only in the case the force acting on the particle is radial. The curves you are looking for are called spherical harmonics and are of great importance for quantum mechanics (think hydrogen orbitals). For classical mechanics, see https://en.wikipedia.org/wiki/Classical_central-force_problem. For $1/r^2$ type of force you get for example the Keplerian trajectories.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 24 at 5:24

























          answered Jan 24 at 5:16









          AndreiAndrei

          12.9k21129




          12.9k21129












          • $begingroup$
            Thanks a lot for the comment. Could you please elaborate more on why the curves are spherical harmonics. For instance, do you mean that the curve is a nodal set of a spherical harmonic?
            $endgroup$
            – Sunghan Kim
            Jan 24 at 6:25










          • $begingroup$
            The angular momentum is conserved. So the trajectory will be an expansion of spherical harmonics (weighted by some radial functions) with the same indices. Obviously, in classical mechanics you can just use simple planar motion. The plane of the orbit is given by the initial velocity direction and the initial position. Since the acceleration is in the plane with the initial velocity, the particle cannot leave the plane.
            $endgroup$
            – Andrei
            Jan 24 at 7:10


















          • $begingroup$
            Thanks a lot for the comment. Could you please elaborate more on why the curves are spherical harmonics. For instance, do you mean that the curve is a nodal set of a spherical harmonic?
            $endgroup$
            – Sunghan Kim
            Jan 24 at 6:25










          • $begingroup$
            The angular momentum is conserved. So the trajectory will be an expansion of spherical harmonics (weighted by some radial functions) with the same indices. Obviously, in classical mechanics you can just use simple planar motion. The plane of the orbit is given by the initial velocity direction and the initial position. Since the acceleration is in the plane with the initial velocity, the particle cannot leave the plane.
            $endgroup$
            – Andrei
            Jan 24 at 7:10
















          $begingroup$
          Thanks a lot for the comment. Could you please elaborate more on why the curves are spherical harmonics. For instance, do you mean that the curve is a nodal set of a spherical harmonic?
          $endgroup$
          – Sunghan Kim
          Jan 24 at 6:25




          $begingroup$
          Thanks a lot for the comment. Could you please elaborate more on why the curves are spherical harmonics. For instance, do you mean that the curve is a nodal set of a spherical harmonic?
          $endgroup$
          – Sunghan Kim
          Jan 24 at 6:25












          $begingroup$
          The angular momentum is conserved. So the trajectory will be an expansion of spherical harmonics (weighted by some radial functions) with the same indices. Obviously, in classical mechanics you can just use simple planar motion. The plane of the orbit is given by the initial velocity direction and the initial position. Since the acceleration is in the plane with the initial velocity, the particle cannot leave the plane.
          $endgroup$
          – Andrei
          Jan 24 at 7:10




          $begingroup$
          The angular momentum is conserved. So the trajectory will be an expansion of spherical harmonics (weighted by some radial functions) with the same indices. Obviously, in classical mechanics you can just use simple planar motion. The plane of the orbit is given by the initial velocity direction and the initial position. Since the acceleration is in the plane with the initial velocity, the particle cannot leave the plane.
          $endgroup$
          – Andrei
          Jan 24 at 7:10


















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