Special curves in geometry and kinematics
$begingroup$
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
asked Jan 24 at 3:56
Sunghan KimSunghan Kim
588
$endgroup$
add a comment |
$begingroup$
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
asked Jan 24 at 3:56
Sunghan KimSunghan Kim
588
$endgroup$
add a comment |
$begingroup$
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
asked Jan 24 at 3:56
Sunghan KimSunghan Kim
588
$endgroup$
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
asked Jan 24 at 3:56
Sunghan KimSunghan Kim
588
asked Jan 24 at 3:56
Sunghan KimSunghan Kim
588
asked Jan 24 at 3:56
Sunghan KimSunghan Kim
588
asked Jan 24 at 3:56
Sunghan KimSunghan Kim
588
asked Jan 24 at 3:56
Sunghan KimSunghan Kim
588
588
add a comment |
add a comment |
1 Answer
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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.
answered Jan 24 at 5:16
AndreiAndrei
12.9k21129
$endgroup$
$begingroup$
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
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
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votes
$begingroup$
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.
answered Jan 24 at 5:16
AndreiAndrei
12.9k21129
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
answered Jan 24 at 5:16
AndreiAndrei
12.9k21129
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
answered Jan 24 at 5:16
AndreiAndrei
12.9k21129
$endgroup$
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.
answered Jan 24 at 5:16
AndreiAndrei
12.9k21129
edited Jan 24 at 5:24
answered Jan 24 at 5:16
AndreiAndrei
12.9k21129
answered Jan 24 at 5:16
AndreiAndrei
12.9k21129
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
add a comment |
$begingroup$
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
add a comment |
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