Toroidal frame system
$begingroup$
*Context:
I'm following the textbook of Barrett O'Neill about differential geometry. In order to calculate the connection forms of the torous I need the relation between a toroidal frame and the natural frame. *
I'm trying to construct a toridal frame system in terms of the natural frame system ${hat{e}_x,,hat{e}_y,,hat{e}_z}$, where $hat{e}_x=(1,,0,,0)$, etc. The same as with an spherical frame ${hat{e}_r,,hat{e}_theta,,hat{e}_phi}$, where:
$hat{e}_r=cos(phi)cos(theta)hat{e}_x+cos(phi)sin(theta)hat{e}_y+sin(phi)hat{e}_z$
$hat{e}_theta=-sin(theta)hat{e}_x+cos(theta)hat{e}_y$
$hat{e}_phi=-sin(phi)cos(theta)hat{e}_x-sin(phi)sin(theta)hat{e}_y+cos(phi)hat{e}_z$
My attempt was to construct parametric curves, then take its derivative to make a tangent vector and finally make it a unit vector (this seems to work fine por cylindrical and spherical frames). For example, i'll take the next curve on the tourus with its standar coordinates
$alpha(phi)=((R+rho_o,cos(phi)),cos(theta_o),,(R+rho_o,cos(phi)),sin(theta_o),,rho_o,\sin(phi))$
where $R$, $rho_o$ and $theta_o$ are fixed (i'll just take a point on the torous with its standar coordinates and fix two of them). Then:
$alpha'(phi)=(-rho_o,sin(phi),cos(theta_o),,-rho_o,sin(phi),sin(theta_o),,rho_o,cos(phi))$
where it's easy to see that $|alpha'(phi)|=rho_o$. Finally:
$hat{e}_phi=-sin(phi)cos(theta)hat{e}_x-sin(phi)sin(theta)hat{e}_y+cos(phi)hat{e}_z$
wich is the same as in spherical frame and this happens with the rest. So, I came to this: Toroidal frame is same as spherical frame...And this jus't feels wrong.
So, I would love and I will appreciate some help here
differential-geometry frenet-frame
asked Dec 4 '18 at 22:35
raulraul
1108
$endgroup$
add a comment |
$begingroup$
*Context:
I'm following the textbook of Barrett O'Neill about differential geometry. In order to calculate the connection forms of the torous I need the relation between a toroidal frame and the natural frame. *
I'm trying to construct a toridal frame system in terms of the natural frame system ${hat{e}_x,,hat{e}_y,,hat{e}_z}$, where $hat{e}_x=(1,,0,,0)$, etc. The same as with an spherical frame ${hat{e}_r,,hat{e}_theta,,hat{e}_phi}$, where:
$hat{e}_r=cos(phi)cos(theta)hat{e}_x+cos(phi)sin(theta)hat{e}_y+sin(phi)hat{e}_z$
$hat{e}_theta=-sin(theta)hat{e}_x+cos(theta)hat{e}_y$
$hat{e}_phi=-sin(phi)cos(theta)hat{e}_x-sin(phi)sin(theta)hat{e}_y+cos(phi)hat{e}_z$
My attempt was to construct parametric curves, then take its derivative to make a tangent vector and finally make it a unit vector (this seems to work fine por cylindrical and spherical frames). For example, i'll take the next curve on the tourus with its standar coordinates
$alpha(phi)=((R+rho_o,cos(phi)),cos(theta_o),,(R+rho_o,cos(phi)),sin(theta_o),,rho_o,\sin(phi))$
where $R$, $rho_o$ and $theta_o$ are fixed (i'll just take a point on the torous with its standar coordinates and fix two of them). Then:
$alpha'(phi)=(-rho_o,sin(phi),cos(theta_o),,-rho_o,sin(phi),sin(theta_o),,rho_o,cos(phi))$
where it's easy to see that $|alpha'(phi)|=rho_o$. Finally:
$hat{e}_phi=-sin(phi)cos(theta)hat{e}_x-sin(phi)sin(theta)hat{e}_y+cos(phi)hat{e}_z$
wich is the same as in spherical frame and this happens with the rest. So, I came to this: Toroidal frame is same as spherical frame...And this jus't feels wrong.
So, I would love and I will appreciate some help here
differential-geometry frenet-frame
asked Dec 4 '18 at 22:35
raulraul
1108
$endgroup$
$begingroup$
How does this happen with the rest? And I'm not sure why you're doing this to find the connection form of the torus. I don't remember precisely what O'Neil does, but don't you have the metric in terms of $phi$ and $theta$? This gives you an orthonormal coframe, and then the structure equations are immediately in play.
$endgroup$
– Ted Shifrin
Dec 4 '18 at 23:19
$begingroup$
@TedShifrin I think I understand my mistake. They are actually share the same expresión and because of that have the same connection forms $omega_{ij}$. The only thing that changes is how they act wit respect functions; i.e. While $hat{e}_theta[f]={1over rcos(phi)}{partial foverpartialtheta}$ in a espherical frame, in a toroidal frame will be $hat{e}_theta[f]={1over R+rhocos(phi)}{partial foverpartialtheta}$
$endgroup$
– raul
Dec 5 '18 at 0:21
add a comment |
$begingroup$
*Context:
I'm following the textbook of Barrett O'Neill about differential geometry. In order to calculate the connection forms of the torous I need the relation between a toroidal frame and the natural frame. *
I'm trying to construct a toridal frame system in terms of the natural frame system ${hat{e}_x,,hat{e}_y,,hat{e}_z}$, where $hat{e}_x=(1,,0,,0)$, etc. The same as with an spherical frame ${hat{e}_r,,hat{e}_theta,,hat{e}_phi}$, where:
$hat{e}_r=cos(phi)cos(theta)hat{e}_x+cos(phi)sin(theta)hat{e}_y+sin(phi)hat{e}_z$
$hat{e}_theta=-sin(theta)hat{e}_x+cos(theta)hat{e}_y$
$hat{e}_phi=-sin(phi)cos(theta)hat{e}_x-sin(phi)sin(theta)hat{e}_y+cos(phi)hat{e}_z$
My attempt was to construct parametric curves, then take its derivative to make a tangent vector and finally make it a unit vector (this seems to work fine por cylindrical and spherical frames). For example, i'll take the next curve on the tourus with its standar coordinates
$alpha(phi)=((R+rho_o,cos(phi)),cos(theta_o),,(R+rho_o,cos(phi)),sin(theta_o),,rho_o,\sin(phi))$
where $R$, $rho_o$ and $theta_o$ are fixed (i'll just take a point on the torous with its standar coordinates and fix two of them). Then:
$alpha'(phi)=(-rho_o,sin(phi),cos(theta_o),,-rho_o,sin(phi),sin(theta_o),,rho_o,cos(phi))$
where it's easy to see that $|alpha'(phi)|=rho_o$. Finally:
$hat{e}_phi=-sin(phi)cos(theta)hat{e}_x-sin(phi)sin(theta)hat{e}_y+cos(phi)hat{e}_z$
wich is the same as in spherical frame and this happens with the rest. So, I came to this: Toroidal frame is same as spherical frame...And this jus't feels wrong.
So, I would love and I will appreciate some help here
differential-geometry frenet-frame
asked Dec 4 '18 at 22:35
raulraul
1108
$endgroup$
*Context:
I'm following the textbook of Barrett O'Neill about differential geometry. In order to calculate the connection forms of the torous I need the relation between a toroidal frame and the natural frame. *
I'm trying to construct a toridal frame system in terms of the natural frame system ${hat{e}_x,,hat{e}_y,,hat{e}_z}$, where $hat{e}_x=(1,,0,,0)$, etc. The same as with an spherical frame ${hat{e}_r,,hat{e}_theta,,hat{e}_phi}$, where:
$hat{e}_r=cos(phi)cos(theta)hat{e}_x+cos(phi)sin(theta)hat{e}_y+sin(phi)hat{e}_z$
$hat{e}_theta=-sin(theta)hat{e}_x+cos(theta)hat{e}_y$
$hat{e}_phi=-sin(phi)cos(theta)hat{e}_x-sin(phi)sin(theta)hat{e}_y+cos(phi)hat{e}_z$
My attempt was to construct parametric curves, then take its derivative to make a tangent vector and finally make it a unit vector (this seems to work fine por cylindrical and spherical frames). For example, i'll take the next curve on the tourus with its standar coordinates
$alpha(phi)=((R+rho_o,cos(phi)),cos(theta_o),,(R+rho_o,cos(phi)),sin(theta_o),,rho_o,\sin(phi))$
where $R$, $rho_o$ and $theta_o$ are fixed (i'll just take a point on the torous with its standar coordinates and fix two of them). Then:
$alpha'(phi)=(-rho_o,sin(phi),cos(theta_o),,-rho_o,sin(phi),sin(theta_o),,rho_o,cos(phi))$
where it's easy to see that $|alpha'(phi)|=rho_o$. Finally:
$hat{e}_phi=-sin(phi)cos(theta)hat{e}_x-sin(phi)sin(theta)hat{e}_y+cos(phi)hat{e}_z$
wich is the same as in spherical frame and this happens with the rest. So, I came to this: Toroidal frame is same as spherical frame...And this jus't feels wrong.
So, I would love and I will appreciate some help here
differential-geometry frenet-frame
differential-geometry frenet-frame
asked Dec 4 '18 at 22:35
raulraul
1108
asked Dec 4 '18 at 22:35
raulraul
1108
edited Dec 4 '18 at 22:43
raul
asked Dec 4 '18 at 22:35
raulraul
1108
asked Dec 4 '18 at 22:35
raulraul
1108
asked Dec 4 '18 at 22:35
raulraul
1108
1108
$begingroup$
How does this happen with the rest? And I'm not sure why you're doing this to find the connection form of the torus. I don't remember precisely what O'Neil does, but don't you have the metric in terms of $phi$ and $theta$? This gives you an orthonormal coframe, and then the structure equations are immediately in play.
$endgroup$
– Ted Shifrin
Dec 4 '18 at 23:19
$begingroup$
@TedShifrin I think I understand my mistake. They are actually share the same expresión and because of that have the same connection forms $omega_{ij}$. The only thing that changes is how they act wit respect functions; i.e. While $hat{e}_theta[f]={1over rcos(phi)}{partial foverpartialtheta}$ in a espherical frame, in a toroidal frame will be $hat{e}_theta[f]={1over R+rhocos(phi)}{partial foverpartialtheta}$
$endgroup$
– raul
Dec 5 '18 at 0:21
add a comment |
$begingroup$
How does this happen with the rest? And I'm not sure why you're doing this to find the connection form of the torus. I don't remember precisely what O'Neil does, but don't you have the metric in terms of $phi$ and $theta$? This gives you an orthonormal coframe, and then the structure equations are immediately in play.
$endgroup$
– Ted Shifrin
Dec 4 '18 at 23:19
$begingroup$
@TedShifrin I think I understand my mistake. They are actually share the same expresión and because of that have the same connection forms $omega_{ij}$. The only thing that changes is how they act wit respect functions; i.e. While $hat{e}_theta[f]={1over rcos(phi)}{partial foverpartialtheta}$ in a espherical frame, in a toroidal frame will be $hat{e}_theta[f]={1over R+rhocos(phi)}{partial foverpartialtheta}$
$endgroup$
– raul
Dec 5 '18 at 0:21
$begingroup$
How does this happen with the rest? And I'm not sure why you're doing this to find the connection form of the torus. I don't remember precisely what O'Neil does, but don't you have the metric in terms of $phi$ and $theta$? This gives you an orthonormal coframe, and then the structure equations are immediately in play.
$endgroup$
– Ted Shifrin
Dec 4 '18 at 23:19
$begingroup$
How does this happen with the rest? And I'm not sure why you're doing this to find the connection form of the torus. I don't remember precisely what O'Neil does, but don't you have the metric in terms of $phi$ and $theta$? This gives you an orthonormal coframe, and then the structure equations are immediately in play.
$endgroup$
– Ted Shifrin
Dec 4 '18 at 23:19
$begingroup$
@TedShifrin I think I understand my mistake. They are actually share the same expresión and because of that have the same connection forms $omega_{ij}$. The only thing that changes is how they act wit respect functions; i.e. While $hat{e}_theta[f]={1over rcos(phi)}{partial foverpartialtheta}$ in a espherical frame, in a toroidal frame will be $hat{e}_theta[f]={1over R+rhocos(phi)}{partial foverpartialtheta}$
$endgroup$
– raul
Dec 5 '18 at 0:21
$begingroup$
@TedShifrin I think I understand my mistake. They are actually share the same expresión and because of that have the same connection forms $omega_{ij}$. The only thing that changes is how they act wit respect functions; i.e. While $hat{e}_theta[f]={1over rcos(phi)}{partial foverpartialtheta}$ in a espherical frame, in a toroidal frame will be $hat{e}_theta[f]={1over R+rhocos(phi)}{partial foverpartialtheta}$
$endgroup$
– raul
Dec 5 '18 at 0:21
add a comment |
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$begingroup$
How does this happen with the rest? And I'm not sure why you're doing this to find the connection form of the torus. I don't remember precisely what O'Neil does, but don't you have the metric in terms of $phi$ and $theta$? This gives you an orthonormal coframe, and then the structure equations are immediately in play.
$endgroup$
– Ted Shifrin
Dec 4 '18 at 23:19
$begingroup$
@TedShifrin I think I understand my mistake. They are actually share the same expresión and because of that have the same connection forms $omega_{ij}$. The only thing that changes is how they act wit respect functions; i.e. While $hat{e}_theta[f]={1over rcos(phi)}{partial foverpartialtheta}$ in a espherical frame, in a toroidal frame will be $hat{e}_theta[f]={1over R+rhocos(phi)}{partial foverpartialtheta}$
$endgroup$
– raul
Dec 5 '18 at 0:21