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Oblate Spheroidal coordinate system graphic representation of ellipse

0

$begingroup$

I am having difficulty understanding how to interpret the coordinate system proposed by Spencer. From his handbook Field Theory Handbooks ISBN 9783540184300 proposed a system with the domains illustrated in the picture.

The domain of eta represents the curvilinear axis represented by the ellipse rotating around the minor axis. What exactly is the characteristic of the ellipse in the picture? Is 'Eta' representing a number describing the minor or mayor axis of the ellipse?

conic-sections coordinate-systems

share|cite|improve this question

edited Mar 26 at 6:30

J. M. is a poor mathematician

61.3k5152291

asked Mar 25 at 18:12

Jose Enrique CalderonJose Enrique Calderon

1177

$endgroup$

add a comment | 

0

$begingroup$

I am having difficulty understanding how to interpret the coordinate system proposed by Spencer. From his handbook Field Theory Handbooks ISBN 9783540184300 proposed a system with the domains illustrated in the picture.

The domain of eta represents the curvilinear axis represented by the ellipse rotating around the minor axis. What exactly is the characteristic of the ellipse in the picture? Is 'Eta' representing a number describing the minor or mayor axis of the ellipse?

conic-sections coordinate-systems

share|cite|improve this question

edited Mar 26 at 6:30

J. M. is a poor mathematician

61.3k5152291

asked Mar 25 at 18:12

Jose Enrique CalderonJose Enrique Calderon

1177

$endgroup$

add a comment | 

$begingroup$

I am having difficulty understanding how to interpret the coordinate system proposed by Spencer. From his handbook Field Theory Handbooks ISBN 9783540184300 proposed a system with the domains illustrated in the picture.

The domain of eta represents the curvilinear axis represented by the ellipse rotating around the minor axis. What exactly is the characteristic of the ellipse in the picture? Is 'Eta' representing a number describing the minor or mayor axis of the ellipse?

conic-sections coordinate-systems

share|cite|improve this question

edited Mar 26 at 6:30

J. M. is a poor mathematician

61.3k5152291

asked Mar 25 at 18:12

Jose Enrique CalderonJose Enrique Calderon

1177

$endgroup$

share|cite|improve this question

edited Mar 26 at 6:30

J. M. is a poor mathematician

61.3k5152291

asked Mar 25 at 18:12

Jose Enrique CalderonJose Enrique Calderon

1177

share|cite|improve this question

edited Mar 26 at 6:30

J. M. is a poor mathematician

61.3k5152291

asked Mar 25 at 18:12

Jose Enrique CalderonJose Enrique Calderon

1177

edited Mar 26 at 6:30

J. M. is a poor mathematician

61.3k5152291

edited Mar 26 at 6:30

J. M. is a poor mathematician

61.3k5152291

asked Mar 25 at 18:12

Jose Enrique CalderonJose Enrique Calderon

1177

asked Mar 25 at 18:12

Jose Enrique CalderonJose Enrique Calderon

1177

2 Answers
2

active

oldest

votes

1

$begingroup$

Neither or both, depending on your point of view. Refer to the illustration: the surfaces with $eta$ held constant are oblate spheriods. From the equations of the transformation to Cartesian coordinates, these spheroids have the parameterization $$begin{align}x &= (acosheta)sinthetacospsi \ y &= (acosheta)sinthetasinpsi \ z &= (asinheta)costheta.end{align}$$ I’ve added parentheses to emphasize the constant quantity in each equation. The half-axis lengths of the spheroid are therefore $acosheta$ and $asinheta$. Similarly, the surfaces with constant $theta$ are hyperboloids of revolution with half-axis lengths $acostheta$ and $asintheta$. The angle $theta$ represents the aperture half-angle of the hyperboloid’s asymptotic cone. Surfaces with constant $psi$ are half-planes that make an angle of $psi$ with the positive $x$-$z$ half-plane.

share|cite|improve this answer

edited Mar 26 at 4:34

answered Mar 25 at 18:41

amdamd

31.9k21053

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks for your clear details of the constant. I could not see this.
  $endgroup$
  – Jose Enrique Calderon
  Mar 25 at 22:32
  

  
  
  
  

add a comment | 

1

$begingroup$

For any $psi=$ constant sectioning plane containing z-axis we have confocal conic sections with the marked $F$ as common focal locus for variable $eta$ value on all ellipses formed by cutting planes$psi=$ constant.

Parametrization for horizontal ellipse sections

$$ r=sqrt{x^2+y^2} = a cosh eta sin theta ;, z = a sinh eta cos theta; $$
$$ big(frac{r}{a cosh eta}big)^2 + big(frac{r}{a sinh eta}big)^2 =1 $$
$$ epsilon^2= 1- frac{sinh^2 eta}{cosh ^2eta}= sech^2 eta $$

So have $epsilon =sech, eta$ geometrically interpreted as eccentricity of horizontal ellipses formed by cutting planes parallel to $x-y$ plane.

Circular conicoids when $eta rightarrow infty.$

share|cite|improve this answer

edited Mar 26 at 6:49

answered Mar 25 at 19:18

NarasimhamNarasimham

21.3k62258

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @ Narasimham From where is derrived the eccentricity= 𝜖=𝑠𝑒𝑐ℎ𝜂 ?
  $endgroup$
  – Jose Enrique Calderon
  Mar 27 at 12:47
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Is the edit clear enough? $ epsilon = sech eta le 1 $
  $endgroup$
  – Narasimham
  Mar 27 at 16:27
  

  
  
  
  

add a comment | 

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1

$begingroup$

Neither or both, depending on your point of view. Refer to the illustration: the surfaces with $eta$ held constant are oblate spheriods. From the equations of the transformation to Cartesian coordinates, these spheroids have the parameterization $$begin{align}x &= (acosheta)sinthetacospsi \ y &= (acosheta)sinthetasinpsi \ z &= (asinheta)costheta.end{align}$$ I’ve added parentheses to emphasize the constant quantity in each equation. The half-axis lengths of the spheroid are therefore $acosheta$ and $asinheta$. Similarly, the surfaces with constant $theta$ are hyperboloids of revolution with half-axis lengths $acostheta$ and $asintheta$. The angle $theta$ represents the aperture half-angle of the hyperboloid’s asymptotic cone. Surfaces with constant $psi$ are half-planes that make an angle of $psi$ with the positive $x$-$z$ half-plane.

share|cite|improve this answer

edited Mar 26 at 4:34

answered Mar 25 at 18:41

amdamd

31.9k21053

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks for your clear details of the constant. I could not see this.
  $endgroup$
  – Jose Enrique Calderon
  Mar 25 at 22:32
  

  
  
  
  

add a comment | 

1

$begingroup$

Neither or both, depending on your point of view. Refer to the illustration: the surfaces with $eta$ held constant are oblate spheriods. From the equations of the transformation to Cartesian coordinates, these spheroids have the parameterization $$begin{align}x &= (acosheta)sinthetacospsi \ y &= (acosheta)sinthetasinpsi \ z &= (asinheta)costheta.end{align}$$ I’ve added parentheses to emphasize the constant quantity in each equation. The half-axis lengths of the spheroid are therefore $acosheta$ and $asinheta$. Similarly, the surfaces with constant $theta$ are hyperboloids of revolution with half-axis lengths $acostheta$ and $asintheta$. The angle $theta$ represents the aperture half-angle of the hyperboloid’s asymptotic cone. Surfaces with constant $psi$ are half-planes that make an angle of $psi$ with the positive $x$-$z$ half-plane.

share|cite|improve this answer

edited Mar 26 at 4:34

answered Mar 25 at 18:41

amdamd

31.9k21053

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks for your clear details of the constant. I could not see this.
  $endgroup$
  – Jose Enrique Calderon
  Mar 25 at 22:32
  

  
  
  
  

add a comment | 

$begingroup$

Neither or both, depending on your point of view. Refer to the illustration: the surfaces with $eta$ held constant are oblate spheriods. From the equations of the transformation to Cartesian coordinates, these spheroids have the parameterization $$begin{align}x &= (acosheta)sinthetacospsi \ y &= (acosheta)sinthetasinpsi \ z &= (asinheta)costheta.end{align}$$ I’ve added parentheses to emphasize the constant quantity in each equation. The half-axis lengths of the spheroid are therefore $acosheta$ and $asinheta$. Similarly, the surfaces with constant $theta$ are hyperboloids of revolution with half-axis lengths $acostheta$ and $asintheta$. The angle $theta$ represents the aperture half-angle of the hyperboloid’s asymptotic cone. Surfaces with constant $psi$ are half-planes that make an angle of $psi$ with the positive $x$-$z$ half-plane.

share|cite|improve this answer

edited Mar 26 at 4:34

answered Mar 25 at 18:41

amdamd

31.9k21053

$endgroup$

share|cite|improve this answer

edited Mar 26 at 4:34

answered Mar 25 at 18:41

amdamd

31.9k21053

answered Mar 25 at 18:41

amdamd

31.9k21053

answered Mar 25 at 18:41

amdamd

31.9k21053

1

$begingroup$

For any $psi=$ constant sectioning plane containing z-axis we have confocal conic sections with the marked $F$ as common focal locus for variable $eta$ value on all ellipses formed by cutting planes$psi=$ constant.

Parametrization for horizontal ellipse sections

$$ r=sqrt{x^2+y^2} = a cosh eta sin theta ;, z = a sinh eta cos theta; $$
$$ big(frac{r}{a cosh eta}big)^2 + big(frac{r}{a sinh eta}big)^2 =1 $$
$$ epsilon^2= 1- frac{sinh^2 eta}{cosh ^2eta}= sech^2 eta $$

So have $epsilon =sech, eta$ geometrically interpreted as eccentricity of horizontal ellipses formed by cutting planes parallel to $x-y$ plane.

Circular conicoids when $eta rightarrow infty.$

share|cite|improve this answer

edited Mar 26 at 6:49

answered Mar 25 at 19:18

NarasimhamNarasimham

21.3k62258

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @ Narasimham From where is derrived the eccentricity= 𝜖=𝑠𝑒𝑐ℎ𝜂 ?
  $endgroup$
  – Jose Enrique Calderon
  Mar 27 at 12:47
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Is the edit clear enough? $ epsilon = sech eta le 1 $
  $endgroup$
  – Narasimham
  Mar 27 at 16:27
  

  
  
  
  

add a comment | 

1

$begingroup$

For any $psi=$ constant sectioning plane containing z-axis we have confocal conic sections with the marked $F$ as common focal locus for variable $eta$ value on all ellipses formed by cutting planes$psi=$ constant.

Parametrization for horizontal ellipse sections

$$ r=sqrt{x^2+y^2} = a cosh eta sin theta ;, z = a sinh eta cos theta; $$
$$ big(frac{r}{a cosh eta}big)^2 + big(frac{r}{a sinh eta}big)^2 =1 $$
$$ epsilon^2= 1- frac{sinh^2 eta}{cosh ^2eta}= sech^2 eta $$

So have $epsilon =sech, eta$ geometrically interpreted as eccentricity of horizontal ellipses formed by cutting planes parallel to $x-y$ plane.

Circular conicoids when $eta rightarrow infty.$

share|cite|improve this answer

edited Mar 26 at 6:49

answered Mar 25 at 19:18

NarasimhamNarasimham

21.3k62258

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @ Narasimham From where is derrived the eccentricity= 𝜖=𝑠𝑒𝑐ℎ𝜂 ?
  $endgroup$
  – Jose Enrique Calderon
  Mar 27 at 12:47
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Is the edit clear enough? $ epsilon = sech eta le 1 $
  $endgroup$
  – Narasimham
  Mar 27 at 16:27
  

  
  
  
  

add a comment | 

$begingroup$

For any $psi=$ constant sectioning plane containing z-axis we have confocal conic sections with the marked $F$ as common focal locus for variable $eta$ value on all ellipses formed by cutting planes$psi=$ constant.

Parametrization for horizontal ellipse sections

$$ r=sqrt{x^2+y^2} = a cosh eta sin theta ;, z = a sinh eta cos theta; $$
$$ big(frac{r}{a cosh eta}big)^2 + big(frac{r}{a sinh eta}big)^2 =1 $$
$$ epsilon^2= 1- frac{sinh^2 eta}{cosh ^2eta}= sech^2 eta $$

So have $epsilon =sech, eta$ geometrically interpreted as eccentricity of horizontal ellipses formed by cutting planes parallel to $x-y$ plane.

Circular conicoids when $eta rightarrow infty.$

share|cite|improve this answer

edited Mar 26 at 6:49

answered Mar 25 at 19:18

NarasimhamNarasimham

21.3k62258

$endgroup$

share|cite|improve this answer

edited Mar 26 at 6:49

answered Mar 25 at 19:18

NarasimhamNarasimham

21.3k62258

answered Mar 25 at 19:18

NarasimhamNarasimham

21.3k62258

answered Mar 25 at 19:18

NarasimhamNarasimham

21.3k62258