Mixing
equation of the sphere
$begingroup$
In Einstein's "The Meaning of Relativity", https://en.wikisource.org/wiki/The_Meaning_of_Relativity/Lecture_1 I don't understand this passage:
"We shall now show briefly that there are geometrical entities which lead to the concept of tensors. Let $P_0$ be the centre of a surface of the second degree, $P$ any point on the surface, and $ξ_ν$ the projections of the interval $P_{0}P$ upon the co-ordinate axes. Then the equation of the surface is"
$a_{mu nu }xi _{mu }xi _{nu }=1$
The quantities a $displaystyle a_{mu nu }$ determine the surface completely, for a given position of the centre, with respect to the chosen system of Cartesian co-ordinates. From the known law of transformation for the $xi _{nu }$, (3a) for linear orthogonal transformations, we easily find the law of transformation for the a $a_{mu nu }$"
I don't understand how $a_{mu nu }$ was brought in, nor the form of the equation, (and by the way 1 is the radius?)
linear-algebra linear-transformations
asked Jan 22 at 10:10
Giuliano MalatestaGiuliano Malatesta
244
$endgroup$
add a comment |
$begingroup$
In Einstein's "The Meaning of Relativity", https://en.wikisource.org/wiki/The_Meaning_of_Relativity/Lecture_1 I don't understand this passage:
"We shall now show briefly that there are geometrical entities which lead to the concept of tensors. Let $P_0$ be the centre of a surface of the second degree, $P$ any point on the surface, and $ξ_ν$ the projections of the interval $P_{0}P$ upon the co-ordinate axes. Then the equation of the surface is"
$a_{mu nu }xi _{mu }xi _{nu }=1$
The quantities a $displaystyle a_{mu nu }$ determine the surface completely, for a given position of the centre, with respect to the chosen system of Cartesian co-ordinates. From the known law of transformation for the $xi _{nu }$, (3a) for linear orthogonal transformations, we easily find the law of transformation for the a $a_{mu nu }$"
I don't understand how $a_{mu nu }$ was brought in, nor the form of the equation, (and by the way 1 is the radius?)
linear-algebra linear-transformations
asked Jan 22 at 10:10
Giuliano MalatestaGiuliano Malatesta
244
$endgroup$
add a comment |
$begingroup$
In Einstein's "The Meaning of Relativity", https://en.wikisource.org/wiki/The_Meaning_of_Relativity/Lecture_1 I don't understand this passage:
"We shall now show briefly that there are geometrical entities which lead to the concept of tensors. Let $P_0$ be the centre of a surface of the second degree, $P$ any point on the surface, and $ξ_ν$ the projections of the interval $P_{0}P$ upon the co-ordinate axes. Then the equation of the surface is"
$a_{mu nu }xi _{mu }xi _{nu }=1$
The quantities a $displaystyle a_{mu nu }$ determine the surface completely, for a given position of the centre, with respect to the chosen system of Cartesian co-ordinates. From the known law of transformation for the $xi _{nu }$, (3a) for linear orthogonal transformations, we easily find the law of transformation for the a $a_{mu nu }$"
I don't understand how $a_{mu nu }$ was brought in, nor the form of the equation, (and by the way 1 is the radius?)
linear-algebra linear-transformations
asked Jan 22 at 10:10
Giuliano MalatestaGiuliano Malatesta
244
$endgroup$
In Einstein's "The Meaning of Relativity", https://en.wikisource.org/wiki/The_Meaning_of_Relativity/Lecture_1 I don't understand this passage:
"We shall now show briefly that there are geometrical entities which lead to the concept of tensors. Let $P_0$ be the centre of a surface of the second degree, $P$ any point on the surface, and $ξ_ν$ the projections of the interval $P_{0}P$ upon the co-ordinate axes. Then the equation of the surface is"
$a_{mu nu }xi _{mu }xi _{nu }=1$
The quantities a $displaystyle a_{mu nu }$ determine the surface completely, for a given position of the centre, with respect to the chosen system of Cartesian co-ordinates. From the known law of transformation for the $xi _{nu }$, (3a) for linear orthogonal transformations, we easily find the law of transformation for the a $a_{mu nu }$"
I don't understand how $a_{mu nu }$ was brought in, nor the form of the equation, (and by the way 1 is the radius?)
linear-algebra linear-transformations
linear-algebra linear-transformations
asked Jan 22 at 10:10
Giuliano MalatestaGiuliano Malatesta
244
asked Jan 22 at 10:10
Giuliano MalatestaGiuliano Malatesta
244
asked Jan 22 at 10:10
Giuliano MalatestaGiuliano Malatesta
244
asked Jan 22 at 10:10
Giuliano MalatestaGiuliano Malatesta
244
asked Jan 22 at 10:10
Giuliano MalatestaGiuliano Malatesta
244
244
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Just because it is easy to draw in 2D consider an ellipse centered at $P_0$
$$
left(frac{xi_1}{a}right)^2 + left(frac{xi_2}{b}right)^2 = 1 tag{1}
$$
for some numbers $a$ and $b$, this is the red ellipse in the sketch below.
Note that this equation can also be written as
$$
a_{11}xi_1xi_1 + a_{22}xi_2xi_2 = 1 ~~mbox{with}~~ a_{11} = 1/a^2, a_{22} = 1/b^2tag{2a}
$$
or as
$$
a_{munu}xi_muxi_nu = 1 ~~mbox{with}~~ a_{11} = 1/a^2, a_{22} = 1/b^2,a_{12} = a_{21} = 0tag{2b}
$$
A slightly more general case is the blue rotated ellipse on the sketch, for this one you can write the set of points that define it as
$$
Axi_1 xi_1 + B xi_1 xi_2 + Cxi_2 xi_2 = 1 tag{3a}
$$
For some appropriate constants $A$, $B$ and $C$. As before you can rewrite it as
$$
a_{munu}xi_muxi_nu = 1 tag{3b}
$$
which is again Eqn (2b). You can see the pattern here. You do not need to limit this to ellipses, you can also include hyperbolas. And this extends naturally to 3D, where you can include spheres, ellipsoids, hyperboloids of one and two sheets (...) The point is that if the surface is quadratic, you can always write as
$$
a_{munu}xi_muxi_nu = 1
$$
answered Jan 22 at 18:24
caveraccaverac
14.8k31130
$endgroup$
$begingroup$
the blue ellipse could be seen as the red one as seen from another frame of reference rotated with respect to the first? So that you obtain (3a) by $a_{sigma tau }'=b_{sigma mu }b_{tau nu }a_{mu nu} $ and then you have $a_{sigma tau }' xi_mu'xi_nu' = 1$ ?
$endgroup$
– Giuliano Malatesta
Jan 24 at 8:58
$begingroup$
@mattfick Correct, don't know if that is part of your reading, but that's basically how tensors trasnform
$endgroup$
– caverac
Jan 24 at 9:47
add a comment |
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1 Answer
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active
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1 Answer
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active
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$begingroup$
Just because it is easy to draw in 2D consider an ellipse centered at $P_0$
$$
left(frac{xi_1}{a}right)^2 + left(frac{xi_2}{b}right)^2 = 1 tag{1}
$$
for some numbers $a$ and $b$, this is the red ellipse in the sketch below.
Note that this equation can also be written as
$$
a_{11}xi_1xi_1 + a_{22}xi_2xi_2 = 1 ~~mbox{with}~~ a_{11} = 1/a^2, a_{22} = 1/b^2tag{2a}
$$
or as
$$
a_{munu}xi_muxi_nu = 1 ~~mbox{with}~~ a_{11} = 1/a^2, a_{22} = 1/b^2,a_{12} = a_{21} = 0tag{2b}
$$
A slightly more general case is the blue rotated ellipse on the sketch, for this one you can write the set of points that define it as
$$
Axi_1 xi_1 + B xi_1 xi_2 + Cxi_2 xi_2 = 1 tag{3a}
$$
For some appropriate constants $A$, $B$ and $C$. As before you can rewrite it as
$$
a_{munu}xi_muxi_nu = 1 tag{3b}
$$
which is again Eqn (2b). You can see the pattern here. You do not need to limit this to ellipses, you can also include hyperbolas. And this extends naturally to 3D, where you can include spheres, ellipsoids, hyperboloids of one and two sheets (...) The point is that if the surface is quadratic, you can always write as
$$
a_{munu}xi_muxi_nu = 1
$$
answered Jan 22 at 18:24
caveraccaverac
14.8k31130
$endgroup$
$begingroup$
the blue ellipse could be seen as the red one as seen from another frame of reference rotated with respect to the first? So that you obtain (3a) by $a_{sigma tau }'=b_{sigma mu }b_{tau nu }a_{mu nu} $ and then you have $a_{sigma tau }' xi_mu'xi_nu' = 1$ ?
$endgroup$
– Giuliano Malatesta
Jan 24 at 8:58
$begingroup$
@mattfick Correct, don't know if that is part of your reading, but that's basically how tensors trasnform
$endgroup$
– caverac
Jan 24 at 9:47
add a comment |
$begingroup$
Just because it is easy to draw in 2D consider an ellipse centered at $P_0$
$$
left(frac{xi_1}{a}right)^2 + left(frac{xi_2}{b}right)^2 = 1 tag{1}
$$
for some numbers $a$ and $b$, this is the red ellipse in the sketch below.
Note that this equation can also be written as
$$
a_{11}xi_1xi_1 + a_{22}xi_2xi_2 = 1 ~~mbox{with}~~ a_{11} = 1/a^2, a_{22} = 1/b^2tag{2a}
$$
or as
$$
a_{munu}xi_muxi_nu = 1 ~~mbox{with}~~ a_{11} = 1/a^2, a_{22} = 1/b^2,a_{12} = a_{21} = 0tag{2b}
$$
A slightly more general case is the blue rotated ellipse on the sketch, for this one you can write the set of points that define it as
$$
Axi_1 xi_1 + B xi_1 xi_2 + Cxi_2 xi_2 = 1 tag{3a}
$$
For some appropriate constants $A$, $B$ and $C$. As before you can rewrite it as
$$
a_{munu}xi_muxi_nu = 1 tag{3b}
$$
which is again Eqn (2b). You can see the pattern here. You do not need to limit this to ellipses, you can also include hyperbolas. And this extends naturally to 3D, where you can include spheres, ellipsoids, hyperboloids of one and two sheets (...) The point is that if the surface is quadratic, you can always write as
$$
a_{munu}xi_muxi_nu = 1
$$
answered Jan 22 at 18:24
caveraccaverac
14.8k31130
$endgroup$
$begingroup$
the blue ellipse could be seen as the red one as seen from another frame of reference rotated with respect to the first? So that you obtain (3a) by $a_{sigma tau }'=b_{sigma mu }b_{tau nu }a_{mu nu} $ and then you have $a_{sigma tau }' xi_mu'xi_nu' = 1$ ?
$endgroup$
– Giuliano Malatesta
Jan 24 at 8:58
$begingroup$
@mattfick Correct, don't know if that is part of your reading, but that's basically how tensors trasnform
$endgroup$
– caverac
Jan 24 at 9:47
add a comment |
$begingroup$
Just because it is easy to draw in 2D consider an ellipse centered at $P_0$
$$
left(frac{xi_1}{a}right)^2 + left(frac{xi_2}{b}right)^2 = 1 tag{1}
$$
for some numbers $a$ and $b$, this is the red ellipse in the sketch below.
Note that this equation can also be written as
$$
a_{11}xi_1xi_1 + a_{22}xi_2xi_2 = 1 ~~mbox{with}~~ a_{11} = 1/a^2, a_{22} = 1/b^2tag{2a}
$$
or as
$$
a_{munu}xi_muxi_nu = 1 ~~mbox{with}~~ a_{11} = 1/a^2, a_{22} = 1/b^2,a_{12} = a_{21} = 0tag{2b}
$$
A slightly more general case is the blue rotated ellipse on the sketch, for this one you can write the set of points that define it as
$$
Axi_1 xi_1 + B xi_1 xi_2 + Cxi_2 xi_2 = 1 tag{3a}
$$
For some appropriate constants $A$, $B$ and $C$. As before you can rewrite it as
$$
a_{munu}xi_muxi_nu = 1 tag{3b}
$$
which is again Eqn (2b). You can see the pattern here. You do not need to limit this to ellipses, you can also include hyperbolas. And this extends naturally to 3D, where you can include spheres, ellipsoids, hyperboloids of one and two sheets (...) The point is that if the surface is quadratic, you can always write as
$$
a_{munu}xi_muxi_nu = 1
$$
answered Jan 22 at 18:24
caveraccaverac
14.8k31130
$endgroup$
Just because it is easy to draw in 2D consider an ellipse centered at $P_0$
$$
left(frac{xi_1}{a}right)^2 + left(frac{xi_2}{b}right)^2 = 1 tag{1}
$$
for some numbers $a$ and $b$, this is the red ellipse in the sketch below.
Note that this equation can also be written as
$$
a_{11}xi_1xi_1 + a_{22}xi_2xi_2 = 1 ~~mbox{with}~~ a_{11} = 1/a^2, a_{22} = 1/b^2tag{2a}
$$
or as
$$
a_{munu}xi_muxi_nu = 1 ~~mbox{with}~~ a_{11} = 1/a^2, a_{22} = 1/b^2,a_{12} = a_{21} = 0tag{2b}
$$
A slightly more general case is the blue rotated ellipse on the sketch, for this one you can write the set of points that define it as
$$
Axi_1 xi_1 + B xi_1 xi_2 + Cxi_2 xi_2 = 1 tag{3a}
$$
For some appropriate constants $A$, $B$ and $C$. As before you can rewrite it as
$$
a_{munu}xi_muxi_nu = 1 tag{3b}
$$
which is again Eqn (2b). You can see the pattern here. You do not need to limit this to ellipses, you can also include hyperbolas. And this extends naturally to 3D, where you can include spheres, ellipsoids, hyperboloids of one and two sheets (...) The point is that if the surface is quadratic, you can always write as
$$
a_{munu}xi_muxi_nu = 1
$$
answered Jan 22 at 18:24
caveraccaverac
14.8k31130
answered Jan 22 at 18:24
caveraccaverac
14.8k31130
answered Jan 22 at 18:24
caveraccaverac
14.8k31130
answered Jan 22 at 18:24
caveraccaverac
14.8k31130
14.8k31130
$begingroup$
the blue ellipse could be seen as the red one as seen from another frame of reference rotated with respect to the first? So that you obtain (3a) by $a_{sigma tau }'=b_{sigma mu }b_{tau nu }a_{mu nu} $ and then you have $a_{sigma tau }' xi_mu'xi_nu' = 1$ ?
$endgroup$
– Giuliano Malatesta
Jan 24 at 8:58
$begingroup$
@mattfick Correct, don't know if that is part of your reading, but that's basically how tensors trasnform
$endgroup$
– caverac
Jan 24 at 9:47
add a comment |
$begingroup$
the blue ellipse could be seen as the red one as seen from another frame of reference rotated with respect to the first? So that you obtain (3a) by $a_{sigma tau }'=b_{sigma mu }b_{tau nu }a_{mu nu} $ and then you have $a_{sigma tau }' xi_mu'xi_nu' = 1$ ?
$endgroup$
– Giuliano Malatesta
Jan 24 at 8:58
$begingroup$
@mattfick Correct, don't know if that is part of your reading, but that's basically how tensors trasnform
$endgroup$
– caverac
Jan 24 at 9:47
$begingroup$
the blue ellipse could be seen as the red one as seen from another frame of reference rotated with respect to the first? So that you obtain (3a) by $a_{sigma tau }'=b_{sigma mu }b_{tau nu }a_{mu nu} $ and then you have $a_{sigma tau }' xi_mu'xi_nu' = 1$ ?
$endgroup$
– Giuliano Malatesta
Jan 24 at 8:58
$begingroup$
the blue ellipse could be seen as the red one as seen from another frame of reference rotated with respect to the first? So that you obtain (3a) by $a_{sigma tau }'=b_{sigma mu }b_{tau nu }a_{mu nu} $ and then you have $a_{sigma tau }' xi_mu'xi_nu' = 1$ ?
$endgroup$
– Giuliano Malatesta
Jan 24 at 8:58
$begingroup$
@mattfick Correct, don't know if that is part of your reading, but that's basically how tensors trasnform
$endgroup$
– caverac
Jan 24 at 9:47
$begingroup$
@mattfick Correct, don't know if that is part of your reading, but that's basically how tensors trasnform
$endgroup$
– caverac
Jan 24 at 9:47
add a comment |
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