Intuition for orientation of tri-vectors in geometric algebra
$begingroup$
I am learning geometric algebra from the MacDonald textbook and it states that the outer product is associative. Letting $bf{u}$, $bf{v}$, and $bf{w}$ be vectors
$$bf{u} wedge bf{v} wedge bf{w}= (bf{u} wedge bf{v}) wedge bf{w} =bf{u} wedge (bf{v} wedge bf{w}).$$
But the book describes the volume created by
$(bf{u} wedge bf{v}) wedge bf{w}$
has orientation given by $bf{u} wedge bf{v}$, while $bf{u} wedge (bf{v} wedge bf{w})$ has orientation given by $bf{v} wedge bf{w}$.
I understand that they are the same volume, but what I'm not getting is the understanding of what the orientation of an oriented volume actually is. Is it saying that $bf{u} wedge bf{v}$ and $bf{v} wedge bf{w}$ have the same orientation even though they are not in parallel planes?
Also, any tips on gaining an overall intuition of orientation of higher dimensional objects would be greatly appreciated.
UPDATE
An updated version of the section in question has been posted here:
http://www.faculty.luther.edu/~macdonal/laga/LAGA%20Section%205.2.pdf
This rewritten section provides a different explanation than above.
geometry vectors geometric-algebras
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This question came from our site for active researchers, academics and students of physics.
add a comment |
$begingroup$
I am learning geometric algebra from the MacDonald textbook and it states that the outer product is associative. Letting $bf{u}$, $bf{v}$, and $bf{w}$ be vectors
$$bf{u} wedge bf{v} wedge bf{w}= (bf{u} wedge bf{v}) wedge bf{w} =bf{u} wedge (bf{v} wedge bf{w}).$$
But the book describes the volume created by
$(bf{u} wedge bf{v}) wedge bf{w}$
has orientation given by $bf{u} wedge bf{v}$, while $bf{u} wedge (bf{v} wedge bf{w})$ has orientation given by $bf{v} wedge bf{w}$.
I understand that they are the same volume, but what I'm not getting is the understanding of what the orientation of an oriented volume actually is. Is it saying that $bf{u} wedge bf{v}$ and $bf{v} wedge bf{w}$ have the same orientation even though they are not in parallel planes?
Also, any tips on gaining an overall intuition of orientation of higher dimensional objects would be greatly appreciated.
UPDATE
An updated version of the section in question has been posted here:
http://www.faculty.luther.edu/~macdonal/laga/LAGA%20Section%205.2.pdf
This rewritten section provides a different explanation than above.
geometry vectors geometric-algebras
$endgroup$
migrated from physics.stackexchange.com Jan 15 at 21:54
This question came from our site for active researchers, academics and students of physics.
add a comment |
$begingroup$
I am learning geometric algebra from the MacDonald textbook and it states that the outer product is associative. Letting $bf{u}$, $bf{v}$, and $bf{w}$ be vectors
$$bf{u} wedge bf{v} wedge bf{w}= (bf{u} wedge bf{v}) wedge bf{w} =bf{u} wedge (bf{v} wedge bf{w}).$$
But the book describes the volume created by
$(bf{u} wedge bf{v}) wedge bf{w}$
has orientation given by $bf{u} wedge bf{v}$, while $bf{u} wedge (bf{v} wedge bf{w})$ has orientation given by $bf{v} wedge bf{w}$.
I understand that they are the same volume, but what I'm not getting is the understanding of what the orientation of an oriented volume actually is. Is it saying that $bf{u} wedge bf{v}$ and $bf{v} wedge bf{w}$ have the same orientation even though they are not in parallel planes?
Also, any tips on gaining an overall intuition of orientation of higher dimensional objects would be greatly appreciated.
UPDATE
An updated version of the section in question has been posted here:
http://www.faculty.luther.edu/~macdonal/laga/LAGA%20Section%205.2.pdf
This rewritten section provides a different explanation than above.
geometry vectors geometric-algebras
$endgroup$
I am learning geometric algebra from the MacDonald textbook and it states that the outer product is associative. Letting $bf{u}$, $bf{v}$, and $bf{w}$ be vectors
$$bf{u} wedge bf{v} wedge bf{w}= (bf{u} wedge bf{v}) wedge bf{w} =bf{u} wedge (bf{v} wedge bf{w}).$$
But the book describes the volume created by
$(bf{u} wedge bf{v}) wedge bf{w}$
has orientation given by $bf{u} wedge bf{v}$, while $bf{u} wedge (bf{v} wedge bf{w})$ has orientation given by $bf{v} wedge bf{w}$.
I understand that they are the same volume, but what I'm not getting is the understanding of what the orientation of an oriented volume actually is. Is it saying that $bf{u} wedge bf{v}$ and $bf{v} wedge bf{w}$ have the same orientation even though they are not in parallel planes?
Also, any tips on gaining an overall intuition of orientation of higher dimensional objects would be greatly appreciated.
UPDATE
An updated version of the section in question has been posted here:
http://www.faculty.luther.edu/~macdonal/laga/LAGA%20Section%205.2.pdf
This rewritten section provides a different explanation than above.
geometry vectors geometric-algebras
geometry vectors geometric-algebras
edited Jan 21 at 19:21
roshoka
asked Jan 14 at 21:35
roshokaroshoka
93
93
migrated from physics.stackexchange.com Jan 15 at 21:54
This question came from our site for active researchers, academics and students of physics.
migrated from physics.stackexchange.com Jan 15 at 21:54
This question came from our site for active researchers, academics and students of physics.
add a comment |
add a comment |
2 Answers
2
active
oldest
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$begingroup$
The orientation of $bf{u} wedge bf{v}$ is given by the orientation of $bf{u}$ on the boundary of $bf{u} wedge bf{v}$. See Fig. 5.5 in my book and the accompanying text.
Similarly, the orientation of $bf{u} wedge bf{v} wedge bf{w}$ is given by the orientation of $bf{u} wedge bf{v}$ (or $bf{v} wedge bf{w}$ or $bf{w} wedge bf{u}$) on the boundary of $bf{u} wedge bf{v} wedge bf{w}$. All are clockwise.
Perhaps this is a better explanation than in my book.
$endgroup$
$begingroup$
Thanks for the reply. Note 3 on pg 75 of your book says that orientation of areas in a plane can only be compared, but can you compare them if they aren't in the same plane?
$endgroup$
– roshoka
Jan 16 at 3:03
1
$begingroup$
Well, this is embarrassing: The orientation aspects of Section 5.2 and my reply yesterday are wrongheaded. I am rewriting the section and will post it as a pdf on the book's website within a few days. My apologies.
$endgroup$
– Alan Macdonald
Jan 16 at 20:40
1
$begingroup$
Orientations of areas not in parallel planes cannot be compared.
$endgroup$
– Alan Macdonald
Jan 16 at 20:41
$begingroup$
Alright, I'll keep an eye out for the update. Thanks
$endgroup$
– roshoka
Jan 17 at 16:53
1
$begingroup$
The outer product is definitely associative. My explanation of this fact was incorrect. There is no problem proceeding with the text if you accept the associativity.
$endgroup$
– Alan Macdonald
Jan 18 at 16:58
|
show 2 more comments
$begingroup$
I'm not familiar with MacDonalds book, but this does not sound right to me; the orientation of a volume element $uwedge v wedge w$ should depend upon all three factors and not just two of them.
The outer product is otherwise known as the wedge or exterior product. It is a signed version of the tensor product, hence used when orientation matters.
They generalise vectors; recall that vectors have direction and magnitude. The outer product of two vectors give a '2d vector', an orientated area element, this ought to be thought of as the generalisation of direction to higher dimensions, where instead of working in higher dimensions per se - for example in a 10d or 20d space - we are working with higher dimensional vectors; the generalisation of magnitude then is the area.
The generalisation to 3d vectors is then obvious: the outer product of three vectors gives an orientated volume element and the magnitude is now the volume.
This explains why the outer product of a vector by itself always vanishes; this is because the area element degenerates into a line element, and hence it's area vanishes; likewise its volume.
The arithmetic of the outer product shows that a generalisation of the triangle law still holds, so long as the two area elements in question have an edge of common magnitude; it's also important to note that the multi-linearity of the outer product means that we can rescale the area element in such a way that it's total area does not change; for example we can magnify one edge by a scale factor $k$ so long as we reduce the other side by $k$ - this has no parallel in the arithmetic of vectors.
$endgroup$
add a comment |
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2 Answers
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$begingroup$
The orientation of $bf{u} wedge bf{v}$ is given by the orientation of $bf{u}$ on the boundary of $bf{u} wedge bf{v}$. See Fig. 5.5 in my book and the accompanying text.
Similarly, the orientation of $bf{u} wedge bf{v} wedge bf{w}$ is given by the orientation of $bf{u} wedge bf{v}$ (or $bf{v} wedge bf{w}$ or $bf{w} wedge bf{u}$) on the boundary of $bf{u} wedge bf{v} wedge bf{w}$. All are clockwise.
Perhaps this is a better explanation than in my book.
$endgroup$
$begingroup$
Thanks for the reply. Note 3 on pg 75 of your book says that orientation of areas in a plane can only be compared, but can you compare them if they aren't in the same plane?
$endgroup$
– roshoka
Jan 16 at 3:03
1
$begingroup$
Well, this is embarrassing: The orientation aspects of Section 5.2 and my reply yesterday are wrongheaded. I am rewriting the section and will post it as a pdf on the book's website within a few days. My apologies.
$endgroup$
– Alan Macdonald
Jan 16 at 20:40
1
$begingroup$
Orientations of areas not in parallel planes cannot be compared.
$endgroup$
– Alan Macdonald
Jan 16 at 20:41
$begingroup$
Alright, I'll keep an eye out for the update. Thanks
$endgroup$
– roshoka
Jan 17 at 16:53
1
$begingroup$
The outer product is definitely associative. My explanation of this fact was incorrect. There is no problem proceeding with the text if you accept the associativity.
$endgroup$
– Alan Macdonald
Jan 18 at 16:58
|
show 2 more comments
$begingroup$
The orientation of $bf{u} wedge bf{v}$ is given by the orientation of $bf{u}$ on the boundary of $bf{u} wedge bf{v}$. See Fig. 5.5 in my book and the accompanying text.
Similarly, the orientation of $bf{u} wedge bf{v} wedge bf{w}$ is given by the orientation of $bf{u} wedge bf{v}$ (or $bf{v} wedge bf{w}$ or $bf{w} wedge bf{u}$) on the boundary of $bf{u} wedge bf{v} wedge bf{w}$. All are clockwise.
Perhaps this is a better explanation than in my book.
$endgroup$
$begingroup$
Thanks for the reply. Note 3 on pg 75 of your book says that orientation of areas in a plane can only be compared, but can you compare them if they aren't in the same plane?
$endgroup$
– roshoka
Jan 16 at 3:03
1
$begingroup$
Well, this is embarrassing: The orientation aspects of Section 5.2 and my reply yesterday are wrongheaded. I am rewriting the section and will post it as a pdf on the book's website within a few days. My apologies.
$endgroup$
– Alan Macdonald
Jan 16 at 20:40
1
$begingroup$
Orientations of areas not in parallel planes cannot be compared.
$endgroup$
– Alan Macdonald
Jan 16 at 20:41
$begingroup$
Alright, I'll keep an eye out for the update. Thanks
$endgroup$
– roshoka
Jan 17 at 16:53
1
$begingroup$
The outer product is definitely associative. My explanation of this fact was incorrect. There is no problem proceeding with the text if you accept the associativity.
$endgroup$
– Alan Macdonald
Jan 18 at 16:58
|
show 2 more comments
$begingroup$
The orientation of $bf{u} wedge bf{v}$ is given by the orientation of $bf{u}$ on the boundary of $bf{u} wedge bf{v}$. See Fig. 5.5 in my book and the accompanying text.
Similarly, the orientation of $bf{u} wedge bf{v} wedge bf{w}$ is given by the orientation of $bf{u} wedge bf{v}$ (or $bf{v} wedge bf{w}$ or $bf{w} wedge bf{u}$) on the boundary of $bf{u} wedge bf{v} wedge bf{w}$. All are clockwise.
Perhaps this is a better explanation than in my book.
$endgroup$
The orientation of $bf{u} wedge bf{v}$ is given by the orientation of $bf{u}$ on the boundary of $bf{u} wedge bf{v}$. See Fig. 5.5 in my book and the accompanying text.
Similarly, the orientation of $bf{u} wedge bf{v} wedge bf{w}$ is given by the orientation of $bf{u} wedge bf{v}$ (or $bf{v} wedge bf{w}$ or $bf{w} wedge bf{u}$) on the boundary of $bf{u} wedge bf{v} wedge bf{w}$. All are clockwise.
Perhaps this is a better explanation than in my book.
answered Jan 16 at 1:18
Alan MacdonaldAlan Macdonald
43625
43625
$begingroup$
Thanks for the reply. Note 3 on pg 75 of your book says that orientation of areas in a plane can only be compared, but can you compare them if they aren't in the same plane?
$endgroup$
– roshoka
Jan 16 at 3:03
1
$begingroup$
Well, this is embarrassing: The orientation aspects of Section 5.2 and my reply yesterday are wrongheaded. I am rewriting the section and will post it as a pdf on the book's website within a few days. My apologies.
$endgroup$
– Alan Macdonald
Jan 16 at 20:40
1
$begingroup$
Orientations of areas not in parallel planes cannot be compared.
$endgroup$
– Alan Macdonald
Jan 16 at 20:41
$begingroup$
Alright, I'll keep an eye out for the update. Thanks
$endgroup$
– roshoka
Jan 17 at 16:53
1
$begingroup$
The outer product is definitely associative. My explanation of this fact was incorrect. There is no problem proceeding with the text if you accept the associativity.
$endgroup$
– Alan Macdonald
Jan 18 at 16:58
|
show 2 more comments
$begingroup$
Thanks for the reply. Note 3 on pg 75 of your book says that orientation of areas in a plane can only be compared, but can you compare them if they aren't in the same plane?
$endgroup$
– roshoka
Jan 16 at 3:03
1
$begingroup$
Well, this is embarrassing: The orientation aspects of Section 5.2 and my reply yesterday are wrongheaded. I am rewriting the section and will post it as a pdf on the book's website within a few days. My apologies.
$endgroup$
– Alan Macdonald
Jan 16 at 20:40
1
$begingroup$
Orientations of areas not in parallel planes cannot be compared.
$endgroup$
– Alan Macdonald
Jan 16 at 20:41
$begingroup$
Alright, I'll keep an eye out for the update. Thanks
$endgroup$
– roshoka
Jan 17 at 16:53
1
$begingroup$
The outer product is definitely associative. My explanation of this fact was incorrect. There is no problem proceeding with the text if you accept the associativity.
$endgroup$
– Alan Macdonald
Jan 18 at 16:58
$begingroup$
Thanks for the reply. Note 3 on pg 75 of your book says that orientation of areas in a plane can only be compared, but can you compare them if they aren't in the same plane?
$endgroup$
– roshoka
Jan 16 at 3:03
$begingroup$
Thanks for the reply. Note 3 on pg 75 of your book says that orientation of areas in a plane can only be compared, but can you compare them if they aren't in the same plane?
$endgroup$
– roshoka
Jan 16 at 3:03
1
1
$begingroup$
Well, this is embarrassing: The orientation aspects of Section 5.2 and my reply yesterday are wrongheaded. I am rewriting the section and will post it as a pdf on the book's website within a few days. My apologies.
$endgroup$
– Alan Macdonald
Jan 16 at 20:40
$begingroup$
Well, this is embarrassing: The orientation aspects of Section 5.2 and my reply yesterday are wrongheaded. I am rewriting the section and will post it as a pdf on the book's website within a few days. My apologies.
$endgroup$
– Alan Macdonald
Jan 16 at 20:40
1
1
$begingroup$
Orientations of areas not in parallel planes cannot be compared.
$endgroup$
– Alan Macdonald
Jan 16 at 20:41
$begingroup$
Orientations of areas not in parallel planes cannot be compared.
$endgroup$
– Alan Macdonald
Jan 16 at 20:41
$begingroup$
Alright, I'll keep an eye out for the update. Thanks
$endgroup$
– roshoka
Jan 17 at 16:53
$begingroup$
Alright, I'll keep an eye out for the update. Thanks
$endgroup$
– roshoka
Jan 17 at 16:53
1
1
$begingroup$
The outer product is definitely associative. My explanation of this fact was incorrect. There is no problem proceeding with the text if you accept the associativity.
$endgroup$
– Alan Macdonald
Jan 18 at 16:58
$begingroup$
The outer product is definitely associative. My explanation of this fact was incorrect. There is no problem proceeding with the text if you accept the associativity.
$endgroup$
– Alan Macdonald
Jan 18 at 16:58
|
show 2 more comments
$begingroup$
I'm not familiar with MacDonalds book, but this does not sound right to me; the orientation of a volume element $uwedge v wedge w$ should depend upon all three factors and not just two of them.
The outer product is otherwise known as the wedge or exterior product. It is a signed version of the tensor product, hence used when orientation matters.
They generalise vectors; recall that vectors have direction and magnitude. The outer product of two vectors give a '2d vector', an orientated area element, this ought to be thought of as the generalisation of direction to higher dimensions, where instead of working in higher dimensions per se - for example in a 10d or 20d space - we are working with higher dimensional vectors; the generalisation of magnitude then is the area.
The generalisation to 3d vectors is then obvious: the outer product of three vectors gives an orientated volume element and the magnitude is now the volume.
This explains why the outer product of a vector by itself always vanishes; this is because the area element degenerates into a line element, and hence it's area vanishes; likewise its volume.
The arithmetic of the outer product shows that a generalisation of the triangle law still holds, so long as the two area elements in question have an edge of common magnitude; it's also important to note that the multi-linearity of the outer product means that we can rescale the area element in such a way that it's total area does not change; for example we can magnify one edge by a scale factor $k$ so long as we reduce the other side by $k$ - this has no parallel in the arithmetic of vectors.
$endgroup$
add a comment |
$begingroup$
I'm not familiar with MacDonalds book, but this does not sound right to me; the orientation of a volume element $uwedge v wedge w$ should depend upon all three factors and not just two of them.
The outer product is otherwise known as the wedge or exterior product. It is a signed version of the tensor product, hence used when orientation matters.
They generalise vectors; recall that vectors have direction and magnitude. The outer product of two vectors give a '2d vector', an orientated area element, this ought to be thought of as the generalisation of direction to higher dimensions, where instead of working in higher dimensions per se - for example in a 10d or 20d space - we are working with higher dimensional vectors; the generalisation of magnitude then is the area.
The generalisation to 3d vectors is then obvious: the outer product of three vectors gives an orientated volume element and the magnitude is now the volume.
This explains why the outer product of a vector by itself always vanishes; this is because the area element degenerates into a line element, and hence it's area vanishes; likewise its volume.
The arithmetic of the outer product shows that a generalisation of the triangle law still holds, so long as the two area elements in question have an edge of common magnitude; it's also important to note that the multi-linearity of the outer product means that we can rescale the area element in such a way that it's total area does not change; for example we can magnify one edge by a scale factor $k$ so long as we reduce the other side by $k$ - this has no parallel in the arithmetic of vectors.
$endgroup$
add a comment |
$begingroup$
I'm not familiar with MacDonalds book, but this does not sound right to me; the orientation of a volume element $uwedge v wedge w$ should depend upon all three factors and not just two of them.
The outer product is otherwise known as the wedge or exterior product. It is a signed version of the tensor product, hence used when orientation matters.
They generalise vectors; recall that vectors have direction and magnitude. The outer product of two vectors give a '2d vector', an orientated area element, this ought to be thought of as the generalisation of direction to higher dimensions, where instead of working in higher dimensions per se - for example in a 10d or 20d space - we are working with higher dimensional vectors; the generalisation of magnitude then is the area.
The generalisation to 3d vectors is then obvious: the outer product of three vectors gives an orientated volume element and the magnitude is now the volume.
This explains why the outer product of a vector by itself always vanishes; this is because the area element degenerates into a line element, and hence it's area vanishes; likewise its volume.
The arithmetic of the outer product shows that a generalisation of the triangle law still holds, so long as the two area elements in question have an edge of common magnitude; it's also important to note that the multi-linearity of the outer product means that we can rescale the area element in such a way that it's total area does not change; for example we can magnify one edge by a scale factor $k$ so long as we reduce the other side by $k$ - this has no parallel in the arithmetic of vectors.
$endgroup$
I'm not familiar with MacDonalds book, but this does not sound right to me; the orientation of a volume element $uwedge v wedge w$ should depend upon all three factors and not just two of them.
The outer product is otherwise known as the wedge or exterior product. It is a signed version of the tensor product, hence used when orientation matters.
They generalise vectors; recall that vectors have direction and magnitude. The outer product of two vectors give a '2d vector', an orientated area element, this ought to be thought of as the generalisation of direction to higher dimensions, where instead of working in higher dimensions per se - for example in a 10d or 20d space - we are working with higher dimensional vectors; the generalisation of magnitude then is the area.
The generalisation to 3d vectors is then obvious: the outer product of three vectors gives an orientated volume element and the magnitude is now the volume.
This explains why the outer product of a vector by itself always vanishes; this is because the area element degenerates into a line element, and hence it's area vanishes; likewise its volume.
The arithmetic of the outer product shows that a generalisation of the triangle law still holds, so long as the two area elements in question have an edge of common magnitude; it's also important to note that the multi-linearity of the outer product means that we can rescale the area element in such a way that it's total area does not change; for example we can magnify one edge by a scale factor $k$ so long as we reduce the other side by $k$ - this has no parallel in the arithmetic of vectors.
answered Jan 15 at 0:35
Mozibur UllahMozibur Ullah
2,7391129
2,7391129
add a comment |
add a comment |
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