Covariant Contravariant Dot product and Length
$begingroup$
So one thing that I find really interesting is that if I have a vector $vec V = V_x hat i + V_y hat j$ its length is just:
$$ |V|^2 = V_x^2 + V_y^2 $$
That is all well and good, but then if I transform the vector into a new basis, I can rewrite the vector in terms of a covariant basis as:
$$ vec V = V^1 vec b_1 + V^2 vec b_2 $$
Now of course, it is obvious that since $vec b_1 $ and $vec b_2$ are not necessarily orthogonormal, that $|V| ne sqrt{(V^1)^2 + (V^2)^2}$, that is all well and good:
Contravariant Basis
Now the usual way this goes is that we then define a new set of basis vectors: we define $b^1$ to be orthogonal to all $b_i$ when $ine 1$ but we define strangely that $b_1 cdot b^1 = 1$. Then we rinse and repeat for all other vectors.
We can then represent v in terms of this new basis directly as:
$$ vec V = V_1 vec b^1 + V_2 vec b^2 $$
Now this is also fine, but then something totally out of the blue happens:
The Dot Product
If we take the dot product of these two representations, we can get an alternative formula for the length:
$$ |V| ^2 = (V^1 vec b_1 + V^2 vec b_2) cdot (V_1 vec b^1 + V_2 vec b^2) = V_x^2 + V_y^2 $$
Now I can verify this by calculation, but I now realise that I have absolutely no understanding of why this should be true.
Any help would be most appreciated :) I don't see the connection here, why does defining this new basis with the rule that $b_j cdot b^k = delta_{jk}$ lead to such an elegant formula for length?
vectors coordinate-systems tensors
asked Jan 12 '18 at 15:21
user2662833user2662833
1,068815
$endgroup$
add a comment |
$begingroup$
So one thing that I find really interesting is that if I have a vector $vec V = V_x hat i + V_y hat j$ its length is just:
$$ |V|^2 = V_x^2 + V_y^2 $$
That is all well and good, but then if I transform the vector into a new basis, I can rewrite the vector in terms of a covariant basis as:
$$ vec V = V^1 vec b_1 + V^2 vec b_2 $$
Now of course, it is obvious that since $vec b_1 $ and $vec b_2$ are not necessarily orthogonormal, that $|V| ne sqrt{(V^1)^2 + (V^2)^2}$, that is all well and good:
Contravariant Basis
Now the usual way this goes is that we then define a new set of basis vectors: we define $b^1$ to be orthogonal to all $b_i$ when $ine 1$ but we define strangely that $b_1 cdot b^1 = 1$. Then we rinse and repeat for all other vectors.
We can then represent v in terms of this new basis directly as:
$$ vec V = V_1 vec b^1 + V_2 vec b^2 $$
Now this is also fine, but then something totally out of the blue happens:
The Dot Product
If we take the dot product of these two representations, we can get an alternative formula for the length:
$$ |V| ^2 = (V^1 vec b_1 + V^2 vec b_2) cdot (V_1 vec b^1 + V_2 vec b^2) = V_x^2 + V_y^2 $$
Now I can verify this by calculation, but I now realise that I have absolutely no understanding of why this should be true.
Any help would be most appreciated :) I don't see the connection here, why does defining this new basis with the rule that $b_j cdot b^k = delta_{jk}$ lead to such an elegant formula for length?
vectors coordinate-systems tensors
asked Jan 12 '18 at 15:21
user2662833user2662833
1,068815
$endgroup$
$begingroup$
I think you lack the correct terminology to deal with these things. Such correct terminology is the machinery of metric tensors. Try having a look at the book of Itskov "Tensor analysis and tensor algebra for engineers".
$endgroup$
– Giuseppe Negro
Jan 12 '18 at 15:25
$begingroup$
Well I am trying to learn this now. But I have found every explanation of this fact to be very hand wavy - and I haven't found a concrete explanation/proof for this fact in particular. I have started looking into the metric tensor, but I don't want to progress beyond this until I fundamentally understand why this should be true in general.
$endgroup$
– user2662833
Jan 12 '18 at 15:30
$begingroup$
I do appreciate the recommendation though; I will try to find it regardless :)
$endgroup$
– user2662833
Jan 12 '18 at 15:31
$begingroup$
Look here and subsequent discussion
$endgroup$
– Giuseppe Negro
Jan 12 '18 at 15:47
$begingroup$
Yep I found it, it goes through it before page 10, but as is the case with all books I've found; it doesn't prove this fact at all; it makes no connection between length and the definition of this dual basis. It just states that there exists a dual basis for every basis and that this formula should hold. What I'm looking for is a connection between the euclidean length and the length in another basis. I can definitely see that this formula is true; but the author here has glossed over why it should be true.
$endgroup$
– user2662833
Jan 12 '18 at 15:52
add a comment |
$begingroup$
So one thing that I find really interesting is that if I have a vector $vec V = V_x hat i + V_y hat j$ its length is just:
$$ |V|^2 = V_x^2 + V_y^2 $$
That is all well and good, but then if I transform the vector into a new basis, I can rewrite the vector in terms of a covariant basis as:
$$ vec V = V^1 vec b_1 + V^2 vec b_2 $$
Now of course, it is obvious that since $vec b_1 $ and $vec b_2$ are not necessarily orthogonormal, that $|V| ne sqrt{(V^1)^2 + (V^2)^2}$, that is all well and good:
Contravariant Basis
Now the usual way this goes is that we then define a new set of basis vectors: we define $b^1$ to be orthogonal to all $b_i$ when $ine 1$ but we define strangely that $b_1 cdot b^1 = 1$. Then we rinse and repeat for all other vectors.
We can then represent v in terms of this new basis directly as:
$$ vec V = V_1 vec b^1 + V_2 vec b^2 $$
Now this is also fine, but then something totally out of the blue happens:
The Dot Product
If we take the dot product of these two representations, we can get an alternative formula for the length:
$$ |V| ^2 = (V^1 vec b_1 + V^2 vec b_2) cdot (V_1 vec b^1 + V_2 vec b^2) = V_x^2 + V_y^2 $$
Now I can verify this by calculation, but I now realise that I have absolutely no understanding of why this should be true.
Any help would be most appreciated :) I don't see the connection here, why does defining this new basis with the rule that $b_j cdot b^k = delta_{jk}$ lead to such an elegant formula for length?
vectors coordinate-systems tensors
asked Jan 12 '18 at 15:21
user2662833user2662833
1,068815
$endgroup$
So one thing that I find really interesting is that if I have a vector $vec V = V_x hat i + V_y hat j$ its length is just:
$$ |V|^2 = V_x^2 + V_y^2 $$
That is all well and good, but then if I transform the vector into a new basis, I can rewrite the vector in terms of a covariant basis as:
$$ vec V = V^1 vec b_1 + V^2 vec b_2 $$
Now of course, it is obvious that since $vec b_1 $ and $vec b_2$ are not necessarily orthogonormal, that $|V| ne sqrt{(V^1)^2 + (V^2)^2}$, that is all well and good:
Contravariant Basis
Now the usual way this goes is that we then define a new set of basis vectors: we define $b^1$ to be orthogonal to all $b_i$ when $ine 1$ but we define strangely that $b_1 cdot b^1 = 1$. Then we rinse and repeat for all other vectors.
We can then represent v in terms of this new basis directly as:
$$ vec V = V_1 vec b^1 + V_2 vec b^2 $$
Now this is also fine, but then something totally out of the blue happens:
The Dot Product
If we take the dot product of these two representations, we can get an alternative formula for the length:
$$ |V| ^2 = (V^1 vec b_1 + V^2 vec b_2) cdot (V_1 vec b^1 + V_2 vec b^2) = V_x^2 + V_y^2 $$
Now I can verify this by calculation, but I now realise that I have absolutely no understanding of why this should be true.
Any help would be most appreciated :) I don't see the connection here, why does defining this new basis with the rule that $b_j cdot b^k = delta_{jk}$ lead to such an elegant formula for length?
vectors coordinate-systems tensors
vectors coordinate-systems tensors
asked Jan 12 '18 at 15:21
user2662833user2662833
1,068815
asked Jan 12 '18 at 15:21
user2662833user2662833
1,068815
asked Jan 12 '18 at 15:21
user2662833user2662833
1,068815
asked Jan 12 '18 at 15:21
user2662833user2662833
1,068815
asked Jan 12 '18 at 15:21
user2662833user2662833
1,068815
1,068815
$begingroup$
I think you lack the correct terminology to deal with these things. Such correct terminology is the machinery of metric tensors. Try having a look at the book of Itskov "Tensor analysis and tensor algebra for engineers".
$endgroup$
– Giuseppe Negro
Jan 12 '18 at 15:25
$begingroup$
Well I am trying to learn this now. But I have found every explanation of this fact to be very hand wavy - and I haven't found a concrete explanation/proof for this fact in particular. I have started looking into the metric tensor, but I don't want to progress beyond this until I fundamentally understand why this should be true in general.
$endgroup$
– user2662833
Jan 12 '18 at 15:30
$begingroup$
I do appreciate the recommendation though; I will try to find it regardless :)
$endgroup$
– user2662833
Jan 12 '18 at 15:31
$begingroup$
Look here and subsequent discussion
$endgroup$
– Giuseppe Negro
Jan 12 '18 at 15:47
$begingroup$
Yep I found it, it goes through it before page 10, but as is the case with all books I've found; it doesn't prove this fact at all; it makes no connection between length and the definition of this dual basis. It just states that there exists a dual basis for every basis and that this formula should hold. What I'm looking for is a connection between the euclidean length and the length in another basis. I can definitely see that this formula is true; but the author here has glossed over why it should be true.
$endgroup$
– user2662833
Jan 12 '18 at 15:52
add a comment |
$begingroup$
I think you lack the correct terminology to deal with these things. Such correct terminology is the machinery of metric tensors. Try having a look at the book of Itskov "Tensor analysis and tensor algebra for engineers".
$endgroup$
– Giuseppe Negro
Jan 12 '18 at 15:25
$begingroup$
Well I am trying to learn this now. But I have found every explanation of this fact to be very hand wavy - and I haven't found a concrete explanation/proof for this fact in particular. I have started looking into the metric tensor, but I don't want to progress beyond this until I fundamentally understand why this should be true in general.
$endgroup$
– user2662833
Jan 12 '18 at 15:30
$begingroup$
I do appreciate the recommendation though; I will try to find it regardless :)
$endgroup$
– user2662833
Jan 12 '18 at 15:31
$begingroup$
Look here and subsequent discussion
$endgroup$
– Giuseppe Negro
Jan 12 '18 at 15:47
$begingroup$
Yep I found it, it goes through it before page 10, but as is the case with all books I've found; it doesn't prove this fact at all; it makes no connection between length and the definition of this dual basis. It just states that there exists a dual basis for every basis and that this formula should hold. What I'm looking for is a connection between the euclidean length and the length in another basis. I can definitely see that this formula is true; but the author here has glossed over why it should be true.
$endgroup$
– user2662833
Jan 12 '18 at 15:52
$begingroup$
I think you lack the correct terminology to deal with these things. Such correct terminology is the machinery of metric tensors. Try having a look at the book of Itskov "Tensor analysis and tensor algebra for engineers".
$endgroup$
– Giuseppe Negro
Jan 12 '18 at 15:25
$begingroup$
I think you lack the correct terminology to deal with these things. Such correct terminology is the machinery of metric tensors. Try having a look at the book of Itskov "Tensor analysis and tensor algebra for engineers".
$endgroup$
– Giuseppe Negro
Jan 12 '18 at 15:25
$begingroup$
Well I am trying to learn this now. But I have found every explanation of this fact to be very hand wavy - and I haven't found a concrete explanation/proof for this fact in particular. I have started looking into the metric tensor, but I don't want to progress beyond this until I fundamentally understand why this should be true in general.
$endgroup$
– user2662833
Jan 12 '18 at 15:30
$begingroup$
Well I am trying to learn this now. But I have found every explanation of this fact to be very hand wavy - and I haven't found a concrete explanation/proof for this fact in particular. I have started looking into the metric tensor, but I don't want to progress beyond this until I fundamentally understand why this should be true in general.
$endgroup$
– user2662833
Jan 12 '18 at 15:30
$begingroup$
I do appreciate the recommendation though; I will try to find it regardless :)
$endgroup$
– user2662833
Jan 12 '18 at 15:31
$begingroup$
I do appreciate the recommendation though; I will try to find it regardless :)
$endgroup$
– user2662833
Jan 12 '18 at 15:31
$begingroup$
Look here and subsequent discussion
$endgroup$
– Giuseppe Negro
Jan 12 '18 at 15:47
$begingroup$
Look here and subsequent discussion
$endgroup$
– Giuseppe Negro
Jan 12 '18 at 15:47
$begingroup$
Yep I found it, it goes through it before page 10, but as is the case with all books I've found; it doesn't prove this fact at all; it makes no connection between length and the definition of this dual basis. It just states that there exists a dual basis for every basis and that this formula should hold. What I'm looking for is a connection between the euclidean length and the length in another basis. I can definitely see that this formula is true; but the author here has glossed over why it should be true.
$endgroup$
– user2662833
Jan 12 '18 at 15:52
$begingroup$
Yep I found it, it goes through it before page 10, but as is the case with all books I've found; it doesn't prove this fact at all; it makes no connection between length and the definition of this dual basis. It just states that there exists a dual basis for every basis and that this formula should hold. What I'm looking for is a connection between the euclidean length and the length in another basis. I can definitely see that this formula is true; but the author here has glossed over why it should be true.
$endgroup$
– user2662833
Jan 12 '18 at 15:52
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The intuition here is as follows: we define the dual basis to "correct for" all the departures from orthonormality of the original basis. So if the angle between two basis vectors in the original basis was acute, the angle in the new basis will be obtuse; if one if the basis vectors was longer in the original basis, it will be shorter in the new basis. Then when we "average out" the two bases by taking the product of the coordinate in one basis with the coordinate in the other basis, all the departures from orthonormality are corrected for, and we get the appropriate length as if we had a single orthonormal basis.
Of course, to describe precisely how this "correction" is done and to prove that it works, you have to go through the math, as, e.g., the book Giuseppe Negro pointed you to does.
Did you have a particular problem with the proof there? Or were you looking for intuition?
answered Jan 12 '18 at 16:33
Y. FormanY. Forman
11.5k523
$endgroup$
add a comment |
$begingroup$
Okay, so I suppose its time I come back and answer my own question here. Its wasn't too
complicated, but thats the beauty of learning isn't it!
I was missing the fact that we define the dual basis as the basis that makes
$bf{e_i} cdot bf{e^j} = delta_i^j$. Given this fact alone, it is possible to find
all of the dual basis vectors $left{bf{e^j}right}$. So using my old notation,
(but bolding instead of adding all those ugly hats), we always know that the squared magnitude equals the dot product of any vector with itself, so:
$$ |V|^2 = bf{v} cdot bf{v} $$
So we can write $bf{v}$ in terms of the original basis, and in terms of the dual basis as I've shown below:
$$ |V| ^2 = (V^1 bf {b_1} + V^2 bf {b_2}) cdot (V_1 bf {b^1} + V_2 bf{b^2})$$
Now if you expand this out, you'll get:
$$ |V| ^2 = V^1 V_1 bf {b_1}cdotbf {b^1} + V^1 V_2 bf {b_1}cdotbf {b^2} +
V^2 V_1 bf {b_2}cdotbf {b^1} + V^2 V_2 bf {b_2}cdotbf {b^2}$$
Now just by using our definition, we can immediately say that $bf{b_1}cdotbf{b^1} = bf{b_2}cdotbf{b^2} = 1$. We have also stated that the dual basis vectors are perpendicular to all original basis vectors (except the one with the same index), so $bf{b_1}cdotbf{b^2} = bf{b_2}cdotbf{b^1} = 0$. If we make these substitutions, we get the simple formula:
$$ |V| ^2 = V^1 V_1 + V^2 V_2 $$
Which I was struggling to understand in the past.
answered Jan 14 at 11:37
user2662833user2662833
1,068815
$endgroup$
add a comment |
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2 Answers
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2 Answers
2
active
oldest
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$begingroup$
The intuition here is as follows: we define the dual basis to "correct for" all the departures from orthonormality of the original basis. So if the angle between two basis vectors in the original basis was acute, the angle in the new basis will be obtuse; if one if the basis vectors was longer in the original basis, it will be shorter in the new basis. Then when we "average out" the two bases by taking the product of the coordinate in one basis with the coordinate in the other basis, all the departures from orthonormality are corrected for, and we get the appropriate length as if we had a single orthonormal basis.
Of course, to describe precisely how this "correction" is done and to prove that it works, you have to go through the math, as, e.g., the book Giuseppe Negro pointed you to does.
Did you have a particular problem with the proof there? Or were you looking for intuition?
answered Jan 12 '18 at 16:33
Y. FormanY. Forman
11.5k523
$endgroup$
add a comment |
$begingroup$
The intuition here is as follows: we define the dual basis to "correct for" all the departures from orthonormality of the original basis. So if the angle between two basis vectors in the original basis was acute, the angle in the new basis will be obtuse; if one if the basis vectors was longer in the original basis, it will be shorter in the new basis. Then when we "average out" the two bases by taking the product of the coordinate in one basis with the coordinate in the other basis, all the departures from orthonormality are corrected for, and we get the appropriate length as if we had a single orthonormal basis.
Of course, to describe precisely how this "correction" is done and to prove that it works, you have to go through the math, as, e.g., the book Giuseppe Negro pointed you to does.
Did you have a particular problem with the proof there? Or were you looking for intuition?
answered Jan 12 '18 at 16:33
Y. FormanY. Forman
11.5k523
$endgroup$
add a comment |
$begingroup$
The intuition here is as follows: we define the dual basis to "correct for" all the departures from orthonormality of the original basis. So if the angle between two basis vectors in the original basis was acute, the angle in the new basis will be obtuse; if one if the basis vectors was longer in the original basis, it will be shorter in the new basis. Then when we "average out" the two bases by taking the product of the coordinate in one basis with the coordinate in the other basis, all the departures from orthonormality are corrected for, and we get the appropriate length as if we had a single orthonormal basis.
Of course, to describe precisely how this "correction" is done and to prove that it works, you have to go through the math, as, e.g., the book Giuseppe Negro pointed you to does.
Did you have a particular problem with the proof there? Or were you looking for intuition?
answered Jan 12 '18 at 16:33
Y. FormanY. Forman
11.5k523
$endgroup$
The intuition here is as follows: we define the dual basis to "correct for" all the departures from orthonormality of the original basis. So if the angle between two basis vectors in the original basis was acute, the angle in the new basis will be obtuse; if one if the basis vectors was longer in the original basis, it will be shorter in the new basis. Then when we "average out" the two bases by taking the product of the coordinate in one basis with the coordinate in the other basis, all the departures from orthonormality are corrected for, and we get the appropriate length as if we had a single orthonormal basis.
Of course, to describe precisely how this "correction" is done and to prove that it works, you have to go through the math, as, e.g., the book Giuseppe Negro pointed you to does.
Did you have a particular problem with the proof there? Or were you looking for intuition?
answered Jan 12 '18 at 16:33
Y. FormanY. Forman
11.5k523
answered Jan 12 '18 at 16:33
Y. FormanY. Forman
11.5k523
answered Jan 12 '18 at 16:33
Y. FormanY. Forman
11.5k523
answered Jan 12 '18 at 16:33
Y. FormanY. Forman
11.5k523
11.5k523
add a comment |
add a comment |
$begingroup$
Okay, so I suppose its time I come back and answer my own question here. Its wasn't too
complicated, but thats the beauty of learning isn't it!
I was missing the fact that we define the dual basis as the basis that makes
$bf{e_i} cdot bf{e^j} = delta_i^j$. Given this fact alone, it is possible to find
all of the dual basis vectors $left{bf{e^j}right}$. So using my old notation,
(but bolding instead of adding all those ugly hats), we always know that the squared magnitude equals the dot product of any vector with itself, so:
$$ |V|^2 = bf{v} cdot bf{v} $$
So we can write $bf{v}$ in terms of the original basis, and in terms of the dual basis as I've shown below:
$$ |V| ^2 = (V^1 bf {b_1} + V^2 bf {b_2}) cdot (V_1 bf {b^1} + V_2 bf{b^2})$$
Now if you expand this out, you'll get:
$$ |V| ^2 = V^1 V_1 bf {b_1}cdotbf {b^1} + V^1 V_2 bf {b_1}cdotbf {b^2} +
V^2 V_1 bf {b_2}cdotbf {b^1} + V^2 V_2 bf {b_2}cdotbf {b^2}$$
Now just by using our definition, we can immediately say that $bf{b_1}cdotbf{b^1} = bf{b_2}cdotbf{b^2} = 1$. We have also stated that the dual basis vectors are perpendicular to all original basis vectors (except the one with the same index), so $bf{b_1}cdotbf{b^2} = bf{b_2}cdotbf{b^1} = 0$. If we make these substitutions, we get the simple formula:
$$ |V| ^2 = V^1 V_1 + V^2 V_2 $$
Which I was struggling to understand in the past.
answered Jan 14 at 11:37
user2662833user2662833
1,068815
$endgroup$
add a comment |
$begingroup$
Okay, so I suppose its time I come back and answer my own question here. Its wasn't too
complicated, but thats the beauty of learning isn't it!
I was missing the fact that we define the dual basis as the basis that makes
$bf{e_i} cdot bf{e^j} = delta_i^j$. Given this fact alone, it is possible to find
all of the dual basis vectors $left{bf{e^j}right}$. So using my old notation,
(but bolding instead of adding all those ugly hats), we always know that the squared magnitude equals the dot product of any vector with itself, so:
$$ |V|^2 = bf{v} cdot bf{v} $$
So we can write $bf{v}$ in terms of the original basis, and in terms of the dual basis as I've shown below:
$$ |V| ^2 = (V^1 bf {b_1} + V^2 bf {b_2}) cdot (V_1 bf {b^1} + V_2 bf{b^2})$$
Now if you expand this out, you'll get:
$$ |V| ^2 = V^1 V_1 bf {b_1}cdotbf {b^1} + V^1 V_2 bf {b_1}cdotbf {b^2} +
V^2 V_1 bf {b_2}cdotbf {b^1} + V^2 V_2 bf {b_2}cdotbf {b^2}$$
Now just by using our definition, we can immediately say that $bf{b_1}cdotbf{b^1} = bf{b_2}cdotbf{b^2} = 1$. We have also stated that the dual basis vectors are perpendicular to all original basis vectors (except the one with the same index), so $bf{b_1}cdotbf{b^2} = bf{b_2}cdotbf{b^1} = 0$. If we make these substitutions, we get the simple formula:
$$ |V| ^2 = V^1 V_1 + V^2 V_2 $$
Which I was struggling to understand in the past.
answered Jan 14 at 11:37
user2662833user2662833
1,068815
$endgroup$
add a comment |
$begingroup$
Okay, so I suppose its time I come back and answer my own question here. Its wasn't too
complicated, but thats the beauty of learning isn't it!
I was missing the fact that we define the dual basis as the basis that makes
$bf{e_i} cdot bf{e^j} = delta_i^j$. Given this fact alone, it is possible to find
all of the dual basis vectors $left{bf{e^j}right}$. So using my old notation,
(but bolding instead of adding all those ugly hats), we always know that the squared magnitude equals the dot product of any vector with itself, so:
$$ |V|^2 = bf{v} cdot bf{v} $$
So we can write $bf{v}$ in terms of the original basis, and in terms of the dual basis as I've shown below:
$$ |V| ^2 = (V^1 bf {b_1} + V^2 bf {b_2}) cdot (V_1 bf {b^1} + V_2 bf{b^2})$$
Now if you expand this out, you'll get:
$$ |V| ^2 = V^1 V_1 bf {b_1}cdotbf {b^1} + V^1 V_2 bf {b_1}cdotbf {b^2} +
V^2 V_1 bf {b_2}cdotbf {b^1} + V^2 V_2 bf {b_2}cdotbf {b^2}$$
Now just by using our definition, we can immediately say that $bf{b_1}cdotbf{b^1} = bf{b_2}cdotbf{b^2} = 1$. We have also stated that the dual basis vectors are perpendicular to all original basis vectors (except the one with the same index), so $bf{b_1}cdotbf{b^2} = bf{b_2}cdotbf{b^1} = 0$. If we make these substitutions, we get the simple formula:
$$ |V| ^2 = V^1 V_1 + V^2 V_2 $$
Which I was struggling to understand in the past.
answered Jan 14 at 11:37
user2662833user2662833
1,068815
$endgroup$
Okay, so I suppose its time I come back and answer my own question here. Its wasn't too
complicated, but thats the beauty of learning isn't it!
I was missing the fact that we define the dual basis as the basis that makes
$bf{e_i} cdot bf{e^j} = delta_i^j$. Given this fact alone, it is possible to find
all of the dual basis vectors $left{bf{e^j}right}$. So using my old notation,
(but bolding instead of adding all those ugly hats), we always know that the squared magnitude equals the dot product of any vector with itself, so:
$$ |V|^2 = bf{v} cdot bf{v} $$
So we can write $bf{v}$ in terms of the original basis, and in terms of the dual basis as I've shown below:
$$ |V| ^2 = (V^1 bf {b_1} + V^2 bf {b_2}) cdot (V_1 bf {b^1} + V_2 bf{b^2})$$
Now if you expand this out, you'll get:
$$ |V| ^2 = V^1 V_1 bf {b_1}cdotbf {b^1} + V^1 V_2 bf {b_1}cdotbf {b^2} +
V^2 V_1 bf {b_2}cdotbf {b^1} + V^2 V_2 bf {b_2}cdotbf {b^2}$$
Now just by using our definition, we can immediately say that $bf{b_1}cdotbf{b^1} = bf{b_2}cdotbf{b^2} = 1$. We have also stated that the dual basis vectors are perpendicular to all original basis vectors (except the one with the same index), so $bf{b_1}cdotbf{b^2} = bf{b_2}cdotbf{b^1} = 0$. If we make these substitutions, we get the simple formula:
$$ |V| ^2 = V^1 V_1 + V^2 V_2 $$
Which I was struggling to understand in the past.
answered Jan 14 at 11:37
user2662833user2662833
1,068815
answered Jan 14 at 11:37
user2662833user2662833
1,068815
answered Jan 14 at 11:37
user2662833user2662833
1,068815
answered Jan 14 at 11:37
user2662833user2662833
1,068815
1,068815
add a comment |
add a comment |
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$begingroup$
I think you lack the correct terminology to deal with these things. Such correct terminology is the machinery of metric tensors. Try having a look at the book of Itskov "Tensor analysis and tensor algebra for engineers".
$endgroup$
– Giuseppe Negro
Jan 12 '18 at 15:25
$begingroup$
Well I am trying to learn this now. But I have found every explanation of this fact to be very hand wavy - and I haven't found a concrete explanation/proof for this fact in particular. I have started looking into the metric tensor, but I don't want to progress beyond this until I fundamentally understand why this should be true in general.
$endgroup$
– user2662833
Jan 12 '18 at 15:30
$begingroup$
I do appreciate the recommendation though; I will try to find it regardless :)
$endgroup$
– user2662833
Jan 12 '18 at 15:31
$begingroup$
Look here and subsequent discussion
$endgroup$
– Giuseppe Negro
Jan 12 '18 at 15:47
$begingroup$
Yep I found it, it goes through it before page 10, but as is the case with all books I've found; it doesn't prove this fact at all; it makes no connection between length and the definition of this dual basis. It just states that there exists a dual basis for every basis and that this formula should hold. What I'm looking for is a connection between the euclidean length and the length in another basis. I can definitely see that this formula is true; but the author here has glossed over why it should be true.
$endgroup$
– user2662833
Jan 12 '18 at 15:52