Anomaly in Curl Computation
$begingroup$
Suppose computing the 3D curl of a vector field involves finding this determinant:
$$
mathrm{det} begin{bmatrix}
mathbf{hat{i}} & frac{partial}{partial x} & 7y^3z^2 \
mathbf{hat{j}} & frac{partial}{partial y} & xyz \
mathbf{hat{k}} & frac{partial}{partial z} & -3x-z
end{bmatrix}
$$
Using cofactor expansion, we should be able to break this into the subdeterminants obtained by excluding any row or column of our choice. However, if we choose the right column, the determinant will incorrectly appear as $(7y^3z^2 - xyz + (-3x - z))mathbf{0} = mathbf{0}$, the zero vector.
This is because the unit vectors have nothing but constant entries, thus their partial derivatives will always equal $mathbf{0}$. Choosing the left column is the sane route, while choosing the middle column appears to work, but choosing the right column leads to a different result. How is this consistent with the typical rules of determinants?
linear-algebra multivariable-calculus vectors determinant cross-product
asked Nov 7 '18 at 4:20
user10478user10478
436211
$endgroup$
add a comment |
$begingroup$
Suppose computing the 3D curl of a vector field involves finding this determinant:
$$
mathrm{det} begin{bmatrix}
mathbf{hat{i}} & frac{partial}{partial x} & 7y^3z^2 \
mathbf{hat{j}} & frac{partial}{partial y} & xyz \
mathbf{hat{k}} & frac{partial}{partial z} & -3x-z
end{bmatrix}
$$
Using cofactor expansion, we should be able to break this into the subdeterminants obtained by excluding any row or column of our choice. However, if we choose the right column, the determinant will incorrectly appear as $(7y^3z^2 - xyz + (-3x - z))mathbf{0} = mathbf{0}$, the zero vector.
This is because the unit vectors have nothing but constant entries, thus their partial derivatives will always equal $mathbf{0}$. Choosing the left column is the sane route, while choosing the middle column appears to work, but choosing the right column leads to a different result. How is this consistent with the typical rules of determinants?
linear-algebra multivariable-calculus vectors determinant cross-product
asked Nov 7 '18 at 4:20
user10478user10478
436211
$endgroup$
1
$begingroup$
It is not. The determinant method for computing the curl is merely a mnemonic and is only meant to be applied via the left column. The curl is not actually a determinant; the fact that the matrix contains entries like $hat i$ and $partial/partial x$, which are not numbers, should give that away.
$endgroup$
– Rahul
Nov 7 '18 at 4:39
$begingroup$
I think that at the least you can also use the middle column.
$endgroup$
– user10478
Nov 7 '18 at 6:40
$begingroup$
Writing the basis vectors in bold and with a hat is overkill.
$endgroup$
– Christian Blatter
Nov 7 '18 at 9:53
add a comment |
$begingroup$
Suppose computing the 3D curl of a vector field involves finding this determinant:
$$
mathrm{det} begin{bmatrix}
mathbf{hat{i}} & frac{partial}{partial x} & 7y^3z^2 \
mathbf{hat{j}} & frac{partial}{partial y} & xyz \
mathbf{hat{k}} & frac{partial}{partial z} & -3x-z
end{bmatrix}
$$
Using cofactor expansion, we should be able to break this into the subdeterminants obtained by excluding any row or column of our choice. However, if we choose the right column, the determinant will incorrectly appear as $(7y^3z^2 - xyz + (-3x - z))mathbf{0} = mathbf{0}$, the zero vector.
This is because the unit vectors have nothing but constant entries, thus their partial derivatives will always equal $mathbf{0}$. Choosing the left column is the sane route, while choosing the middle column appears to work, but choosing the right column leads to a different result. How is this consistent with the typical rules of determinants?
linear-algebra multivariable-calculus vectors determinant cross-product
asked Nov 7 '18 at 4:20
user10478user10478
436211
$endgroup$
Suppose computing the 3D curl of a vector field involves finding this determinant:
$$
mathrm{det} begin{bmatrix}
mathbf{hat{i}} & frac{partial}{partial x} & 7y^3z^2 \
mathbf{hat{j}} & frac{partial}{partial y} & xyz \
mathbf{hat{k}} & frac{partial}{partial z} & -3x-z
end{bmatrix}
$$
Using cofactor expansion, we should be able to break this into the subdeterminants obtained by excluding any row or column of our choice. However, if we choose the right column, the determinant will incorrectly appear as $(7y^3z^2 - xyz + (-3x - z))mathbf{0} = mathbf{0}$, the zero vector.
This is because the unit vectors have nothing but constant entries, thus their partial derivatives will always equal $mathbf{0}$. Choosing the left column is the sane route, while choosing the middle column appears to work, but choosing the right column leads to a different result. How is this consistent with the typical rules of determinants?
linear-algebra multivariable-calculus vectors determinant cross-product
linear-algebra multivariable-calculus vectors determinant cross-product
asked Nov 7 '18 at 4:20
user10478user10478
436211
asked Nov 7 '18 at 4:20
user10478user10478
436211
edited Dec 8 '18 at 4:48
user10478
asked Nov 7 '18 at 4:20
user10478user10478
436211
asked Nov 7 '18 at 4:20
user10478user10478
436211
asked Nov 7 '18 at 4:20
user10478user10478
436211
436211
1
$begingroup$
It is not. The determinant method for computing the curl is merely a mnemonic and is only meant to be applied via the left column. The curl is not actually a determinant; the fact that the matrix contains entries like $hat i$ and $partial/partial x$, which are not numbers, should give that away.
$endgroup$
– Rahul
Nov 7 '18 at 4:39
$begingroup$
I think that at the least you can also use the middle column.
$endgroup$
– user10478
Nov 7 '18 at 6:40
$begingroup$
Writing the basis vectors in bold and with a hat is overkill.
$endgroup$
– Christian Blatter
Nov 7 '18 at 9:53
add a comment |
1
$begingroup$
It is not. The determinant method for computing the curl is merely a mnemonic and is only meant to be applied via the left column. The curl is not actually a determinant; the fact that the matrix contains entries like $hat i$ and $partial/partial x$, which are not numbers, should give that away.
$endgroup$
– Rahul
Nov 7 '18 at 4:39
$begingroup$
I think that at the least you can also use the middle column.
$endgroup$
– user10478
Nov 7 '18 at 6:40
$begingroup$
Writing the basis vectors in bold and with a hat is overkill.
$endgroup$
– Christian Blatter
Nov 7 '18 at 9:53
1
1
$begingroup$
It is not. The determinant method for computing the curl is merely a mnemonic and is only meant to be applied via the left column. The curl is not actually a determinant; the fact that the matrix contains entries like $hat i$ and $partial/partial x$, which are not numbers, should give that away.
$endgroup$
– Rahul
Nov 7 '18 at 4:39
$begingroup$
It is not. The determinant method for computing the curl is merely a mnemonic and is only meant to be applied via the left column. The curl is not actually a determinant; the fact that the matrix contains entries like $hat i$ and $partial/partial x$, which are not numbers, should give that away.
$endgroup$
– Rahul
Nov 7 '18 at 4:39
$begingroup$
I think that at the least you can also use the middle column.
$endgroup$
– user10478
Nov 7 '18 at 6:40
$begingroup$
I think that at the least you can also use the middle column.
$endgroup$
– user10478
Nov 7 '18 at 6:40
$begingroup$
Writing the basis vectors in bold and with a hat is overkill.
$endgroup$
– Christian Blatter
Nov 7 '18 at 9:53
$begingroup$
Writing the basis vectors in bold and with a hat is overkill.
$endgroup$
– Christian Blatter
Nov 7 '18 at 9:53
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
If you're going to use this method, you have to be careful to preserve the order of the terms at all times, because the "multiplication" is not commutative in this context! You must have a unit vector, followed by a partial derivative, followed by one of the polynomials in the right column. The partial derivative applies only to the polynomial, not to the unit vector. The unit vector just comes along for the ride. In particular, the following is incorrect:
This is because the unit vectors have nothing but constant entries, thus their partial derivatives will always equal the 0 vector.
Let's say we exclude the right column, as you did. Then, for the $7x^2y^2$ term, the correct computation is:
$$j frac{partial}{partial z} 7x^2y^2 - k frac{partial}{partial y} 7x^2y^2$$
which equals $-k (14x^2y)$.
Now repeat for the other two factors in the third column.
answered Nov 7 '18 at 4:37
TedTed
21.5k13260
$endgroup$
$begingroup$
Aren't the columns required to be commutative up to the determinant's sign? If we swapped columns 1 and 2, then the partial differential operator would be eating the unit vectors once again. How can the order be decided systematically if we don't already know what the solution should look like?
$endgroup$
– user10478
Nov 7 '18 at 6:30
$begingroup$
No, in this non-commutative context, the usual rule for swapping columns (or rows) in the determinant doesn't apply. The original formula is a correct method for computing the curl only if the columns are ordered in this way.
$endgroup$
– Ted
Nov 7 '18 at 7:12
add a comment |
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$begingroup$
If you're going to use this method, you have to be careful to preserve the order of the terms at all times, because the "multiplication" is not commutative in this context! You must have a unit vector, followed by a partial derivative, followed by one of the polynomials in the right column. The partial derivative applies only to the polynomial, not to the unit vector. The unit vector just comes along for the ride. In particular, the following is incorrect:
This is because the unit vectors have nothing but constant entries, thus their partial derivatives will always equal the 0 vector.
Let's say we exclude the right column, as you did. Then, for the $7x^2y^2$ term, the correct computation is:
$$j frac{partial}{partial z} 7x^2y^2 - k frac{partial}{partial y} 7x^2y^2$$
which equals $-k (14x^2y)$.
Now repeat for the other two factors in the third column.
answered Nov 7 '18 at 4:37
TedTed
21.5k13260
$endgroup$
$begingroup$
Aren't the columns required to be commutative up to the determinant's sign? If we swapped columns 1 and 2, then the partial differential operator would be eating the unit vectors once again. How can the order be decided systematically if we don't already know what the solution should look like?
$endgroup$
– user10478
Nov 7 '18 at 6:30
$begingroup$
No, in this non-commutative context, the usual rule for swapping columns (or rows) in the determinant doesn't apply. The original formula is a correct method for computing the curl only if the columns are ordered in this way.
$endgroup$
– Ted
Nov 7 '18 at 7:12
add a comment |
$begingroup$
If you're going to use this method, you have to be careful to preserve the order of the terms at all times, because the "multiplication" is not commutative in this context! You must have a unit vector, followed by a partial derivative, followed by one of the polynomials in the right column. The partial derivative applies only to the polynomial, not to the unit vector. The unit vector just comes along for the ride. In particular, the following is incorrect:
This is because the unit vectors have nothing but constant entries, thus their partial derivatives will always equal the 0 vector.
Let's say we exclude the right column, as you did. Then, for the $7x^2y^2$ term, the correct computation is:
$$j frac{partial}{partial z} 7x^2y^2 - k frac{partial}{partial y} 7x^2y^2$$
which equals $-k (14x^2y)$.
Now repeat for the other two factors in the third column.
answered Nov 7 '18 at 4:37
TedTed
21.5k13260
$endgroup$
$begingroup$
Aren't the columns required to be commutative up to the determinant's sign? If we swapped columns 1 and 2, then the partial differential operator would be eating the unit vectors once again. How can the order be decided systematically if we don't already know what the solution should look like?
$endgroup$
– user10478
Nov 7 '18 at 6:30
$begingroup$
No, in this non-commutative context, the usual rule for swapping columns (or rows) in the determinant doesn't apply. The original formula is a correct method for computing the curl only if the columns are ordered in this way.
$endgroup$
– Ted
Nov 7 '18 at 7:12
add a comment |
$begingroup$
If you're going to use this method, you have to be careful to preserve the order of the terms at all times, because the "multiplication" is not commutative in this context! You must have a unit vector, followed by a partial derivative, followed by one of the polynomials in the right column. The partial derivative applies only to the polynomial, not to the unit vector. The unit vector just comes along for the ride. In particular, the following is incorrect:
This is because the unit vectors have nothing but constant entries, thus their partial derivatives will always equal the 0 vector.
Let's say we exclude the right column, as you did. Then, for the $7x^2y^2$ term, the correct computation is:
$$j frac{partial}{partial z} 7x^2y^2 - k frac{partial}{partial y} 7x^2y^2$$
which equals $-k (14x^2y)$.
Now repeat for the other two factors in the third column.
answered Nov 7 '18 at 4:37
TedTed
21.5k13260
$endgroup$
If you're going to use this method, you have to be careful to preserve the order of the terms at all times, because the "multiplication" is not commutative in this context! You must have a unit vector, followed by a partial derivative, followed by one of the polynomials in the right column. The partial derivative applies only to the polynomial, not to the unit vector. The unit vector just comes along for the ride. In particular, the following is incorrect:
This is because the unit vectors have nothing but constant entries, thus their partial derivatives will always equal the 0 vector.
Let's say we exclude the right column, as you did. Then, for the $7x^2y^2$ term, the correct computation is:
$$j frac{partial}{partial z} 7x^2y^2 - k frac{partial}{partial y} 7x^2y^2$$
which equals $-k (14x^2y)$.
Now repeat for the other two factors in the third column.
answered Nov 7 '18 at 4:37
TedTed
21.5k13260
answered Nov 7 '18 at 4:37
TedTed
21.5k13260
answered Nov 7 '18 at 4:37
TedTed
21.5k13260
answered Nov 7 '18 at 4:37
TedTed
21.5k13260
21.5k13260
$begingroup$
Aren't the columns required to be commutative up to the determinant's sign? If we swapped columns 1 and 2, then the partial differential operator would be eating the unit vectors once again. How can the order be decided systematically if we don't already know what the solution should look like?
$endgroup$
– user10478
Nov 7 '18 at 6:30
$begingroup$
No, in this non-commutative context, the usual rule for swapping columns (or rows) in the determinant doesn't apply. The original formula is a correct method for computing the curl only if the columns are ordered in this way.
$endgroup$
– Ted
Nov 7 '18 at 7:12
add a comment |
$begingroup$
Aren't the columns required to be commutative up to the determinant's sign? If we swapped columns 1 and 2, then the partial differential operator would be eating the unit vectors once again. How can the order be decided systematically if we don't already know what the solution should look like?
$endgroup$
– user10478
Nov 7 '18 at 6:30
$begingroup$
No, in this non-commutative context, the usual rule for swapping columns (or rows) in the determinant doesn't apply. The original formula is a correct method for computing the curl only if the columns are ordered in this way.
$endgroup$
– Ted
Nov 7 '18 at 7:12
$begingroup$
Aren't the columns required to be commutative up to the determinant's sign? If we swapped columns 1 and 2, then the partial differential operator would be eating the unit vectors once again. How can the order be decided systematically if we don't already know what the solution should look like?
$endgroup$
– user10478
Nov 7 '18 at 6:30
$begingroup$
Aren't the columns required to be commutative up to the determinant's sign? If we swapped columns 1 and 2, then the partial differential operator would be eating the unit vectors once again. How can the order be decided systematically if we don't already know what the solution should look like?
$endgroup$
– user10478
Nov 7 '18 at 6:30
$begingroup$
No, in this non-commutative context, the usual rule for swapping columns (or rows) in the determinant doesn't apply. The original formula is a correct method for computing the curl only if the columns are ordered in this way.
$endgroup$
– Ted
Nov 7 '18 at 7:12
$begingroup$
No, in this non-commutative context, the usual rule for swapping columns (or rows) in the determinant doesn't apply. The original formula is a correct method for computing the curl only if the columns are ordered in this way.
$endgroup$
– Ted
Nov 7 '18 at 7:12
add a comment |
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$begingroup$
It is not. The determinant method for computing the curl is merely a mnemonic and is only meant to be applied via the left column. The curl is not actually a determinant; the fact that the matrix contains entries like $hat i$ and $partial/partial x$, which are not numbers, should give that away.
$endgroup$
– Rahul
Nov 7 '18 at 4:39
$begingroup$
I think that at the least you can also use the middle column.
$endgroup$
– user10478
Nov 7 '18 at 6:40
$begingroup$
Writing the basis vectors in bold and with a hat is overkill.
$endgroup$
– Christian Blatter
Nov 7 '18 at 9:53