Formula involving cross products $nabla times (F times G)$?
$begingroup$
I was knowing this formula for vectors $a,b,c$:
$a times (b times c) = (acdot c)b - (acdot b)c$
But I was thinking of the differential operator as a vector and hence I thought the above formula would also work for
$nabla times (F times G) $
but it did not rather we have the following:
$nabla times (F times G)= (nabla cdot G)F - (Fcdot nabla)G + (Gcdot nabla)F - (nablacdot F)G$
multivariable-calculus vector-analysis
edited Dec 21 '18 at 23:17
Ruslan
3,72721633
asked Feb 18 '18 at 2:57
BAYMAXBAYMAX
2,91621125
$endgroup$
add a comment |
$begingroup$
I was knowing this formula for vectors $a,b,c$:
$a times (b times c) = (acdot c)b - (acdot b)c$
But I was thinking of the differential operator as a vector and hence I thought the above formula would also work for
$nabla times (F times G) $
but it did not rather we have the following:
$nabla times (F times G)= (nabla cdot G)F - (Fcdot nabla)G + (Gcdot nabla)F - (nablacdot F)G$
multivariable-calculus vector-analysis
edited Dec 21 '18 at 23:17
Ruslan
3,72721633
asked Feb 18 '18 at 2:57
BAYMAXBAYMAX
2,91621125
$endgroup$
add a comment |
$begingroup$
I was knowing this formula for vectors $a,b,c$:
$a times (b times c) = (acdot c)b - (acdot b)c$
But I was thinking of the differential operator as a vector and hence I thought the above formula would also work for
$nabla times (F times G) $
but it did not rather we have the following:
$nabla times (F times G)= (nabla cdot G)F - (Fcdot nabla)G + (Gcdot nabla)F - (nablacdot F)G$
multivariable-calculus vector-analysis
edited Dec 21 '18 at 23:17
Ruslan
3,72721633
asked Feb 18 '18 at 2:57
BAYMAXBAYMAX
2,91621125
$endgroup$
I was knowing this formula for vectors $a,b,c$:
$a times (b times c) = (acdot c)b - (acdot b)c$
But I was thinking of the differential operator as a vector and hence I thought the above formula would also work for
$nabla times (F times G) $
but it did not rather we have the following:
$nabla times (F times G)= (nabla cdot G)F - (Fcdot nabla)G + (Gcdot nabla)F - (nablacdot F)G$
multivariable-calculus vector-analysis
multivariable-calculus vector-analysis
edited Dec 21 '18 at 23:17
Ruslan
3,72721633
asked Feb 18 '18 at 2:57
BAYMAXBAYMAX
2,91621125
edited Dec 21 '18 at 23:17
Ruslan
3,72721633
asked Feb 18 '18 at 2:57
BAYMAXBAYMAX
2,91621125
edited Dec 21 '18 at 23:17
Ruslan
3,72721633
edited Dec 21 '18 at 23:17
Ruslan
3,72721633
edited Dec 21 '18 at 23:17
Ruslan
3,72721633
3,72721633
asked Feb 18 '18 at 2:57
BAYMAXBAYMAX
2,91621125
asked Feb 18 '18 at 2:57
BAYMAXBAYMAX
2,91621125
asked Feb 18 '18 at 2:57
BAYMAXBAYMAX
2,91621125
2,91621125
add a comment |
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
To see why this fails, examine the Wikipedia proof of the triple product expansion. Writing their work in more detail, we find:
begin{align}
(utimes (vtimes w))_x &= u_y(v_xw_y - v_yw_x) - u_z(v_zw_x - v_xw_z)\
&=u_yv_xw_y - u_yv_yw_x - u_zv_zw_x +u_zv_xw_z \
&=u_ycolor{red}{v_x}w_y + u_zcolor{red}{v_x}w_z - u_yv_ycolor{magenta}{w_x}-u_zv_zcolor{magenta}{w_x} \
&= color{red}{v_x}(u_yw_y + u_zw_z) - color{magenta}{w_x}(u_yv_y + u_zv_z)
end{align}
You can see that they took advantage of the commutativity of products of scalars. But operators generally do not commute with other things! As an easy example, $frac{d}{dx}f(x)g(x) neq f(x)frac{d}{dx}g(x)$. You will notice that when $u$ is replaced with the operator $nabla$, we will instead obtain:
begin{align}
(nablatimes (vtimes w))_x &= partial_y (v_xw_y - v_yw_x) - partial_z(v_zw_x - v_xw_z)\
&=partial_y(v_xw_y)-partial_y(v_yw_x) - partial_z(v_zw_x) + partial_z(v_xw_z)\
&= (partial_yv_x) w_y+v_x(partial_y w_y) - (partial_yv_y)w_x-v_y(partial_yw_x)-(partial_zv_z)w_x - v_z(partial_zw_x)\&qquadqquad+(partial_zv_x)w_z + v_x(partial_zw_z) \
&= v_x(partial_y w_y) + v_x(partial_zw_z) - (partial_yv_y)w_x - (partial_zv_z)w_x + (partial_yv_x) w_y + (partial_zv_x)w_z\&qquadqquad -v_y(partial_yw_x)-v_z(partial_zw_x) \
&=(partial_yw_y+partial_zw_z)v_x - (partial_yv_y+partial_zv_z)w_x+(w_ypartial_y + w_zpartial_z)v_x - (v_ypartial_y +v_zpartial_z)w_x \
&vdots\
&=(nablacdot w)v_x - (nablacdot v)w_x+(wcdotnabla)v_x - (vcdotnabla)w_x
end{align}
and similarly for the other components.
answered Feb 18 '18 at 5:13
bamesbames
1,9311315
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
To see why this fails, examine the Wikipedia proof of the triple product expansion. Writing their work in more detail, we find:
begin{align}
(utimes (vtimes w))_x &= u_y(v_xw_y - v_yw_x) - u_z(v_zw_x - v_xw_z)\
&=u_yv_xw_y - u_yv_yw_x - u_zv_zw_x +u_zv_xw_z \
&=u_ycolor{red}{v_x}w_y + u_zcolor{red}{v_x}w_z - u_yv_ycolor{magenta}{w_x}-u_zv_zcolor{magenta}{w_x} \
&= color{red}{v_x}(u_yw_y + u_zw_z) - color{magenta}{w_x}(u_yv_y + u_zv_z)
end{align}
You can see that they took advantage of the commutativity of products of scalars. But operators generally do not commute with other things! As an easy example, $frac{d}{dx}f(x)g(x) neq f(x)frac{d}{dx}g(x)$. You will notice that when $u$ is replaced with the operator $nabla$, we will instead obtain:
begin{align}
(nablatimes (vtimes w))_x &= partial_y (v_xw_y - v_yw_x) - partial_z(v_zw_x - v_xw_z)\
&=partial_y(v_xw_y)-partial_y(v_yw_x) - partial_z(v_zw_x) + partial_z(v_xw_z)\
&= (partial_yv_x) w_y+v_x(partial_y w_y) - (partial_yv_y)w_x-v_y(partial_yw_x)-(partial_zv_z)w_x - v_z(partial_zw_x)\&qquadqquad+(partial_zv_x)w_z + v_x(partial_zw_z) \
&= v_x(partial_y w_y) + v_x(partial_zw_z) - (partial_yv_y)w_x - (partial_zv_z)w_x + (partial_yv_x) w_y + (partial_zv_x)w_z\&qquadqquad -v_y(partial_yw_x)-v_z(partial_zw_x) \
&=(partial_yw_y+partial_zw_z)v_x - (partial_yv_y+partial_zv_z)w_x+(w_ypartial_y + w_zpartial_z)v_x - (v_ypartial_y +v_zpartial_z)w_x \
&vdots\
&=(nablacdot w)v_x - (nablacdot v)w_x+(wcdotnabla)v_x - (vcdotnabla)w_x
end{align}
and similarly for the other components.
answered Feb 18 '18 at 5:13
bamesbames
1,9311315
$endgroup$
add a comment |
$begingroup$
To see why this fails, examine the Wikipedia proof of the triple product expansion. Writing their work in more detail, we find:
begin{align}
(utimes (vtimes w))_x &= u_y(v_xw_y - v_yw_x) - u_z(v_zw_x - v_xw_z)\
&=u_yv_xw_y - u_yv_yw_x - u_zv_zw_x +u_zv_xw_z \
&=u_ycolor{red}{v_x}w_y + u_zcolor{red}{v_x}w_z - u_yv_ycolor{magenta}{w_x}-u_zv_zcolor{magenta}{w_x} \
&= color{red}{v_x}(u_yw_y + u_zw_z) - color{magenta}{w_x}(u_yv_y + u_zv_z)
end{align}
You can see that they took advantage of the commutativity of products of scalars. But operators generally do not commute with other things! As an easy example, $frac{d}{dx}f(x)g(x) neq f(x)frac{d}{dx}g(x)$. You will notice that when $u$ is replaced with the operator $nabla$, we will instead obtain:
begin{align}
(nablatimes (vtimes w))_x &= partial_y (v_xw_y - v_yw_x) - partial_z(v_zw_x - v_xw_z)\
&=partial_y(v_xw_y)-partial_y(v_yw_x) - partial_z(v_zw_x) + partial_z(v_xw_z)\
&= (partial_yv_x) w_y+v_x(partial_y w_y) - (partial_yv_y)w_x-v_y(partial_yw_x)-(partial_zv_z)w_x - v_z(partial_zw_x)\&qquadqquad+(partial_zv_x)w_z + v_x(partial_zw_z) \
&= v_x(partial_y w_y) + v_x(partial_zw_z) - (partial_yv_y)w_x - (partial_zv_z)w_x + (partial_yv_x) w_y + (partial_zv_x)w_z\&qquadqquad -v_y(partial_yw_x)-v_z(partial_zw_x) \
&=(partial_yw_y+partial_zw_z)v_x - (partial_yv_y+partial_zv_z)w_x+(w_ypartial_y + w_zpartial_z)v_x - (v_ypartial_y +v_zpartial_z)w_x \
&vdots\
&=(nablacdot w)v_x - (nablacdot v)w_x+(wcdotnabla)v_x - (vcdotnabla)w_x
end{align}
and similarly for the other components.
answered Feb 18 '18 at 5:13
bamesbames
1,9311315
$endgroup$
add a comment |
$begingroup$
To see why this fails, examine the Wikipedia proof of the triple product expansion. Writing their work in more detail, we find:
begin{align}
(utimes (vtimes w))_x &= u_y(v_xw_y - v_yw_x) - u_z(v_zw_x - v_xw_z)\
&=u_yv_xw_y - u_yv_yw_x - u_zv_zw_x +u_zv_xw_z \
&=u_ycolor{red}{v_x}w_y + u_zcolor{red}{v_x}w_z - u_yv_ycolor{magenta}{w_x}-u_zv_zcolor{magenta}{w_x} \
&= color{red}{v_x}(u_yw_y + u_zw_z) - color{magenta}{w_x}(u_yv_y + u_zv_z)
end{align}
You can see that they took advantage of the commutativity of products of scalars. But operators generally do not commute with other things! As an easy example, $frac{d}{dx}f(x)g(x) neq f(x)frac{d}{dx}g(x)$. You will notice that when $u$ is replaced with the operator $nabla$, we will instead obtain:
begin{align}
(nablatimes (vtimes w))_x &= partial_y (v_xw_y - v_yw_x) - partial_z(v_zw_x - v_xw_z)\
&=partial_y(v_xw_y)-partial_y(v_yw_x) - partial_z(v_zw_x) + partial_z(v_xw_z)\
&= (partial_yv_x) w_y+v_x(partial_y w_y) - (partial_yv_y)w_x-v_y(partial_yw_x)-(partial_zv_z)w_x - v_z(partial_zw_x)\&qquadqquad+(partial_zv_x)w_z + v_x(partial_zw_z) \
&= v_x(partial_y w_y) + v_x(partial_zw_z) - (partial_yv_y)w_x - (partial_zv_z)w_x + (partial_yv_x) w_y + (partial_zv_x)w_z\&qquadqquad -v_y(partial_yw_x)-v_z(partial_zw_x) \
&=(partial_yw_y+partial_zw_z)v_x - (partial_yv_y+partial_zv_z)w_x+(w_ypartial_y + w_zpartial_z)v_x - (v_ypartial_y +v_zpartial_z)w_x \
&vdots\
&=(nablacdot w)v_x - (nablacdot v)w_x+(wcdotnabla)v_x - (vcdotnabla)w_x
end{align}
and similarly for the other components.
answered Feb 18 '18 at 5:13
bamesbames
1,9311315
$endgroup$
To see why this fails, examine the Wikipedia proof of the triple product expansion. Writing their work in more detail, we find:
begin{align}
(utimes (vtimes w))_x &= u_y(v_xw_y - v_yw_x) - u_z(v_zw_x - v_xw_z)\
&=u_yv_xw_y - u_yv_yw_x - u_zv_zw_x +u_zv_xw_z \
&=u_ycolor{red}{v_x}w_y + u_zcolor{red}{v_x}w_z - u_yv_ycolor{magenta}{w_x}-u_zv_zcolor{magenta}{w_x} \
&= color{red}{v_x}(u_yw_y + u_zw_z) - color{magenta}{w_x}(u_yv_y + u_zv_z)
end{align}
You can see that they took advantage of the commutativity of products of scalars. But operators generally do not commute with other things! As an easy example, $frac{d}{dx}f(x)g(x) neq f(x)frac{d}{dx}g(x)$. You will notice that when $u$ is replaced with the operator $nabla$, we will instead obtain:
begin{align}
(nablatimes (vtimes w))_x &= partial_y (v_xw_y - v_yw_x) - partial_z(v_zw_x - v_xw_z)\
&=partial_y(v_xw_y)-partial_y(v_yw_x) - partial_z(v_zw_x) + partial_z(v_xw_z)\
&= (partial_yv_x) w_y+v_x(partial_y w_y) - (partial_yv_y)w_x-v_y(partial_yw_x)-(partial_zv_z)w_x - v_z(partial_zw_x)\&qquadqquad+(partial_zv_x)w_z + v_x(partial_zw_z) \
&= v_x(partial_y w_y) + v_x(partial_zw_z) - (partial_yv_y)w_x - (partial_zv_z)w_x + (partial_yv_x) w_y + (partial_zv_x)w_z\&qquadqquad -v_y(partial_yw_x)-v_z(partial_zw_x) \
&=(partial_yw_y+partial_zw_z)v_x - (partial_yv_y+partial_zv_z)w_x+(w_ypartial_y + w_zpartial_z)v_x - (v_ypartial_y +v_zpartial_z)w_x \
&vdots\
&=(nablacdot w)v_x - (nablacdot v)w_x+(wcdotnabla)v_x - (vcdotnabla)w_x
end{align}
and similarly for the other components.
answered Feb 18 '18 at 5:13
bamesbames
1,9311315
edited Feb 18 '18 at 5:24
answered Feb 18 '18 at 5:13
bamesbames
1,9311315
answered Feb 18 '18 at 5:13
bamesbames
1,9311315
answered Feb 18 '18 at 5:13
bamesbames
1,9311315
1,9311315
add a comment |
add a comment |
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