Calculating the cross product of a cross product
up vote
1
down vote
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so I really can't see what I am doing wrong. I want to use this formula:
$atimes (btimes c) = b(acdot c) - c(acdot b)$
Calculate the rotation of $v(x,y,z)=(x,y,z)^T times omega$ with $omega in mathbb R^3$
Solution:
$acdot c=nablacdot omega=0$
and
$acdot b = nabla cdot (x,y,z)^T= partial_x x + partial_y y + partial_z z = 3$
so we get $-omega 3=-3omega$
The actual solution (which I do get by direct calculation) is: $-2omega$
cross-product curl
asked Oct 23 at 18:16
xotix
36829
add a comment |
up vote
1
down vote
favorite
so I really can't see what I am doing wrong. I want to use this formula:
$atimes (btimes c) = b(acdot c) - c(acdot b)$
Calculate the rotation of $v(x,y,z)=(x,y,z)^T times omega$ with $omega in mathbb R^3$
Solution:
$acdot c=nablacdot omega=0$
and
$acdot b = nabla cdot (x,y,z)^T= partial_x x + partial_y y + partial_z z = 3$
so we get $-omega 3=-3omega$
The actual solution (which I do get by direct calculation) is: $-2omega$
cross-product curl
asked Oct 23 at 18:16
xotix
36829
1
You can't just replace $a$ with $nabla$ like that, you also need to keep track of what is integrated. See here: math.stackexchange.com/questions/836041/…
– Hans Lundmark
Oct 23 at 18:32
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
so I really can't see what I am doing wrong. I want to use this formula:
$atimes (btimes c) = b(acdot c) - c(acdot b)$
Calculate the rotation of $v(x,y,z)=(x,y,z)^T times omega$ with $omega in mathbb R^3$
Solution:
$acdot c=nablacdot omega=0$
and
$acdot b = nabla cdot (x,y,z)^T= partial_x x + partial_y y + partial_z z = 3$
so we get $-omega 3=-3omega$
The actual solution (which I do get by direct calculation) is: $-2omega$
cross-product curl
asked Oct 23 at 18:16
xotix
36829
so I really can't see what I am doing wrong. I want to use this formula:
$atimes (btimes c) = b(acdot c) - c(acdot b)$
Calculate the rotation of $v(x,y,z)=(x,y,z)^T times omega$ with $omega in mathbb R^3$
Solution:
$acdot c=nablacdot omega=0$
and
$acdot b = nabla cdot (x,y,z)^T= partial_x x + partial_y y + partial_z z = 3$
so we get $-omega 3=-3omega$
The actual solution (which I do get by direct calculation) is: $-2omega$
cross-product curl
cross-product curl
asked Oct 23 at 18:16
xotix
36829
asked Oct 23 at 18:16
xotix
36829
asked Oct 23 at 18:16
xotix
36829
asked Oct 23 at 18:16
xotix
36829
asked Oct 23 at 18:16
xotix
36829
36829
1
You can't just replace $a$ with $nabla$ like that, you also need to keep track of what is integrated. See here: math.stackexchange.com/questions/836041/…
– Hans Lundmark
Oct 23 at 18:32
add a comment |
1
You can't just replace $a$ with $nabla$ like that, you also need to keep track of what is integrated. See here: math.stackexchange.com/questions/836041/…
– Hans Lundmark
Oct 23 at 18:32
1
1
You can't just replace $a$ with $nabla$ like that, you also need to keep track of what is integrated. See here: math.stackexchange.com/questions/836041/…
– Hans Lundmark
Oct 23 at 18:32
You can't just replace $a$ with $nabla$ like that, you also need to keep track of what is integrated. See here: math.stackexchange.com/questions/836041/…
– Hans Lundmark
Oct 23 at 18:32
add a comment |
2 Answers
2
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up vote
1
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If you want to generalise formulae with vectors having commuting components, make sure to write the usual formula so as to preserve the order in products. For example, here the $i$th component of the RHS is $sum_j a_j (b_i c_j-b_j c_i)$, once we impose the $abc$ order of the LHS. Now you know the generalisation off the top of your head. For $a_i=partial_i$, the result's $i$th component is $$sum_j partial_j (b_i c_j-b_j c_i)=b_inablacdot c+ccdotnabla b_i-(nablacdot b) c-bcdotnabla c_i.$$The vector, in other words, is $$vec{b}(nablacdotvec{c})-(nablacdotvec{b})vec{c}+(vec{c}cdotnabla)vec{b}-(vec{b}cdotnabla)vec{c}.$$(I've swapped the middle terms to mirror @md2perpe's quoted result, but the letters $b,,c$ still need to be changed to $A,,B$).
answered Oct 23 at 20:23
J.G.
21.2k21933
add a comment |
up vote
1
down vote
On Wikipedia you can see that the formula for the curl of a cross product is given by
$$
nabla times (mathbf{A} times mathbf{B}) = mathbf{A} (nabla cdot mathbf{B}) - mathbf{B} (nabla cdot mathbf{A}) + (mathbf{B} cdot nabla) mathbf{A} - (mathbf{A} cdot nabla) mathbf{B}
.
$$
Applying this on your case gives
$$begin{align}
nabla times (mathbf{x} times mathbf{omega})
&= mathbf{x} (nabla cdot mathbf{omega}) - mathbf{omega} (nabla cdot mathbf{x}) + (mathbf{omega} cdot nabla) mathbf{x} - (mathbf{x} cdot nabla) mathbf{omega} \
&= mathbf{x} 0 - mathbf{omega} 3 + mathbf{omega} - mathbf{0}
= -2 mathbf{omega}
.
end{align}$$
answered Oct 23 at 20:07
md2perpe
7,33811027
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
If you want to generalise formulae with vectors having commuting components, make sure to write the usual formula so as to preserve the order in products. For example, here the $i$th component of the RHS is $sum_j a_j (b_i c_j-b_j c_i)$, once we impose the $abc$ order of the LHS. Now you know the generalisation off the top of your head. For $a_i=partial_i$, the result's $i$th component is $$sum_j partial_j (b_i c_j-b_j c_i)=b_inablacdot c+ccdotnabla b_i-(nablacdot b) c-bcdotnabla c_i.$$The vector, in other words, is $$vec{b}(nablacdotvec{c})-(nablacdotvec{b})vec{c}+(vec{c}cdotnabla)vec{b}-(vec{b}cdotnabla)vec{c}.$$(I've swapped the middle terms to mirror @md2perpe's quoted result, but the letters $b,,c$ still need to be changed to $A,,B$).
answered Oct 23 at 20:23
J.G.
21.2k21933
add a comment |
up vote
1
down vote
accepted
If you want to generalise formulae with vectors having commuting components, make sure to write the usual formula so as to preserve the order in products. For example, here the $i$th component of the RHS is $sum_j a_j (b_i c_j-b_j c_i)$, once we impose the $abc$ order of the LHS. Now you know the generalisation off the top of your head. For $a_i=partial_i$, the result's $i$th component is $$sum_j partial_j (b_i c_j-b_j c_i)=b_inablacdot c+ccdotnabla b_i-(nablacdot b) c-bcdotnabla c_i.$$The vector, in other words, is $$vec{b}(nablacdotvec{c})-(nablacdotvec{b})vec{c}+(vec{c}cdotnabla)vec{b}-(vec{b}cdotnabla)vec{c}.$$(I've swapped the middle terms to mirror @md2perpe's quoted result, but the letters $b,,c$ still need to be changed to $A,,B$).
answered Oct 23 at 20:23
J.G.
21.2k21933
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
If you want to generalise formulae with vectors having commuting components, make sure to write the usual formula so as to preserve the order in products. For example, here the $i$th component of the RHS is $sum_j a_j (b_i c_j-b_j c_i)$, once we impose the $abc$ order of the LHS. Now you know the generalisation off the top of your head. For $a_i=partial_i$, the result's $i$th component is $$sum_j partial_j (b_i c_j-b_j c_i)=b_inablacdot c+ccdotnabla b_i-(nablacdot b) c-bcdotnabla c_i.$$The vector, in other words, is $$vec{b}(nablacdotvec{c})-(nablacdotvec{b})vec{c}+(vec{c}cdotnabla)vec{b}-(vec{b}cdotnabla)vec{c}.$$(I've swapped the middle terms to mirror @md2perpe's quoted result, but the letters $b,,c$ still need to be changed to $A,,B$).
answered Oct 23 at 20:23
J.G.
21.2k21933
If you want to generalise formulae with vectors having commuting components, make sure to write the usual formula so as to preserve the order in products. For example, here the $i$th component of the RHS is $sum_j a_j (b_i c_j-b_j c_i)$, once we impose the $abc$ order of the LHS. Now you know the generalisation off the top of your head. For $a_i=partial_i$, the result's $i$th component is $$sum_j partial_j (b_i c_j-b_j c_i)=b_inablacdot c+ccdotnabla b_i-(nablacdot b) c-bcdotnabla c_i.$$The vector, in other words, is $$vec{b}(nablacdotvec{c})-(nablacdotvec{b})vec{c}+(vec{c}cdotnabla)vec{b}-(vec{b}cdotnabla)vec{c}.$$(I've swapped the middle terms to mirror @md2perpe's quoted result, but the letters $b,,c$ still need to be changed to $A,,B$).
answered Oct 23 at 20:23
J.G.
21.2k21933
edited Nov 22 at 19:40
answered Oct 23 at 20:23
J.G.
21.2k21933
answered Oct 23 at 20:23
J.G.
21.2k21933
answered Oct 23 at 20:23
J.G.
21.2k21933
21.2k21933
add a comment |
add a comment |
up vote
1
down vote
On Wikipedia you can see that the formula for the curl of a cross product is given by
$$
nabla times (mathbf{A} times mathbf{B}) = mathbf{A} (nabla cdot mathbf{B}) - mathbf{B} (nabla cdot mathbf{A}) + (mathbf{B} cdot nabla) mathbf{A} - (mathbf{A} cdot nabla) mathbf{B}
.
$$
Applying this on your case gives
$$begin{align}
nabla times (mathbf{x} times mathbf{omega})
&= mathbf{x} (nabla cdot mathbf{omega}) - mathbf{omega} (nabla cdot mathbf{x}) + (mathbf{omega} cdot nabla) mathbf{x} - (mathbf{x} cdot nabla) mathbf{omega} \
&= mathbf{x} 0 - mathbf{omega} 3 + mathbf{omega} - mathbf{0}
= -2 mathbf{omega}
.
end{align}$$
answered Oct 23 at 20:07
md2perpe
7,33811027
add a comment |
up vote
1
down vote
On Wikipedia you can see that the formula for the curl of a cross product is given by
$$
nabla times (mathbf{A} times mathbf{B}) = mathbf{A} (nabla cdot mathbf{B}) - mathbf{B} (nabla cdot mathbf{A}) + (mathbf{B} cdot nabla) mathbf{A} - (mathbf{A} cdot nabla) mathbf{B}
.
$$
Applying this on your case gives
$$begin{align}
nabla times (mathbf{x} times mathbf{omega})
&= mathbf{x} (nabla cdot mathbf{omega}) - mathbf{omega} (nabla cdot mathbf{x}) + (mathbf{omega} cdot nabla) mathbf{x} - (mathbf{x} cdot nabla) mathbf{omega} \
&= mathbf{x} 0 - mathbf{omega} 3 + mathbf{omega} - mathbf{0}
= -2 mathbf{omega}
.
end{align}$$
answered Oct 23 at 20:07
md2perpe
7,33811027
add a comment |
up vote
1
down vote
up vote
1
down vote
On Wikipedia you can see that the formula for the curl of a cross product is given by
$$
nabla times (mathbf{A} times mathbf{B}) = mathbf{A} (nabla cdot mathbf{B}) - mathbf{B} (nabla cdot mathbf{A}) + (mathbf{B} cdot nabla) mathbf{A} - (mathbf{A} cdot nabla) mathbf{B}
.
$$
Applying this on your case gives
$$begin{align}
nabla times (mathbf{x} times mathbf{omega})
&= mathbf{x} (nabla cdot mathbf{omega}) - mathbf{omega} (nabla cdot mathbf{x}) + (mathbf{omega} cdot nabla) mathbf{x} - (mathbf{x} cdot nabla) mathbf{omega} \
&= mathbf{x} 0 - mathbf{omega} 3 + mathbf{omega} - mathbf{0}
= -2 mathbf{omega}
.
end{align}$$
answered Oct 23 at 20:07
md2perpe
7,33811027
On Wikipedia you can see that the formula for the curl of a cross product is given by
$$
nabla times (mathbf{A} times mathbf{B}) = mathbf{A} (nabla cdot mathbf{B}) - mathbf{B} (nabla cdot mathbf{A}) + (mathbf{B} cdot nabla) mathbf{A} - (mathbf{A} cdot nabla) mathbf{B}
.
$$
Applying this on your case gives
$$begin{align}
nabla times (mathbf{x} times mathbf{omega})
&= mathbf{x} (nabla cdot mathbf{omega}) - mathbf{omega} (nabla cdot mathbf{x}) + (mathbf{omega} cdot nabla) mathbf{x} - (mathbf{x} cdot nabla) mathbf{omega} \
&= mathbf{x} 0 - mathbf{omega} 3 + mathbf{omega} - mathbf{0}
= -2 mathbf{omega}
.
end{align}$$
answered Oct 23 at 20:07
md2perpe
7,33811027
answered Oct 23 at 20:07
md2perpe
7,33811027
answered Oct 23 at 20:07
md2perpe
7,33811027
answered Oct 23 at 20:07
md2perpe
7,33811027
7,33811027
add a comment |
add a comment |
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You can't just replace $a$ with $nabla$ like that, you also need to keep track of what is integrated. See here: math.stackexchange.com/questions/836041/…
– Hans Lundmark
Oct 23 at 18:32