Helmholtz decomposition proof
$begingroup$
Reading derivation of Helmholtz decomposition in wikipedia, there are two points I do not understand. First one appears in the step:
$$
begin{align}
dots
&=-frac{1}{4pi}left[nablaleft(nablacdotint_Vfrac{mathbf{F}(mathbf{r}')}{left|mathbf{r}-mathbf{r}'right|}mathrm{d}V'right)-nablatimesleft(nablatimesint_Vfrac{mathbf{F}(mathbf{r}')}{left|mathbf{r}-mathbf{r}'right|}mathrm{d}V'right)right] \
&= -frac{1}{4pi} left[nablaleft(int_Vmathbf{F}(mathbf{r}')cdotnablafrac{1}{left|mathbf{r}-mathbf{r}'right|}mathrm{d}V'right)+nablatimesleft(int_Vmathbf{F}(mathbf{r}')timesnablafrac{1}{left|mathbf{r}-mathbf{r}'right|}mathrm{d}V'right)right] \
&= dots
end{align}
$$
where sign of second term jumps from "-" to "+" at same time than curl is inserted in the integral, something I do not see why must change the sign.
The second doubts appears a few lines after, where I can read:
$$
mathbf{a}timesnablapsi =psi(nablatimesmathbf{a})-nabla times (psimathbf{a})
$$
because, starting from product rule of curl over scalar and vector:
$$
nabla times (psimathbf{a}) = psi(nablatimesmathbf{a}) + mathbf{a}timesnablapsi
$$
I expected:
$$
mathbf{a}timesnablapsi = -psi(nablatimesmathbf{a})+nabla times (psimathbf{a})
$$
Could be two obvious errors from my side, but I can not find them. Any help is welcome, thanks.
calculus curl
asked Jan 23 at 19:23
pasaba por aquipasaba por aqui
429315
$endgroup$
add a comment |
$begingroup$
Reading derivation of Helmholtz decomposition in wikipedia, there are two points I do not understand. First one appears in the step:
$$
begin{align}
dots
&=-frac{1}{4pi}left[nablaleft(nablacdotint_Vfrac{mathbf{F}(mathbf{r}')}{left|mathbf{r}-mathbf{r}'right|}mathrm{d}V'right)-nablatimesleft(nablatimesint_Vfrac{mathbf{F}(mathbf{r}')}{left|mathbf{r}-mathbf{r}'right|}mathrm{d}V'right)right] \
&= -frac{1}{4pi} left[nablaleft(int_Vmathbf{F}(mathbf{r}')cdotnablafrac{1}{left|mathbf{r}-mathbf{r}'right|}mathrm{d}V'right)+nablatimesleft(int_Vmathbf{F}(mathbf{r}')timesnablafrac{1}{left|mathbf{r}-mathbf{r}'right|}mathrm{d}V'right)right] \
&= dots
end{align}
$$
where sign of second term jumps from "-" to "+" at same time than curl is inserted in the integral, something I do not see why must change the sign.
The second doubts appears a few lines after, where I can read:
$$
mathbf{a}timesnablapsi =psi(nablatimesmathbf{a})-nabla times (psimathbf{a})
$$
because, starting from product rule of curl over scalar and vector:
$$
nabla times (psimathbf{a}) = psi(nablatimesmathbf{a}) + mathbf{a}timesnablapsi
$$
I expected:
$$
mathbf{a}timesnablapsi = -psi(nablatimesmathbf{a})+nabla times (psimathbf{a})
$$
Could be two obvious errors from my side, but I can not find them. Any help is welcome, thanks.
calculus curl
asked Jan 23 at 19:23
pasaba por aquipasaba por aqui
429315
$endgroup$
add a comment |
$begingroup$
Reading derivation of Helmholtz decomposition in wikipedia, there are two points I do not understand. First one appears in the step:
$$
begin{align}
dots
&=-frac{1}{4pi}left[nablaleft(nablacdotint_Vfrac{mathbf{F}(mathbf{r}')}{left|mathbf{r}-mathbf{r}'right|}mathrm{d}V'right)-nablatimesleft(nablatimesint_Vfrac{mathbf{F}(mathbf{r}')}{left|mathbf{r}-mathbf{r}'right|}mathrm{d}V'right)right] \
&= -frac{1}{4pi} left[nablaleft(int_Vmathbf{F}(mathbf{r}')cdotnablafrac{1}{left|mathbf{r}-mathbf{r}'right|}mathrm{d}V'right)+nablatimesleft(int_Vmathbf{F}(mathbf{r}')timesnablafrac{1}{left|mathbf{r}-mathbf{r}'right|}mathrm{d}V'right)right] \
&= dots
end{align}
$$
where sign of second term jumps from "-" to "+" at same time than curl is inserted in the integral, something I do not see why must change the sign.
The second doubts appears a few lines after, where I can read:
$$
mathbf{a}timesnablapsi =psi(nablatimesmathbf{a})-nabla times (psimathbf{a})
$$
because, starting from product rule of curl over scalar and vector:
$$
nabla times (psimathbf{a}) = psi(nablatimesmathbf{a}) + mathbf{a}timesnablapsi
$$
I expected:
$$
mathbf{a}timesnablapsi = -psi(nablatimesmathbf{a})+nabla times (psimathbf{a})
$$
Could be two obvious errors from my side, but I can not find them. Any help is welcome, thanks.
calculus curl
asked Jan 23 at 19:23
pasaba por aquipasaba por aqui
429315
$endgroup$
Reading derivation of Helmholtz decomposition in wikipedia, there are two points I do not understand. First one appears in the step:
$$
begin{align}
dots
&=-frac{1}{4pi}left[nablaleft(nablacdotint_Vfrac{mathbf{F}(mathbf{r}')}{left|mathbf{r}-mathbf{r}'right|}mathrm{d}V'right)-nablatimesleft(nablatimesint_Vfrac{mathbf{F}(mathbf{r}')}{left|mathbf{r}-mathbf{r}'right|}mathrm{d}V'right)right] \
&= -frac{1}{4pi} left[nablaleft(int_Vmathbf{F}(mathbf{r}')cdotnablafrac{1}{left|mathbf{r}-mathbf{r}'right|}mathrm{d}V'right)+nablatimesleft(int_Vmathbf{F}(mathbf{r}')timesnablafrac{1}{left|mathbf{r}-mathbf{r}'right|}mathrm{d}V'right)right] \
&= dots
end{align}
$$
where sign of second term jumps from "-" to "+" at same time than curl is inserted in the integral, something I do not see why must change the sign.
The second doubts appears a few lines after, where I can read:
$$
mathbf{a}timesnablapsi =psi(nablatimesmathbf{a})-nabla times (psimathbf{a})
$$
because, starting from product rule of curl over scalar and vector:
$$
nabla times (psimathbf{a}) = psi(nablatimesmathbf{a}) + mathbf{a}timesnablapsi
$$
I expected:
$$
mathbf{a}timesnablapsi = -psi(nablatimesmathbf{a})+nabla times (psimathbf{a})
$$
Could be two obvious errors from my side, but I can not find them. Any help is welcome, thanks.
calculus curl
calculus curl
asked Jan 23 at 19:23
pasaba por aquipasaba por aqui
429315
asked Jan 23 at 19:23
pasaba por aquipasaba por aqui
429315
edited Jan 23 at 19:28
pasaba por aqui
asked Jan 23 at 19:23
pasaba por aquipasaba por aqui
429315
asked Jan 23 at 19:23
pasaba por aquipasaba por aqui
429315
asked Jan 23 at 19:23
pasaba por aquipasaba por aqui
429315
429315
add a comment |
add a comment |
1 Answer
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In both cases: the sign CHANGES if you reverse order of cross product.
$${bf a}times {bf b}=-{bf b}times{bf a}$$
There is a reverse of order in cross product from $nablatimes {bf F}f({bf r})$ to $ {bf F}times nabla f({bf r})$. Note that the derivative works on $bf r$, not $bf r'$ so ${bf F}$ acts as a constant. So, the minus sign is correct, because the reverse of order happened.
In the second one, you are again reversing the order without paying attention. The product you wrote is wrong, it should be:
$$nablatimes (psi {bf a})=psi(nablatimes {bf a})+(nablapsi)times{bf a}$$
Notice how ${bf a}$ which gives the vector direction to the right operand of $times$, stays on the right, and $nabla$ which acts as a vector on the left, stays on the left.
answered Jan 23 at 19:38
orionorion
13.6k11837
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1 Answer
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1 Answer
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$begingroup$
In both cases: the sign CHANGES if you reverse order of cross product.
$${bf a}times {bf b}=-{bf b}times{bf a}$$
There is a reverse of order in cross product from $nablatimes {bf F}f({bf r})$ to $ {bf F}times nabla f({bf r})$. Note that the derivative works on $bf r$, not $bf r'$ so ${bf F}$ acts as a constant. So, the minus sign is correct, because the reverse of order happened.
In the second one, you are again reversing the order without paying attention. The product you wrote is wrong, it should be:
$$nablatimes (psi {bf a})=psi(nablatimes {bf a})+(nablapsi)times{bf a}$$
Notice how ${bf a}$ which gives the vector direction to the right operand of $times$, stays on the right, and $nabla$ which acts as a vector on the left, stays on the left.
answered Jan 23 at 19:38
orionorion
13.6k11837
$endgroup$
add a comment |
$begingroup$
In both cases: the sign CHANGES if you reverse order of cross product.
$${bf a}times {bf b}=-{bf b}times{bf a}$$
There is a reverse of order in cross product from $nablatimes {bf F}f({bf r})$ to $ {bf F}times nabla f({bf r})$. Note that the derivative works on $bf r$, not $bf r'$ so ${bf F}$ acts as a constant. So, the minus sign is correct, because the reverse of order happened.
In the second one, you are again reversing the order without paying attention. The product you wrote is wrong, it should be:
$$nablatimes (psi {bf a})=psi(nablatimes {bf a})+(nablapsi)times{bf a}$$
Notice how ${bf a}$ which gives the vector direction to the right operand of $times$, stays on the right, and $nabla$ which acts as a vector on the left, stays on the left.
answered Jan 23 at 19:38
orionorion
13.6k11837
$endgroup$
add a comment |
$begingroup$
In both cases: the sign CHANGES if you reverse order of cross product.
$${bf a}times {bf b}=-{bf b}times{bf a}$$
There is a reverse of order in cross product from $nablatimes {bf F}f({bf r})$ to $ {bf F}times nabla f({bf r})$. Note that the derivative works on $bf r$, not $bf r'$ so ${bf F}$ acts as a constant. So, the minus sign is correct, because the reverse of order happened.
In the second one, you are again reversing the order without paying attention. The product you wrote is wrong, it should be:
$$nablatimes (psi {bf a})=psi(nablatimes {bf a})+(nablapsi)times{bf a}$$
Notice how ${bf a}$ which gives the vector direction to the right operand of $times$, stays on the right, and $nabla$ which acts as a vector on the left, stays on the left.
answered Jan 23 at 19:38
orionorion
13.6k11837
$endgroup$
In both cases: the sign CHANGES if you reverse order of cross product.
$${bf a}times {bf b}=-{bf b}times{bf a}$$
There is a reverse of order in cross product from $nablatimes {bf F}f({bf r})$ to $ {bf F}times nabla f({bf r})$. Note that the derivative works on $bf r$, not $bf r'$ so ${bf F}$ acts as a constant. So, the minus sign is correct, because the reverse of order happened.
In the second one, you are again reversing the order without paying attention. The product you wrote is wrong, it should be:
$$nablatimes (psi {bf a})=psi(nablatimes {bf a})+(nablapsi)times{bf a}$$
Notice how ${bf a}$ which gives the vector direction to the right operand of $times$, stays on the right, and $nabla$ which acts as a vector on the left, stays on the left.
answered Jan 23 at 19:38
orionorion
13.6k11837
edited Jan 23 at 19:42
answered Jan 23 at 19:38
orionorion
13.6k11837
answered Jan 23 at 19:38
orionorion
13.6k11837
answered Jan 23 at 19:38
orionorion
13.6k11837
13.6k11837
add a comment |
add a comment |
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