$L= left( frac1rsintheta fracpartialpartial phi hatphi - frac1rfracpartial partial theta hattheta right) $
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Surely I'm making a trivial calculation error but I can not find it.
Let:
$$L(r,theta,phi)= frachbari r left( frac1rsintheta fracpartialpartial phi hatphi - frac1rfracpartial partial theta hattheta right)$$
$$L^2=Lcdot L= -hbar^2 r^2 left( frac1rsintheta fracpartialpartial phi hatphi - frac1rfracpartial partial theta hattheta right)^2=$$
$$= -hbar^2 r^2 left( frac1rsintheta fracpartialpartial phi hatphi - frac1rfracpartial partial theta hattheta right) cdot left( frac1rsintheta fracpartialpartial phi hatphi - frac1rfracpartial partial theta hattheta right)=$$
$$-hbar^2 left( frac1sin^2(theta) fracpartial^2partial phi^2 + fracpartial^2partial theta^2 right)$$
But I know that the correct result is:
$$L^2=-hbar^2 left( frac1sin^2(theta) fracpartial^2partial phi^2 + fracpartial^2partial theta^2 + cottheta fracpartialpartial theta right)$$
multivariable-calculus partial-derivative vector-analysis
edited Aug 28 at 15:40
Andrei
7,8802923
asked Aug 28 at 13:56
Stefano Barone
391110
add a comment |Â
up vote
1
down vote
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Surely I'm making a trivial calculation error but I can not find it.
Let:
$$L(r,theta,phi)= frachbari r left( frac1rsintheta fracpartialpartial phi hatphi - frac1rfracpartial partial theta hattheta right)$$
$$L^2=Lcdot L= -hbar^2 r^2 left( frac1rsintheta fracpartialpartial phi hatphi - frac1rfracpartial partial theta hattheta right)^2=$$
$$= -hbar^2 r^2 left( frac1rsintheta fracpartialpartial phi hatphi - frac1rfracpartial partial theta hattheta right) cdot left( frac1rsintheta fracpartialpartial phi hatphi - frac1rfracpartial partial theta hattheta right)=$$
$$-hbar^2 left( frac1sin^2(theta) fracpartial^2partial phi^2 + fracpartial^2partial theta^2 right)$$
But I know that the correct result is:
$$L^2=-hbar^2 left( frac1sin^2(theta) fracpartial^2partial phi^2 + fracpartial^2partial theta^2 + cottheta fracpartialpartial theta right)$$
multivariable-calculus partial-derivative vector-analysis
edited Aug 28 at 15:40
Andrei
7,8802923
asked Aug 28 at 13:56
Stefano Barone
391110
2
The vectors $hatphi$ and $hattheta$ are also coordinate dependent.
– HBR
Aug 28 at 14:34
1
Also, you would be better off writing the diff'l operators 'last', e.g, $$ L= left( hatphifrac1rsintheta fracpartialpartial phi - hatthetafrac1rfracpartial partial theta right).$$ The way it was written originally, formally it looks like the operators are applied to the vectors. Reordering will make clearer that the product rule comes into play in the question.
– peter a g
Aug 28 at 14:57
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Surely I'm making a trivial calculation error but I can not find it.
Let:
$$L(r,theta,phi)= frachbari r left( frac1rsintheta fracpartialpartial phi hatphi - frac1rfracpartial partial theta hattheta right)$$
$$L^2=Lcdot L= -hbar^2 r^2 left( frac1rsintheta fracpartialpartial phi hatphi - frac1rfracpartial partial theta hattheta right)^2=$$
$$= -hbar^2 r^2 left( frac1rsintheta fracpartialpartial phi hatphi - frac1rfracpartial partial theta hattheta right) cdot left( frac1rsintheta fracpartialpartial phi hatphi - frac1rfracpartial partial theta hattheta right)=$$
$$-hbar^2 left( frac1sin^2(theta) fracpartial^2partial phi^2 + fracpartial^2partial theta^2 right)$$
But I know that the correct result is:
$$L^2=-hbar^2 left( frac1sin^2(theta) fracpartial^2partial phi^2 + fracpartial^2partial theta^2 + cottheta fracpartialpartial theta right)$$
multivariable-calculus partial-derivative vector-analysis
edited Aug 28 at 15:40
Andrei
7,8802923
asked Aug 28 at 13:56
Stefano Barone
391110
Surely I'm making a trivial calculation error but I can not find it.
Let:
$$L(r,theta,phi)= frachbari r left( frac1rsintheta fracpartialpartial phi hatphi - frac1rfracpartial partial theta hattheta right)$$
$$L^2=Lcdot L= -hbar^2 r^2 left( frac1rsintheta fracpartialpartial phi hatphi - frac1rfracpartial partial theta hattheta right)^2=$$
$$= -hbar^2 r^2 left( frac1rsintheta fracpartialpartial phi hatphi - frac1rfracpartial partial theta hattheta right) cdot left( frac1rsintheta fracpartialpartial phi hatphi - frac1rfracpartial partial theta hattheta right)=$$
$$-hbar^2 left( frac1sin^2(theta) fracpartial^2partial phi^2 + fracpartial^2partial theta^2 right)$$
But I know that the correct result is:
$$L^2=-hbar^2 left( frac1sin^2(theta) fracpartial^2partial phi^2 + fracpartial^2partial theta^2 + cottheta fracpartialpartial theta right)$$
multivariable-calculus partial-derivative vector-analysis
edited Aug 28 at 15:40
Andrei
7,8802923
asked Aug 28 at 13:56
Stefano Barone
391110
edited Aug 28 at 15:40
Andrei
7,8802923
edited Aug 28 at 15:40
Andrei
7,8802923
edited Aug 28 at 15:40
Andrei
7,8802923
7,8802923
asked Aug 28 at 13:56
Stefano Barone
391110
asked Aug 28 at 13:56
Stefano Barone
391110
asked Aug 28 at 13:56
Stefano Barone
391110
391110
2
The vectors $hatphi$ and $hattheta$ are also coordinate dependent.
– HBR
Aug 28 at 14:34
1
Also, you would be better off writing the diff'l operators 'last', e.g, $$ L= left( hatphifrac1rsintheta fracpartialpartial phi - hatthetafrac1rfracpartial partial theta right).$$ The way it was written originally, formally it looks like the operators are applied to the vectors. Reordering will make clearer that the product rule comes into play in the question.
– peter a g
Aug 28 at 14:57
add a comment |Â
2
The vectors $hatphi$ and $hattheta$ are also coordinate dependent.
– HBR
Aug 28 at 14:34
1
Also, you would be better off writing the diff'l operators 'last', e.g, $$ L= left( hatphifrac1rsintheta fracpartialpartial phi - hatthetafrac1rfracpartial partial theta right).$$ The way it was written originally, formally it looks like the operators are applied to the vectors. Reordering will make clearer that the product rule comes into play in the question.
– peter a g
Aug 28 at 14:57
2
2
The vectors $hatphi$ and $hattheta$ are also coordinate dependent.
– HBR
Aug 28 at 14:34
The vectors $hatphi$ and $hattheta$ are also coordinate dependent.
– HBR
Aug 28 at 14:34
1
1
Also, you would be better off writing the diff'l operators 'last', e.g, $$ L= left( hatphifrac1rsintheta fracpartialpartial phi - hatthetafrac1rfracpartial partial theta right).$$ The way it was written originally, formally it looks like the operators are applied to the vectors. Reordering will make clearer that the product rule comes into play in the question.
– peter a g
Aug 28 at 14:57
Also, you would be better off writing the diff'l operators 'last', e.g, $$ L= left( hatphifrac1rsintheta fracpartialpartial phi - hatthetafrac1rfracpartial partial theta right).$$ The way it was written originally, formally it looks like the operators are applied to the vectors. Reordering will make clearer that the product rule comes into play in the question.
– peter a g
Aug 28 at 14:57
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
The problem is that if you change $theta$, the direction of $hatphi$ changes. You should be able to prove that $$fracpartial hatthetapartial phi=hatphicostheta$$
If you plug this into your equation, you will get the missing term.
answered Aug 28 at 14:42
Andrei
7,8802923
Much easier to calculate $L_x$, $L_y$, and $L_z$ in terms of spherical coordinates
– Andrei
Aug 28 at 14:43
And also the coefficient of $hat phi$ depends on $theta$. The partial wrt $theta$ will act on that too (i.e., I mean it cannot be treated as a constant wrt $theta$).
– peter a g
Aug 28 at 14:45
Correct, but then you are left with $hatthetacdothatphi=0$ for that term
– Andrei
Aug 28 at 14:47
Yes - but the OP should notice that!
– peter a g
Aug 28 at 14:47
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
The problem is that if you change $theta$, the direction of $hatphi$ changes. You should be able to prove that $$fracpartial hatthetapartial phi=hatphicostheta$$
If you plug this into your equation, you will get the missing term.
answered Aug 28 at 14:42
Andrei
7,8802923
Much easier to calculate $L_x$, $L_y$, and $L_z$ in terms of spherical coordinates
– Andrei
Aug 28 at 14:43
And also the coefficient of $hat phi$ depends on $theta$. The partial wrt $theta$ will act on that too (i.e., I mean it cannot be treated as a constant wrt $theta$).
– peter a g
Aug 28 at 14:45
Correct, but then you are left with $hatthetacdothatphi=0$ for that term
– Andrei
Aug 28 at 14:47
Yes - but the OP should notice that!
– peter a g
Aug 28 at 14:47
add a comment |Â
up vote
2
down vote
accepted
The problem is that if you change $theta$, the direction of $hatphi$ changes. You should be able to prove that $$fracpartial hatthetapartial phi=hatphicostheta$$
If you plug this into your equation, you will get the missing term.
answered Aug 28 at 14:42
Andrei
7,8802923
Much easier to calculate $L_x$, $L_y$, and $L_z$ in terms of spherical coordinates
– Andrei
Aug 28 at 14:43
And also the coefficient of $hat phi$ depends on $theta$. The partial wrt $theta$ will act on that too (i.e., I mean it cannot be treated as a constant wrt $theta$).
– peter a g
Aug 28 at 14:45
Correct, but then you are left with $hatthetacdothatphi=0$ for that term
– Andrei
Aug 28 at 14:47
Yes - but the OP should notice that!
– peter a g
Aug 28 at 14:47
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
The problem is that if you change $theta$, the direction of $hatphi$ changes. You should be able to prove that $$fracpartial hatthetapartial phi=hatphicostheta$$
If you plug this into your equation, you will get the missing term.
answered Aug 28 at 14:42
Andrei
7,8802923
The problem is that if you change $theta$, the direction of $hatphi$ changes. You should be able to prove that $$fracpartial hatthetapartial phi=hatphicostheta$$
If you plug this into your equation, you will get the missing term.
answered Aug 28 at 14:42
Andrei
7,8802923
answered Aug 28 at 14:42
Andrei
7,8802923
answered Aug 28 at 14:42
Andrei
7,8802923
answered Aug 28 at 14:42
Andrei
7,8802923
7,8802923
Much easier to calculate $L_x$, $L_y$, and $L_z$ in terms of spherical coordinates
– Andrei
Aug 28 at 14:43
And also the coefficient of $hat phi$ depends on $theta$. The partial wrt $theta$ will act on that too (i.e., I mean it cannot be treated as a constant wrt $theta$).
– peter a g
Aug 28 at 14:45
Correct, but then you are left with $hatthetacdothatphi=0$ for that term
– Andrei
Aug 28 at 14:47
Yes - but the OP should notice that!
– peter a g
Aug 28 at 14:47
add a comment |Â
Much easier to calculate $L_x$, $L_y$, and $L_z$ in terms of spherical coordinates
– Andrei
Aug 28 at 14:43
And also the coefficient of $hat phi$ depends on $theta$. The partial wrt $theta$ will act on that too (i.e., I mean it cannot be treated as a constant wrt $theta$).
– peter a g
Aug 28 at 14:45
Correct, but then you are left with $hatthetacdothatphi=0$ for that term
– Andrei
Aug 28 at 14:47
Yes - but the OP should notice that!
– peter a g
Aug 28 at 14:47
Much easier to calculate $L_x$, $L_y$, and $L_z$ in terms of spherical coordinates
– Andrei
Aug 28 at 14:43
Much easier to calculate $L_x$, $L_y$, and $L_z$ in terms of spherical coordinates
– Andrei
Aug 28 at 14:43
And also the coefficient of $hat phi$ depends on $theta$. The partial wrt $theta$ will act on that too (i.e., I mean it cannot be treated as a constant wrt $theta$).
– peter a g
Aug 28 at 14:45
And also the coefficient of $hat phi$ depends on $theta$. The partial wrt $theta$ will act on that too (i.e., I mean it cannot be treated as a constant wrt $theta$).
– peter a g
Aug 28 at 14:45
Correct, but then you are left with $hatthetacdothatphi=0$ for that term
– Andrei
Aug 28 at 14:47
Correct, but then you are left with $hatthetacdothatphi=0$ for that term
– Andrei
Aug 28 at 14:47
Yes - but the OP should notice that!
– peter a g
Aug 28 at 14:47
Yes - but the OP should notice that!
– peter a g
Aug 28 at 14:47
add a comment |Â
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2
The vectors $hatphi$ and $hattheta$ are also coordinate dependent.
– HBR
Aug 28 at 14:34
1
Also, you would be better off writing the diff'l operators 'last', e.g, $$ L= left( hatphifrac1rsintheta fracpartialpartial phi - hatthetafrac1rfracpartial partial theta right).$$ The way it was written originally, formally it looks like the operators are applied to the vectors. Reordering will make clearer that the product rule comes into play in the question.
– peter a g
Aug 28 at 14:57