Del operator apply directly to orthogonal curvilinear coordinate does not match












0












$begingroup$


I understand that $nabla$ in general orthogonal coordinate $(u_1,u_2,u_3)$ as follows:



$$
nabla=mathbf{a}_{u_1}frac{partial}{h_1partial u_1}+mathbf{a}_{u_2}frac{partial}{h_2partial u_2}+mathbf{a}_{u_3}frac{partial}{h_3partial u_3}tag{1}
$$



It is also given that for general curvilinear coordinates $(u_1,u_2,u_3)$



$$
nablacdotmathbf{A}=frac1{h_1h_2h_3}left[frac{partial}{partial u_1}(h_2h_3A_1)+frac{partial}{partial u_2}(h_1h_3A_2)+frac{partial}{partial u_3}(h_1h_2A_3)right]tag{2}
$$



Assuming $mathbf{A}$ is in curvilinear coordinates, apply directly equation (1) to $mathbf{A}$ does not give me equation (2) .... or am I missing anything?




Clarification on what I am trying to do.... please point out any mistake..



I am taking that
A= $mathbf{a}_{u_1}$$A_1$ + $mathbf{a}_{u_2}$$A_2$ + $mathbf{a}_{u_3}$$A_3$



To make thing really simple, let assume A has only 1 term : A=$mathbf{a}_{u_1}$$A_1$



so $nablacdotmathbf{A}$ = ($mathbf{a}_{u_1}frac{partial}{h_1partial u_1}$). ($mathbf{a}_{u_1}$$A_1$)



which end up as $nablacdotmathbf{A}$ = ($frac{partial A_1}{h_1partial u_1}$)



But the equation 2 first term shows $frac1{h_1h_2h_3}[frac{partial}{partial u_1}(h_2h_3A_1) ] $ .. somthing is wrong ..






curvilinear-coordinates







share|cite|improve this question











$endgroup$












  • $begingroup$
    See Maxim's comment here.
    $endgroup$
    – user10354138
    Dec 9 '18 at 14:00










  • $begingroup$
    hi @user10354138 , i have seem Maxim comment , but not quite understand , i have make some edit on the post to make it clearer... kindly advise...
    $endgroup$
    – lau
    Dec 9 '18 at 15:38










  • $begingroup$
    hi @user10354138 , sorrry for my weak understanding , i thinking that i am dealing with orthogonal coordinates , $mathbf{a}_{u_1}$ . $mathbf{a}_{u_1}$ should give you 1 , given they are unit vectors. So you are saying that it is not true ?
    $endgroup$
    – lau
    Dec 9 '18 at 16:20










  • $begingroup$
    Your $mathbf{a}_{u_i}$ is just the unit direction of $nabla u_i$. However, they are not constant (e.g., polar coordinates where the $mathbf{e}_r,mathbf{e}_theta$ varies with the angular coordinate $theta$), so $frac{partialmathbf{a}_{u_1}}{partial u_i}$ are not necessarily zero and thus you cannot just consider the $mathbf{a}_{u_1}$-component of $nabla$, and also $frac{partial}{partial u_1}(mathbf{a}_{u_1}A_1)$ need not be $mathbf{a}_{u_1}frac{partial A_1}{partial u_1}$.
    $endgroup$
    – user10354138
    Dec 9 '18 at 16:55


















0












$begingroup$


I understand that $nabla$ in general orthogonal coordinate $(u_1,u_2,u_3)$ as follows:



$$
nabla=mathbf{a}_{u_1}frac{partial}{h_1partial u_1}+mathbf{a}_{u_2}frac{partial}{h_2partial u_2}+mathbf{a}_{u_3}frac{partial}{h_3partial u_3}tag{1}
$$



It is also given that for general curvilinear coordinates $(u_1,u_2,u_3)$



$$
nablacdotmathbf{A}=frac1{h_1h_2h_3}left[frac{partial}{partial u_1}(h_2h_3A_1)+frac{partial}{partial u_2}(h_1h_3A_2)+frac{partial}{partial u_3}(h_1h_2A_3)right]tag{2}
$$



Assuming $mathbf{A}$ is in curvilinear coordinates, apply directly equation (1) to $mathbf{A}$ does not give me equation (2) .... or am I missing anything?




Clarification on what I am trying to do.... please point out any mistake..



I am taking that
A= $mathbf{a}_{u_1}$$A_1$ + $mathbf{a}_{u_2}$$A_2$ + $mathbf{a}_{u_3}$$A_3$



To make thing really simple, let assume A has only 1 term : A=$mathbf{a}_{u_1}$$A_1$



so $nablacdotmathbf{A}$ = ($mathbf{a}_{u_1}frac{partial}{h_1partial u_1}$). ($mathbf{a}_{u_1}$$A_1$)



which end up as $nablacdotmathbf{A}$ = ($frac{partial A_1}{h_1partial u_1}$)



But the equation 2 first term shows $frac1{h_1h_2h_3}[frac{partial}{partial u_1}(h_2h_3A_1) ] $ .. somthing is wrong ..






curvilinear-coordinates







share|cite|improve this question











$endgroup$












  • $begingroup$
    See Maxim's comment here.
    $endgroup$
    – user10354138
    Dec 9 '18 at 14:00










  • $begingroup$
    hi @user10354138 , i have seem Maxim comment , but not quite understand , i have make some edit on the post to make it clearer... kindly advise...
    $endgroup$
    – lau
    Dec 9 '18 at 15:38










  • $begingroup$
    hi @user10354138 , sorrry for my weak understanding , i thinking that i am dealing with orthogonal coordinates , $mathbf{a}_{u_1}$ . $mathbf{a}_{u_1}$ should give you 1 , given they are unit vectors. So you are saying that it is not true ?
    $endgroup$
    – lau
    Dec 9 '18 at 16:20










  • $begingroup$
    Your $mathbf{a}_{u_i}$ is just the unit direction of $nabla u_i$. However, they are not constant (e.g., polar coordinates where the $mathbf{e}_r,mathbf{e}_theta$ varies with the angular coordinate $theta$), so $frac{partialmathbf{a}_{u_1}}{partial u_i}$ are not necessarily zero and thus you cannot just consider the $mathbf{a}_{u_1}$-component of $nabla$, and also $frac{partial}{partial u_1}(mathbf{a}_{u_1}A_1)$ need not be $mathbf{a}_{u_1}frac{partial A_1}{partial u_1}$.
    $endgroup$
    – user10354138
    Dec 9 '18 at 16:55
















0












0








0





$begingroup$


I understand that $nabla$ in general orthogonal coordinate $(u_1,u_2,u_3)$ as follows:



$$
nabla=mathbf{a}_{u_1}frac{partial}{h_1partial u_1}+mathbf{a}_{u_2}frac{partial}{h_2partial u_2}+mathbf{a}_{u_3}frac{partial}{h_3partial u_3}tag{1}
$$



It is also given that for general curvilinear coordinates $(u_1,u_2,u_3)$



$$
nablacdotmathbf{A}=frac1{h_1h_2h_3}left[frac{partial}{partial u_1}(h_2h_3A_1)+frac{partial}{partial u_2}(h_1h_3A_2)+frac{partial}{partial u_3}(h_1h_2A_3)right]tag{2}
$$



Assuming $mathbf{A}$ is in curvilinear coordinates, apply directly equation (1) to $mathbf{A}$ does not give me equation (2) .... or am I missing anything?




Clarification on what I am trying to do.... please point out any mistake..



I am taking that
A= $mathbf{a}_{u_1}$$A_1$ + $mathbf{a}_{u_2}$$A_2$ + $mathbf{a}_{u_3}$$A_3$



To make thing really simple, let assume A has only 1 term : A=$mathbf{a}_{u_1}$$A_1$



so $nablacdotmathbf{A}$ = ($mathbf{a}_{u_1}frac{partial}{h_1partial u_1}$). ($mathbf{a}_{u_1}$$A_1$)



which end up as $nablacdotmathbf{A}$ = ($frac{partial A_1}{h_1partial u_1}$)



But the equation 2 first term shows $frac1{h_1h_2h_3}[frac{partial}{partial u_1}(h_2h_3A_1) ] $ .. somthing is wrong ..






curvilinear-coordinates







share|cite|improve this question











$endgroup$




I understand that $nabla$ in general orthogonal coordinate $(u_1,u_2,u_3)$ as follows:



$$
nabla=mathbf{a}_{u_1}frac{partial}{h_1partial u_1}+mathbf{a}_{u_2}frac{partial}{h_2partial u_2}+mathbf{a}_{u_3}frac{partial}{h_3partial u_3}tag{1}
$$



It is also given that for general curvilinear coordinates $(u_1,u_2,u_3)$



$$
nablacdotmathbf{A}=frac1{h_1h_2h_3}left[frac{partial}{partial u_1}(h_2h_3A_1)+frac{partial}{partial u_2}(h_1h_3A_2)+frac{partial}{partial u_3}(h_1h_2A_3)right]tag{2}
$$



Assuming $mathbf{A}$ is in curvilinear coordinates, apply directly equation (1) to $mathbf{A}$ does not give me equation (2) .... or am I missing anything?




Clarification on what I am trying to do.... please point out any mistake..



I am taking that
A= $mathbf{a}_{u_1}$$A_1$ + $mathbf{a}_{u_2}$$A_2$ + $mathbf{a}_{u_3}$$A_3$



To make thing really simple, let assume A has only 1 term : A=$mathbf{a}_{u_1}$$A_1$



so $nablacdotmathbf{A}$ = ($mathbf{a}_{u_1}frac{partial}{h_1partial u_1}$). ($mathbf{a}_{u_1}$$A_1$)



which end up as $nablacdotmathbf{A}$ = ($frac{partial A_1}{h_1partial u_1}$)



But the equation 2 first term shows $frac1{h_1h_2h_3}[frac{partial}{partial u_1}(h_2h_3A_1) ] $ .. somthing is wrong ..






curvilinear-coordinates




curvilinear-coordinates






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 9 '18 at 15:36







lau

















asked Dec 9 '18 at 13:47









laulau

11




11












  • $begingroup$
    See Maxim's comment here.
    $endgroup$
    – user10354138
    Dec 9 '18 at 14:00










  • $begingroup$
    hi @user10354138 , i have seem Maxim comment , but not quite understand , i have make some edit on the post to make it clearer... kindly advise...
    $endgroup$
    – lau
    Dec 9 '18 at 15:38










  • $begingroup$
    hi @user10354138 , sorrry for my weak understanding , i thinking that i am dealing with orthogonal coordinates , $mathbf{a}_{u_1}$ . $mathbf{a}_{u_1}$ should give you 1 , given they are unit vectors. So you are saying that it is not true ?
    $endgroup$
    – lau
    Dec 9 '18 at 16:20










  • $begingroup$
    Your $mathbf{a}_{u_i}$ is just the unit direction of $nabla u_i$. However, they are not constant (e.g., polar coordinates where the $mathbf{e}_r,mathbf{e}_theta$ varies with the angular coordinate $theta$), so $frac{partialmathbf{a}_{u_1}}{partial u_i}$ are not necessarily zero and thus you cannot just consider the $mathbf{a}_{u_1}$-component of $nabla$, and also $frac{partial}{partial u_1}(mathbf{a}_{u_1}A_1)$ need not be $mathbf{a}_{u_1}frac{partial A_1}{partial u_1}$.
    $endgroup$
    – user10354138
    Dec 9 '18 at 16:55




















  • $begingroup$
    See Maxim's comment here.
    $endgroup$
    – user10354138
    Dec 9 '18 at 14:00










  • $begingroup$
    hi @user10354138 , i have seem Maxim comment , but not quite understand , i have make some edit on the post to make it clearer... kindly advise...
    $endgroup$
    – lau
    Dec 9 '18 at 15:38










  • $begingroup$
    hi @user10354138 , sorrry for my weak understanding , i thinking that i am dealing with orthogonal coordinates , $mathbf{a}_{u_1}$ . $mathbf{a}_{u_1}$ should give you 1 , given they are unit vectors. So you are saying that it is not true ?
    $endgroup$
    – lau
    Dec 9 '18 at 16:20










  • $begingroup$
    Your $mathbf{a}_{u_i}$ is just the unit direction of $nabla u_i$. However, they are not constant (e.g., polar coordinates where the $mathbf{e}_r,mathbf{e}_theta$ varies with the angular coordinate $theta$), so $frac{partialmathbf{a}_{u_1}}{partial u_i}$ are not necessarily zero and thus you cannot just consider the $mathbf{a}_{u_1}$-component of $nabla$, and also $frac{partial}{partial u_1}(mathbf{a}_{u_1}A_1)$ need not be $mathbf{a}_{u_1}frac{partial A_1}{partial u_1}$.
    $endgroup$
    – user10354138
    Dec 9 '18 at 16:55


















$begingroup$
See Maxim's comment here.
$endgroup$
– user10354138
Dec 9 '18 at 14:00




$begingroup$
See Maxim's comment here.
$endgroup$
– user10354138
Dec 9 '18 at 14:00












$begingroup$
hi @user10354138 , i have seem Maxim comment , but not quite understand , i have make some edit on the post to make it clearer... kindly advise...
$endgroup$
– lau
Dec 9 '18 at 15:38




$begingroup$
hi @user10354138 , i have seem Maxim comment , but not quite understand , i have make some edit on the post to make it clearer... kindly advise...
$endgroup$
– lau
Dec 9 '18 at 15:38












$begingroup$
hi @user10354138 , sorrry for my weak understanding , i thinking that i am dealing with orthogonal coordinates , $mathbf{a}_{u_1}$ . $mathbf{a}_{u_1}$ should give you 1 , given they are unit vectors. So you are saying that it is not true ?
$endgroup$
– lau
Dec 9 '18 at 16:20




$begingroup$
hi @user10354138 , sorrry for my weak understanding , i thinking that i am dealing with orthogonal coordinates , $mathbf{a}_{u_1}$ . $mathbf{a}_{u_1}$ should give you 1 , given they are unit vectors. So you are saying that it is not true ?
$endgroup$
– lau
Dec 9 '18 at 16:20












$begingroup$
Your $mathbf{a}_{u_i}$ is just the unit direction of $nabla u_i$. However, they are not constant (e.g., polar coordinates where the $mathbf{e}_r,mathbf{e}_theta$ varies with the angular coordinate $theta$), so $frac{partialmathbf{a}_{u_1}}{partial u_i}$ are not necessarily zero and thus you cannot just consider the $mathbf{a}_{u_1}$-component of $nabla$, and also $frac{partial}{partial u_1}(mathbf{a}_{u_1}A_1)$ need not be $mathbf{a}_{u_1}frac{partial A_1}{partial u_1}$.
$endgroup$
– user10354138
Dec 9 '18 at 16:55






$begingroup$
Your $mathbf{a}_{u_i}$ is just the unit direction of $nabla u_i$. However, they are not constant (e.g., polar coordinates where the $mathbf{e}_r,mathbf{e}_theta$ varies with the angular coordinate $theta$), so $frac{partialmathbf{a}_{u_1}}{partial u_i}$ are not necessarily zero and thus you cannot just consider the $mathbf{a}_{u_1}$-component of $nabla$, and also $frac{partial}{partial u_1}(mathbf{a}_{u_1}A_1)$ need not be $mathbf{a}_{u_1}frac{partial A_1}{partial u_1}$.
$endgroup$
– user10354138
Dec 9 '18 at 16:55












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