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A Formula of Orthogonal Curvilinear Coordinate Systems
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Let $(u_1,u_2,u_3)$ represent the three coordinates in a general curvilinear system in $mathbbR^3$, and let $mathbfe_i$ be the unit vector that points the direction of increasing $u_i$. Let $mathbfe_i cdot mathbfe_j=delta_ij$, and $mathbfe_itimesmathbfe_j=varepsilon_ijkmathbfe_k$ for $i,j,k=1,2,3$. For any $mathbfr$ in $mathbbR^3$, let $fracpartial mathbfrpartial u_i=h_imathbfe_i$, where $h_i$ is the scale vector. We then have $dr^2=(sum_i=1^3 h_idu_i)^2$. This is called orthogonal curvilinear coordinate system. Moreover, by making use of $nabla u_i=h_i^-1mathbfe_i$, we can write the operator $nabla$ as $nabla=sum_i=1^3 mathbfe_i h_i^-1fracpartialpartial u_i$. Then, the gradient formula for the orthogonal curvilinear system is obvious.
We have some tricks to obtain the divergence and curl formulas in the orthognal curvilinear system. For example, we may use the relation $nablacdotleft(fracmathbfe_1h_2h_3right)=nablacdot(nabla mathbfe_2timesnablamathbfe_3)=0$ to obtain the divergence formula. However, we may consider a direct approach: Compute the derivative of a unit vector against a coordinate, i.e., compute the value of $fracpartial mathbfe_jpartial u_i$. If $i=j$, by noticing $fracpartial mathbfe_ipartial u_i=varepsilon_ijkfracpartial(mathbfe_jtimesmathbfe_k)partial u_i$, it suffices to compute $fracpartial mathbfe_jpartial u_i$ for $jneq i$. Actually, there is a simple formula about it: For $jneq i$, $fracpartial mathbfe_jpartial u_i=h_j^-1fracpartial h_ipartial u_jmathbfe_i$. My question is how to derive this formula.
vector-analysis
asked Aug 5 at 22:03
user581944
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Let $(u_1,u_2,u_3)$ represent the three coordinates in a general curvilinear system in $mathbbR^3$, and let $mathbfe_i$ be the unit vector that points the direction of increasing $u_i$. Let $mathbfe_i cdot mathbfe_j=delta_ij$, and $mathbfe_itimesmathbfe_j=varepsilon_ijkmathbfe_k$ for $i,j,k=1,2,3$. For any $mathbfr$ in $mathbbR^3$, let $fracpartial mathbfrpartial u_i=h_imathbfe_i$, where $h_i$ is the scale vector. We then have $dr^2=(sum_i=1^3 h_idu_i)^2$. This is called orthogonal curvilinear coordinate system. Moreover, by making use of $nabla u_i=h_i^-1mathbfe_i$, we can write the operator $nabla$ as $nabla=sum_i=1^3 mathbfe_i h_i^-1fracpartialpartial u_i$. Then, the gradient formula for the orthogonal curvilinear system is obvious.
We have some tricks to obtain the divergence and curl formulas in the orthognal curvilinear system. For example, we may use the relation $nablacdotleft(fracmathbfe_1h_2h_3right)=nablacdot(nabla mathbfe_2timesnablamathbfe_3)=0$ to obtain the divergence formula. However, we may consider a direct approach: Compute the derivative of a unit vector against a coordinate, i.e., compute the value of $fracpartial mathbfe_jpartial u_i$. If $i=j$, by noticing $fracpartial mathbfe_ipartial u_i=varepsilon_ijkfracpartial(mathbfe_jtimesmathbfe_k)partial u_i$, it suffices to compute $fracpartial mathbfe_jpartial u_i$ for $jneq i$. Actually, there is a simple formula about it: For $jneq i$, $fracpartial mathbfe_jpartial u_i=h_j^-1fracpartial h_ipartial u_jmathbfe_i$. My question is how to derive this formula.
vector-analysis
asked Aug 5 at 22:03
user581944
11
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $(u_1,u_2,u_3)$ represent the three coordinates in a general curvilinear system in $mathbbR^3$, and let $mathbfe_i$ be the unit vector that points the direction of increasing $u_i$. Let $mathbfe_i cdot mathbfe_j=delta_ij$, and $mathbfe_itimesmathbfe_j=varepsilon_ijkmathbfe_k$ for $i,j,k=1,2,3$. For any $mathbfr$ in $mathbbR^3$, let $fracpartial mathbfrpartial u_i=h_imathbfe_i$, where $h_i$ is the scale vector. We then have $dr^2=(sum_i=1^3 h_idu_i)^2$. This is called orthogonal curvilinear coordinate system. Moreover, by making use of $nabla u_i=h_i^-1mathbfe_i$, we can write the operator $nabla$ as $nabla=sum_i=1^3 mathbfe_i h_i^-1fracpartialpartial u_i$. Then, the gradient formula for the orthogonal curvilinear system is obvious.
We have some tricks to obtain the divergence and curl formulas in the orthognal curvilinear system. For example, we may use the relation $nablacdotleft(fracmathbfe_1h_2h_3right)=nablacdot(nabla mathbfe_2timesnablamathbfe_3)=0$ to obtain the divergence formula. However, we may consider a direct approach: Compute the derivative of a unit vector against a coordinate, i.e., compute the value of $fracpartial mathbfe_jpartial u_i$. If $i=j$, by noticing $fracpartial mathbfe_ipartial u_i=varepsilon_ijkfracpartial(mathbfe_jtimesmathbfe_k)partial u_i$, it suffices to compute $fracpartial mathbfe_jpartial u_i$ for $jneq i$. Actually, there is a simple formula about it: For $jneq i$, $fracpartial mathbfe_jpartial u_i=h_j^-1fracpartial h_ipartial u_jmathbfe_i$. My question is how to derive this formula.
vector-analysis
asked Aug 5 at 22:03
user581944
11
Let $(u_1,u_2,u_3)$ represent the three coordinates in a general curvilinear system in $mathbbR^3$, and let $mathbfe_i$ be the unit vector that points the direction of increasing $u_i$. Let $mathbfe_i cdot mathbfe_j=delta_ij$, and $mathbfe_itimesmathbfe_j=varepsilon_ijkmathbfe_k$ for $i,j,k=1,2,3$. For any $mathbfr$ in $mathbbR^3$, let $fracpartial mathbfrpartial u_i=h_imathbfe_i$, where $h_i$ is the scale vector. We then have $dr^2=(sum_i=1^3 h_idu_i)^2$. This is called orthogonal curvilinear coordinate system. Moreover, by making use of $nabla u_i=h_i^-1mathbfe_i$, we can write the operator $nabla$ as $nabla=sum_i=1^3 mathbfe_i h_i^-1fracpartialpartial u_i$. Then, the gradient formula for the orthogonal curvilinear system is obvious.
We have some tricks to obtain the divergence and curl formulas in the orthognal curvilinear system. For example, we may use the relation $nablacdotleft(fracmathbfe_1h_2h_3right)=nablacdot(nabla mathbfe_2timesnablamathbfe_3)=0$ to obtain the divergence formula. However, we may consider a direct approach: Compute the derivative of a unit vector against a coordinate, i.e., compute the value of $fracpartial mathbfe_jpartial u_i$. If $i=j$, by noticing $fracpartial mathbfe_ipartial u_i=varepsilon_ijkfracpartial(mathbfe_jtimesmathbfe_k)partial u_i$, it suffices to compute $fracpartial mathbfe_jpartial u_i$ for $jneq i$. Actually, there is a simple formula about it: For $jneq i$, $fracpartial mathbfe_jpartial u_i=h_j^-1fracpartial h_ipartial u_jmathbfe_i$. My question is how to derive this formula.
vector-analysis
asked Aug 5 at 22:03
user581944
11
asked Aug 5 at 22:03
user581944
11
asked Aug 5 at 22:03
user581944
11
asked Aug 5 at 22:03
user581944
11
11
add a comment |Â
add a comment |Â
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