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Mathematics
Differential Geometry
LeviCivita in Orthogonal Curvilinear Coordinate System: "Cross Product Matrix
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[QUOTE="ltkach2015, post: 5759765, member: 520445"] [COLOR=#ff0000]If given a position vector defined for a orthogonal curvilinear coordinate system HOW would the matrices that make up the Levi Civita 3x3x3 matrix remain the same? [/COLOR] "Levi Civita 3x3x3 is said to be independent of any coordinate system or metric tensor"([URL]https://en.wikipedia.org/wiki/Levi-Civita_symbol[/URL])[B][U][COLOR=#0000ff]Position Vector in a cartesian coordinate system[/COLOR] [/U] R = [/B][x;y;z] = x*[B]i [/B]+ y*[B]j[/B] + z*[B]k [/B] = x*[1;0;0] + y*[0;1;0] + z*[0;0;1] [COLOR=#0000ff][B][U]Cartesian Unit Vectors [/U][/B][/COLOR] [COLOR=#000000][B]i = [/B][1;0;0], [B]j = [/B][0;1;0], [B]k = [/B][0;0;1][/COLOR] [B][U][COLOR=#0000ff]Angular Velocity Vector[/COLOR][/U] w = [/B][w1;w2;w3] [B][U][COLOR=#0000ff]Velocity Vector[/COLOR][/U] V = [/B]cross([B]w[/B],[B]R[/B]) = [B]w [/B]x [B]R [/B]= [U][B]CPM[/B]([B]w[/B])[/U][B]*R[/B] [U][B][COLOR=#0000ff]Cross Product Matrix (CPM) Derivation[/COLOR][/B] [/U] [INDENT][U][COLOR=rgb(0, 0, 255)][B]Matrices that make up each 'page' of the 3x3x3 alternating tensor/symbol/Levi-Civita symbol[/B][/COLOR][/U] [B][U]I[/U][/B] = [B] j*[/B]transpose([B]k[/B]) - [B]k[/B]*transpose([B]j[/B]) = [0 0 0; 0 0 1; 0 -1 0] [U][B]J[/B][/U] = [B] -[/B]([B]i*[/B]transpose([B]k[/B]) - [B]k[/B]*transpose([B]i[/B])) = [0 0 -1; 0 0 0; 1 0 0] [U][B]K[/B][/U] = [B]i*[/B]transpose([B]j[/B]) - [B]j[/B]*transpose([B]i[/B]) = [0 1 0; -1 0 0; 0 0 0][/INDENT] [INDENT][U][B][COLOR=#0000ff]LeviCivita here is a 3x3x3 matrix[/COLOR][/B][/U] [B][I][U]LeviCivita[/U][/I][/B](all rows, all columns, page1) = [B][U]I[/U][/B] [I][B][U]LeviCivita[/U][/B][/I](all rows, all columns, page2)[B] = [U]J[/U][/B] [U][I][B]LeviCivita[/B][/I][/U](all rows, all columns, page3) = [U][B]K[/B][/U][/INDENT] [INDENT][B][U]CPM[/U] [/B]= transpose([B]w[/B])*[I][B][U]LeviCivita[/U][/B][/I] = [transpose([B]w[/B])*[B][I][U]LeviCivita[/U][/I][/B](all rows, all columns, page1) ; transpose([B]w[/B])*[B][I]LeviCivita[/I][/B](all rows, all columns, page2); ...transpose([B]w[/B])*[B][I][U]LeviCivita[/U][/I][/B](all rows, all columns, page3) ] = [transpose([B]w[/B])*[U][B]I[/B][/U] ; transpose([B]w[/B])*[B][U]J[/U][/B] ; transpose([B]w[/B])*[B][U]K[/U][/B] ][/INDENT] [COLOR=#ff0000][ 0 -w3 w2 ] [ w3 0 -w1 ] = [U][B]CPM[/B]([/U][B]w[/B]) [ -w2 w1 0 ][/COLOR] And, as expected [B]V [/B]= [B]w[/B] x [B]R[/B] = [U][B]CPM[/B]([/U][B]w[/B])*[B]R[/B] = [w2*z - w3*y; w3*x - w1*z; w1*y - w2*x] -------------------------------------------------------------------------------------------------------------------------------------- -------------------------------------------------------------------------------------------------------------------------------------- Again how would the Levi Civita 3x3x3 matrix remain the same in an orthogonal curvilinear coordinate system? For example given the position vector defined in spherical coordinates. Coordinates: r = radius, o = theta, p = phi [INDENT][B]R [/B]= [r*cos(o)*sin(p); r*sin(o)*sin(p); r*cos(p)] = r*[B]normalized_r normalized_r[/B] is the normalized unit vector for the r direction[/INDENT] Should one assume the following to arrive at the matrices that form the Levi Civita symbol: [INDENT][B][U][SIZE=7]r[/SIZE][/U][/B] = [B] o*[/B]transpose([B]p[/B]) - [B]o[/B]*transpose([B]p[/B]) [SIZE=7][B][U]o[/U][/B] [/SIZE]= [B] -[/B]([B]r*[/B]transpose([B]p[/B]) - [B]p[/B]*transpose([B]r[/B])) [SIZE=7][B][U]p[/U][/B] [/SIZE]= [B]r*[/B]transpose([B]o[/B]) - [B]o[/B]*transpose([B]r[/B])[/INDENT] where the unit vectors are found from: [B]r = gradient[/B](x) = [partial_dx/partial_dr; partial_dx/partial_do; partial_dx/partial_dp] [B]o = gradient[/B](y) = [partial_dy/partial_dr; partial_dy/partial_do; partial_dy/partial_dp] [B]p = gradient[/B](z) = [partial_dz/partial_dr; partial_dz/partial_do; partial_dz/partial_dp]OR Should one use the normalized unit vectors to form the matrices instead? [INDENT][B][U][SIZE=7]r[/SIZE][/U][/B] = [B] normalized_o*[/B]transpose([B]normalized_p[/B]) - [B]normalized_o[/B]*transpose([B]normalized_o[/B]) [SIZE=7][B][U]o[/U][/B] [/SIZE]= [B] -[/B]([B]normalized_r*[/B]transpose([B]normalized_p[/B]) - [B]normalized_p[/B]*transpose([B]normalized_r[/B])) [SIZE=7][B][U]p[/U][/B] [/SIZE]= [B]normalized_r*[/B]transpose([B]normalized_o[/B]) - [B]normalized_o[/B]*transpose([B]normalized_r)[/B][/INDENT] None of these resulted in the matrices that comprised the Levi Civita 3x3x3 matrix. This spherical coordinate system is not right handed. Should one change the order from r, o, p to r, p, o? [URL]https://math.stackexchange.com/questions/243142/what-is-the-general-formula-for-calculating-dot-and-cross-products-in-spherical[/URL] [/QUOTE]
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LeviCivita in Orthogonal Curvilinear Coordinate System: "Cross Product Matrix
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