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the 3-dimensional metric tensor \"g\" for Cylindrical and \ Spherical Coordinates. \nHint: This is a good assignment to do with ", StyleBox["Mathematica", FontSlant->"Italic"], " or equivalent, therefore, I have shown you some ", StyleBox["Mathematica", FontSlant->"Italic"], " code below to get you started if you like to do it this way. " }], "Text", CellFrame->True, Background->GrayLevel[1]], Cell[BoxData[ \(Clear["\"]\)], "Input", Background->GrayLevel[1]], Cell["<GrayLevel[1], CellTags->"S5.16.1"], Cell[CellGroupData[{ Cell["a) Cylindrical Coordinates:", "Subsection", Background->GrayLevel[1]], Cell[CellGroupData[{ Cell[BoxData[ \(\(?CoordinatesToCartesian\)\)], "Input", Background->GrayLevel[1]], Cell[BoxData[ \("CoordinatesToCartesian[pt] gives the Cartesian coordinates of the \ point pt given in the default coordinate system. CoordinatesToCartesian[pt, \ coordsys] gives the Cartesian coordinates of the point given in the \ coordinate system coordsys."\)], "Print", Background->GrayLevel[1]] }, Closed]], Cell[CellGroupData[{ Cell[BoxData[ \(x2rRule = \ Thread[\ {x, y, z} -> CoordinatesToCartesian[{r, \[Theta], z}, Cylindrical]\ ]\)], "Input",\ Background->GrayLevel[1]], Cell[BoxData[ \({x \[Rule] r\ Cos[\[Theta]], y \[Rule] r\ Sin[\[Theta]], z \[Rule] z}\)], "Output", Background->GrayLevel[1]] }, Closed]], Cell[CellGroupData[{ Cell[BoxData[{ \(\(vector = {r, \[Theta], z};\)\), "\[IndentingNewLine]", \(\(tmp = \ Outer[d, {x, y, z}, vector];\)\), "\[IndentingNewLine]", \(tmp\ // MatrixForm\)}], "Input", Background->GrayLevel[1]], Cell[BoxData[ TagBox[ RowBox[{"(", "\[NoBreak]", GridBox[{ {\(d[x, r]\), \(d[x, \[Theta]]\), \(d[x, z]\)}, {\(d[y, r]\), \(d[y, \[Theta]]\), \(d[y, z]\)}, {\(d[z, r]\), \(d[z, \[Theta]]\), \(d[z, z]\)} }], "\[NoBreak]", ")"}], (MatrixForm[ #]&)]], "Output", Background->GrayLevel[1]] }, Closed]], Cell[BoxData[{ \(\(gMetric = Transpose[tmp] . tmp;\)\), "\[IndentingNewLine]", \(gMetric // MatrixForm\)}], "Input", Background->GrayLevel[1]], Cell[BoxData[{ \(\(gMetric = \(gMetric\ /. x2rRule\)\ /. {d \[Rule] D}\ // Simplify;\)\), "\[IndentingNewLine]", \(gMetric // MatrixForm\)}], "Input", Background->GrayLevel[1]], Cell[BoxData[{ \(\(dvector = {dr, d\[Theta], dz};\)\), "\[IndentingNewLine]", \(dvector . gMetric . dvector\)}], "Input", Background->GrayLevel[1]] }, Closed]], Cell[CellGroupData[{ Cell["b) Spherical Coordinates:", "Subsection", Background->GrayLevel[1]], Cell[CellGroupData[{ Cell[BoxData[ \(\(?CoordinatesToCartesian\)\)], "Input", Background->GrayLevel[1]], Cell[BoxData[ \("CoordinatesToCartesian[pt] gives the Cartesian coordinates of the \ point pt given in the default coordinate system. CoordinatesToCartesian[pt, \ coordsys] gives the Cartesian coordinates of the point given in the \ coordinate system coordsys."\)], "Print", Background->GrayLevel[1]] }, Closed]], Cell[BoxData[ \(x2rRule = \ Thread[\ {x, y, z} -> CoordinatesToCartesian[{r, \[Theta], \[Phi]}, Spherical]\ ]\)], "Input", Background->GrayLevel[1]], Cell[BoxData[{ \(\(vector = {r, \[Theta], \[Phi]};\)\), "\[IndentingNewLine]", \(\(tmp = \ Outer[d, {x, y, z}, vector];\)\), "\[IndentingNewLine]", \(tmp\ // MatrixForm\)}], "Input", Background->GrayLevel[1]], Cell[BoxData[{ \(\(gMetric = Transpose[tmp] . tmp;\)\), "\[IndentingNewLine]", \(gMetric // MatrixForm\)}], "Input", Background->GrayLevel[1]], Cell[BoxData[{ \(\(gMetric = \(gMetric\ /. x2rRule\)\ /. {d \[Rule] D}\ // Simplify;\)\), "\[IndentingNewLine]", \(gMetric // MatrixForm\)}], "Input", Background->GrayLevel[1]], Cell[BoxData[{ \(\(dvector = {dr, d\[Theta], d\[Phi]};\)\), "\[IndentingNewLine]", \(dvector . gMetric . dvector\)}], "Input", Background->GrayLevel[1]] }, Closed]] }, Closed]], Cell[CellGroupData[{ Cell["\<\ Problem #2: Gaussian Curvature for Cylindrical and Spherical Coordinates: 2-D part only:\ \ \>", "Section", Background->GrayLevel[1]], Cell[TextData[{ "We will take 2-dimensional sub-spaces of Cylindrical and Spherical \ Coordinates,and then compute the Gaussian Curvature for these sub-spaces. \n\ Since there are 3-coordinates, and we take them 2 at a time, there are a \ total for 3-combinations for each coordinate system. \n\nFor Cylindrical {r,\ \[Theta],z}, we have: {r,\[Theta]},{r,z}, and {\[Theta],z}\nFor Spherical {r,\ \[Theta],\[Phi]}, we have: {r,\[Theta]},{r,,\[Phi]} and {\[Theta],\[Phi]}\n\n\ For each pair or coordinates, compute the Gaussian curvature. Make a sketch \ of each 2-dimensional surface, and explain intuitively why you expect the \ Gaussian curvature is zero or not (as the case may be).\n\n\nHint: This is a \ good assignment to do with ", StyleBox["Mathematica", FontSlant->"Italic"], " or equivalent, therefore, I have shown you some ", StyleBox["Mathematica", FontSlant->"Italic"], " code below to get you started if you like to do it this way. I'll show \ you an example with Cartesian coordinates. " }], "Text", CellFrame->True, Background->GrayLevel[1]], Cell[BoxData[ \(Clear["\"]\)], "Input", Background->GrayLevel[1]], Cell["<GrayLevel[1], CellTags->"S5.16.1"], Cell[CellGroupData[{ Cell["a) Cartesian Coordinates: {x,y} ", "Subsection", Background->GrayLevel[1]], Cell[BoxData[ RowBox[{ RowBox[{"gMetric", "=", TagBox[ RowBox[{"(", "\[NoBreak]", GridBox[{ {"1", "0"}, {"0", "1"} }], "\[NoBreak]", ")"}], (MatrixForm[ #]&)]}], ";"}]], "Input", Background->GrayLevel[1]], Cell[CellGroupData[{ Cell[BoxData[{ \(\(vector = {x, y};\)\), "\[IndentingNewLine]", \(\(dvector = {dx, dy};\)\), "\[IndentingNewLine]", \(dvector . gMetric . dvector\)}], "Input", Background->GrayLevel[1]], Cell[BoxData[ \(dx\^2 + dy\^2\)], "Output"] }, Closed]], Cell[BoxData[ \(\(curvature = \ \(1\/\(2\ g11\ g22\)\) \((\[IndentingNewLine]\(-d[ g11, {x2, 2}]\)\ - \ d[g22, {x1, 2}]\ \[IndentingNewLine] + \ \(1\/\(2\ g11\)\) \((d[g11, x1]\ d[g22, x1]\ + \ d[g11, x2]^2)\)\ + \ \ \(+\ \(1\/\(2\ g22\)\)\) \((d[ g11, x2]\ d[g22, x1]\ + \ d[g22, x1]^2)\))\);\)\)], "Input", Background->GrayLevel[1]], Cell[CellGroupData[{ Cell[BoxData[ \(rule1 = Thread[\ {x1, x2} \[Rule] \ vector\ ]\)], "Input", Background->GrayLevel[1]], Cell[BoxData[ \({x1 \[Rule] x, x2 \[Rule] y}\)], "Output"] }, Closed]], Cell[CellGroupData[{ Cell[BoxData[ \(rule2 = \ \ \({{g11, g12}, {g21, g22}} \[Rule] \ gMetric\ // \ Map[Flatten, #\ \ \ , 1] &\)\ // Thread\)], "Input", Background->GrayLevel[1]], Cell[BoxData[ \({g11 \[Rule] 1, g12 \[Rule] 0, g21 \[Rule] 0, g22 \[Rule] 1}\)], "Output"] }, Closed]], Cell[CellGroupData[{ Cell[BoxData[ \(\(\(curvature\ /. rule1\)\ /. rule2\)\ /. {d \[Rule] D}\)], "Input", Background->GrayLevel[1]], Cell[BoxData[ \(0\)], "Output"] }, Closed]], Cell["\<\ Discussion: The {x,y} sub-space is simply a flat 2-dimensional \ plane, like a sheet of flat paper. Therefore, we expect the curvature should \ be zero as computed above. \ \>", "Text", CellFrame->True, Background->GrayLevel[1]] }, Closed]], Cell["\<\ b) Cylindrical Coordinates: {r,\[Theta]} \ \>", "Subsection", Background->GrayLevel[1]], Cell["c) Cylindrical Coordinates: {r,z} ", "Subsection", Background->GrayLevel[1]], Cell["\<\ d) Cylindrical Coordinates: {\[Theta],z} \ \>", "Subsection", Background->GrayLevel[1]], Cell["\<\ e) Spherical Coordinates: {r,\[Theta]} \ \>", "Subsection", Background->GrayLevel[1]], Cell["\<\ f) Spherical Coordinates: {r,\[Phi]} \ \>", "Subsection", Background->GrayLevel[1]], Cell["\<\ g) Spherical Coordinates: {\[Theta],\[Phi]} \ \>", "Subsection", Background->GrayLevel[1]] }, Closed]], Cell[CellGroupData[{ Cell["\<\ Problem #3: Gaussian Curvature for Generalized Coordinates: \ \>", "Section", Background->GrayLevel[1]], Cell[TextData[{ "This is a variation of the above problem. We will work with coordinates \ {r,\[Phi]}, but assume a more general metric form, as shown below. \na) \ Compute the Gaussian Curvature, K.\nb) Solve for f[r] in terms of K and \ f'[r]=df/dr.\nc) Solve the differential equation in part b) to obtain f[r]. \n\ (See me if you have never solved such an equation before. This is a simple \ 1'st order equation which can be solved by the age-old method of guessing.)\n\ d) Now that you have obtained f[r], substitue back in to obtain: ", Cell[BoxData[ \(ds\^2 = d\[Phi]\^2\ r\^2 + dr\^2\ f[r]\)]], "." }], "Text", CellFrame->True, Background->GrayLevel[1]], Cell[BoxData[ \(Clear["\"]\)], "Input", Background->GrayLevel[1]], Cell["<GrayLevel[1], CellTags->"S5.16.1"], Cell[CellGroupData[{ Cell["\<\ a) Generalized Coordinates: {r,\[Phi]}\ \>", "Subsection", Background->GrayLevel[1]], Cell[CellGroupData[{ Cell[BoxData[{ \(\(gMetric = DiagonalMatrix[{f[r], r\^2}];\)\), "\[IndentingNewLine]", \(gMetric\ // MatrixForm\)}], "Input", Background->GrayLevel[1]], Cell[BoxData[ TagBox[ RowBox[{"(", "\[NoBreak]", GridBox[{ {\(f[r]\), "0"}, {"0", \(r\^2\)} }], "\[NoBreak]", ")"}], (MatrixForm[ #]&)]], "Output", Background->GrayLevel[1]] }, Closed]], Cell[BoxData[{ \(\(vector = {r, \[Phi]};\)\), "\[IndentingNewLine]", \(\(dvector = {dr, d\[Phi]};\)\), "\[IndentingNewLine]", \(dvector . gMetric . dvector\)}], "Input", Background->GrayLevel[1]] }, Closed]] }, Closed]], Cell[CellGroupData[{ Cell["\<\ Problem #4: Metric Tensors for Spherical and Cylindrical Coordinates: 4- D\ \>", "Section",\ Background->GrayLevel[1]], Cell[TextData[{ "By any means, generalize the 3-dimensional metrics \"g\" you computed in \ Problem #1 to the 4-dimensional space-time case. Order your variables as \ follows: {t,x,y,z}.\nAssume the signs of the Cartesian metric \"g\" are ", Cell[BoxData[ TagBox[ RowBox[{"(", "\[NoBreak]", GridBox[{ {"1", "0", "0", "0"}, {"0", \(-1\), "0", "0"}, {"0", "0", \(-1\), "0"}, {"0", "0", "0", \(-1\)} }], "\[NoBreak]", ")"}], (MatrixForm[ #]&)]]], "so that ", Cell[BoxData[ \(TraditionalForm\`ds\^2 = dt\^2 - \((dx\^2 + dy\^2 + dz\^2)\)\)]], ".\nExplain the reasoning you used to getyour result. " }], "Text", CellFrame->True, Background->GrayLevel[1]], Cell[BoxData[ \(Clear["\"]\)], "Input", Background->GrayLevel[1]], Cell["<GrayLevel[1], CellTags->"S5.16.1"], Cell[CellGroupData[{ Cell["a) Cylindrical Coordinates:", "Subsection", Background->GrayLevel[1]], Cell[CellGroupData[{ Cell[BoxData[ \(\(?CoordinatesToCartesian\)\)], "Input", Background->GrayLevel[1]], Cell[BoxData[ \("CoordinatesToCartesian[pt] gives the Cartesian coordinates of the \ point pt given in the default coordinate system. 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