(************** Content-type: application/mathematica ************** CreatedBy='Mathematica 5.2' Mathematica-Compatible Notebook This notebook can be used with any Mathematica-compatible application, such as Mathematica, MathReader or Publicon. The data for the notebook starts with the line containing stars above. To get the notebook into a Mathematica-compatible application, do one of the following: * Save the data starting with the line of stars above into a file with a name ending in .nb, then open the file inside the application; * Copy the data starting with the line of stars above to the clipboard, then use the Paste menu command inside the application. Data for notebooks contains only printable 7-bit ASCII and can be sent directly in email or through ftp in text mode. Newlines can be CR, LF or CRLF (Unix, Macintosh or MS-DOS style). 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Travaux pratiques : \ corrig\[EAcute]\ \>", "Subtitle", FormatType->TextForm, FontSize->14, CellTags->"i:1"], Cell[CellGroupData[{ Cell[TextData[{ "Notons (", StyleBox["x", FontSlant->"Italic"], ", ", StyleBox["t", FontSlant->"Italic"], ")", " les coordonn\[EAcute]es d'un \[EAcute]v\[EAcute]nement dans le syst\ \[EGrave]me de r\[EAcute]f\[EAcute]rence \[Sum] et (", Cell[BoxData[ \(TraditionalForm\`\(\(\ \)\(x', \ t'\)\)\)]], ") les coordonn\[EAcute]es du m\[EHat]me \[EAcute]v\[EAcute]nement dans le \ syst\[EGrave]me de r\[EAcute]f\[EAcute]rence \[Sum]'. ", StyleBox["v", FontSlant->"Italic"], " est la vitesse relative des syst\[EGrave]mes selon ", StyleBox["Ox", FontSlant->"Italic"], ". Admettons que la transformation permettant d'exprimer les coordonn\ \[EAcute]es dans \[Sum]' \[AGrave] partir de celles dans \[Sum] \ s'\[EAcute]crit :" }], "Text", TextAlignment->Left, TextJustification->0, FontSize->12, CellTags->"i:1"], Cell[TextData[{ Cell[BoxData[ FormBox[ RowBox[{"(", "\[NegativeThinSpace]", GridBox[{ { FormBox[\(x'\), "TextForm"]}, { RowBox[{ FormBox["t", "TextForm"], "'"}]} }], "\[NegativeThinSpace]", ")"}], TraditionalForm]]], " = ", Cell[BoxData[ FormBox[ RowBox[{ RowBox[{"(", "\[NegativeThinSpace]", GridBox[{ {"a", "b"}, {"d", "e"} }], "\[NegativeThinSpace]", ")"}], RowBox[{"(", "\[NegativeThinSpace]", GridBox[{ {"x"}, {"t"} }], "\[NegativeThinSpace]", ")"}]}], TraditionalForm]]] }], "Text", TextAlignment->Center, TextJustification->0, FontSize->12, CellTags->"i:1"], Cell[TextData[{ StyleBox["D\[EAcute]finition des deux \[EAcute]v\[EAcute]nements utilis\ \[EAcute]s", FontWeight->"Bold"], StyleBox["\n", FontColor->RGBColor[1, 0, 0]], StyleBox["Lorsque les origines des deux syst\[EGrave]mes de r\[EAcute]f\ \[EAcute]rence co\[IDoubleDot]ncident, il se produit, \[AGrave] l'origine des \ syst\[EGrave]mes, l'\[EAcute]v\[EAcute]nement ", FontColor->RGBColor[0, 0, 1]], Cell[BoxData[ \(TraditionalForm\`E\_0\)], FontColor->RGBColor[0, 0, 1]], StyleBox[". Au temps ", FontColor->RGBColor[0, 0, 1]], StyleBox["t ", FontSlant->"Italic", FontColor->RGBColor[0, 0, 1]], StyleBox["(temps mesur\[EAcute] en m dans \[Sum]), il se produit, \[AGrave] \ l'origine de \[Sum], l'\[EAcute]v\[EAcute]nement ", FontColor->RGBColor[0, 0, 1]], Cell[BoxData[ \(TraditionalForm\`E\_1\)], FontColor->RGBColor[0, 0, 1]], StyleBox[". Rappel : le temps ", FontColor->RGBColor[0, 0, 1]], StyleBox["t", FontSlant->"Italic", FontColor->RGBColor[0, 0, 1]], StyleBox[" mesur\[EAcute] en m est donn\[EAcute] par ", FontColor->RGBColor[0, 0, 1]], StyleBox["t", FontSlant->"Italic", FontColor->RGBColor[0, 0, 1]], StyleBox[" = ", FontColor->RGBColor[0, 0, 1]], StyleBox["c", FontSlant->"Italic", FontColor->RGBColor[0, 0, 1]], StyleBox[" ", FontColor->RGBColor[0, 0, 1]], Cell[BoxData[ FormBox[ StyleBox[\(t\_s\), FontSlant->"Italic"], TextForm]], FontColor->RGBColor[0, 0, 1]], "\n\n", StyleBox["Questions et r\[EAcute]ponses\n", FontWeight->"Bold"], "1. Les coordonn\[EAcute]es d'un \[EAcute]v\[EAcute]nement, comme les \ coordonn\[EAcute]es d'un point, peuvent \[EHat]tre \ consid\[EAcute]r\[EAcute]es comme les composantes d'un vecteur. Comment \ faut-il \[EAcute]crire les composantes d'un \[EAcute]v\[EAcute]nement pour \ que le carr\[EAcute] de la norme du vecteur qui le repr\[EAcute]sente donne \ (", Cell[BoxData[ FormBox[ SuperscriptBox["t", StyleBox["2", FontSize->10]], TraditionalForm]]], "-", Cell[BoxData[ FormBox[ SuperscriptBox["x", StyleBox["2", FontSize->9]], TraditionalForm]]], ") ?" }], "Text", CellMargins->{{Inherited, 1}, {Inherited, Inherited}}, TextAlignment->Left, TextJustification->0, FontSize->12, CellTags->"i:1"], Cell[CellGroupData[{ Cell[BoxData[ \(event = {\[ImaginaryI]*x, \ t}; \ \ (*\ temps\ mesur\[EAcute]\ en\ m\[EGrave]tre\ \ *) \[IndentingNewLine]\(\(event\)\(.\)\(event\)\(\ \ \)\( (*\ produit\ scalaire\ du\ vecteur\ par\ lui - m\[EHat]me\ *) \)\)\)], "Input", FontSize->10, CellTags->"i:1"], Cell[BoxData[ \(t\^2 - x\^2\)], "Output", CellTags->"i:1"] }, Open ]], Cell[TextData[{ "2", StyleBox[". Ecrivez les coordonn\[EAcute]s des \[EAcute]v\[EAcute]nements ", FontVariations->{"CompatibilityType"->0}], Cell[BoxData[ \(TraditionalForm\`E\_0\)]], " et ", Cell[BoxData[ \(TraditionalForm\`E\_1\)]], StyleBox["dans les deux syst\[EGrave]mes ", FontVariations->{"CompatibilityType"->0}], "\[Sum] et \[Sum]'." }], "Text", FontSize->12, CellTags->"i:1"], Cell[CellGroupData[{ Cell[BoxData[{ \(\(e0 = {0, 0};\)\ (*\ coordonn\[EAcute]es\ des\ \[EAcute]v\[EAcute]nements\ dans\ \[Sum]\ *) \ \), "\[IndentingNewLine]", \(e1 = {0, t}\)}], "Input", FontSize->10, CellTags->"i:1"], Cell[BoxData[ \({0, t}\)], "Output", CellTags->"i:1"], Cell[CellGroupData[{ Cell[BoxData[ \(e0prime = {0, 0}; \ \ (*\ coordonn\[EAcute]es\ des\ \[EAcute]v\[EAcute]nements\ dans\ \(\[Sum]\^'\ \)\ *) \[IndentingNewLine]e1prime = {\[ImaginaryI]*xprime, tprime}\)], "Input"], Cell[BoxData[ \({\[ImaginaryI]\ xprime, tprime}\)], "Output"] }, Open ]], Cell[TextData[{ "3. Exprimez ", Cell[BoxData[ \(TraditionalForm\`x'\)]], " \[AGrave] l'aide de ", StyleBox["v", FontSlant->"Italic"], " de ", StyleBox["c ", FontSlant->"Italic"], "et de ", StyleBox["t' ", FontSlant->"Italic"], "puis utilisez l'invariance de l'intervalle (", Cell[BoxData[ FormBox[ SuperscriptBox["t", StyleBox["2", FontSize->9]], TraditionalForm]]], "-", Cell[BoxData[ FormBox[ SuperscriptBox["x", StyleBox["2", FontSize->9]], TraditionalForm]]], ") = (", Cell[BoxData[ FormBox[ SuperscriptBox[\(\(t'\)\(\ \)\), StyleBox["2", FontSize->9]], TraditionalForm]]], "-", Cell[BoxData[ FormBox[ SuperscriptBox[\(\(x'\)\(\ \)\), StyleBox["2", FontSize->9]], TraditionalForm]]], ") pour obtenir ", Cell[BoxData[ \(TraditionalForm\`\(\(t'\)\(\ \)\)\)]], "en fonction de ", StyleBox["v", FontSlant->"Italic"], " de ", StyleBox["c", FontSlant->"Italic"], " et de ", Cell[BoxData[ \(TraditionalForm\`t\)]], "." }], "Text", FontSize->12, CellTags->"i:1"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[{ \(\(e1prime = {\[ImaginaryI]*v/c*tprime, tprime};\)\), "\[IndentingNewLine]", \(sol = Solve[e1 . e1 == e1prime . e1prime, tprime]\)}], "Input", FontSize->10, CellTags->"i:1"], Cell[BoxData[ \({{tprime \[Rule] \(-\(t\/\@\(1 - v\^2\/c\^2\)\)\)}, {tprime \[Rule] t\/\@\(1 - v\^2\/c\^2\)}}\)], "Output", CellTags->"i:1"] }, Open ]], Cell[TextData[{ StyleBox["4. Exprimez ", FontVariations->{"CompatibilityType"->0}], "ensuite", StyleBox[" x' ", FontSlant->"Italic"], "\[AGrave] l'aide de la relation ", Cell[BoxData[ \(TraditionalForm\`x'\)]], " = ", Cell[BoxData[ FormBox[ FractionBox[ StyleBox["v", FontSize->10], StyleBox["c", FontSize->10]], TraditionalForm]], FontSize->18], StyleBox["t'", FontSlant->"Italic"] }], "Text", FontSize->12, CellTags->"i:1"], Cell[CellGroupData[{ Cell[BoxData[ \(v/c*sol[\([2, 1, 2]\)]\)], "Input", FontSize->10, CellTags->"i:1"], Cell[BoxData[ \(\(t\ v\)\/\(c\ \@\(1 - v\^2\/c\^2\)\)\)], "Output", CellTags->"i:1"] }, Open ]], Cell[TextData[{ "5. Exprimez les coordonn\[EAcute]es ", StyleBox["x' e", FontSlant->"Italic"], "t ", StyleBox["t' ", FontSlant->"Italic"], "de l'\[EAcute]v\[EAcute]nement ", Cell[BoxData[ \(TraditionalForm\`E\_1\)]], " en fonction de ses coordonn\[EAcute]es dans \[Sum] \[AGrave] partir de :" }], "Text", FontSize->12, CellTags->"i:1"], Cell[TextData[{ Cell[BoxData[ FormBox[ RowBox[{"(", "\[NegativeThinSpace]", GridBox[{ { FormBox[\(x'\), "TextForm"]}, { RowBox[{ FormBox["t", "TextForm"], "'"}]} }], "\[NegativeThinSpace]", ")"}], TraditionalForm]]], " = ", Cell[BoxData[ FormBox[ RowBox[{ RowBox[{"(", "\[NegativeThinSpace]", GridBox[{ {"a", "b"}, {"d", "e"} }], "\[NegativeThinSpace]", ")"}], RowBox[{"(", "\[NegativeThinSpace]", GridBox[{ {"x"}, {"t"} }], "\[NegativeThinSpace]", ")"}]}], TraditionalForm]]] }], "Text", TextAlignment->Center, TextJustification->0, FontSize->12, CellTags->"i:1"], Cell[CellGroupData[{ Cell[BoxData[ \({{a, b}, {d, e}} . {0, t}\)], "Input", FontSize->10, CellTags->"i:1"], Cell[BoxData[ \({b\ t, e\ t}\)], "Output", CellTags->"i:1"] }, Open ]], Cell[TextData[{ "6. Comparez les valeurs obtenues pour ", StyleBox["x'", FontSlant->"Italic"], " et ", StyleBox["t'", FontSlant->"Italic"], " sous 5. \[AGrave] celles obtenues respectivement sous 4. et 3. \ D\[EAcute]duisez la valeur des coefficients ", StyleBox["b", FontSlant->"Italic"], " et ", StyleBox["e", FontSlant->"Italic"], " de cette comparaison." }], "Text", FontSize->12, CellTags->"i:1"], Cell[CellGroupData[{ Cell[BoxData[ \(Solve[b*t == \(t\ v\)\/\(c\ \@\(1 - v\^2\/c\^2\)\), b]\)], "Input", FontSize->10, CellTags->"i:1"], Cell[BoxData[ \({{b \[Rule] v\/\(c\ \@\(\(c\^2 - v\^2\)\/c\^2\)\)}}\)], "Output", CellTags->"i:1"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(Solve[e*t == t\/\@\(1 - v\^2\/c\^2\), e]\)], "Input", FontSize->10, CellTags->"i:1"], Cell[BoxData[ \({{e \[Rule] 1\/\@\(\(c\^2 - v\^2\)\/c\^2\)}}\)], "Output", CellTags->"i:1"] }, Open ]], Cell[TextData[{ "7. Exprimez les coordonn\[EAcute]es ", StyleBox["x", FontSlant->"Italic"], " et ", StyleBox["t", FontSlant->"Italic"], " d'un \[EAcute]v\[EAcute]nement quelconque ", Cell[BoxData[ \(TraditionalForm\`\(\(E\)\(\ \)\)\)]], "en fonction de ses coordonn\[EAcute]es dans \[Sum]'" }], "Text", FontSize->12, CellTags->"i:1"], Cell[CellGroupData[{ Cell[BoxData[ \({xprime, tprime} = {{a, b}, {d, e}} . {x, t}\)], "Input", FontSize->10, CellTags->"i:1"], Cell[BoxData[ \({b\ t + a\ x, e\ t + d\ x}\)], "Output", CellTags->"i:1"] }, Open ]], Cell[TextData[{ "8. Calculez, \[AGrave] partir des coordonn\[EAcute]es (", StyleBox["x", FontSlant->"Italic"], ", ", StyleBox["t", FontSlant->"Italic"], ") obtenues sous 7. l'intervalle (", Cell[BoxData[ FormBox[ SuperscriptBox["t", StyleBox["2", FontSize->9]], TraditionalForm]]], "-", Cell[BoxData[ FormBox[ SuperscriptBox["x", StyleBox["2", FontSize->9]], TraditionalForm]]], ")" }], "Text", FontSize->12, CellTags->"i:1"], Cell[CellGroupData[{ Cell[BoxData[ \(intervalle = Expand[{\[ImaginaryI]*xprime, tprime} . {\[ImaginaryI]*xprime, tprime}]\)], "Input", FontSize->10, CellTags->"i:1"], Cell[BoxData[ \(\(-b\^2\)\ t\^2 + e\^2\ t\^2 - 2\ a\ b\ t\ x + 2\ d\ e\ t\ x - a\^2\ x\^2 + d\^2\ x\^2\)], "Output", CellTags->"i:1"] }, Open ]], Cell[TextData[{ "9. Identifiez le coefficient de ", StyleBox["xt", FontSlant->"Italic"], " qui doit \[EHat]tre nul car (", Cell[BoxData[ FormBox[ SuperscriptBox["t", StyleBox["2", FontSize->9]], TraditionalForm]]], "-", Cell[BoxData[ FormBox[ SuperscriptBox["x", StyleBox["2", FontSize->9]], TraditionalForm]]], ") = (", Cell[BoxData[ FormBox[ SuperscriptBox[\(\(t'\)\(\ \)\), StyleBox["2", FontSize->9]], TraditionalForm]]], "-", Cell[BoxData[ FormBox[ SuperscriptBox[\(\(x'\)\(\ \)\), StyleBox["2", FontSize->9]], TraditionalForm]]], ")" }], "Text", FontSize->12, CellTags->"i:1"], Cell[CellGroupData[{ Cell[BoxData[ \(cxt = Coefficient[intervalle, x*t]\)], "Input", FontSize->10, CellTags->"i:1"], Cell[BoxData[ \(\(-2\)\ a\ b + 2\ d\ e\)], "Output", CellTags->"i:1"] }, Open ]], Cell[TextData[{ "10. Identifiez le coefficient de ", Cell[BoxData[ FormBox[ SuperscriptBox[\(\(x\)\(\ \)\), StyleBox["2", FontSize->9]], TraditionalForm]], CellMargins->{{5.6875, Inherited}, {Inherited, Inherited}}], " qui doit \[EHat]tre \[EAcute]gal \[AGrave] -1." }], "Text", CellMargins->{{2, Inherited}, {Inherited, Inherited}}, FontSize->12, CellTags->"i:1"], Cell[CellGroupData[{ Cell[BoxData[ \(cx2 = Coefficient[intervalle, x^2]\)], "Input", FontSize->10, CellTags->"i:1"], Cell[BoxData[ \(\(-a\^2\) + d\^2\)], "Output", CellTags->"i:1"] }, Open ]], Cell[TextData[{ "11. Identifiez le coefficient de ", Cell[BoxData[ FormBox[ SuperscriptBox[\(\(t\)\(\ \)\), StyleBox["2", FontSize->9]], TraditionalForm]]], " qui doit \[EHat]tre \[EAcute]gal \[AGrave] 1." }], "Text", FontSize->12, CellTags->"i:1"], Cell[CellGroupData[{ Cell[BoxData[ \(ct2 = Coefficient[intervalle, t^2]\)], "Input", FontSize->10, CellTags->"i:1"], Cell[BoxData[ \(\(-b\^2\) + e\^2\)], "Output", CellTags->"i:1"] }, Open ]], Cell[TextData[{ "12. 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