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There are at least three \ ways to solve this question. See if you can find two."] }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData[{ "A: (a) Let the inverse of the frequency be the time ", Cell[BoxData[ \(t\_o\)], "Input"], ". Redshift it!" }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[CellGroupData[{ Cell[BoxData[ \(Simplify[ \((\[Gamma][\[Beta]]\ + \ \[Beta]\ \[Gamma][\[Beta]])\) q[t\_o, \ t\_o, 0, 0]\ .\ {1, 0, 0, 0}]\)], "Input"], Cell[BoxData[ \({\(\((1 + \[Beta])\)\ t\_o\)\/\@\(1 - \[Beta]\^2\), \(\((1 + \[Beta])\)\ t\_o\)\/\@\(1 - \[Beta]\^2\), 0, 0}\)], "Output"] }, Open ]], Cell[TextData[{ "The frequency is the inverse of the time component, or \n", Cell[BoxData[ RowBox[{"\[Nu]", " ", "=", " ", FormBox[ \(\[Nu]\_o\ \@\(\((1\ - \ \[Beta])\)\/\(1\ + \ \[Beta]\[AliasDelimiter]\)\)\ \), "TraditionalForm"]}]], "Input"] }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell["(b) We need another redshift of exactly the same size. ", "Text", Evaluatable->False, AspectRatioFixed->True], Cell[CellGroupData[{ Cell[BoxData[ \(Simplify[ \((\[Gamma][\[Beta]]\ + \ \[Beta]\ \[Gamma][\[Beta]])\)\^2\ q[to, to, 0, 0]\ .\ {1, 0, 0, 0}]\)], "Input"], Cell[BoxData[ \({\(-\(\(to\ \((1 + \[Beta])\)\)\/\(\(-1\) + \[Beta]\)\)\), \(-\(\(to\ \((1 + \[Beta])\)\)\/\(\(-1\) + \[Beta]\)\)\), 0, 0}\)], "Output"] }, Open ]], Cell[TextData[{ "The frequency it the inverse of the time component, so ", Cell[BoxData[ RowBox[{"\[Nu]", " ", "=", " ", FormBox[\(\[Nu]\_o\ \((1\ - \ \[Beta])\)\/\(1\ + \ \[Beta]\)\), "TraditionalForm"]}]], "Input"], "." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell["\<\ (b') Another approach is to boost the initial event with a speed \ equal to the two boosts, which by the addition of velocity formula is shown \ below.\ \>", "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(\(\[Beta]\[Beta]\ = \ \(2\ \[Beta]\)\/\(1\ + \ \[Beta]\^2\)\ ; \)\)], "Input"], Cell["Redshift with this velocity and try to simplify.", "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(PowerExpand[ Cancel[\n\t Simplify[\n\t\t Expand[\n\t\t\t \((\[Gamma][\[Beta]\[Beta]]\ + \ \[Beta]\[Beta]\ \[Gamma][\[Beta]\[Beta]])\)\ q[to, to, 0, 0]\ .\ {1, 0, 0, 0}]]]]\)], "Input"], Cell[BoxData[ \({\(to\ \((\(-1\) + \[Beta]\^2)\)\)\/\((\(-1\) + \[Beta])\)\^2, \(to\ \((\(-1\) + \[Beta]\^2)\)\)\/\((\(-1\) + \[Beta])\)\^2, 0, 0}\)], "Output"] }, Open ]], Cell["A step away from the previous result.", "Text", Evaluatable->False, AspectRatioFixed->True] }, Open ]], Cell[CellGroupData[{ Cell[TextData[StyleBox[" French: 5-10"]], "Subsection", CellMargins->{{0, Inherited}, {Inherited, Inherited}}, Evaluatable->False, AspectRatioFixed->True, CellTags->"f5-10"], Cell["\<\ Q: A pulsed radar source is at rest at the point x = 0. A large \ meteorite moves with constant velocity v toward the source; it is at the \ point x = -d at t = 0. A first radar pulse is emitted by the source at t = \ 0, and a second pulse at t = to (to << d/c). The pulses are reflected by the \ meteorite and return to the source. (a) Draw in spacetime graph (1) the \ source, (2) the meteorite, (3) the two outgoing pulses, (4) the reflected \ pulses. (b) Evaluate the time interval between the arrivals at x = 0 of the \ two reflected pulses. (c) Evaluate the time interval between the arrivals as \ the meteorite of the two outgoing pulses, as measured in the rest frame of \ the meteorite. Answer (b) and (c) first with a well-chosen Lorentz transformation. Then \ answer again, this time using the Doppler effect and the results of the above \ problem (French 5-9).\ \>", "Text", Evaluatable->False, AspectRatioFixed->True], Cell["\<\ A: (a) This spacetime graph of the meteorite was constructed in \ the program \"Spacetime\" by Prof. Edwin F. 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gRQoo`09Kg/8@W>LOom_NgooMkd007Nm00Eoo`03Nmioogoo019oo`00:7oo 00EkgWooOomoo`0002]oo`04125cW7ooMkd3Ool00`00Oomg_@02Ool00g_N Mkeoo`0DOol002Yoo`9g_@035:E_Ngoo029oo`03Nmioogoo00Aoo`03001o ogoo00Aoo`03001oogoo009oo`04Nmioof]J2482Ool00g_NOomg_@0>Ool0 02aoo`04F]IoogooNmhQOol017_NOomoogNm0Woo00=_N`@QNmh017_N00Ao o`@QOomkgP9oo`06MkeoogooJeYoog_N0goo00=kgWooOol02goo001AOol0 0gNmOomoo`02Ool02Ukg5:D00124Kg]oof]JOomkgP0017oo0W_N0Woo00Mg _GooOomW>GooMkekgP0"], "Graphics", ImageSize->{123, 173}, ImageMargins->{{0, 0}, {0, 0}}, ImageRegion->{{0, 1}, {0, 1}}, ImageCacheContents->"Empty"], Cell["\<\ (b) Chose the frame of the Earth. The world line of the first \ pulse is q[t, -t, 0,0]. The worldline of the meteorite is q[t, v t/c - d,0,0]. Solve \ for the time when the distances are the same.\ \>", "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(Solve[\n\t\t \(q[t, \ \[Beta]\ t\ - \ d, 0, 0]\)[\([2, 1]\)]\ == \ \n\t\ \ \(q[t, \ \(-t\), 0, 0]\)[\([2, 1]\)], \ t]\)], "Input"], Cell[BoxData[ \({{t \[Rule] d\/\(1 + \[Beta]\)}}\)], "Output"] }, Open ]], Cell["\<\ The distance traveled is the same, so it arrives back at the Earth \ at\ \>", "Text"], Cell[BoxData[ \(\(pulse1\_back\ = \ \(2\ d\)\/\(1\ + \ \[Beta]\); \)\)], "Input"], Cell["\<\ Find the time at which the second pulse arrives at the meteorite. \ The only change is the departure of the pulse.\ \>", "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(pulse2time\_met\ = \ Solve[\n\t\t \(q[t, \ \[Beta]\ t\ - \ d, 0, 0]\)[\([2, 1]\)]\ == \ \n\t\ \ \(q[t, \ \(-t\)\ + \ to, 0, 0]\)[\([2, 1]\)], \ t]\)], "Input"], Cell[BoxData[ \({{t \[Rule] \(-\(\(\(-d\) - to\)\/\(1 + \[Beta]\)\)\)}}\)], "Output"] }, Open ]], Cell["Use the time to find pulse 2's position.", "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(pulse2pos\_met\ = \ \[Beta]\ t\ - \ L\ \ /. \ pulse2time\_met\)], "Input"], Cell[BoxData[ \({\(-L\) - \(\((\(-d\) - to)\)\ \[Beta]\)\/\(1 + \[Beta]\)}\)], "Output"] }, Open ]], Cell["Add these together to find the return time.", "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(pulse2\_back\ = \ Simplify[\(d\ + \ to\)\/\(1\ + \ \[Beta]\)\ + \ d\ - \ \(\[Beta] \((d\ + \ to)\)\)\/\(1\ + \ \[Beta]\)]\)], "Input"], Cell[BoxData[ \(\(2\ d + to - to\ \[Beta]\)\/\(1 + \[Beta]\)\)], "Output"] }, Open ]], Cell["\<\ Examine the interval between the arrival of the two pulses back to \ the Earth.\ \>", "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(Simplify[pulse2\_back\ - \ pulse1\_back]\)], "Input"], Cell[BoxData[ \(\(to - to\ \[Beta]\)\/\(1 + \[Beta]\)\)], "Output"] }, Open ]], Cell[TextData[{ "The interval between the arrival of the two pulses is shifted by a factor \ ", Cell[BoxData[ \(t'\ = \ \(1\ - \ \[Beta]\)\/\(1\ + \ \[Beta]\)\ t\_o\)], "Input"], "." }], "Text"], Cell["\<\ (c) Choose the rest frame of the meteorite. Boost the emission \ event to this frame.\ \>", "Text"], Cell[BoxData[ \(\(pulse1\_source\ = \ q[\[Gamma][\[Beta]], \ \[Beta]\ \[Gamma][\[Beta]], 0, 0]\ .\ q[0, d, 0, 0]; \)\)], "Input"], Cell["\<\ Add the time after being boosted together with the position needed \ to travel to the meteorite to get the time of pulse 1 at the meteorite.\ \>", "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(pulse1\_met\ = \ pulse1\_source[\([1, 1]\)]\ + \ pulse1\_source[\([2, 1]\)]\)], "Input"], Cell[BoxData[ \(d\/\@\(1 - \[Beta]\^2\) - \(d\ \[Beta]\)\/\@\(1 - \[Beta]\^2\)\)], "Output"] }, Open ]], Cell["Repeat this process for pulse 2.", "Text"], Cell[BoxData[ \(\(pulse2\_source\ = \ Simplify[L[to, \ d, \ \[Beta]]\ .\ q[to, d, 0, 0]]; \)\)], "Input"], Cell[CellGroupData[{ Cell[BoxData[ \(pulse2\_met\ = \ pulse2\_source[\([1, 1]\)]\ + \ pulse2\_source[\([2, 1]\)]\)], "Input"], Cell[BoxData[ \(\(to - d\ \[Beta]\)\/\@\(1 - \[Beta]\^2\) + \(d - to\ \[Beta]\)\/\@\(1 - \[Beta]\^2\)\)], "Output"] }, Open ]], Cell[TextData[StyleBox[ "Examine the interval between the arrival of the two pulses at the \ meteorite."]], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(Simplify[pulse2\_met\ - \ pulse1\_met]\)], "Input"], Cell[BoxData[ \(\(to - to\ \[Beta]\)\/\@\(1 - \[Beta]\^2\)\)], "Output"] }, Open ]], Cell[TextData[{ "The interval between the arrival of the two pulses at the meteorite is \ shifted by a factor ", Cell[BoxData[ \(t'\ = \ \@\(\(1\ - \ \[Beta]\)\/\(1\ + \ \[Beta]\)\)\ t\_o\)], "Input"], "." }], "Text"], Cell["\<\ (b') From the reference frame of the meteorite, the pulse of light \ would be blueshifted from the source, and blueshifted to the receiver. Use \ the result from 5-9 (b).\ \>", "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(Simplify[ \((\[Gamma][\(-\[Beta]\)]\ - \ \[Beta]\ \[Gamma][\(-\[Beta]\)])\)\^2\ q[to, to, 0, 0]\ .\ {1, 0, 0, 0}]\)], "Input"], Cell[BoxData[ \({\(-\(\(to\ \((\(-1\) + \[Beta])\)\)\/\(1 + \[Beta]\)\)\), \(-\(\(to\ \((\(-1\) + \[Beta])\)\)\/\(1 + \[Beta]\)\)\), 0, 0}\)], "Output"] }, Open ]], Cell[TextData[{ "The time interval between pulses is ", Cell[BoxData[ \(t'\ = \ \(1\ - \ \[Beta]\)\/\(1\ + \ \[Beta]\)\ t\_o\)], "Input"], "." }], "Text"], Cell["\<\ (c') As stated above, the pulse of light from the source is \ blueshifted, so using a modified answer from 5-9 (a).\ \>", "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(Simplify[ \((\[Gamma][\(-\[Beta]\)]\ - \ \[Beta]\ \[Gamma][\(-\[Beta]\)])\) q[to, to, 0, 0]\ .\ {1, 0, 0, 0}]\)], "Input"], Cell[BoxData[ \({\(-\(\(to\ \((\(-1\) + \[Beta])\)\)\/\@\(1 - \[Beta]\^2\)\)\), \(-\(\(to\ \((\(-1\) + \[Beta])\)\)\/\@\(1 - \[Beta]\^2\)\)\), 0, 0}\)], "Output"] }, Open ]], Cell[TextData[{ "The time interval between pulse at the meteorite is ", Cell[BoxData[ \(t'\ = \ \@\(\(1\ - \ \[Beta]\)\/\(1\ + \ \[Beta]\)\)\ t\_o\)], "Input"], "." }], "Text"] }, Open ]], Cell[CellGroupData[{ Cell[TextData[StyleBox[" French: 5-11"]], "Subsection", CellMargins->{{-2, Inherited}, {Inherited, Inherited}}, Evaluatable->False, AspectRatioFixed->True, CellTags->"f5-11"], Cell["\<\ Q: An astronaut moves radially away from the Earth at a constant \ acceleration (as measured in the Earth's reference frame) of 9.8 m/s^2. How \ long will it be before the redshift makes the red glare of the neon signs of \ Earth invisible to his human eyesight?\ \>", "Text", Evaluatable->False, AspectRatioFixed->True], Cell["\<\ A: Solve for the velocity which redshifts the wavelength of neon (~600 nm) to invisible (~700 nm).\ \>", "Text", Evaluatable->False, AspectRatioFixed->True], Cell[CellGroupData[{ Cell[BoxData[ \(Solve[\n\t \((\[Gamma][\[Beta]]\ + \ \[Beta]\ \[Gamma][\[Beta]])\)\ \(q[600, \ 600, 0, 0]\)\ [\([2, 1]\)]\ == \ 700, \ \[Beta]]\)], "Input"], Cell[BoxData[ \({{\[Beta] \[Rule] 13\/85}}\)], "Output"] }, Open ]], Cell["The time required at a constant acceleration is t = v/a.", "Text", Evaluatable->False, AspectRatioFixed->True], Cell[CellGroupData[{ Cell[BoxData[ \(\(\(13\ c\)\/85\) \(1\/\(9.8\ m/s\^2\)\) min\/\(60\ s\)\ hr\/\(60\ min\)\ day\/\(24\ hr\)\)], "Input"], Cell[BoxData[ \(54.1883420034680538847`\ day\)], "Output"] }, Open ]], Cell["\<\ After 54 days, the neon lights become invisible to the astronaut's \ eyes.\ \>", "Text", Evaluatable->False, AspectRatioFixed->True] }, Open ]], Cell[CellGroupData[{ Cell[TextData[StyleBox[" French: 5-12"]], "Subsection", CellMargins->{{0, Inherited}, {Inherited, Inherited}}, Evaluatable->False, AspectRatioFixed->True, CellTags->"f5-12"], Cell["\<\ Q: There is a spaceship shuttle service from the Earth to Mars. \ Each spaceship is equipped with two identical lights, one at the front and \ one at the rear. The spaceships normally travel at a speed vo, relative to \ the Earth, such that the headlight of a spaceship approaching Earth appears \ green (500 nm) and the taillight of a departing spaceship appears red (600 \ nm). (a) what is the value of vo/c? (b) One spaceship accelerates to \ overtake the spaceship ahead of it. At what speed must the overtaking \ spaceship travel relative to the Earth so that the taillight of the \ Mars-bound spaceship ahead of it looks like a headlight (500 nm green)?\ \>", "Text", Evaluatable->False, AspectRatioFixed->True], Cell["\<\ A: (a) Solve for the velocity that reverses the shifts to the same \ wavelength (i.e., redshift the headlight's blueshifted light to the \ wavelength of the taillight's blueshifted redshifted light ; )\ \>", "Text", Evaluatable->False, AspectRatioFixed->True], Cell[CellGroupData[{ Cell[BoxData[ \(Solve[\n\t \((\[Gamma][\[Beta]]\ + \ \[Beta]\ \[Gamma][\[Beta]])\)\ \(q[500\ nm, \ 500\ nm, 0, 0]\)[\([2, 1]\)]\ == \n\ \ \((\[Gamma][\[Beta]]\ - \ \[Beta]\ \[Gamma][\[Beta]])\)\ \(q[600\ nm, \ 600\ nm, 0, 0]\)[\([2, 1]\)], \ \[Beta]]\)], "Input"], Cell[BoxData[ \({{\[Beta] \[Rule] 1\/11}}\)], "Output"] }, Open ]], Cell["The spaceships travel at vo/c = 1/11.", "Text", Evaluatable->False, AspectRatioFixed->True], Cell["\<\ (b) Solve for the velocity needed to shift the wavelength from 600 \ to 500 nm.\ \>", "Text", Evaluatable->False, AspectRatioFixed->True], Cell[CellGroupData[{ Cell[BoxData[ \(Solve[\n\t \((\[Gamma][\[Beta]]\ - \ \[Beta]\ \[Gamma][\[Beta]])\)\ \(q[600\ nm, \ 600\ nm, 0, 0]\)[\([2, 1]\)]\ == \ 500\ nm, \ \[Beta]]\)], "Input"], Cell[BoxData[ \({{\[Beta] \[Rule] 11\/61}}\)], "Output"] }, Open ]], Cell["The required velocity is v/c = 11/61.", "Text", Evaluatable->False, AspectRatioFixed->True] }, Open ]], Cell[CellGroupData[{ Cell[TextData[StyleBox[ " R & H: 2-68 A Doppler shift revealed as a color change"]], "Subsection", CellMargins->{{0, 28}, {Inherited, Inherited}}, Evaluatable->False, AspectRatioFixed->True, CellTags->"2-68"], Cell["\<\ Q: A spaceship is receding from the Earth at a speed of 0.20c. A \ light on the rear of the ship appears blue (450 nm) to the passengers on the \ ship. What color would it appear to an observer on the Earth?\ \>", "Text", Evaluatable->False, AspectRatioFixed->True], Cell["A: Redshift the light at 450 nm by 0.20c.", "Text", Evaluatable->False, AspectRatioFixed->True], Cell[CellGroupData[{ Cell[BoxData[ \(\((\[Gamma][0.2]\ + \ 0.2\ \[Gamma][0.2])\)\ q[450\ nm, 450\ nm, 0, 0]\ .\ {1, 0, 0, 0}\)], "Input"], Cell[BoxData[ \({551.135192126215072089`\ nm, 551.135192126215072089`\ nm, 0, 0}\)], "Output"] }, Open ]], Cell["The light appears at 551 nm, yellow.", "Text", Evaluatable->False, AspectRatioFixed->True] }, Open ]], Cell[CellGroupData[{ Cell[TextData[StyleBox[ " R & H: 2-71 The Ives-Stillwell experiment"]], "Subsection", CellMargins->{{0, Inherited}, {Inherited, Inherited}}, Evaluatable->False, AspectRatioFixed->True, CellTags->"2-71"], Cell["\<\ Q: (long) Neutral hydrogen atoms are moving along the axis of an \ evacuated tube with a speed of 2.0 x 10^6 m/s. A spectrometer is arranged to \ receive light emitted by these atoms in the direction of their forward \ motion. This light, if emitted from resting hydrogen atoms, would have a \ measured (proper) wavelength of 486.13 nm. (a) Calculate the expected \ wavelength for light emitted from the forward-moving (approaching) atoms, \ using the exact relativistic formula. (b) By using a mirror this same \ spectrometer can also measure the wavelength of light emitted by these moving \ atoms in the direction opposite to their motion. What wavelength is expected \ under this arrangement? (c) Calculate the difference between the average of \ the two wavelengths found in (a) and (b) and the unshifted (proper) \ wavelength. Find the second order dependence on beta. By this technique, \ Ives and Stillwell were able to distinguish between the predictions of the \ classical and the relativistic Doppler formulas.\ \>", "Text", Evaluatable->False, AspectRatioFixed->True], Cell["\<\ A: (a) & (b) Red- and blue shift the light at 486 nm +/- the speed \ of the moving hydrogen.\ \>", "Text", Evaluatable->False, AspectRatioFixed->True], Cell[CellGroupData[{ Cell[BoxData[ \(\[Beta]\_H\ = \ 2\ 10\^6\ \(m/s\)\ /c\)], "Input"], Cell[BoxData[ \(1\/150\)], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(\((\[Gamma][\[Beta]\_H]\ + \ \[Beta]\_H\ \[Gamma][\[Beta]\_H])\)\ q[486.133, \ 486.133, 0, 0]\ .\ {1, 0, 0, 0}\)], "Input"], Cell[BoxData[ \({489.384762004438527652`, 489.384762004438527652`, 0.`, 0.`}\)], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(\((\[Gamma][\[Beta]\_H]\ - \ \[Beta]\_H\ \[Gamma][\[Beta]\_H])\)\ q[486.133, \ 486.133, 0, 0]\ .\ {1, 0, 0, 0}\)], "Input"], Cell[BoxData[ \({482.902844626896295522`, 482.902844626896295522`, 0.`, 0.`}\)], "Output"] }, Open ]], Cell["\<\ The light is redshifted to 489.4 nm and blueshifted to 482.9 \ nm.\ \>", "Text", Evaluatable->False, AspectRatioFixed->True], Cell["\<\ (c) We can measure the average of these two shifted wavelengths, \ or their average difference.\ \>", "Text", Evaluatable->False, AspectRatioFixed->True], Cell[CellGroupData[{ Cell[BoxData[ \(\((489.385\ + \ 482.903)\)/2\ - \ 486.133\)], "Input"], Cell[BoxData[ \(0.0110000000000000097699`\)], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(\((489.385\ - \ 482.903)\)/2\)], "Input"], Cell[BoxData[ \(3.24100000000000000588`\)], "Output"] }, Open ]], Cell["\<\ According to classical theory, if the observer is fixed, and the \ source moves, there is no second order dependence on beta.\ \>", "Text"], Cell[BoxData[ \(\[Lambda]\ = \ \[Lambda]\_o\ \((1\ - \ \[Beta])\)\)], "Input"], Cell["If the source is fixed but the observer moves, then", "Text"], Cell[BoxData[ \(\[Lambda]\ = \ \[Lambda]\_o\ \((1\ - \ \[Beta]\ + \ \[Beta]\^2)\)\)], "Input"], Cell["\<\ Special relativity predicts a coefficient of +0.5 for the beta \ squared term, the one measured in the lab.\ \>", "Text", Evaluatable->False, AspectRatioFixed->True] }, Open ]], Cell[CellGroupData[{ Cell[TextData[StyleBox[" R&H: 2-83 The headlight effect"]], "Subsection", CellMargins->{{-1, Inherited}, {Inherited, Inherited}}, Evaluatable->False, AspectRatioFixed->True, CellTags->"2-83"], Cell["\<\ Q: A source of light, at rest in the S' frame, emits uniformly in \ all directions. The source is viewed from frame S, the relative speed \ parameter relating the two frames being beta. (a) Show that at high speeds, \ the forward-pointing cone into which the source emits half of its radiation \ has a half angle given closely, in radian measure, by \ \>", "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(\[Theta]\_\(1/2\)\ = \ \@\(2 \((1 - \[Beta])\)\)\)], "Input"], Cell["\<\ (b) What value of the half angle is predicted for the gamma \ radiation emitted by a beam of energetic neutral pions, for which v/c = \ 0.993? (c) At what speed would a light source have to move toward an \ observer to have half of its radiation concentrated into a narrow forward \ cone of half angle 5.0\[Degree]?\ \>", "Text"], Cell["\<\ A: This problem requires a boost quaternion that works with \ nonzero values for t, x, y, and z. See the last problem set in the addition \ of velocities section for the derivation of the following boost quaternion, \ or the notebook on \"Alternative algebra for boosts\":\ \>", "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(\(\[CapitalLambda][t_, \ x_, \ y_, \ z_, \ \[Beta]_]\ := \n 1/\((t^2\ + \ x^2\ + \ y^2\ + \ z^2)\)\ *\n q[\[Gamma][\[Beta]]\ t^2\ - \ 2\ \[Beta]\ \[Gamma][\[Beta]]\ t\ x\ + \ \[Gamma][\[Beta]]\ x^2\ + \ y^2\ + \ z^2, \n \(-\ \[Beta]\)\ \[Gamma][\[Beta]]\ \((t^2\ - \ x^2)\), \n t\ y\ - \ x\ z\ + \ \[Gamma][\[Beta]]\ \((\(-t\)\ + \ \[Beta]\ x)\)\ y\ + \ \[Gamma][\[Beta]]\ \((\(-\[Beta]\)\ t\ + \ x)\)\ z, \n x\ y\ + \ t\ z\ + \ \[Gamma][\[Beta]]\ \((\ \[Beta]\ t\ - \ x)\)\ y\ + \ \[Gamma][\[Beta]]\ \((\(-t\)\ + \ \[Beta]\ x)\)\ z]\n\)\)], "Input"], Cell["\<\ Boost a spherically symmetric velocity quaternion, normalizing to \ the resulting gamma so the resulting quaternion still characterizes \ velocities.\ \>", "Text", Evaluatable->False, AspectRatioFixed->True], Cell[CellGroupData[{ Cell[BoxData[ \(\[Beta]\_sphere = \ Simplify[\n\t\t \[CapitalLambda][1, 1, 1, 1, \(-\[Beta]\)]\ .\ q[1, 1, 1, 1]\ . \ {1, 0, 0, 0}\ /\n\t \((\[CapitalLambda][1, 1, 1, 1, \(-\[Beta]\)]\ .\ q[1, 1, 1, 1])\)[ \([1, 1]\)]\ ]\)], "Input"], Cell[BoxData[ \({1, 1, \@\(1 - \[Beta]\^2\)\/\(1 + \[Beta]\), \@\(1 - \[Beta]\^2\)\/\(1 + \[Beta]\)}\)], "Output"] }, Open ]], Cell["Calculate the angle directly from the ratio of speeds.", "Text", Evaluatable->False, AspectRatioFixed->True], Cell[CellGroupData[{ Cell[BoxData[ \(\[Theta]\_0.5\ = \ \((\[Beta]\_sphere[\([3]\)]\ + \ \[Beta]\_sphere[\([4]\)])\)/ \[Beta]\_sphere[\([2]\)]\)], "Input"], Cell[BoxData[ \(\(2\ \@\(1 - \[Beta]\^2\)\)\/\(1 + \[Beta]\)\)], "Output"] }, Open ]], Cell[TextData[{ "As beta approaches 1, this angle approaches ", Cell[BoxData[ \(TraditionalForm\`\@\(2 \((1\ - \ \[Beta])\)\)\)]], "." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData[{ "(b) Let beta", StyleBox[" ->", FontFamily->"Symbol"], " 0.993." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[CellGroupData[{ Cell[BoxData[ \(\[Theta]\_0.5\ 180\/\[Pi] /. \ \[Beta]\ -> \ 0.993\)], "Input"], Cell[BoxData[ \(6.79122311443450636441`\)], "Output"] }, Open ]], Cell["The predicted half angle for the gamma rays is 6.79\[Degree].", "Text", Evaluatable->False, AspectRatioFixed->True], Cell["(c) Solve for beta, given the angle.", "Text", Evaluatable->False, AspectRatioFixed->True], Cell[CellGroupData[{ Cell[BoxData[ \(Solve[\[Theta]\_0.5\ == \ 5.\ \[Pi]\/180, \[Beta]]\)], "Input"], Cell[BoxData[ \({{\[Beta] \[Rule] 0.996199517834361406454`}}\)], "Output"] }, Open ]], Cell["\<\ A light source would have to travel at 0.9962c to concentrate its \ radiation in a forward cone of half angle 5\[Degree].\ \>", "Text", Evaluatable->False, AspectRatioFixed->True] }, Open ]] }, Open ]], Cell[CellGroupData[{ Cell[TextData[{ StyleBox[" ", FontWeight->"Plain", FontVariations->{"Underline"->False}], StyleBox["Four-Vector Invariants", FontWeight->"Plain"] }], "Section", CellMargins->{{-1, Inherited}, {Inherited, Inherited}}, Evaluatable->False, AspectRatioFixed->True, CellTags->"4 vectors"], Cell[CellGroupData[{ Cell[TextData[StyleBox[ " Baranger: Decay of a particle - timelike or spacelike?"]], "Subsection", CellMargins->{{1, Inherited}, {Inherited, Inherited}}, Evaluatable->False, AspectRatioFixed->True, CellTags->"decay"], Cell["\<\ Q: In event 1, an unstable particle is produced in the target of \ an accelerator. In event 2, this particle decays 5 meters away. Is the \ interval between these two events timelike or spacelike? Why?\ \>", "Text", Evaluatable->False, AspectRatioFixed->True], Cell["\<\ A: The speed of the particle must be less than one, or x/t < 1. \ If event 1 is at the origin and event 2 has a spatial position of 5m, it must \ have a time of 5m + a (a>0). Calculate the square of the interval by \ squaring the quaternion.\ \>", "Text", Evaluatable->False, AspectRatioFixed->True], Cell[CellGroupData[{ Cell[BoxData[ \(Simplify[ \((q[a\ + \ 5, \ 5, 0, 0]\ .\ q[a\ + \ 5, \ 5, 0, 0])\)[\([1, 1]\)]\ ] \)], "Input"], Cell[BoxData[ \(a\ \((10 + a)\)\)], "Output"] }, Open ]], Cell["\<\ The square of the interval a^2 + 10a is always positive, so the \ interval is timelike in the future.\ \>", "Text", Evaluatable->False, AspectRatioFixed->True] }, Open ]], Cell[CellGroupData[{ Cell[TextData[StyleBox[ " R&H: 2-42 The interval is invariant - check it out"]], "Subsection", CellMargins->{{-1, Inherited}, {Inherited, Inherited}}, Evaluatable->False, AspectRatioFixed->True, CellTags->"2-42"], Cell["\<\ Q: Two events occur on the x axis of reference frame S, their \ spacetime coordinates being event1 = q[5 us, 720 m,0,0 ] and event 2 = [2 us, \ 1200 m,0,0 ]. (a) What is the square of the spacetime interval for these two \ events? (b) What are the coordinates of these events in a frame S' that \ moves at speed 0.60c in the direction of increasing x? Calculate the square \ of the interval in this frame and compare it to the value calculated for \ frame S. (c) What are the coordinates of these events in a frame S\" that \ moves at a speed of 0.95c in the direction of decreasing x? Again calculate \ the square of the spacetime interval and compare it with the values found in \ (a) and (b). Do your calculations bear out the invariance of the spacetime \ interval?\ \>", "Text", Evaluatable->False, AspectRatioFixed->True], Cell["\<\ A: (a) The square of the spacetime interval between events 1 and 2 \ is the first term of difference between the quaternions squared.\ \>", "Text",\ Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(\(event1\ = \ q[5.\ 10\^\(-6\)\ s\ c, \ 720\ m, 0, 0]; \)\)], "Input"], Cell[BoxData[ \(\(event2\ = \ q[2\ 10\^\(-6\)\ s\ c, \ 1200\ m, 0, 0]; \)\)], "Input"], Cell[CellGroupData[{ Cell[BoxData[ \(\((\((event1\ - \ event2)\)\ .\((event1\ - \ event2)\))\)[\([1, 1]\)] \)], "Input"], Cell[BoxData[ \(579600.000000000000009`\ m\^2\)], "Output"] }, Open ]], Cell["\<\ The square of the interval between event 1 and 2 is 5.8 x 10^5 m^2.\ \ \>", "Text", Evaluatable->False, AspectRatioFixed->True], Cell["(b) Boost the quaternions and then square them.", "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(\(e1b6\ = \ L[5.\ 10\^\(-6\)\ s\ c, \ 720\ m, \ 0.6]\ .\ event1; \)\)], "Input"], Cell[BoxData[ \(\(e2b6\ = \ L[2.\ 10\^\(-6\)\ s\ c, \ 1200\ m, \ 0.6]\ .\ event2; \)\)], "Input"], Cell[CellGroupData[{ Cell[BoxData[ \(\((\((e1b6\ - \ e2b6)\)\ .\ \((e1b6\ - \ e2b6)\))\)[\([1, 1]\)]\)], "Input"], Cell[BoxData[ \(579600.000000000000009`\ m\^2\)], "Output"] }, Open ]], Cell["\<\ The square of the interval between the boosted events is the \ same.\ \>", "Text", Evaluatable->False, AspectRatioFixed->True], Cell["(c) Repeat the exercise with a new value for beta.", "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(\(e1b95\ = \ L[5.\ 10\^\(-6\)\ s\ c, \ 720\ m, \ \(-0.95\)]\ .\ event1; \)\)], "Input"], Cell[BoxData[ \(\(e2b95\ = \ L[2.\ 10\^\(-6\)\ s\ c, \ 1200\ m, \ \(-0.95\)]\ .\ event2; \)\)], "Input"], Cell[CellGroupData[{ Cell[BoxData[ \(\((\((e1b95\ - \ e2b95)\)\ .\ \((e1b95\ - \ e2b95)\))\)[\([1, 1]\)] \)], "Input"], Cell[BoxData[ \(579600.000000000000919`\ m\^2\)], "Output"] }, Open ]], Cell["\<\ Again, the square of the interval between the boosted events is the \ same. The first term of the square of a quaternion is identical to the first \ term of a square of a boosted quaternion.\ \>", "Text", Evaluatable->False, AspectRatioFixed->True] }, Open ]], Cell[CellGroupData[{ Cell[TextData[StyleBox[ " R&H: 2-43 An event pair - timelike or spacelike?"]], "Subsection", CellMargins->{{0, Inherited}, {Inherited, Inherited}}, Evaluatable->False, AspectRatioFixed->True, CellTags->"2-43"], Cell["\<\ Q: Two events occur on the x axis of reference frame S, their \ spacetime coordinates being event1 = q[5 us, 200 m,0,0] and event 2 = [2 us, \ 1200 m,0,0]. (a) What is the square of the spacetime interval for these two \ events? (b) What is the proper distance interval between them? (c) If two \ events possess a (mathematically real) proper distance interval, it should be \ possible to find a frame S' in which these events would be seen to occur \ simultaneously. Find this frame. (d) Can you calculate a (mathematically \ real) proper time interval for this pair of events? (e) Would you describe \ this pair of events as timelike? Spacelike? Lightlike?\ \>", "Text", Evaluatable->False, AspectRatioFixed->True], Cell["\<\ A: (a) The square of the spacetime interval between events 1 and 2 \ is the first term of difference between the quaternions squared.\ \>", "Text",\ Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(\(event1\ = \ q[5.\ 10\^\(-6\)\ s\ c, \ \ \ 200\ m, 0, 0]; \)\)], "Input"], Cell[BoxData[ \(\(event2\ = \ q[2.\ \(10\^\(-6\)\) s\ c, \ 1200\ m, 0, 0]; \)\)], "Input"], Cell[CellGroupData[{ Cell[BoxData[ \(\((\((event2\ - \ event1)\).\((event2\ - \ event1)\))\)[\([1, 1]\)] \)], "Input"], Cell[BoxData[ \(\(-190000.`\)\ m\^2\)], "Output"] }, Open ]], Cell["\<\ The square of the interval between event 1 and 2 is -1.9 x 10^5 m^2\ \ \>", "Text", Evaluatable->False, AspectRatioFixed->True], Cell["\<\ (b) The proper distance interval is the square root of the negative \ of this number.\ \>", "Text", Evaluatable->False, AspectRatioFixed->True], Cell[CellGroupData[{ Cell[BoxData[ \(\@\(1.9\ \ 10\^5\ m\^2\)\)], "Input"], Cell[BoxData[ \(435.889894354067355215`\ \@m\^2\)], "Output"] }, Open ]], Cell["The proper distance is 436 m.", "Text", Evaluatable->False, AspectRatioFixed->True], Cell["\<\ (c) Boost both event quaternions by beta, set the time components \ equal to each other, and solve for beta.\ \>", "Text", Evaluatable->False, AspectRatioFixed->True], Cell[CellGroupData[{ Cell[BoxData[ \(Solve[\n\t \((L[5.\ 10\^\(-6\)\ s\ c, \ \ \ 200\ m, \ \[Beta]]\ .\ event1)\)[ \([1, 1]\)]\ == \n\t \((L[2.\ 10\^\(-6\)\ s\ c, \ 1200\ m, \ \[Beta]]\ .\ event2)\)[ \([1, 1]\)], \ \[Beta]]\)], "Input"], Cell[BoxData[ \({{\[Beta] \[Rule] \(-0.900000000000000000086`\)}}\)], "Output"] }, Open ]], Cell[TextData[{ "In frame S', the events will appear simultaneous for v/c", StyleBox[" ", FontFamily->"Symbol"], "= 0.9 in the direction of decreasing x." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell["\<\ (d) & (e) For events that are spacelike separated, there is no \ meaningful measure of proper time.\ \>", "Text", Evaluatable->False, AspectRatioFixed->True] }, Open ]], Cell[CellGroupData[{ Cell[TextData[StyleBox[ " R&H: 2-44 An event pair - spacelike or timelike? "]], "Subsection", CellMargins->{{0, Inherited}, {Inherited, Inherited}}, Evaluatable->False, AspectRatioFixed->True, CellTags->"2-44"], Cell["\<\ Q: Two events occur on the x axis of reference frame S, their \ spacetime coordinates being event1 = q[5 us, 720 m,0,0] and event 2 = [2 us, \ 1200 m,0,0]. (a) Using the data from problem 2-42 above, calculate the \ proper time interval for this pair of events. The proper time interval that \ you have calculated should be smaller than any of the actual time intervals \ in the three given frames of problem 2-42. Is it? (b) If two events possess \ a (mathematically real) proper time interval, it should be possible to find a \ frame S' in which these events would be seen to occur at the same place. \ Find this frame. (c) Can you calculate a (mathematically real) proper \ distance interval for this pair of events? (d) Would you describe this pair \ of events as timelike? Spacelike? Lightlike?\ \>", "Text", Evaluatable->False, AspectRatioFixed->True], Cell["\<\ A: (a) To make this question more of a challenge, let's define a \ quaternion \"Ltau\" which maps an arbitrary timelike quaternion to its proper \ time: \tLtau . q[t,x,y,z] = q[tau, 0, 0, 0] . To find Ltau, multiply the above equation on the right by the inverse of \ q[t,x,y,z].\ \>", "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(Simplify[\n q[\@\((q[t, x, y, z]\ .\ q[t, x, y, z])\)[\([1, 1]\)], 0, 0, 0]\ .\n\t\t Transpose[q[t, x, y, z]]\ .{1, 0, 0, 0}/\n\t\t \((t^2\ + \ x^2\ + \ y^2\ + \ z^2)\)]\)], "Input"], Cell[BoxData[ \({\(t\ \@\(t\^2 - x\^2 - y\^2 - z\^2\)\)\/\(t\^2 + x\^2 + y\^2 + z\^2\), \(-\(\(x\ \@\(t\^2 - x\^2 - y\^2 - z\^2\)\)\/\(t\^2 + x\^2 + y\^2 + z\^2\)\)\), \n\t \(-\(\(y\ \@\(t\^2 - x\^2 - y\^2 - z\^2\)\)\/\(t\^2 + x\^2 + y\^2 + z\^2\)\)\), \(-\(\(z\ \@\(t\^2 - x\^2 - y\^2 - z\^2\)\)\/\(t\^2 + x\^2 + y\^2 + z\^2\)\)\)}\)], "Input"], Cell[BoxData[ \(L\_tau[tlq_]\ := \n\n\t Transpose[tlq]\ *\n\t \@\(tlq[\([1, 1]\)]\^2\ - \ tlq[\([2, 1]\)]\^2\ - \ tlq[\([3, 1]\)]\^2\ - \ tlq[\([4, 1]\)]\^2\)\/\(tlq[\([1, 1]\)]\^2\ + \ tlq[\([2, 1]\)]\^2\ + \ tlq[\([3, 1]\)]\^2\ + \ tlq[\([4, 1]\)]\^2\)\)], "Input"], Cell[CellGroupData[{ Cell[BoxData[ \(L\_tau[q[32., 2., 5, \(-4.3\)]]\ .\ q[32, 2, 5, \(-4.3\)]\ . \ {1, 0, 0, 0}\)], "Input"], Cell[BoxData[ \({31.2491599887100965295`, 0.`, \(-1.084202172485504434`*^-19\), 0.`}\)], "Output"] }, Open ]], Cell["Works to within default accuracy.", "Text", Evaluatable->False, AspectRatioFixed->True], Cell["\<\ Now on to the question. Map the given quaternion to its proper \ time interval.\ \>", "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(\(intq\ = \ q[2.\ 10\^\(-6\)\ s, \ 1200\ m/c, 0, 0]\ - \ q\ [5.\ 10\^\(-6\)\ s, \ 720\ m/c, 0, 0]; \)\)], "Input"], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{ RowBox[{"(", RowBox[{ RowBox[{ SubscriptBox[ StyleBox["L", FontWeight->"Plain"], "tau"], StyleBox["[", FontFamily->"Symbol", FontWeight->"Plain"], StyleBox["intq", FontWeight->"Plain"], StyleBox["]", FontFamily->"Symbol", FontWeight->"Plain"]}], StyleBox[" ", FontFamily->"Symbol", FontWeight->"Plain"], StyleBox[".", FontFamily->"Symbol", FontWeight->"Plain"], StyleBox[" ", FontFamily->"Symbol", FontWeight->"Plain"], StyleBox["intq", FontWeight->"Plain"]}], StyleBox[")", FontWeight->"Plain"]}], StyleBox["[", FontWeight->"Plain"], StyleBox[\([1, 1]\), FontWeight->"Plain"], StyleBox["]", FontWeight->"Plain"]}]], "Input"], Cell[BoxData[ \(2.53771550808990407595`*^-6\ \@s\^2\)], "Output"] }, Open ]], Cell["\<\ The proper time is 2.54 microseconds. This is less than the time \ of 3 microseconds observed in this frame.\ \>", "Text", Evaluatable->False, AspectRatioFixed->True], Cell["Boost the interval up 0.6c, & repeat the cycle.", "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(\(intqb6\ = \n\t L[intq[\([1, 1]\)], \ intq[\([2, 1]\)], \ 0.6]\ .\ intq; \)\)], "Input"], Cell[CellGroupData[{ Cell[BoxData[ \(\((L\_tau[intqb6]\ .\ intqb6)\)[\([1, 1]\)]\)], "Input"], Cell[BoxData[ \(2.53771550808990407552`*^-6\ \@s\^2\)], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(intqb6[\([1, 1]\)]\)], "Input"], Cell[BoxData[ \(\(-4.94999999999999999982`*^-6\)\ s\)], "Output"] }, Open ]], Cell[BoxData[ \(\ \)], "Input"], Cell[TextData[{ "The interval 2.53 microseconds is the same, less that 4.95", StyleBox[" ", FontFamily->"Symbol"], "microseconds observed." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(\(intqb95\ = \n\t L[intq[\([1, 1]\)], \ intq[\([2, 1]\)], \ \(-0.95\)]\ .\ intq; \)\)], "Input"], Cell[CellGroupData[{ Cell[BoxData[ \(\((L\_tau[intqb95]\ .\ intqb95)\)[\([1, 1]\)]\)], "Input"], Cell[BoxData[ \(2.53771550808990407639`*^-6\ \@s\^2\)], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(intqb95[\([1, 1]\)]\)], "Input"], Cell[BoxData[ \(\(-4.7397933526305791505`*^-6\)\ s\)], "Output"] }, Open ]], Cell["\<\ The interval is the same, less that 4.74 microseconds \ observed.\ \>", "Text", Evaluatable->False, AspectRatioFixed->True], Cell["\<\ (b) Boost both event quaternions by beta, set the space components \ equal to each other, and solve for beta.\ \>", "Text", Evaluatable->False, AspectRatioFixed->True], Cell[CellGroupData[{ Cell[BoxData[ \(Solve[\n\t \((L[5.\ \(10\^\(-6\)\) s\ c, \ \ 720\ m, \[Beta]]\ .\ q[5.\ \(10\^\(-6\)\) s\ c, \ \ \ 720\ m, 0, 0])\)[\([2, 1]\)]\ == \n\t\(( L[2.\ \(10\^\(-6\)\) s\ c, 1200\ m, \[Beta]]\ .\ q[2.\ \(10\^\(-6\)\) s\ c, \ 1200\ m, 0, 0])\)[\([2, 1]\)], \ \[Beta]]\)], "Input"], Cell[BoxData[ \({{\[Beta] \[Rule] \(-0.533333333333333333347`\)}}\)], "Output"] }, Open ]], Cell["\<\ The frame must move a speed 0.53c in the direction of decreasing x. \ \ \>", "Text", Evaluatable->False, AspectRatioFixed->True], Cell["\<\ (c) & (d) The interval is timelike. It is not meaningful to search \ for a proper distance between these two events.\ \>", "Text", Evaluatable->False, AspectRatioFixed->True] }, Open ]] }, Open ]], Cell[CellGroupData[{ Cell[TextData[{ StyleBox[" ", FontWeight->"Plain", FontVariations->{"Underline"->False}], StyleBox["The Twin Paradox", FontWeight->"Plain"] }], "Section", CellMargins->{{0, Inherited}, {Inherited, Inherited}}, Evaluatable->False, AspectRatioFixed->True, CellTags->"twins"], Cell[CellGroupData[{ Cell[TextData[StyleBox[" The tortoise & the hare"]], "Subsection", CellMargins->{{0, Inherited}, {Inherited, Inherited}}, Evaluatable->False, AspectRatioFixed->True, CellTags->"tort"], Cell["\<\ Q: The tortoise challenges the hare to a race in the woods. The \ hare laughs hysterically saying \"Surely, M'am, you are not serious?\" But \ the tortoise is serious; she gets on the course and starts running(?) right \ away. The course is a closed loop beginning and ending at the same tree. \ While the tortoise is running, the hare continues telling jokes with his \ friends. But when he sees that she has almost gotten back to the finish, he \ decides that it is time to teach her a lesson, and he dashes on the course as \ quick as he can to catch up with her. Alas, he miscalculated slightly and he \ returns to the tree just barely behind her! QUESTION: Assuming that the two animals were of the same age before the race, \ which one is older at the end of it? Justify your answer with quantitative \ arguments!\ \>", "Text", Evaluatable->False, AspectRatioFixed->True], Cell["\<\ A: Let the hare run the fraction f of the tortoise's proper time \ t. Calculate the tortoise's squared interval in terms of this fraction.\ \>", "Text", Evaluatable->False, AspectRatioFixed->True], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{ SubscriptBox[ StyleBox["tau", FontWeight->"Plain"], "tort"], StyleBox[" ", FontWeight->"Plain"], StyleBox["=", FontWeight->"Plain"], StyleBox[" ", FontWeight->"Plain"], "\n", " ", StyleBox[ \(\((q[f\ t, 0, 0, 0]\ .\ q[f\ t, 0, 0, 0]\ + \ \n\t\t q[\((1 - f)\)\ t, 0, 0, 0]\ .\ q[\((1 - f)\)\ t, 0, 0, 0])\)[ \([1, 1]\)]\), FontWeight->"Plain"]}]], "Input"], Cell[BoxData[ \(\((1 - f)\)\^2\ t\^2 + f\^2\ t\^2\)], "Output"] }, Open ]], Cell["\<\ In the tortoise's reference frame, the hare initially travels away \ from the tortoise at the slow Btort speed for time t. Then the hare starts \ traveling toward the tortoise at Bhare speed for a time (1-f) t. Calculate \ the hare's squared interval.\ \>", "Text", Evaluatable->False, AspectRatioFixed->True], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{ SubscriptBox[ StyleBox["tau", FontWeight->"Plain"], "hare"], StyleBox[" ", FontWeight->"Plain"], StyleBox["=", FontWeight->"Plain"], StyleBox[" ", FontWeight->"Plain"], "\n", "\t", RowBox[{ RowBox[{ StyleBox["(", FontWeight->"Plain"], RowBox[{ RowBox[{ RowBox[{ StyleBox["q", FontWeight->"Plain"], StyleBox["[", FontWeight->"Plain"], RowBox[{ StyleBox[\(f\ t\), FontWeight->"Plain"], StyleBox[",", FontWeight->"Plain"], StyleBox[" ", FontWeight->"Plain"], RowBox[{ SubscriptBox[ StyleBox["\[Beta]", FontWeight->"Plain"], "tort"], " ", "f", " ", "t"}], ",", "0", ",", "0"}], "]"}], " ", ".", " ", RowBox[{"q", "[", RowBox[{\(f\ t\), ",", StyleBox[ RowBox[{" ", StyleBox[" ", FontWeight->"Plain"]}]], StyleBox[ RowBox[{ SubscriptBox[ StyleBox["\[Beta]", FontWeight->"Plain"], "tort"], " ", "f", " ", "t"}], FontWeight->"Plain"], StyleBox[",", FontWeight->"Plain"], StyleBox["0", FontWeight->"Plain"], StyleBox[",", FontWeight->"Plain"], StyleBox["0", FontWeight->"Plain"]}], StyleBox["]", FontWeight->"Plain"]}]}], StyleBox[" ", FontWeight->"Plain"], StyleBox["+", FontWeight->"Plain"], StyleBox[" ", FontWeight->"Plain"], "\n", "\t\t", RowBox[{ RowBox[{ StyleBox["q", FontWeight->"Plain"], StyleBox["[", FontWeight->"Plain"], RowBox[{ StyleBox[\(\((1 - f)\) t\), FontWeight->"Plain"], StyleBox[",", FontWeight->"Plain"], StyleBox[" ", FontWeight->"Plain"], RowBox[{ RowBox[{ StyleBox["-", FontWeight->"Plain"], SubscriptBox[ StyleBox["\[Beta]", FontWeight->"Plain"], "hare"]}], " ", "f", " ", "t"}], ",", "0", ",", "0"}], "]"}], " ", ".", " ", RowBox[{"q", "[", StyleBox[ RowBox[{\(\((1 - f)\)\ t\), ",", " ", RowBox[{ RowBox[{"-", SubscriptBox[ StyleBox["\[Beta]", FontWeight->"Plain"], "hare"]}], " ", "f", " ", "t"}], ",", "0", ",", "0"}], FontWeight->"Plain"], StyleBox["]", FontWeight->"Plain"]}]}]}], StyleBox[")", FontWeight->"Plain"]}], StyleBox["[", FontWeight->"Plain"], StyleBox[\([1, 1]\), FontWeight->"Plain"], StyleBox["]", FontWeight->"Plain"]}]}]], "Input"], Cell[BoxData[ \(f\^2\ t\^2 + \((t - f\ t)\)\^2 - f\^2\ t\^2\ \[Beta]\_hare\%2 - f\^2\ t\^2\ \[Beta]\_tort\%2\)], "Output"] }, Open ]], Cell["Look at the difference.", "Text", Evaluatable->False, AspectRatioFixed->True], Cell[CellGroupData[{ Cell[BoxData[ \(Simplify[tau\_hare\ - \ tau\_tort]\)], "Input"], Cell[BoxData[ \(\(-f\^2\)\ t\^2\ \((\[Beta]\_hare\%2 + \[Beta]\_tort\%2)\)\)], "Output"] }, Open ]], Cell["\<\ Since this term is always negative, the hare is necessarily \ younger than the tortoise.\ \>", "Text", Evaluatable->False, AspectRatioFixed->True] }, Open ]], Cell[CellGroupData[{ Cell[TextData[StyleBox[ " R&H: B2-2 Einstein and the clock \"paradox\""]], "Subsection", CellMargins->{{0, Inherited}, {Inherited, Inherited}}, Evaluatable->False, AspectRatioFixed->True, CellTags->"b2-2"], Cell["\<\ Q: Einstein, in his first paper on the special theory of \ relativity, wrote the following: \"If one of the two synchronous clocks at A \ is moved in a closed curve with constant velocity until it returns to A, the \ journey lasting t seconds, then by the clock that has remained at rest the \ travelled clock on its arrival at A will be t v^2/2 c^2 seconds slow.\" \ Prove this statement. (Note: Elsewhere in his paper Einstein indicated that \ this result is an approximation, valid only for v << c.)\ \>", "Text", Evaluatable->False, AspectRatioFixed->True], Cell["\<\ A: Compare the intervals of the two clocks, one that has move, the \ other that has remained.\ \>", "Text", Evaluatable->False, AspectRatioFixed->True], Cell[CellGroupData[{ Cell[BoxData[ \(\@\((q[t, \ \[Beta]\ t, 0, 0]\ .\ q[t, \ \[Beta]\ t, 0, 0])\)[ \([1, 1]\)]\ - \ t\)], "Input"], Cell[BoxData[ \(\(-t\) + \@\(t\^2 - t\^2\ \[Beta]\^2\)\)], "Output"] }, Open ]], Cell["If beta << 1, calculate the series expansion.", "Text", Evaluatable->False, AspectRatioFixed->True], Cell[CellGroupData[{ Cell[BoxData[ \(Simplify[PowerExpand[\n\t\tSeries[%, \ {\[Beta], 0, 2}]]]\)], "Input"], Cell[BoxData[ InterpretationBox[ RowBox[{\(-\(\(t\ \[Beta]\^2\)\/2\)\), "+", InterpretationBox[\(O[\[Beta]]\^3\), SeriesData[ \[Beta], 0, {}, 2, 3, 1]]}], SeriesData[ \[Beta], 0, { Times[ Rational[ -1, 2], t]}, 2, 3, 1]]], "Output"] }, Open ]], Cell["The moving clock is t v^2/2 c^2 slower than the one at rest.", "Text", Evaluatable->False, AspectRatioFixed->True] }, Open ]], Cell[CellGroupData[{ Cell[TextData[StyleBox[" R&H B2-12: Getting Younger"]], "Subsection", CellMargins->{{0, Inherited}, {Inherited, Inherited}}, Evaluatable->False, AspectRatioFixed->True, CellTags->"b2-12"], Cell["\<\ Q: Can you think of any way to use space travel to reverse the \ aging process, that is, to get younger? Could you send your parents out on a \ long space voyage and have them be younger than you are when they get back?\ \ \>", "Text", Evaluatable->False, AspectRatioFixed->True], Cell["\<\ A: There are actually two questions here. Starting with the last \ question first, with a HUGE investment of energy for the parents, time will \ appear to run at a slower rate than the clocks back at home. The energy \ investment is the critical parameter to determine if the clocks will run at \ different enough rates to have the parents return younger than their \ children.\ \>", "Text", Evaluatable->False, AspectRatioFixed->True], Cell["\<\ The second question concerns reversing the aging process. The \ aging process will appear to procede in the same manor for both parent and \ child. Why is this not reversable? Find the quaternion that reverses time. \tLTimeRev q[t, x, y, z] = q[-t, x, y, z] \t Compute LTimeRev by multiplying on the right by the inverse of \ q[t,x,y,z].\ \>", "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(Simplify[ q[\(-t\), x, y, z]\ .\ \n\ \ Transpose[q[t, x, y, z]]\ .\ {1, 0, 0, 0}/ \n\t\t\ \((t^2\ + \ x^2\ + \ y^2\ + \ z^2)\)]\)], "Input"], Cell[BoxData[ \({\(\(-t\^2\) + x\^2 + y\^2 + z\^2\)\/\(t\^2 + x\^2 + y\^2 + z\^2\), \(2\ t\ x\)\/\(t\^2 + x\^2 + y\^2 + z\^2\), \n\t \(2\ t\ y\)\/\(t\^2 + x\^2 + y\^2 + z\^2\), \(2\ t\ z\)\/\(t\^2 + x\^2 + y\^2 + z\^2\)}\)], "Input"], Cell["\<\ Aboard the spaceship, or on the Earth, t >>> x, y and z, so the \ time reversal quaternion is approximately\ \>", "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(LTimeRevBsmall[t_, \ x_, \ y_, \ z_]\ := \ \n\t\n\t q[\(-1\), \ 2\ x/t, \ 2\ y/t, \ 2\ z/t]\)], "Input"], Cell["\<\ Test that this works for someone moving a meter per second in the \ x direction, 0.5 m/s in the y.\ \>", "Text", Evaluatable->False, AspectRatioFixed->True], Cell[CellGroupData[{ Cell[BoxData[ \(LTimeRevBsmall[1.\ s, \ 1.\ m/c, \ 0.5\ m/c, 0]\ .\ \n\t\t\t\ \n\t\t\t\ q[1.\ s, \ 1.\ m/c, \ 0.5\ m/c, 0]\ .\ {1, 0, 0, 0}\)], "Input"], Cell[BoxData[ \({\(-1.00000000000000002775`\)\ s, 3.33333333333333333347`*^-9\ s, 1.66666666666666666673`*^-9\ s, 0.`\ s}\)], "Output"] }, Open ]], Cell["\<\ The proposed quaternion does reverse time in the classical regime. \ Note that it is predominantly a scalar, almost q[-1,0,0,0] However, it is \ not _exactly_ the identity. If we think about time reversal for two nearby \ worldlines, they will not commute by the small factor found in the second \ through fourth terms. This observation may lead to a new justification of \ the second law of thermodynamics.\ \>", "Text", Evaluatable->False, AspectRatioFixed->True] }, Open ]], Cell[CellGroupData[{ Cell[TextData[StyleBox[" French: 5-20 Signals from twins"]], "Subsection", CellMargins->{{0, Inherited}, {Inherited, Inherited}}, Evaluatable->False, AspectRatioFixed->True, CellTags->"f5-20"], Cell["\<\ Q: A and B are twins. A goes on a trip to Alpha Centauri (4 \ light-years away) and back again. He travels at speed 0.6c with respect to \ the Earth both ways, and transmits a radio signal every 0.01 year in his \ frame. His twin B similarly sends a signal every 0.01 years in his own rest \ frame. (a) how many signals emitted by A before he turns around does B \ receive? (b) How many signals does A receive before he turns around? (c) \ What is the total number of signals each twin receives from the other? (d) \ Who is younger at the end of the trip, and by how much? Show that the twins \ both agree on this result.\ \>", "Text", Evaluatable->False, AspectRatioFixed->True], Cell["\<\ A: Start out by drawing the signals sent and received from B's \ frame of reference.\ \>", "Text", Evaluatable->False, AspectRatioFixed->True], Cell[GraphicsData["Bitmap", "\<\ CF5dJ6E]HGAYHf4PAg9QL6QYHgOol00g_NOomoo`03Ool2Nmh00gooNmio o`02Ool01G_NOomoogooMkd01Goo00=kgWooOol0:7oo001=Ool00g_NMkeo o`2@Ool00g_NOomoo`02Ool3Nmh3Ool2Nmh3Ool00g>LOomoo`03Ool00g_N Oomoo`02Nmh017NmNmikgWNm0goo00Mg_GooMkekgWooMkeoo`02Nmh2Ool0 17_NOomkgWoo0g_N1Woo00=g_G_NMkd01Goo0W_N0Woo00EkgWooMkeoogNm 009oo`9kgRYoo`00BWoo00=kgWNmNmh00Woo00=kgWooOol0V7oo0W_N27oo 0WNm0Woo00Ag_GooOomkgP=oo`9kgP9oo`04LiaoogooLi`4Ool017_NOomo og_N1goo0WNm00=kgWooNmh00goo0W_N0goo00=kgWooOol01Woo00=g_Goo Ool0:Woo0019Ool00g_NOomoo`02Mkd00g_NOomoo`2AOol00g_NOomoo`04 Ool2Nmh2Ool00g_NMkekgP05Ool2Nmh2Ool00gNmNmikgP02Ool037_NOomo ogooMkekgWooMkeoogNmOomkgP9oo`03Mkeoogoo00ioo`9kgPMoo`9kgP03 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\>"], "Graphics", ImageSize->{345, 147}, ImageMargins->{{0, 0}, {0, 0}}, ImageRegion->{{0, 1}, {0, 1}}, ImageCacheContents->"Empty"], Cell["\<\ (a) The signals from A out are redshifted and received at B for a \ time of ...\ \>", "Text", Evaluatable->False, AspectRatioFixed->True], Cell[CellGroupData[{ Cell[BoxData[ \(time\_BgetsAout\ = \ 2\ 4\ yr\ /\ 0.6\ - \ \((4\ yr\ /\ 0.6\ - \ 4\ yr)\)\)], "Input"], Cell[BoxData[ \(10.666666666666666666`\ yr\)], "Output"] }, Open ]], Cell["\<\ The signals are sent at a rate of 100/year as viewed by the sender. \ This rate is lowered by the redshifting, so the total number of signals is \ the lower rate times the amount of time the signals are received.\ \>", "Text",\ Evaluatable->False, AspectRatioFixed->True], Cell[CellGroupData[{ Cell[BoxData[ \(\((\[Gamma][0.6]\ - \ 0.6\ \[Gamma][0.6])\)\ q[100/yr, \ 100/yr, 0, 0]\ .\ {1, 0, 0, 0}\ tBgetsAout\)], "Input"], Cell[BoxData[ \({533.333333333333333304`, 533.333333333333333304`, 0, 0}\)], "Output"] }, Open ]], Cell["B receives 533 redshifted signals from A.", "Text", Evaluatable->False, AspectRatioFixed->True], Cell["\<\ (b) The signals from B are not shifted in B's frame, but are \ received at A for a time of...\ \>", "Text", Evaluatable->False, AspectRatioFixed->True], Cell[CellGroupData[{ Cell[BoxData[ \(time\_AgetBout\ = \ \((4\ yr\ /\ 0.6\ - \ 4\ yr)\)\)], "Input"], Cell[BoxData[ \(2.66666666666666666652`\ yr\)], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(q[100/yr, \ 100/yr, 0, 0]\ .{1, 0, 0, 0}\ 2.667\ yr\)], "Input"], Cell[BoxData[ \({266.700000000000000022`, 266.700000000000000022`, 0, 0}\)], "Output"] }, Open ]], Cell["\<\ A receives 266 signals from B during A's trip to Alpha \ Centauri.\ \>", "Text", Evaluatable->False, AspectRatioFixed->True], Cell["\<\ (c) It is easiest to calculate the number of signals received by A \ since the rate with B as a reference frame is constant.\ \>", "Text", Evaluatable->False, AspectRatioFixed->True], Cell[CellGroupData[{ Cell[BoxData[ \(A\_total\ = \ q[100/yr, \ 100/yr, 0, 0]\ .\ {1, 0, 0, 0}\ 2\ 4\ yr/0.6\)], "Input"], Cell[BoxData[ \({1333.33333333333333326`, 1333.33333333333333326`, 0, 0}\)], "Output"] }, Open ]], Cell["A receives a total of 1333 signals from B.", "Text", Evalu