Special Relativity solved examples using an Electrical Analog Circuit

Transcription

1 Specil Reltivity solved exmples using n Electricl Anlog Circuit Mourici Shchter mourici@gmil.com mourici@wll.co.il ISRAE, HOON Introduction In this pper, I develop simple nlog electricl circuit. And I use this circuit to solve some clssicl prolems in specil reltivity. Definitions m, m Rest mss nd moving mss c speed of light v mss velocity reltive to the l v slip c X Inductnce α Inductor qulity fctor 1//11 - mourici@gmil.com, mourici@wll.co.il Isrel tel

2 Step 1 he electricl nlog circuit for Reltivistic Mss And hidden prmeters in Specil Reltivity According to Specil Reltivity, elementry prticles hve rest mss m. When the prticle moves with "slip" it's mss m is growing ccording to m 1] m Becuse of the squre root in the eqution, the "slip" is limited to 1 < < 1 which mens tht the prticle velocity is lwys less then the speed of light Now I multiply oth side of eq.1 y α nd get rid of the squre root α m α m α m 1 1 ] α m Eq. in contrry to eq.1, is "continuous function" with only two "d" points ± 1 According to eq. we cn pss the speed of light, Step Deriving n Electricl Equivlent Circuit tht helps understnd Specil Reltivity et go ck to eq. nd write it s follows 3] m 1 1 m m m α α α + α Eqution 3 is divided into two prts tht now on must e treted seprtely α m 1 α m 1 4] α m 1 α m+ 1+ he totl reltivistic mss is computed using Pythgors 5] α m α m + α m + //11 - mourici@gmil.com, mourici@wll.co.il Isrel tel

3 he following grph is for m ginst Step 3 Mking the eqution it more physicl In eq.1 or eq. we dont see ny fmilir physicl phenomen. And the ide tht speed of light is the sme in ll directions is not intuitive So, I m going to turn those equtions into conventionl not weird nd intuitive physicl eqution tht every student meets during his studies. o do so, et define 1 π j 1, f to e ny frequency, ω π f,, k f λ 6] 1 α m 1 1 1, α m, α m, α m + + let sustitute those definitions in eq. 4, nd divide ech eqution y non zero term jω I define some new terms such s: Z,,, Z Z+ Z nd put them in eq. 3 nd eq. 4 7] α α m 1 m 1 α m 1 1 α m 1 Z jω jω jω jω α m+ 1 1 α m α m 1 1 α m 1 Z jω jω jω jω Wht we get is something more fmilir to electricl engineers 3//11 - mourici@gmil.com, mourici@wll.co.il Isrel tel

4 8] α m 1 1 α m α m + α m α m α m α m + + Z Z Z jω jω jω Z Z Z jω jω jω jω 1 jω 1 + nd from the lst line in eq 8 9] According to electricl engineering the lst eqution descries the equivlent vlue of two inductors nd + in prllel. But it is wrong to compute prllel vlue of inductors tht rect with different frequencies. 1] jω jω jω j ω j ω [ 1 ] [ 1 + ] + eqution 9 nd 1 re correct numericlly ut wrong physiclly ecuse the frequencies re not the sme in nd + his led to the conclusion tht professor Alert Einstein missed something tht will e explined now Every electricl engineer will sy tht,, +, re electricl inductors. And he impednce of n inductor is given in generl y 11] Z jω And the current psses through the inductor is given y Ohm's lw 1] I V Z V jω we found in eq. 8 tht ] + Z Z Z+ ut this eqution is wrong ecuse of the different frequencies in ech inductor. he reson for tht is tht the frequency of the current nd voltge must e the sme nd the sme frequency must lso pper in the impednce of the inductor his is ecuse for n inductor 14] di v dt 4//11 - mourici@gmil.com, mourici@wll.co.il Isrel tel

5 So we must compute the currents in ech inductor seprtely ccording to the electricl circuits in the picture elow nd eq. 15 I jω V Ve ω j 1 t I jω 1 V Ve ω I + j 1 + t + jω 1 + [15] where I I I + ω t ω 1 t j V V V Ve Z jω jω 1 jω 1 ω 1 + t V+ V+ V+ Ve Z jω jω 1 + jω V V Ve j Z jω jω j V Ve V Ve V Ve jωt jω 1 t jω 1 + t 16] + he conclusion is tht when the mss is moving nd we hve two frequencies So 16] I I + I+ he "current" I Is superposition of two different currents I nd I+ with different frequencies nd mplitudes? Before continuing with the theory et solve some clssicl exmples ime Diltion Consider pendulum clock sitting sttionry in the frme S the pendulum ticks t intervls of. his mens tht the tick events in frme S occur t the sme plce. In moving frme S, the intervl etween ticks is due to time diltion 5//11 - mourici@gmil.com, mourici@wll.co.il Isrel tel

6 And every tick occurs t different plce. If 1 then A moving clock runs more slowly. he correct interprettion is tht time itself runs more slowly in moving frmes. Muons ife ime ime diltion is tested most ccurtely is in prticle ccelertors where elementry prticles routinely rech speeds close to c. he effect of time diltion is prticulrly vivid on unstle prticles which live much longer in the l frme thn in their own rest frme. For exmple, he muons re unstle prticles. hey decy into n electron, together with couple of neutrinos, 6 with hlf-life of 1 s Muons re creted when cosmic rys hit the tmosphere, nd susequently rin down on Erth. 6 Yet to mke it down to se level, it tkes out 7 1 s, somewht longer thn their lifetime. Given this, why re there ny muons detected on Erth t ll? Surely they should hve decyed. he reson tht they do not decy is ecuse the muons re trveling t speed close to the speed of light,. From the muon s 7 perspective, the journey tkes only 7 1 s, nd is less thn their lifetime. he eqution of time diltion re represented in more intuitive form squring nd dividing y jω α α where 1 α α α jω jω j jω + + ω 1 1+ jω , α α nd therefore the two inductors in prllel re jω jω 1 jω 1+ nd the inductors with their voltge source ω ω + jωt j 1 t j 1 t V V V + jω jω 1 jω 1+ 6//11 - mourici@gmil.com, mourici@wll.co.il Isrel tel

7 Numericl Exmple he verge lifetime of µ mesons with speed of.95c is mesured to e s. Compute the verge lifetime of µ mesons in system in which they re t rest 6 1 s s But if we use the electric nlog circuit 6 6 α α α jω j jω + + ω 1 1+ jω , α α ω 1 ω 1+ jωt j t j t V V V + jω jω 1 jω 1+ jω 1 t jω 1+ jωt V V V + jω jω 1 jω 1+ t jωα jωα α Of course we get the sme result 6 s 7//11 - mourici@gmil.com, mourici@wll.co.il Isrel tel

8 Conclusion For prticles tht move close to the speed of light ime diltion cn e computed with 1, α α or jω jωt t V V jω jω hose who lerned electricity understnd etter the mening of my equtions ength Contrction ength or more ccurtely "proper length" is defined s the distnce etween two coordintes mesured t the sme time. It is mesured t the sme time in the moving # frme S nd the rest frme S According to orentz trnsformtion t B t A mesured t the sme time x x v t t v # # B A B A Γ Γ Γ xb xa Γ x x B A It is cler tht If 1 then Γ Γ Professor Alert Einstein elieved tht ength contrction cn not e mesured dder Prdox he solution of prolem with length contrction looks the sme s prolem with time diltion 8//11 - mourici@gmil.com, mourici@wll.co.il Isrel tel

9 Reltivistic dynmic A completely inelstic collision et s pply the conservtion of energy nd momentum to specific cse. A ll of ule gum with rest mss 16 kg, nd nother ll of ule gum with rest mss 9 kg, speeds towrd ech other s shown: Before collision m kg, m 9 kg,, 5 5 After collision: he two lls stick together. he totl horizontl momentum nd the totl energy efore nd fter collision re s follow don t miss ny line ecuse the solution is not conventionl mc E 4 4 m c m c 1 1 E m m c 1 1 E c c c 1 1 E c c m c 1 1 E E + E 4c + 5c 65c 5 c m 5 fterc he momentum is computed twice, so keep pying ttention 3 4 m 16 9 c c 5 mc c p 1, 5 c p 1 c m m + m kg p p + p m + c m m c 5 c 1 1 p c c he totl momentum fter collision is zero. herefore fter collision msses doesn t move 9//11 - mourici@gmil.com, mourici@wll.co.il Isrel tel

10 Now I solve the sme exmple using electricl nlog circuit α α m c α m c , p α m c α p *************************************************** ssume α c p c c α α m c α m c p c 1 1 p c 1 c ********************************************************** he sum nd diference of the momentum re hiden in the next expressions + α p α p c c c c α p α p α p α p We know lso tht energy nd momentum depend ccording to E p c + E where E m c E p c + E 4c p c + 16c p c 4c 56c 144 c p 1c E p c + E 5c p c + 9c p c 5c 81c 144 c p 1c 1//11 - mourici@gmil.com, mourici@wll.co.il Isrel tel

11 And if we compute the sum nd difference of energy nd momentum we get gin previous results E p c + E, E m c E p c + E, E p c + E nd the energy sum is; E p c + E so + E p c + E E + E p + p c + E + E E p c + E 4c + 5c p + p c + 56c + 81c c p + p c + 337c c p + p c 4 p + p 88c p 1 c, p 1 c s efore ****************************************************** *** Now, let find the energy nd momentum difference E E E, p p p E p c + E c E E E p p c E E p p p p c E E E 4 5c 175c p p c 56c 81c p p p p c 175c p p p + p c p p Conclusions It is well understood tht the electricl nlog circuit help solving prolems in specil reltivity. he nlog electricl circuit lso explin specil reltivity not the wy professor Alert Einstein did o see detil of this explntion you hve to red short pper Specil Reltivity Electromgnetic nd Grvittion Comined Into One heory 11//11 - mourici@gmil.com, mourici@wll.co.il Isrel tel