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RELATIVISTIC MOMENTUM AND ENERGY We have derived the addition of velocity equation for motion parallel to the motion of the moving frame u—v x ux— vux 2 c Now we need the equation for motion perpendicular to the direction of motion of the moving frame. From the Lorentz-Einstein Equation we have y,=y =y(t-) 1 Differentiate to get dy’=dy vdx\ dt’=y(dt— j 2 c Divide to get dy’ dt’ dy y(dt_ vdx’ c2) Or uy y(1 —“‘ C2J 2 w Di (‘I r) D r-I (IJ = . CD Di CD Di (DO Di CD r+(D ‘0 r1 ZCD CD(D . CD (DO II II CD Di (J CD -1 0 -‘ t7 ( t) x -• )1 :. ... • ...:.,.... 1Ue1i )I N — UTXS — b1 b 4ê — I II —. *b b 411 b 1I Ix I N N I I ii .4- / / :aidwexa joj iaqo tpea o padsa qM UiAOW ae sJaAJasqo OM atu ‘IeD!UaP! ae seq at.j! ‘eq e Lfl!M pea SJaAJaSqO OM Jap!SUOJ Or using the railroad car example. Looking down on the railroad tracks: 0’ ystcin [hrow uL C (utdi .z:ft C’ + I i12 lb () ‘rtirow arid catch (a) As seen by 0 5 N) II %4J -fl 0 CD CD (F) -. C O C D CD (DO O —I, 0< Di D CD (DO 0 Di C (DO r+ - C (F) CDD CD Di r) -‘ 0 r-I EO = < C) DCD CD CD t-I -‘ -‘ - (F) r+ -‘ r+t CD DiCD n CD CD II —. C,) CD r+O CD D CD Di 0-1 CTD Cn CD CD 0 -‘CD Di CD CD Di CO Cr, Di Di —.2L II 0 r ) CD cr Cr, Momentum is not conserved if u * Butz4 is uy y(1x) ux=o Thus U y Y Not Momentum is not conserved. 7 co -, - - -‘ CD C 0 —. D - 3 0 3 0 —‘ ‘I I, CD CD “3 CD 33 0 CD 3 0 —0)0 CD Oo CD 0) ‘ QD CD-00 r) - ,, 0 ii I, -‘ CD CD CD ‘ CTCD D CD ,, 33 0) II —0 CD (I) D ri r+ 0) D CT CD —hCD < 0 P 0 CD cz: 3 : • 5 0(1 CD CD CU CD < DD) CD Iiii III- II CD -‘ CD II 0 CD 3 CD -‘ Cr, 0) CD -‘ CD CD CD 3 i CD 0)0 5•3 D CDD 3 0 3 38 —1 0° CD We will show it from a different point of view and do what some other books do. Consider the equation p=ymu We can interpret this as the mass increasing with velocity. p = (relativistic mass)u = (ym)u Then 9 0’ %y%tc;n I hrow (utch :E::: g :z:: !:If;: I , LJ/2 0 fhrow dnd Lat(.h - I (a)AsseenbyO For the observer in rest frame Change in momentum for ball in rest frame 2 (mestuy) Change in momentum for ball in moving frame 10 2 (mojngu) = 2 mmovjng (i ux=o 2 (mmojgu) = 2 [mmoving 2 (mmojngu) = 0)1 2 m movjng Conservation of momentum the two changes must be equal 11 2 (mrestuy) mmo - = 2 m mov ing = ymres t This states that a moving mass has a larger mass than the same object when moving. The conservation of momentum holds if we have a different definition of momentum p=ymu Or Have a different definition for mass mmovjng = ymrest 12 Either interpretation is proved with the experiment of accelerating electrons and measuring the momentum. 5 E 4 V 3 E C 2 E V 1 0 0 0.2 0.4 0.6 0.8 1.0 1.2 v/c 2GtJ6 Brooks!Ce 13 KINETIC ENERGY WHAT WE DID IN BASIC MECHANICS KE WORK TO BRING OBJECT FROM REST TO STATE OF MOTION WITH VELOCITY = K = CALCULATE THIS WORK AND IT IS THE KINETIC ENERGY WORK = FORCE X DISTANCE Fds AND 14 dv F=m dt so dv I ds K=m ds= I m dv dt dt J OR K =fmvdv If the object goes from 0 to some speed v K = Vmvdv 0 f = mfVvdv = m 15 - z -n -n rn rn z m o - rn rn II —1 C z rn 0 r) - rn C -n - C —1 0 -‘ START WITH NEWTON’S SECOND LAW d F= p dt Relativistic momentum p=ymu d F= ymu dt Kinetic Energy, K, then is force times distance K=W= fFds=(ymu)ds s=ut 17 ds=udt So K = f = K = d(yu) mfu(yu)dt mfud(yu) = udy + ydu 1 I dy=d 1— 3 dy= if \ 2 u 2C2) 2u )du 2 (_ 18 2 1_ d= ) 2 d(yu)=u u ( \ 2 C u / 2 2 u 1_ d(u)= ) 2 du 2 u C2) —3’ /2 dul+Ydu u 2 ) 2 du+(1_ _1/2 Find common denominator 19 d(yu) du = ( C) Put in KE Equation du K=mf U (‘\ 3 K=mc I F I 3 K=mc C) udu 2 (c — 1 — ) 2 u 3’ /2 i 2j 20 Put in the limits 1 3 K=mc 1 2 2 e 2 K=mc [C [2 2 K=mc C — C 2 U 1 F —1 2 me — / C 2 me C K= 2 ymc 2 — me Or K= (y—1)mc 2 21 2 K=(y—1)mc K = 2 ymc 2 ymc = — 2 mc 2 K + mc ALL ENERGY UNITS TOTAL ENERGY = KE TOTAL ENERGY = E = + MASS ENERGY 2 ymc 2 )fmc