Relativity III. Review: Kinetic Energy. Example: He beam from THIA K = 300keV v =? Exact vs non-relativistic calculations Q.37-3.

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1 Relatiity III Today: Time dilation eamples The Lorentz Transformation Four-dimensional spaetime The inariant interal Eamples Reiew: Kineti Energy General relation for total energy: Rest energy, 0: Kineti energy: Momentum: K E γ m E m E m γ p γ m ( m Relation between momentum and energy: E ( m + ( p Eat s non-relatiisti alulations Last time we saw that for the Stanford LINC, a nonrelatiisti alulation was terribly wrong (3 m s miles. That was a ase where K >> m. (ER ase Now let s look at the ase K << m. (NR ase Eample: He beam from THI K 300k? m 4 G 4 0 K So non-relatiisti alulation should be OK. K m K m K m m / s 3.67 mm / ns Compare with eat relatiisti answer K ( γ m K γ + m γ ( / γ Compare with non-relatiisti approimation: % error ( Q.37-3 n eletron (m 0.5 M moes with a speed 0.4 so that γ 3. What is its kineti energy?. 0. M. 0.5 M 3..0 M 4..0 M M

2 Q.37-3 m 0.5 M, γ 3: K? K E m ( γ m. 0. M. 0.5 M 3..0 M 4..0 M M (3 0.5 M.0 M Time Dilation lab obserer ompares two stationary loks against a lok moing with speed, as it passes first one then the other. Lab loks gie t, moing lok t 0. t γ t 0 t 0 Time measured in lab ( t is greater than proper time t 0 measured by o-moing obserer. Moing loks run slow. γ ( / Eample: Problem 37- lok moes along the ais at speed 0.6 and reads zero as it passes the origin. What time does the lok read as it passes 80 m? Lab time: t 80 m s m / s µ 8 Gamma fator: γ Proper time: t µ s t µ s γ.5 Q.37-4 lok moes relatie to a laboratory, at speed suh that γ5. During the time taken for the moing lok to adane 0 ns, how muh time elapses aording to the lab loks? ( 0.5 ns ( ns (3 5 ns (4 0 ns (5 50 ns Q.37-4 lok moes relatie to a laboratory, at speed suh that γ5. During the time taken for the moing lok to adane 0 ns, how muh time elapses aording to the lab loks? Solution: t γ t 5 0 ns 50 ns ( 0.5 ns ( ns (3 5 ns (4 0 ns (5 50 ns Galileo: Lorentz transformation Einstein found that the old self-eident laws for transformations between inertial frames must be replaed by new ones. t t Einstein: use the Lorentz transformation: γ ( t γ ( t / γ where ( /

3 Eample: Problem 37- Two flashbulbs triggered simultaneously. 30 km lso iewed from moing frame. γ ( t γ ( t / For eah eent, knowing,t, alulate, km.5 Coordinates in lab frame: t t 0, 0, L 30 km Coordinates in moing frame: 0, 0,? Eample ontinued γ γ ( t / m / s equation for time of eent in moing frame. ( / (.5 γ L / m 5.8 µ s ut 0. So happens before as seen in moing frame! Eample summarized Times in lab frame: t t Times in moing frame: µ s Eents are not simultaneous in moing frame. is first, then is 5.8 miroseonds later. The spaetime interal Our tet doesn t stress this point, but there is another way of epressing laws of speial relatiity whih is often simpler than using the Lorentz transformation equations or going bak to the two postulates. This is the inariant spaetime interal.

4 The spaetime ontinuum nother way of epressing laws of speial relatiity whih is often simpler than using the Lorentz transformation equations. Instead of thinking of spae and time separately, think of a four-dimensional spaetime. The points in this spaetime are really eents. Then the distane between eents is alled the spaetime interal. Now relatiity follows from the fundamental assumption that the spaetime interal is inariant: the same for all inertial obserers. The spaetime interal Gien two eents (,t and (,t. s seen by inertial obserer O they are separated by interals t t t s seen by another obserer O' these interals are different (relatie. ut the spae-time interal is the same (inariant: s' t t The spaetime diagram n alternatie to the Lorentz transformation equations is the inariant spaetime interal: t ( t ( ' t Unlike ordinary (Eulidean geometry, this interal an be either positie (timelike Two eents or negatie (spaelike. Eample Two eents our at the same plae: Derie time dilation using '. 0, t Same eents seen by obserer moing with speed : s', ( ' ( ( ( ' ( ( ' ( / So now use inariane of spaetime interal: gies t / ( / γ t ll Obserers are Equialent In a moing referene frame, lengths ontrat, time dilates. Moing rods are short and moing loks are slow! How is this possible if all inertial obserers are equialent? If moes with speed relatie to, then s loks are slow as measured by. ut aording to, it s that s moing, so s loks are slow as measured by. Is this possible? Yes, beause they also disagree about how the loks are originally synhronized. areful analysis shows that the effet is perfetly symmetrial: Eah obserer thinks the other is using slow loks! Simple Problem to Show Symmetry Obserers O and O with relatie speed. y C L C y C L C

5 Use inariant spaetime interal y C y L C C L C Four eents: (a C passes C, (b C passes C, ( C passes C, (d C passes C For eah pair of eents figure out the inariant interal t Now apply the inariane s' Result: Find eah obserer thinks the other s loks are slow and improperly synhronized! Eample: Problem Can a long limo fit in a short garage temporarily? (ont d (a Length of ar aording to Garageman (ont d (e Time between eents aording to Carman (b Coordinates of eent aording to Garageman ut tc 0 so tc s ( Car spends this time inside aording to G. (d Length of garage aording to Carman So to Carman, eent ours first, and the ar is neer entirely inside the garage.