8.022pset6solution

# 8.022pset6solution - Massachusetts Institute of Technology...

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Massachusetts Institute of Technology Department of Physics 8.022 Spring 2004 Solution 6:Special relativity; Magnetic fields. TA: Yi Mao, [email protected] 1. Invariant interval: A quantity that is left unchanged by Lorentz transformations is a called a “Lorentz invariant”. Consider two events described in the laboratory frame by ( t 1 , x 1 , y 1 , z 1 ) and ( t 2 , x 2 , y 2 , z 2 ). Show that Δ s 2 = - ( c Δ t ) 2 + (Δ x ) 2 + (Δ y ) 2 + (Δ z ) 2 . (1) is a Lorentz invariant. Let’s label events 1 and 2 in the laboratory frame by ( t 1 , x 1 , y 1 , z 1 ) and ( t 2 , x 2 , y 2 , z 2 ), and in the boosted frame by ( t 0 1 , x 0 1 , y 0 1 , z 0 1 ) and ( t 0 2 , x 0 2 , y 0 2 , z 0 2 ). For simplicity suppose boosted direction is along positive x-axis and the boost frame is of relative velocity v to the laboratory frame. The transformation law for event 1 reads ct 0 1 = γ ( ct 1 - βx 1 ) x 0 1 = γ ( x 1 - βct 1 ) y 0 1 = y 1 z 0 1 = z 1 , (2) where β = v/c and γ = (1 - β 2 ) - 1 / 2 . For event 2 the transformation is simply to rewrite all subscripts “1” in eq.(2) to subscript “2”. The spacetime interval Δ s 2 in boosted frame then becomes Δ s 0 2 = - ( c Δ t 0 ) 2 + (Δ x 0 ) 2 + (Δ y 0 ) 2 + (Δ z 0 ) 2 = - ( ct 0 2 - ct 0 1 ) 2 + ( x 0 2 - x 0 1 ) 2 + ( y 0 2 - y 0 1 ) 2 + ( z 0 2 - z 0 1 ) 2 = - γ 2 [( ct 2 - ct 1 ) - β ( x 2 - x 1 )] 2 + γ 2 [( x 2 - x 1 ) - β ( ct 2 - ct 1 )] 2 +( y 2 - y 1 ) 2 + ( z 2 - z 1 ) 2 = - γ 2 (1 - β 2 )( ct 2 - ct 1 ) 2 + γ 2 (1 - β 2 )( x 2 - x 1 ) 2 + ( y 2 - y 1 ) 2 + ( z 2 - z 1 ) 2 = - ( c Δ t ) 2 + (Δ x ) 2 + (Δ y ) 2 + (Δ z ) 2 = Δ s 2 . (3) So Δ s 2 is a Lorentz invariant. 2. Galilean tranformations: Prior to special relativity, people related coordinates between different frames with the “Galilean transformation” – clocks in different reference frame tick at the same rate, spatial positions are shifted by a term that depends on the relative velocity just as you would expect. For example, for frames that are moving with respect to each other in the x direction, we would have t 0 = t x 0 = x - vt (4)

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Using the binomial expansion on γ , show that for small v/c the Lorentz tranformations reduce to the Galilean tranformations. At what value of v does the next term in the expansion change the x transformation by 1%? The Lorentz transformation reads in eq.(2) without subscripts “1” for the sake of generality. Note that as β = v/c ¿ 1, γ = (1 - β 2 ) - 1 / 2 1 + 1 2 β 2 + · · · , (5) where “ · · · ” denotes terms in the order O ( β 4 ). Then Lorentz transformation reduces to be ct 0 = ( ct - βx )(1 + 1 2 β 2 ) = ct + O ( β ); x 0 = ( x - βct )(1 + 1 2 β 2 ) = x - vt + O ( β 2 ) . (6) or t 0 = t , x 0 = x - vt . This is exactly Galilean transformation. The next term in the expansion of x transformation is (1 / 2) 2 , so it changes by a rate ((1 / 2) 2 ) /x = (1 / 2) β 2 = 1%. This gives v = 0 . 14 c 4 × 10 7 m/s , beyond which, that means, the relativistic effect cannot be ignored and Newton Mechanics is not quite valid anymore. 3. Transforming velocities: A bullet is fired with velocity ~ u 0 in the ( x 0 , y 0 ) plane of a mov- ing frame F 0 . Frame F 0 moves with speed v in the + x direction with respect to the laboratory frame F .
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