8.022pset6solution

8.022pset6solution - Massachusetts Institute of Technology...

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Unformatted text preview: 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 1 , x 1 , y 1 , z 1 ) and ( t 2 , x 2 , y 2 , z 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 1 = γ ( ct 1- βx 1 ) x 1 = γ ( x 1- βct 1 ) y 1 = y 1 z 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 2 =- ( c Δ t ) 2 + (Δ x ) 2 + (Δ y ) 2 + (Δ z ) 2 =- ( ct 2- ct 1 ) 2 + ( x 2- x 1 ) 2 + ( y 2- y 1 ) 2 + ( z 2- z 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 = t x = x- vt (4) 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 = ( ct- βx )(1 + 1 2 β 2 ) = ct + O ( β ); x = ( x- βct )(1 + 1 2 β 2 ) = x- vt + O ( β 2 ) . (6) or t = t , x = x- vt . This is exactly Galilean transformation. The next term in the expansion of x transformation is (1 / 2) xβ 2 , so it changes by a rate ((1 / 2) xβ 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.relativistic effect cannot be ignored and Newton Mechanics is not quite valid anymore....
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This homework help was uploaded on 04/11/2008 for the course PHYSICS 8.022 taught by Professor Hughes during the Spring '08 term at MIT.

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8.022pset6solution - Massachusetts Institute of Technology...

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