This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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 xaxis 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....
View
Full
Document
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.
 Spring '08
 Hughes
 Mass, Special Relativity

Click to edit the document details