Assignment 7

Assignment 7 - PHYS 230 Assignment 7 Fall 2007 1 In class...

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PHYS 230 Fall 2007 Assignment 7 1. In class, as in the textbook, we “proved” that ( c Δ t ) 2 - x ) 2 = ( c Δ τ ) 2 = ( c Δ t ± ) 2 - x ± ) 2 using a geometric construction. Prove it from the Lorentz transformation equations using the following steps: (a) Show that ( ct ) 2 - ( x ) 2 = ( ct ± ) 2 - ( x ± ) 2 Use the Lorentz transformation equations: ct = γ ( ct ± + βx ± ) and x = γ ( x ± + βct ± ) and substitute into the LHS. We obtain γ 2 [( ct ± + βx ± ) 2 - ( x ± + βct ± ) 2 ] or γ 2 (1 - β 2 )[( ct ± ) 2 - ( x ± ) 2 ] since the cross terms cancel. But, γ 2 (1 - β 2 ) is just 1 , so the identity is proved. (b) Differentiate, use an appropriate expansion, and show therefore that ( c Δ t ) 2 - x ) 2 = ( c Δ τ ) 2 = ( c Δ t ± ) 2 - x ± ) 2 The simplest way is to differentiate the Lorentz transformation equations directly, with respect to τ , the proper time, to obtain dx ± = γ ( dx - βcdt ) and cdt ± = γ ( cdt - βdx ) Thus ( dx ± ) 2 - ( cdt ± ) 2 = γ 2 [( dx - βcdt ) 2 - ( cdt - βdx ) 2 ] or ( dx ± ) 2 - ( cdt ± ) 2 = γ 2 (1 - β 2 )[( dx ) 2 - ( cdt ) 2 ] + γ 2 [ - 2 βdxdt + 2 βdxdt ] Since, as before, γ 2 (1 - β 2 ) = 1 , this gives ( dx ± ) 2 - ( cdt ± ) 2 = ( dx ) 2 - ( cdt ) 2 1
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as required. An alternative way is to replace x by x + Δ x , and ct by ct + c Δ t on the LHS, to multiply out, and then show that, order by order, the expression is equal to the RHS. Thus, show that ( x + Δ x ) 2 - ( ct + c Δ t ) 2 = ( x ± + Δ x ± ) 2 - ( ct ± + c Δ t ± ) 2 Multiplying out, the zero order terms are equal from part (a). The first order terms give 2 x Δ x - 2 c 2 t Δ t = 2 x ± Δ x ± - 2 c 2 t ± Δ t ± and the second order terms give what is required. So, we must show that the first order terms are equivalent. Use the Lorentz trans- formations again, and substitute on the LHS to give γ 2 x ± + βc Δ t ± )( x ± + βct ± ) - γ 2 ( c Δ t ± + β Δ x ± )( ct ± + βx ± ) The cross terms cancel, and, once again, γ 2 (1 - β 2 ) = 1 . QED. 2. French 7.1. A K -meson traveling through the laboratory breaks up into two π -mesons. One of the two π -mesons is left at rest. What was the kinetic energy of the K ? What is the kinetic energy of the remaining π -meson? (Note: Mass in MeV is just a measure of Mc 2 , momentum in MeV is a measure of
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Assignment 7 - PHYS 230 Assignment 7 Fall 2007 1 In class...

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