As in Figure (a), reference frame S \' passes reference frame S with a certain v
ID: 2242887 • Letter: A
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As in Figure (a), reference frame S' passes reference frame S with a certain velocity. Events 1 and 2 are to have a certain temporal separation ?t' according to the S' observer. However, their spatial separation ?x' according to that observer has not been set yet. Figure (b) gives their temporal separation ?t according to the S observer as a function of ?x' for a range of ?x' values.The vertical axis scale is set by ?ta = 6.00
As in Figure (a), reference frame S' passes reference frame S with a certain velocity. Events 1 and 2 are to have a certain temporal separation ?t' according to the S' observer. However, their spatial separation ?x' according to that observer has not been set yet. Figure (b) gives their temporal separation ?t according to the S observer as a function of ?x' for a range of ?x' values. The vertical axis scale is set by ?ta = 6.00 mu s. What is ?t'?Explanation / Answer
we have:
dt = gamma dt' + gamma (v/c^2) dx'
therefore the slope of the diagram is "gamma (v/c^2)"
==> slope = (6us - 2us)/400 = 4e-6/400 = 1e-8
==> gamma (v/c^2) = 1e-8
==> (1/(1 - (v/c)^2)^0.5) (v/c^2) = 1e-8
==> (1/(1 - (v/3e8)^2)^0.5) * (v/(3e8)^2) = 1e-8
==> v = 2.84607e8 m/s
therefore at dx' = 0:
dt = gamma dt' + gamma (v/c^2) dx'
==> 2 = gamma dt' + gamma (v/c^2) * 0
==> 2 = gamma dt'
==> 2 = (1/(1 - (v/3e8)^2)^0.5) dt'
==> 2 = (1/(1 - (2.84607e8/3e8)^2)^0.5) dt'
==> 2 = (1/(1 - (2.84607e8/3e8)y2)y0.5) dt'
==> dt' = 0.632 x 10^-6 s
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