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Special Relativity 4. Topics. Paradoxes Twin Temporal The Lorentz Transformation Velocity Transformation The Interval Summary. Betty’s Now in 2022. Ann’s Now in 2030. 2030. E-mail sent by Betty in 2016, Betty’s time, received by Ann in 2028, Ann’s time. 2028. Betty’s Now in 2016.
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Topics • Paradoxes • Twin • Temporal • The Lorentz Transformation • Velocity Transformation • The Interval • Summary
Betty’s Now in 2022 Ann’s Now in 2030 2030 E-mail sent by Betty in 2016, Betty’s time, received by Ann in 2028, Ann’s time. 2028 Betty’s Now in 2016 Betty’s Now in 2010 2020 Ann’s Now in 2020 E-mail sent by Ann in 2012, Ann’s time, received by Betty in 2016, Betty’s time. 2013.6 2012 2010 Ann’s Now in 2010 8 light years Twin Paradox Betty 8 years younger! Puzzle: Why does Betty become younger?
Send super-luminal signal to star A in 2010, arriving in 2016 according to Betty. But for Ann, signal sent in 2010 arrives in 2007! Betty’s Now in 2010 Betty’s Now in 2016 8 light years 2008 Ann and Betty’s Now in 2007.2 2007.2 A Super-luminal signal sent from A arrives in 2008, preventing signal sent in 2010! Temporal Paradox 2020 Ann’s Now in 2020 Ann’s Worldline 2013.6 8 light years Ann’s Now in 2010 2010
t t' P x' C Q x O A B The Lorentz Transformation Define t = tBC + tCP x = xOA + xAB What is the map between the two different frames of reference? x' = xOQ t' = tQP tCP = tQP / g tBC = v x / c2 xOA = xOQ / g xAB = v t
t t' P x' C Q x O A B Lorentz Transformation Lorentz transformation (t,x,y,z) → (t’,x’,y’,z’) t = g (t – v x/c2) x’ = g (x – v t) y’ = y z’ = z tCP = tQP / g tBC = v x / c2 xOA = xOQ / g xAB = v t
t t' P x' Q x O B Velocity Transformation Suppose point P is moving at velocity u = (ux, uy, uz) in the S frame. To find its velocity in the S’ frame we start with: S’ frame S frame
t t' P x' Q x O B Velocity Transformation Example: Consider u’x S’ frame S frame
t t' P S’ frame x' x O B Q S frame Velocity Transformation Similarly, we arrive at Review Example 1-4
P A O B The Interval The distance between O and P is independent of the components used to calculate it: OB2 + BP2 = OP2 = OA2 + AP2 OP2 is said to be invariant. The formula for computing it is called the metric. In 1908, Herman Minkowski introduced a similar idea for spacetime Hermann Minkowski (1864 -1909)
t t' P x' Q x O B The Interval Try OP2 = (ct)2 + x2 + y2 + z2 What is the spacetime distance, i.e., the interval, between event O and event P? This does not work! Minkowski showed that the correct formula is OP2= (ct)2– (x2 + y2 + z2) = (ct’)2– (x’2 + y’2 + z’2) Extra Credit: Show that OP2 is invariant
The Interval In general, the interval Ds2 between two events is either timelike Ds2 = c2Dt2 – Ds2 or spacelike Ds2 = Ds2 – c2Dt2 or null Ds2 = c2Dt2 - Ds2 = 0 depending on whether the temporal difference cDt or spatial difference Ds dominates
The Interval Since the interval Ds2 is an invariant we can immediately deduce the following: when the spatial separation between two events Ds = 0 their temporal separation is invariant: Dt = Ds / c Proper Time when the temporal separation between two events Dt = 0 their spatial separation is invariant: Ds= Ds Proper Distance
Summary • Super-luminal travel would lead to temporal paradoxes • The temporal separation of events depends on the path taken between them and is greatest for an inertial, i.e., non-accelerating, path yields the longest temporal separation • The distance between events in spacetime is computed from the interval Ds2 = c2Dt2 – Ds2