LightTime dilation, Length contraction and Relativistic velocity-additionTime dilation is the phenomenon predicted by the theory ofspecial relativity that says time in two inertial reference frames movingwith respect to each other will appear to be different. In particular, thetime intervals on a speeding rocket will appear to be longer to you than toAtothe people on the ship (figure 1). It is expressed in the formula At =STFigure 1LightObserverSample problem for time dilationThe Hypersonic Technology Vehicle 2 (HTV-2) is an experimentalrocket vehicle capable of traveling at 21,000 km/h (5830 m/s). If anelectronic clock in the HTV-2 measures a time interval of exactly 1-sduration, what would observers on Earth measure the time interval tobe?Special relativity also tells you thatlengths get contracted at high speeds. Theobject is actually contracted in length as seenfrom the stationary reference time frame.Length contraction can be calculated usingthe equation L=Lo1- where L is thelength measured by an observer who is inmotion with respect to the distance beingmeasured, Lo is the length measured by anobserver at rest with respect to the distancebeing measured, v is the relative speed of thetwo observers, and c is the speed of light in aIn this case, Ato is 1.0 seconds (recall that Ato is the time measured inthe same reference frame as the event being measured).Given:Ato 1 s1= 1.0000000004 s(5.830 x 10³ m/s)3.8x108 m/sor 1s + 4.0 x 10-10sv = 5830 m/sc = 3.0x108 m/s(constant)Solution:At =Significance=vacuum.√1-(1.9433 x 10-5) ²1s√.9999999996=The very high speed ofthe HTV-2 is still only 105 times the speed of light.Relativistic effects for theHTV-2 are negligible foralmost all purposes, butare not zero.1=5830m/s3.8x108 m/s1s.999999998The second postulate of relativity saysthat classical velocity addition (velocities addlike regular numbers in one dimensionalmotion) doesn't apply to light. The correctformula for one-dimensional relativistic velocityv+u'addition is u = where u is the velocity of an1+Sample problem for length contractionA crew member of a spaceship measures the ship's length to be100 m. The ship flies past Earth at a speed of 0.900 times thespeed of light. If observers on Earth measure the length of theship, what would they measure?object relative to one observer, u' is the velocityrelative to the other observer, and c is the speed oflight in a vacuum.Solution:Given:ΔΙ = ΔΙ 1-2Al = (100 m)√1- (0.900)²= (100 m)√10.810Al = 100 mv = 0.900 cΔΙ = ?= (100 m)√0.190= (100 m)(0.436)ACTIVITY: Problem solving. Show your solutions.Al = (100 m) 1- (0.900) Al = 43.6 mSample problem for relativistic velocitySuppose a spaceship heading directly toward Earth at half thespeed of light sends a signal to us on a laser-produced beam oflight. Given that the light leaves the ship at speed c as observedfrom the ship, How fast does a light signal approach Earth ofsent from a spaceship traveling at 0.500 c?Given:Solution:v +u'=U=.2. A spaceship, 195 m long as seem on board,moves by the Earth at 0.850 c. What is its lengthas measured by an Earth-bound observer?v = 0.500 cu' = cvu1+C²0.500 + c0.500c2C²(0.500 + 1)c0.5001+&1.500c=1 +0.500= C=1+SignificanceRelativistic velocity gives the correct result. The speed of light isindependent of the relative motion of source and observer, whetherthe observer is on the ship or earthbound.

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ACTIVITY: Problem solving. Show your solutions. 2. A spaceship, 195 m long as seem on board, moves by the Earth at 0.850 c. What is its length as measured by an Earth-bound observer?