Lorentz TransformationA2290-06 1The Lorentz TransformationRelativity and AstrophysicsLecture 06Terry HerterA2290-06 Lorentz Transformation 2Outline Coordinate transformations Lorentz Transformation Statement Proof Addition of velocities Partial proof Examples of velocity addition Proof of contraction along the direction of propagation Reading Spacetime Physics: Chapter 3 & Special Topic: Lorentz Transformation Homework: (due Wed. 9/16/09) 3-1, 3-7, and 3-10 (maybe more on Friday)Lorentz TransformationA2290-06 2A2290-06 Lorentz Transformation 3Moving between inertial frames Events and intervals between events define the physical world. This defines the physics We define isolated events using a latticework (inertial reference frame) of recording clocks. But we will need to move from “clock-lattice” frame to another (for instance, lab to rocket frame and visa-versa) Lorentz transformation Name for the translation between inertial frames It may useful because we may want to tag the location of a number of events in our lab The transformation allows to compute the space and time separations between events in different inertial framesA2290-06 Lorentz Transformation 4Speed example Suppose I travel in a rocket that you observe to be traveling at vrel= 4/5c. I fire a bullet that I observe to fly forward at 4/5c. What velocity do you see for the bullet? The velocity is not 4/5 + 4/5 = 1.6! It is 40/41 which is determine via the Lorentz transformation. Suppose the bullet strikes a target 4 meters away from me and my clock measures the time of flight to be 5 meters.  What do you (in the lab frame) measure for the space and time coordinates of the two events? We use the Lorentz transformation to figure this out.vrel= 4/5v′ = 4/5Lorentz TransformationA2290-06 3A2290-06 Lorentz Transformation 5The Spacetime Interval is not enough Rocket Frame: x′ = 4 m and t′ = 5 m  Bullet Frame: x′′ = 0 m, use spacetime interval to get the time So that, t′′ = 3 m (proper time) Lab Frame: Can’t use interval because it is not sufficient to determine xor t separately The Lorentz transformation allows us to determine these separately  2222xtxt=>2222m4m50 tprimed = rocket framedouble primed = bullet frameunprimed = lab frame222m3 xtA2290-06 Lorentz Transformation 6Lorentz Transformation The Lorentz transformation allows us to move between inertial reference frames x , t are in the lab frame x′, t′ are in the rocket frame vrelis the relative velocity between the rocket and lab frames For convenience let the positive x-axis be along the direction of motion of the rocket. The transformation equations are The LT is powerful but is not fundamental in that it doesn’t expose deep new features of spacetime But it is useful – sometimes want to know the length of a yacht, but at other times you would like the positions of the bow and sternrelative to north 21relrelvxvtt21relrelvtvxxyyzzLorentz TransformationA2290-06 4A2290-06 Lorentz Transformation 7Proof of LT – part 1 Requirements of the transformation The linearity, that is, space and time coordinate must appear the first power, not squared or cubed. - Since we require that we can choose any event as the zero of space and time. Must preserve spacetime interval between two events Let us define x, t are in the lab frame x′, t′ are in the rocket frame vrelis the relative velocity between the rocket and lab frames Step 1: There will be no change in the transverse direction Step 2: Consider a sparkplug that sits at the origin of the rocket frame and emits a spark at time t′.  What are x and t in our lab frame?  Spark must occur at location of sparkplug so that Since x = 0 at t = 0 by the way we chose the frames. tvxrelyyzzA2290-06 Lorentz Transformation 8Proof of LT – part 2 Now use the spacetime interval to get the relation between t and t′. Thus Let us define, , a quantity which occurs often in Lorentz transformations Then we have These equations allow us to move from space and time coordinates in one inertial frame to another but only apply when x′ = 0.  We need to extend to the more general case when x′  0.tt211relvtvxrel2/121relvtt 21relvttor  22220txt22xt 222tvtrelandLorentz TransformationA2290-06 5A2290-06 Lorentz Transformation 9Proof of LT – part 3 Since the Lorentz transformation must be linear the general form should look like: We wish now to find B, D, G, and H.  These coefficients should depend upon the rocket speed but not the coordinates of a particular event. The transformation must agree with our previous result for x′ = 0. B and G will be set by requiring that the Spacetime interval is the same in the rocket and lab frames See pages 101-102 of Spacetime PhysicstxBttvxGxreltDxBttHxGxA2290-06 Lorentz Transformation 10Lorentz Transformation This gives the transformation equations The inverse transformation from (x,t) to (x′,t′) Transforms coordinates the other way.  To derive we note that the laboratory moves with the same speed but opposite sign (negative-x direction) We get the reverse transformation by changing the sign of vreland swapping t and t′. See pages 102-103 of Spacetime Physics for long proof.21relrelvxvtt21relrelvtvxxyyzz21relrelvxvtt21relrelvtvxxyyzzLorentz TransformationA2290-06 6A2290-06 Lorentz Transformation 11Addition of Velocities We can derive how velocities add up from the Lorentz transformation. Writing the LT using , Taking the differential of both equations Now dividing the two This is call the Law of Addition of Velocities See page 105 of Spacetime Physics for a non-calculus derivation Your textbook prefers the term Law of Combination of Velocities For our earlier example using vrel= 4/5 and v′ = 4/5 we get v = 40/41 =