PH300 Modern Physics SP11 Last time: • Time dilation and length contraction Today: • Spacetime • Addition of velocities • Lorentz transformations Thursday: • Relativistic momentum and energy “The only reason for time is so that HW03 due, beginning of class; HW04 assigned everything doesn’t happen at once.” 2/1 Day 6: Next week: - Albert Einstein Questions? Intro to quantum Spacetime Thursday: Exam I (in class) Addition of Velocities Relativistic Momentum & Energy Lorentz Transformations 1 2 Spacetime Diagrams (1D in space) Spacetime Diagrams (1D in space) c · t In PHYS I: v In PH300: x x x x Δx Δx v = /Δt Δt t t Recall: Lucy plays with a fire cracker in the train. (1D in space) Spacetime Diagrams Ricky watches the scene from the track. c· t In PH300: object moving with 0<v<c. ‘Worldline’ of the object L R -2 -1 0 1 2 x object moving with 0>v>-c v c·t c·t Lucy object at rest object moving with v = -c. at x=1 x=0 at time t=0 -2 -1 0 1 2 x -2 -1 0 1 2 x Ricky 1 Example: Ricky on the tracks Example: Lucy in the train ct ct Light reaches both walls at the same time. Light travels to both walls Ricky concludes: Light reaches left side first. x x L R L R Lucy concludes: Light reaches both sides at the same time In Ricky’s frame: Walls are in motion In Lucy’s frame: Walls are at rest S Frame S’ as viewed from S ... -3 -2 -1 0 1 2 3 ... This is the time axis ct S’ v=0.5c These angles ct’ of the frame S’ are equal ... -3 -2 -1 0 1 2 3 ... Frame S’ is moving to the right at v = 0.5c. The origins of x’ S and S’ coincide at t=t’=0. Which shows the world line of the origin of S’ as viewed in S? This is the space axis x of the frame S’ A ct B ct C ct D ct x x x x Frame S’ as viewed from S ct ct’ In S: (x=3,ct=3) In S’: (x’=1.8,ct’=2) x’ Spacetime Interval x Both frames are adequate for describing events – but will give different spacetime coordinates for these events, in general. 2 Distance in Galilean Relativity Remember Lucy? The distance between the blue and the red ball h is: Lucy Event 1 – firecracker explodes Event 2 – light reaches detector If the two balls are not Distance between events is h moving relative to each other, we find that the distance between them is “invariant” under Galileo transformations. Remember Ricky? Spacetime interval Say we have two events: (x1,y1,z1,t1) and (x2,y2,z2,t2). Define the spacetime interval (sort of the "distance") cΔt’ h between two events as: Δx’ Event 1 – firecracker explodes With: Δx = x2 − x1 Event 2 – light reaches detector Spacetime interval Distance between events is cΔt’ Δy = y2 − y1 But distance between x-coordinates is Δx’ Δz = z − z and: (cΔt’)2 = (Δx’)2 + h2 2 1 Δt = t2 − t1 And Lucy got We can write The spacetime interval has the same value in all reference frames! I.e. Δs2 is “invariant” under Lorentz since transformations. Spacetime interval Spacetime ct Here is an event in spacetime. Any light signal that 2 2 2 2 2 passes through this event = − − − (Δs') (cΔt ') (Δx') (Δy') (Δz') has the dashed world x lines. These identify the 2 2 (Δs') = (Δs) ‘light cone’ of this event. The spacetime interval has the same value in all reference frames! I.e. Δs2 is “invariant” under Lorentz transformations. 3 Spacetime Spacetime ct ct Here is an event in Here is an event in spacetime. spacetime. The blue area is the future The yellow area is the on this event. A “elsewhere” of the event. No physical x The pink is its past. x signal can travel from the event to its elsewhere! Spacetime Spacetime Now we have two events ct A and B as shown on the ct D 2 left. C (Δs) >0: Time-like events (A è D) B B The space-time interval (Δs)2 <0: Space-like events (Δs)2 of these two events A A (A è B) is: 2 x x (Δs) =0: Light-like events A) Positive (A è C) B) Negative C) Zero If (Δs)2 is negative in one frame of reference it is also 2 negative in any other inertial frame! ((Δs) is invariant under (Δs)2 is invariant under Lorentz transformations. Lorentz transformation). à Causality is fulfilled in SR. Example: Wavefront of a flash z y t=0 t>0 x Wavefront = Surface of a sphere with radius ct: (ct)2 - x2 - y2 - z2 = 0 Spacetime interval for light-like event: (Δs)2 = 0 Einstein: 'c' is the same in all inertial systems. Therefore: (ct')2 - x'2 - y'2 - z'2 = 0 in all inertial systems! (Here we assumed that the origins of S and S' overlapped at t=0) 4 Velocity transformation (1D) Velocity transformation (1D) , B A , where Δx=x2 – x1, Δx’=x’2 – x’1 γ S Δx Use Lorentz: x’ = (x-vt) . x x . ... -3 -2 -1 0 1 2... 2 1 S’ v ... -3 -2 -1 0 1 2 3 ... Galilean result An object moves from event A=(x1,t1) to event B =(x2,t2). As seen from S, its speed is with: Δx = x2 – x1 New in special relativity Δt = t2 – t1 As seen from S’, its speed is with: Δx’ = x’2 – x’1 Δt’ = t’2 – t’1 Velocity transformation in 3D Velocity transformation (3D) The velocity u=(u , u , u ) measured in S is given by: y y' x y z u u =Δx/ , u =Δy/ , u =Δz/ S (x,y,z,t) x Δt y Δt z Δt , where Δx=x2-x1 … (x',y',z',t') v S' x x' To find the corresponding velocity components u’x, u’y, u’z in the frame S’, which is moving along the x-axes in S with the z z' velocity v, we use again the Lorentz transformation: x’1=γ(x1-vt1), and so on… In a more general case we want to transform a velocity u 2 t’1=γ(t1-vx1/c ), and so on… (measured in frame S) to u’ in frame S’. Note that u can point in any arbitrary direction, but v still points along the x-axes. Algebra Velocity transformation (3D) (aka. “Velocity-Addition formula”) Some applications 5 Relativistic transformations Velocity transformation: Which coordinates are primed? u is what we were looking for! (i.e. velocity measured in S) y y' Suppose a spacecraft travels at speed v=0.5c relative to (x,y,z,t) u the Earth. It launches a missile at speed 0.5c relative to S (x',y',z',t') v the spacecraft in its direction of motion. How fast is the S' missile moving relative to Earth? (Hint: Remember which Earth x x' Spacecraft coordinates are the primed ones. And: Does your answer z z' make sense?) a) 0.8 c b) 0.5 c c) c d) 0.25 c e) 0 Transformations If S’ is moving with speed v in the positive x direction relative to S, and the origin of S and S' overlap at t=0, then the coordinates of the same event in the two frames are related by: The “object” could be light, too! Suppose a spacecraft travels at speed v=0.5c relative to Lorentz transformation Velocity transformation the Earth. It shoots a beam of light out in its direction of (relativistic) (relativistic) motion. How fast is the light moving relative to the Earth? (Get your answer using the formula). a) 1.5c b) 1.25c c) c d) 0.75c e) 0.5c A note of caution: Application: Lorentz transformation The way the Lorentz and Galileo transformations are presented v v ? here assumes the following: … An observer in S would like to express an event (x,y,z,t) (in A B A B his frame S) with the coordinates of the frame S', i.e. he t0 = 0 t1 = 1s wants to find the corresponding event (x',y',z',t') in S'. The frame S' is moving along the x-axes of the frame S with the Two clocks (one at A and one at B) are synchronized. A third clock flies past A at a velocity v. The moment it passes A all velocity v (measured relative to S) and we assume that the three clocks show the same time t0 = 0 (viewed by observers in origins of both frames overlap at the time t=0. A and B. See left image.) y y' What time does the third clock show (as seen by an observer (x,y,z,t) at B) at the moment it passes the clock in B? The clock at B is S (x',y',z',t') v showing t1 = 1s at that moment. à Use Lorentz transformation! S' 2 2 2 2 2 x x' A) γ · (t1-t0) B) γ (t1-t0)(1 – v/c ) C) γ (t1-t0)(1 + v /c ) 2 D) (t1-t0) / γ E) γ(t1-t0)(1 + vx'/c ) z z' 6 Hint: Use the following frames: Hint: Use the following systems: v v v v x' ? x' ? x … x … A B A B A B A B t0 = 0 t1 = 1s t0 = 0 t1 = 1s Two clocks (one at A and one at B) are synchronized.