Announcement Next Quiz on Wednesday 7 - PowerPoint Presentation

1. AnnouncementNext Quiz on Wednesday 7th September.Only three lectures before Midsem.Shall start Quantum Ideas after Midsem.

2. RecapitulateWe need a new definition of momentum.In the process we introduced the concept of Four Vectors.

3. Minkowski Space and Four Vectors

4. Dot Product of two four vectorsThis dot product is a four scalar implying that its value does not change when we change frame from S to S’.

5. Example of a four vectorx, y, z and ict are components of a four vector.

6. Magnitude of DisplacementThe dot product referring to two events is given by

7. Proper Time IntervalThe Proper time interval Δτ is defined as follows, which is a four scalar.

8. Proper Time IntervalIs consistent with earlier definition of proper time interval.Is imaginary for space like intervals, where it is not possible to find a frame in which the events occur at the same position.Is more general definition of proper time interval.

9. Example 6: Proper Time IntervalConsider two events in S.E1: x1=150 m, t1=0.3 s E2: x2=210 m, t2=0.4 s Δx = 60mc Δt=3x108x.1x10-6 =30 mSpace Like

10. Proper Time Interval

11. Same Events in S’Let v=0.6c, so γ= 1.25

12. Proper Time Interval

13. Example 7: Vertical FallAn object is moving in S frame in negative y direction with a constant velocity and travels a fixed tower of height 288 m in 1.2  s. Find the proper time interval for crossing the tower. Verify the same for a frame S’ which is moving in +x direction with a speed 0.6c.

14. Vertical FallSvS’H

15. In SE1: x1=0, y1= 288 m, t1=0E2: x2=0, y2=0, t2= 1.2 μs

16. In S’

17. Proper Time Evaluated in S’

18. Question?We note . Is time dilation applicable here? Is any of the time interval truly proper?Let us go to object frame. Assume that the object is in a gravity free space and moves with a constant velocity.

19. Transformation Equations in Object Frame

20. Time interval in Object FrameΔx=0, Δy=288 m, Δt=1.2 μs

21. Using Time Dilation FormulaObjet Frame to SThis is as expected.Objet Frame to S’. Find relative velocity

22. Velocity of Object in S’ Frame

23.

24. Using time dilation formulaThus we get the expected result for S’ frame

25. SvS’0.72 μs1.5 μs1.2 μsv=0.8c γ=5/3v=0.6c γ=1.25v=2.63x108 γ=1/0.48

26. Velocity Four VectorImagine two events are related to the displacement of a particle.

27. Calculation of Proper Time in S

28. Velocity Four Vector in SThe Four Components are thus given by

29. To obtain velocity transformation

30. Expanding

31. Fourth Equation

32. Substitute in First Equation

33. Substitute in Second Equation

34. The Last Equation

35.

36. Length square (velocity Four vector)

37. Example 8: Velocity Four VectorTake the numbers from the Example 7. The object velocity components are as follows.In S:ux=0, uy=0.8cIn S’ moving with v=0.6c:ux=-0.6c, uy=0.8c/γ. γ=1.25

38. SvS’uy=0.8c γu=5/3v=0.6c γ=1.25ux=-0.6c,uy=0.8c/1.25 γu’=1/0.48

39. Components in S

40. Components in S’

41. The Transformation Equation

42. Verification

43. Momentum Four VectorWe need that irrespective of v, β or γ the conservation of momentum should be valid. What is the fourth component?

44. Rest or Proper Mass, A four scalarWe define a proper or rest mass as mo. Then following would be a valid four vector components with dimension of momentum

45. New SymbolsThus the momentum four vector can be written as follows.

46. Fourth ComponentTaking clues from classical mechanics and our discussion so far, let us define force vector and four vector as follows.

47. Using Force Four Vector

48. Force Velocity Dot ProductLet us obtain a four scalar by taking dot product of force and velocity Four Vectors.

49. Time Derivative of m

50. A frame in which the instantaneous velocity is zeroIn all frames E = mc2