2. The twin paradox H.4.1Describe how the concept of time dilation leads to the twin paradox. H.4.2Discuss the Hafele-Keating experiment. Option H: Relativity H4 Consequences of special relativity

3. The twin paradox Describe how time dilation leads to the paradox.  This has already been done in Topic H3. Option H: Relativity H4 Consequences of special relativity EXAMPLE: The twin paradox: Suppose Einstein has a twin brother who stays on Earth while Einstein travels at great speed in a spaceship. When he returns to Earth Einstein finds that his twin has aged more than himself! Explain why this is so. SOLUTION: Since Einstein is in the moving spaceship, his clock ticks more slowly. But his twin’s ticks at its Earth rate. The twin is thus older than Einstein on his return! By the way, this is NOT the paradox. The paradox is on the next slide…

4. EXAMPLE: The twin paradox: From Einstein’s perspective Einstein is standing still, but his twin is moving (with Earth) in the opposite direction. Thus Einstein’s twin should be the one to age more slowly. Why doesn’t he? SOLUTION: The “paradox” is resolved by general relativity (which is the relativity of non- inertial reference frames). It turns out that because Einstein’s spaceship is the reference frame that actually accelerates, his is the one that “ages” more slowly. The twin paradox Describe how time dilation leads to the paradox.  This has already been done in Topic H3. Option H: Relativity H4 Consequences of special relativity

5. The twin paradox Discuss the Hafele-Keating experiment.  From Wikipedia: “The Hafele–Keating experiment was a test of the theory of relativity. In 1971, Hafele, a physicist, and Keating, an astronomer, took four atomic clocks aboard commercial airliners.”  “They flew twice around the world, first eastward, then westward, and compared the clocks against others that remained at the United States Naval Observatory.”  “When reunited, the three sets of clocks were found to disagree with one another, and their differences were consistent with the predictions of special and general relativity.” Option H: Relativity H4 Consequences of special relativity This experiment obviously tested time dilation.

6. Velocity addition H.4.3Solve one-dimensional problems involving the relativistic addition of velocities. Option H: Relativity H4 Consequences of special relativity

7. Velocity addition Solve one-dimensional problems involving the relativistic addition of velocities.  Suppose you are in Car B, shown below.  Suppose car A is moving at u x = +20 m s -1 and your car B is moving at v = +40 m s -1 as shown.  As far as you are concerned, the velocity u x ’ of car A relative to you is -20 m s -1, because A is moving backwards relative to you. Option H: Relativity H4 Consequences of special relativity A B velocity of A relative to B u x ’ = u x - v FYI  This is the Galilean transformation which we learned way back in Topic 2.1.

8. Velocity addition Solve one-dimensional problems involving the relativistic addition of velocities.  Now consider two spaceships leaving Earth in opposite directions:  Suppose u x = 0.75c and v = 0.50c (to the left).  Then according to the Galilean transformation, the u x ’ (the velocity of A relative to B) would be u x ’ = u x – v = 0.80c - - 0.50c = 1.3c.  This is in contradiction to the tenet that c is the maximum speed anything can have. Option H: Relativity H4 Consequences of special relativity FYI  The Galilean transformation for addition of velocities fails at relativistic speeds.

9. Velocity addition Solve one-dimensional problems involving the relativistic addition of velocities.  At relativistic speeds we will use: Option H: Relativity H4 Consequences of special relativity relativistic velocity addition u x ’ = u x - v 1 - u v c 2 x Where u x and v are signed quantities. EXAMPLE: Spaceship B is travel- ing leftward at 0.50c wrt Earth. Spaceship A is traveling right- ward at 0.80c wrt Earth. What is the speed of A relative to B? SOLUTION: v = -0.50c, u x = +0.80c: ● u x ’ = (u x - v) / (1 - u x v/c 2 ) = (0.8c – - 0.5c) / (1 - - 0.5c  0.8c / c 2 ) = 1.3c / (1 + 0.4) = 0.93c.

10. Velocity addition Solve one-dimensional problems involving the relativistic addition of velocities. Option H: Relativity H4 Consequences of special relativity ●Use u x ’ = (u x – v)/(1 – u x v/c 2 ), where u x = 0.8c and v = -0.8c (since they are traveling in opposite directions. ●Then u x ’ = (0.8c – - 0.8c)/[1 – (0.8c)(-0.8c)/c 2 ]. u x ’ = 1.6c/[1 + 0.64]. u x ’ = 0.98c. FYI  Make sure you are consistent with your signs and that you get an answer that is LESS THAN c!

11. Velocity addition Solve one-dimensional problems involving the relativistic addition of velocities. Option H: Relativity H4 Consequences of special relativity ●A Galilean transformation assumes that time and length are absolute (the same in all reference frames) and fails under conditions of relativistic speeds. ●A Galilean transformation simply adds velocities in a straight-forward manner according to “intuition.” ●EXAMPLE: u x ’ = u x – v.

12. Velocity addition Solve one-dimensional problems involving the relativistic addition of velocities. Option H: Relativity H4 Consequences of special relativity where u x = 0.9800c and v = -0.9800c (since they are traveling in opposite directions). ●Use u x ’ = u x – v, ●Then u x ’ = 0.98c – - 0.98c = 1.960c.

13. Velocity addition Solve one-dimensional problems involving the relativistic addition of velocities. Option H: Relativity H4 Consequences of special relativity ●Use u x ’ = (u x – v)/(1 – u x v/c 2 ), where u x = 0.98c and v = -0.98c (since they are traveling in opposite directions. u x ’ = (0.98c – - 0.98c)/[1 – (0.98c)( - 0.98c)/c 2 ] = 1.96c/[1 + 0.98 2 ] = 0.9998c.

14. Velocity addition Solve one-dimensional problems involving the relativistic addition of velocities. Option H: Relativity H4 Consequences of special relativity ●In (b)(i) v > c which is not possible. ●Thus the Galilean transformation is not applicable to this problem.

15. Mass and energy H.4.4State the formula representing the equivalence of mass and energy. H.4.5Define rest mass. H.4.6Distinguish between the energy of a body at rest and its total energy when it is moving. H.4.7Explain why no object can ever attain the speed of light in a vacuum H.4.8Determine the total energy of an accelerated particle. Option H: Relativity H4 Consequences of special relativity

16. Mass and energy State the formula representing the equivalence of mass and energy. ●Recall E = mc 2. It has been slightly changed: ●We used this formula when we looked at mass defect in nuclear energy problems. It is the energy of a mass m 0 in its rest frame. Option H: Relativity H4 Consequences of special relativity Equivalence of mass and energy E 0 = m 0 c 2 PRACTICE: A nuclear power plant converts about 30 kg of matter into energy each year. How many joules is this? How many watts? SOLUTION: ●E 0 = m 0 c 2 = 30(3  10 8 ) 2 = 2.7  10 18 J. ●P = E 0 /t = 2.7  10 18 /[365  24  3600] = 8.6  10 10 W. Much of this energy is wasted in conversion to electrical power.

17. Mass and energy Define rest mass. ●It is beyond the scope of this course, but not only do time and length change with speed, but so does mass! ●We call m the relativistic mass. Recall that  is the Lorentz factor, and it is instrumental in solving special relativity problems. ●We call m 0 the rest mass. This is the mass of the object as measured in a reference frame in which it is at rest. Option H: Relativity H4 Consequences of special relativity relativistic mass m =  m 0 where  = 1 1 - v 2 /c 2 FYI  Note that as v  c that   . ●Thus as v  c we see that m  .

18. Mass and energy Define rest mass. Option H: Relativity H4 Consequences of special relativity relativistic mass m =  m 0 where  = 1 1 - v 2 /c 2 PRACTICE: At CERN a proton can be accelerated to a speed such that its relativistic mass is that of U 238. How fast is it going? SOLUTION: ●First, find the Lorentz factor: m =  m 0 238m p =  m p  = 238. ●Then solve for v: (1 – v 2 /c 2 ) 1/2 = 1/238 1 – v 2 /c 2 = 1/238 2 1 – 1/238 2 = v 2 /c 2 0.9999823 = v 2 /c 2  v = 0.9999912c. Make sure you can reproduce both of these graphs.

19. Mass and energy Distinguish between the energy of a body at rest and its total energy when it is moving. ●Since mass increases with velocity according to m =  m 0, clearly the total energy of a moving mass is E =  m 0 c 2. Option H: Relativity H4 Consequences of special relativity relativistic energy E 0 = m 0 c 2 where  = 1 1 - v 2 /c 2 E =  m 0 c 2 PRACTICE: Show that the relativistic energy E reduces to the rest energy E 0 when v = 0. SOLUTION: ●If v = 0 then  = 1/(1 - 0 2 /c 2 ) 1/2 = 1/1 1/2 = 1. ●Then E =  m 0 c 2 = 1m 0 c 2 = m 0 c 2 = E 0. FYI  Thus E =  m 0 c 2 is the total energy of the object, whether the object is moving or not.

20. Mass and energy Explain why no object can ever attain the speed of light in a vacuum ●Various arguments can be presented to show that no object with a non-zero rest mass can attain the speed of light. Option H: Relativity H4 Consequences of special relativity EXAMPLE: Argument 1: m   as v  c. ●If v = c then  = 1/(1 - c 2 /c 2 ) = 1/0 = . ●Since m =  m 0 then m   also. ●But there is not even an infinite amount of mass in the universe. (Reductio ad absurdum). Argument 2: E   as v  c. ●Since E =  m 0 c 2 then as    so does E. ●But there is not even an infinite amount of energy in the universe. (Reductio ad absurdum).

21. Mass and energy Determine the total energy of an accelerated particle. ●Recall that the acceleration of a charge e through a potential difference V produces a kinetic energy given by E K = eV. ●The total energy E of a particle of rest mass m 0 is just the sum of its rest energy E 0 and its kinetic energy E K = eV. Option H: Relativity H4 Consequences of special relativity total energy of an accelerated particle E = E 0 + E K where m =  m 0 mc 2 = m 0 c 2 + eV FYI  We can not use (1/2)mv 2 = eV at relativistic speeds to find v because it assumes all of the energy eV is going into the velocity. But the mass also changes at large speeds.

22. Mass and energy Determine the total energy of an accelerated particle. Option H: Relativity H4 Consequences of special relativity ●Use E =  m 0 c 2 =  E 0. Then 3E 0 =  E 0   = 3. ●Since  = 1/(1 – v 2 /c 2 ) 1/2 = 3, then (1 – v 2 /c 2 ) 1/2 = 1/3 1 – v 2 /c 2 = 1/9 v 2 /c 2 = 8/9 v = 0.94c.

23. Mass and energy Determine the total energy of an accelerated particle. Option H: Relativity H4 Consequences of special relativity ●The mass as measured by an observer at rest with respect to the mass. ●The mass as measured by an observer in the rest frame of the mass. ●From the formula we see that v   V. ●Thus if V is large enough, v > c, which cannot happen.

24. Mass and energy Determine the total energy of an accelerated particle. Option H: Relativity H4 Consequences of special relativity ●Use E = E 0 + E K.  mc 2 = eV  m = eV/c 2  m = e(5.0  10 6 V)/c 2  m = 5.0 MeV c -2  m = (1.6  10 -19 )(5  10 6 )/(3  10 8 ) 2  m = 8.9  10 -30 kg. ●Alternate method… ●Then mc 2 = m 0 c 2 + eV. mc 2 - m 0 c 2 = eV.

25. Mass and energy Determine the total energy of an accelerated particle. Option H: Relativity H4 Consequences of special relativity ●Use E = E 0 + eV. Then mc 2 = m 0 c 2 + eV.  m 0 c 2 = m 0 c 2 + eV.  = 1 + eV/(m 0 c 2 )

26. Mass and energy Determine the total energy of an accelerated particle. Option H: Relativity H4 Consequences of special relativity ●Use  = 1 + eV/(m 0 c 2 ).  = 1 + e(500 MV)/(938 MeVc -2 c 2 )  = 1 + 500 MeV/(938 MeV)  = 1 + 500/938 = 1.53  = 1/(1 – v 2 /c 2 ) -1/2 = 1.53 1 – v 2 /c 2 = 1/1.53 2 = 0.427 1 - 0.427 = v 2 /c 2 v 2 = 0.573c 2 v = 0.76c.

27. Mass and energy Determine the total energy of an accelerated particle. Option H: Relativity H4 Consequences of special relativity ●As v  c,   . ●Since E =  m 0 c 2 we see that ●As   , E  . ●Since there is not an infinite amount of energy in the universe, you cannot accelerate an object with a rest mass to the speed of light.

28. Mass and energy Determine the total energy of an accelerated particle. Option H: Relativity H4 Consequences of special relativity E = E 0 + E K = 0.51 MeV + 6.00 MeV = 6.51 MeV. E K = eV = e(6.00  10 6 ) V = 6.00 MeV. (For an electron, E 0 = m 0 c 2 = 0.51 MeV.)

29. Mass and energy Determine the total energy of an accelerated particle. Option H: Relativity H4 Consequences of special relativity ●E =  m 0 c 2 =  (0.51 MeV) = 6.51 MeV ●  = 6.51 MeV / 0.51 MeV = 12.17.  = 1/(1 – v 2 /c 2 ) -1/2 = 12.17 1 – v 2 /c 2 = 1/12.17 2 = 1/12.17 2 1 - 0.0067 = v 2 /c 2 v 2 = 0.9933c 2 v = 0.997c.

30. Mass and energy Determine the total energy of an accelerated particle. Option H: Relativity H4 Consequences of special relativity ●Rest mass energy is E 0 = m 0 c 2 and is the energy that a particle has in its rest frame. ●Total energy is E = m 0 c 2 + E K and is the sum of the rest mass energy and the kinetic energy E K = eV. ●For these problems we always assume there is no potential energy.

31. Mass and energy Determine the total energy of an accelerated particle. Option H: Relativity H4 Consequences of special relativity ●E 0 = m 0 c 2 = (938 MeV c -2 )c 2 = 938 MeV.

32. Mass and energy Determine the total energy of an accelerated particle. Option H: Relativity H4 Consequences of special relativity ●  = 1/(1 – v 2 /c 2 ) -1/2 = 1/(1 – 0.980 2 c 2 /c 2 ) -1/2 = 5.03. ●E =  m 0 c 2 = m 0 c 2 + eV eV = (  - 1)m 0 c 2 V = (  - 1)m 0 c 2 /e V = (5.03 - 1)(938 MeV)/e V = 3780 MeV.