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Relativity. Lesson 3 Velocity addition Rest mass and relativistic mass Use of E = mc 2 as total energy Acceleration of electrons by a p.d. Use of MeVc -2 and MeVc -1. Galilean Velocity addition. S. v. train stationary. ground stationary. Easy!. What if approaches c ?.
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Relativity Lesson 3 Velocity addition Rest mass and relativistic mass Use of E = mc2 as total energy Acceleration of electrons by a p.d. Use of MeVc-2 and MeVc-1 IB Physics – Relativity
Galilean Velocity addition S v train stationary ground stationary Easy! IB Physics – Relativity
What if approaches c ? ….and my special theory of relativity does not allow this u would be bigger than c IB Physics – Relativity
Relativistic velocity formulae Just a small correction needed prove the inverse formulae from this; Notice the similarity to Galileo's formulae with a small correction IB Physics – Relativity
Now try this rocket ground • An electron has a speed of 2.00 x 108 ms-1 relative to a rocket, which itself moves at a speed of 1.00 x 108 ms-1 with respect to the ground. What is the speed of the electron with respect to the ground. Answer; u = 2.45 x 108 ms-1 Tsokos, 2005, p567 The trick is to clearly sort out your reference frames. IB Physics – Relativity
And this; a bit harder 0.8c 0.9c A B 2. Two rockets move away from each other with speeds of of 0.9c to right and 0.8c to the left with respect to the ground. What is the relative speed of each rocket as measured from the other. Answer ± 0.988c Tsokos, 2005, p567 Again the trick is to be very careful with your frames of reference IB Physics – Relativity
Learn this solution rocket A ground rocket B u is then the velocity of A with respect to B. Write down then check the velocity of B with respect to A IB Physics – Relativity
Apply the same approach for the question in Kirk on p147 By now you should have realised that at low velocities the relativistic formulae approximate to the Galilean equations. Check this out. Show also that the relativistic formulae do not allow objects to have velocities greater than c ............. but F = ma allows velocities > c so we might have to look at mass again!!! IB Physics – Relativity
Galileo and Newton allow v > c v constant acceleration c decreasing acceleration as speed approaches c but I don’t allow this t IB Physics – Relativity
Inertial mass Defined as m = F/a For a constant force Einstein predicted that as v approaches c the acceleration will decrease to zero. This ensures that speed can never exceed the speed of light This means the inertial mass must be approaching infinite. IB Physics – Relativity
Rest Mass m/m0 The rest mass, m0, of an object is the mass as measured in a frame where the object is at rest If mass is measured in a frame moving with respect to the object; m = m0 IB Physics – Relativity
Try these • Find the speed of a particle whose relativistic mass is double its rest mass. Answer = 0.866c • A constant force, F, is applied to a particle at rest. Find the acceleration in terms of the rest mass, m0. Find the acceleration again if the same force is applied to the same particle when it moves at a speed 0.8c. • Find the rest energy of an electron and its total energy when it moves at a speed equal to 0.8c. Give values in MeV. Tsokos, 2005, p571 IB Physics – Relativity
Mass - Energy The energy needed to create a particle at rest is called the rest energy given by; E0 = m0c2 If this particle accelerates it gains kinetic energy and the mass also will increase so total energy; E = mc2 or E = m0c2 IB Physics – Relativity
Proper energy; mc2 IB Physics – Relativity
Kinetic energy (T) and momentum (p) The formulae; Ek = ½ mv2 and p = mv no longer work at high speeds. Total energy of a particle; E = Ek + m0c2 So total energy is the kinetic energy plus the rest mass energy IB Physics – Relativity
Now try this ***??** An electron is accelerated through a p.d. of 1.0 x 106V. Calculate its velocity; (use the data book to find the electron rest mass and charge). Answer = 0.94c Hint; First calculate the energy gained then find the rest mass energy. This will give you a value for . Kirk, 2003, p148 IB Physics – Relativity
Relativistic formulae At high speeds you must use the following formulae; p = m0v for momentum E = Ek + m0c2 for kinetic energy E2 = p2c2 + m02c4 for total energy E = m0c2 the derivations of these are not required. IB Physics – Relativity
Useful units The rest mass of an electron = 9.11 x 10-31 Kg This is equivalent to energy E0 = m0c2 = 9.11 x 10-31 x 9 x 1016 = 8.199 x 10-14 J This is more conveniently quoted in MeV. so E0 = 8.199 x 10-14 / 1.6 x 10-19 = 0.512 MeV Working backwards we know E0 /c2 = m0 So mass can be measured in MeVc-2 electron rest mass = 0.512 MeVc-2 Similarly KeVc-2 and GeVc-2 etc. may be used IB Physics – Relativity
and problems are much easier to solve. Similarly MeVc-1 may be used for momentum IB Physics – Relativity
So try these using the new units • Find the momentum of a pion (rest mass 135 MeVc-2) whose speed is 0.80c. Answer 180 MeVc-1 • Find the speed of a muon (rest mass = 105 MeVc-2) whose momentum is 228 MeVc-1. Answer = 0.91c • Find the kinetic energy of an electron (rest mass 0.511 MeVc-2) whose momentum is 1.5MeVc-1. Answer = 1.07 MeV Tsokos, 2005, p581 Work through the example in Kirk on p.150 IB Physics – Relativity