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1 Conservation of energy-momentum In terms of relativistic momentum, the relativistic total energy can be expressed as followed Recap

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2 Reduction of relativistic kinetic energy in the classical limit The expression of the relativistic kinetic energy must reduce to that of classical one in the limit u –> 0 when compared with c, i.e.

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3 Expand with binomial expansion For u << c, we can always expand in terms of (u/c) 2 as i.e., the relativistic kinetic energy reduces to classical expression in the u << c limit

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4 Example An electron moves with speed u = 0.85c. Find its total energy and kinetic energy in eV. CERNs picture: the circular accelerator accelerates electron almost the speed of light

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5 Due to mass-energy equivalence, sometimes we express the mass of an object in unit of energy Electron has rest mass m 0 = 9.1 x kg The rest mass of the electron can be expressed as energy equivalent, via m 0 c 2 = 9.1 x kg x (3 x 10 8 m/s) 2 = 8.19 x J = 8.19 x x (1.6x ) -1 eV = x 10 3 eV = MeV

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6 Solution First, find the Lorentz factor, = 1.89 The rest mass of electron, m 0 c 2, is 0.5 MeV Hence the total energy is E = mc 2 = m 0 c 2 )= 1.89 x 0.5 MeV = 0.97 MeV Kinetic energy is the difference between the total relativistic energy and the rest mass, K = E - m 0 c 2 = (0.97 – 0.51)MeV = 0.46 MeV

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7 Mass-energy equivalence: potential energy In some special case, a system has no potential energy nor kinetic energy, e.g. two nucleons at rest and separated far apart According to mass-energy equivalence E = mc 2 = K + E 0 However, in general, U and K for a system are not zero, e.g. two nucleons fused into one nucleus – potential energy will come into the play in such a nucleus and cannot be ignored In fact, not only does kinetic energy contribute to the relativistic mass, m, to the system, but potential energy too Hence, more generally, E = mc 2 = K + U + E 0 Hence, generally the mass of a system, m, would have a contribution not only from its kinetic energy but also from potential energy Its relativistic mass m and the rest mass m 0 will be different by an amount mc 2 = (m - m 0 )c 2 = K + U (in previous lecture we have temporarily ignore the role of potential energy for the sake of simplicity)

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8 Example: a compressed spring Consider an uncompressed spring system with rest mass m 0 (We shall ignore the kinetic energy as we consider only rest spring) Uncompressed, Rest mass = m 0, total relativistic energy E = E 0 =m 0 c 2 (rest energy only) Compression caused by external force, F ext Now, the spring is compressed by x by some external force The work done by the external force will be converted into the potential energy of the spring, according to conservation of mechanical energy, F ext x = U. U will add to the total relativistic energy of the spring system: E = m 0 c 2 --–> E = m 0 c 2 + U

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9 E.g. if the spring constant k = 100N/m, and is compressed by 10 cm, potential energy stored = U = kx 2 /2 = 0.5 J This will contribute to the total relativistic mass of the spring by a amount m = U/c 2 = 5.56 x kg – what a tiny amount As the total energy, E, increases due to external U, the relativistic mass of the spring will increase by some small amount m due to mass-energy equivalence the increase in the mass is simply mc 2 = U

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10 Binding energy The nucleus of a deuterium comprises of one neutron and one proton. Both nucleons are bounded within the deuterium nucleus Neutron, m n proton, m p Deuterium, m d Nuclear fusion Initially, the total Energy = (m n + m n )c 2 After fusion, the total energy = m d c 2 + U U Analogous to exothermic process in chemistry

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11 U is the energy that will be released when a proton and a neutron is fused in a nuclear reaction. The same amount of energy is required if we want to separate the proton from the neutron in a deuterium nucleus U is called the binding energy

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12 U can be explained in terms energy-mass equivalence relation, as followed For the following argument, we will ignore KE for simplicity sake Experimentally, we finds that m n + m p > m d By conservation of energy-momentum, E(before) = E(after) m n c 2 + m p c = m d c 2 + U Hence, U = (m p + m n )c 2 - m d c 2 = mc 2 The difference in mass between deuterium and the sum of (m n + m n )c 2 is converted into the binding energy that binds the proton to the neutron together (opposite to the case of a compressed spring – in deuterium its mass decreases)

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13 Example m n = u; m p = u; m d = u; u = standard atomic unit = mass of 1/12 of the mass of a 12 C nucleus = 1.66 x kg = 1.66 x x c 2 J = x J = x /(1.6x ) eV = x 10 6 eV = x MeV Hence the binding energy U = mc 2 = (m p + m n )c 2 - m d c 2 = u = 2.23 MeV

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14 SR finishes here… We will go to the next topic … Mainly on the quantum picture of light and matter

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