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In terms of relativistic momentum, the relativistic total energy can be expressed as followed

Conservation of energy-momentum. In terms of relativistic momentum, the relativistic total energy can be expressed as followed. Recap . Reduction of relativistic kinetic energy in the classical limit. The expression of the relativistic kinetic energy.

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In terms of relativistic momentum, the relativistic total energy can be expressed as followed

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

  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.

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

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

  5. Due to mass-energy equivalence, sometimes we express the mass of an object in unit of energy Electron has rest mass m0 = 9.1 x 10-31kg The rest mass of the electron can be expressed as energy equivalent, via m0 c2 = 9.1 x 10-31kg x (3 x 108m/s)2 = 8.19 x 10-14 J = 8.19 x 10-14 x (1.6x10-19)-1 eV = 511.88 x 103 eV = 0.511 MeV

  6. Solution • First, find the Lorentz factor, g = 1.89 • The rest mass of electron, m0c2, is 0.5 MeV • Hence the total energy is E = mc2 = g (m0c2)= 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 -m0c2 = (0.97 – 0.51)MeV = 0.46 MeV

  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 = mc2 = K + E0 • 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 = mc2 = K + U + E0 • 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 m0 will be different by an amount • Dmc2 = (m - m0)c2 = K + U • (in previous lecture we have temporarily ignore the role of potential energy for the sake of simplicity)

  8. Consider an uncompressed spring system with rest mass m0(We shall ignore the kinetic energy as we consider only rest spring) Compression caused by external force, Fext Example: a compressed spring Uncompressed, Rest mass = m0, total relativistic energy E = E0 =m0c2 (rest energy only) Now, the spring is compressed by Dx 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, Fext Dx = U. U will add to the total relativistic energy of the spring system: E = m0c2 --–> E = m0c2 + U

  9. As the total energy, E, increases due to external U, the relativistic mass of the spring will increase by some small amount Dm due to mass-energy equivalence the increase in the mass is simply Dmc2 = U E.g. if the spring constant k = 100N/m, and is compressed by 10 cm, potential energy stored = U = kx2/2 = 0.5 J This will contribute to the total relativistic mass of the spring by a amount Dm = U/c2 = 5.56 x 10-18 kg – what a tiny amount

  10. proton, mp Neutron, mn U Deuterium, md Binding energy • The nucleus of a deuterium comprises of one neutron and one proton. Both nucleons are bounded within the deuterium nucleus Initially, the total Energy = (mn+ mn)c2 Nuclear fusion After fusion, the total energy = mdc2 + U Analogous to exothermic process in chemistry

  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

  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 mn + mp > md • By conservation of energy-momentum, • E(before) = E(after) • mnc2 + mpc2 + 0 = mdc2 + U • Hence, U = (mp + mn)c2 - mdc2 = Dmc2 • The difference in mass between deuterium and the sum of (mn + mn)c2 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)

  13. Example • mn= 1.008665u; mp= 1.007276u; • md= 2.013553u; • u = standard atomic unit = mass of 1/12 of the mass of a 12C nucleus = 1.66 x 10-27kg = 1.66 x 10-27 x c2 J = 1.494 x 10-10 J = 1.494 x 10-10/(1.6x10-19) eV = 933.75 x 106 eV = 933.75 x MeV • Hence the binding energy • U = Dmc2 = (mp + mn)c2 - mdc2 • =0.002388u = 2.23 MeV

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

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