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1 Relativity H6: Relativistic momentum and energy

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2 Two basic physical quantities need to be modified Two basic physical quantities need to be modified Length Length Time Time Other physical quantities will need to be modified as well Other physical quantities will need to be modified as well

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3 Relativistic momentum To properly describe the motion of particles within special relativity, Newtons laws of motion and the definitions of momentum and energy need to be generalized To properly describe the motion of particles within special relativity, Newtons laws of motion and the definitions of momentum and energy need to be generalized These generalized definitions reduce to the classical ones when the speed is much less than c These generalized definitions reduce to the classical ones when the speed is much less than c

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4 Relativisitc momentum To account for conservation of momentum in all inertial frames, the definition must be modified To account for conservation of momentum in all inertial frames, the definition must be modified v is the speed of the particle, m is its mass as measured by an observer at rest with respect to the mass v is the speed of the particle, m is its mass as measured by an observer at rest with respect to the mass When v << c, the denominator approaches 1 and so p approaches mv When v << c, the denominator approaches 1 and so p approaches mv

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5 Problem: particle decay An unstable particle at rest breaks up into two fragments of unequal mass. The mass of the lighter fragment is 2.50 × 10– 28 kg, and that of the heavier fragment is 1.67 × 10–27 kg. If the lighter fragment has a speed of 0.893c after the breakup, what is the speed of the heavier fragment? An unstable particle at rest breaks up into two fragments of unequal mass. The mass of the lighter fragment is 2.50 × 10– 28 kg, and that of the heavier fragment is 1.67 × 10–27 kg. If the lighter fragment has a speed of 0.893c after the breakup, what is the speed of the heavier fragment?

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6 An unstable particle at rest breaks up into two fragments of unequal mass. The mass of the lighter fragment is 2.50 × 10 –28 kg, and that of the heavier fragment is 1.67 × 10 –27 kg. If the lighter fragment has a speed of 0.893c after the breakup, what is the speed of the heavier fragment? Given: v 1 = 0.8 c m 1 =2.50×10 –28 kg m 2 =1.67×10 –27 kg Find: Find: v 2 = ? v 2 = ? Momentum must be conserved, so the momenta of the two fragments must add to zero. Thus, their magnitudes must be equal, or For the heavier fragment, which reduces to and yields

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7 Relativistic energy The definition of kinetic energy requires modification in relativistic mechanics The definition of kinetic energy requires modification in relativistic mechanics KE = mc 2 – mc 2 The term mc 2 is called the rest energy of the object and is independent of its speed The term mc 2 is called the rest energy of the object and is independent of its speed The term mc 2 is the total energy, E, of the object and depends on its speed and its rest energy The term mc 2 is the total energy, E, of the object and depends on its speed and its rest energy

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8 A particle has energy by virtue of its mass alone A particle has energy by virtue of its mass alone A stationary particle with zero kinetic energy has an energy proportional to its inertial mass A stationary particle with zero kinetic energy has an energy proportional to its inertial mass E = mc 2 E = mc 2 The mass of a particle may be completely convertible to energy and pure energy may be converted to particles The mass of a particle may be completely convertible to energy and pure energy may be converted to particles

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9 Energy and Relativistic Momentum It is useful to have an expression relating total energy, E, to the relativistic momentum, p It is useful to have an expression relating total energy, E, to the relativistic momentum, p E 2 = p 2 c 2 + (mc 2 ) 2 E 2 = p 2 c 2 + (mc 2 ) 2 When the particle is at rest, p = 0 and E = mc 2 When the particle is at rest, p = 0 and E = mc 2 Massless particles (m = 0) have E = pc Massless particles (m = 0) have E = pc This is also used to express masses in energy units This is also used to express masses in energy units mass of an electron = 9.11 x 10 -31 kg = 0.511 MeV mass of an electron = 9.11 x 10 -31 kg = 0.511 MeV Conversion: 1 u = 929.494 MeV/c 2 Conversion: 1 u = 929.494 MeV/c 2

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