Relativistic momentum and kinetic energy, and E =
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IOP PUBLISHING Eur. J. Phys. 30 (2009) 325–330
EUROPEAN JOURNAL OF PHYSICS
Relativistic momentum and kinetic energy, and E = mc2
Ben Yu-Kuang Hu
Department of Physics, University of Akron, Akron, OH 44325-4001, USA E-mail: email@example.com
Received 12 May 2008, in ?nal form 24 November 2008 Published 9 February 2009 Online at stacks.iop.org/EJP/30/325
Based on relativistic velocity addition and the conservation of momentum and energy, I present simple derivations of the expressions for the relativistic momentum and kinetic energy of a particle, and for the formula E = mc2 . (Some ?gures in this article are in colour only in the electronic version)
The standard formal way that upper-level introductory undergraduate textbooks  obtain the expressions for the relativistic momentum p, the relativistic kinetic energy T and the mass– energy relationship E = mc2 is by ?rst introducing Lorentz transformations and 4-vectors, and then de?ning the 4-momentum vector pμ = m dx μ /dτ (μ = 0, 1, 2, 3), where τ is the proper time. The temporal and spatial components of pμ reduce in the non-relativistic limit to mc2 + 1 mv 2 and mv, respectively, and are therefore postulated to be the relativistic 2 generalizations of the total energy and momentum. It is then asserted that in an isolated system, all the components of pμ are conserved. In this approach, the student is ?rst confronted with an unfamiliar quantity, the 4-vector, followed by an assertion of the conservation of its components which must be taken ‘on faith’, since there is no way a priori to justify it. This approach is pedagogically rather unsatisfactory. As a result, there have been many papers describing derivations of the relativistic expressions that are based on more physical grounds . This paper describes new simple and concise derivations of the relativistic forms of p and T, and E = mc2 , all based on (i) the conservation of momentum and energy in the collisions of two particles and (ii) the velocity addition rules. Momentum and energy conservation should be concepts familiar to students, and the velocity addition rules can be quite simply derived from the constancy of the speed of light in all inertial reference frames . To simplify the algebra, velocities in this paper are expressed in units of c, the speed of light. Hence velocities are dimensionless, and c = 1. To obtain the standard dimensional expressions, replace all velocities in the expressions given here by v → v/c and multiply all
0143-0807/09/020325+06$30.00 c 2009 IOP Publishing Ltd Printed in the UK 325
B Y-K Hu
masses by c2 in order to obtain energy. Also, in this paper primes on variables denote ‘after collision’.
2. The derivations
In all three derivations, collisions are analysed in the centre-of-momentum frame of reference, Scm , in which both particles have momenta that are equal in magnitude and opposite in direction, and in the laboratory frame of reference, Slab , in which one of the particles is initially at rest. It is self-evident that the appropriate conservation laws are obeyed in Scm . Imposition of conservation laws in Slab gives the desired expressions.
2.1. Relativistic velocity transformations
? Let us recall the relativistic velocity transformation rules. Let S be an inertial frame moving with velocity (u, 0) with respect to frame S. If a particle has velocity (vx , vy ) in frame S, the ? components of its velocity in frame S are [1, 3] vx ? u , 1 ? vx u √ vy 1 ? u2 . vy = ? 1 ? vx u vx = ?
2.2. Relativistic momentum
From dimensional analysis and the vector1 nature of momentum, the momentum of a particle of mass m travelling with velocity v must have the form p = mγ (v)v, (2) where γ (v) is a function of v ≡ |v| that is to be determined. Since p = mv for non-relativistic velocities, γ (0) = 1. Consider the case where the particles are identical; hence, m1 = m2 = m. Let the motion of the particles be in the x–y plane and their initial velocities in Scm be ±(v, 0). Assume that the particles barely graze each other, so that in the collision each particle picks up a very small y-component of the velocity of magnitude δv in Scm . (See ?gure 1(a).) Their speeds in Scm do not change because the collision is elastic, and hence their velocities after the collision are ±( v 2 ? (δv)2 , δv) ≈ ±(v, δv), to ?rst order in δv. Because δv is assumed to be very small, we ignore all terms of order (δv)2 and higher. Now consider the collision in the laboratory frame of reference Slab that moves with velocity (?v, 0) with respect to Scm . (See ?gure 1(b).) The pre-collision velocities of the particles in Slab , using (1a) and (1b) on the Scm velocities ±(v, 0), are v1,lab = (w, 0), where 2v w= , (3) 1 + v2 and v2,lab = (0, 0). After the collision, transforming the post-collision Scm velocities ±(v, δv) to the Slab frame we obtain, to the ?rst order in δv, √ δv 1 ? v 2 v1,lab ≈ w, , (4a) 1 + v2
Here, the terms ‘vector’ and ‘scalar’ are used in the non-relativistic (i.e. not the 4-vector) sense.
Relativistic momentum and kinetic energy, and E = mc2
Figure 1. Grazing collision between two particles of equal mass in (a) centre of momentum and (b) laboratory frames of reference. Dashed and solid lines indicate before and after the collision, respectively.
v2,lab ≈ 0, ?
√ δv 1 ? v 2 1 ? v2
The y-component of the total momentum before the collision is zero, and hence by conservation of momentum, after the collision (p1,lab + p2,lab )y = mγ (v1,lab )v1,lab,y + mγ (v2,lab )v2,lab,y = 0. To the ?rst order in δv, (5) together with (4a) and (4b) gives γ (w) γ (0) δv = 0, (6) ? 2 1+v 1 ? v2 which implies that the term in the square parentheses vanishes. This together with (3) and γ (0) = 1 (the non-relativistic limit) gives the desired result, 1 + v2 2v γ (w) = = 1? 1 ? v2 1 + v2
2.3. Relativistic kinetic energy
= (1 ? w 2 )?1/2 .
Dimensional analysis and the scalar2 property of kinetic energy imply that its form is T = mG(v), (8) where m is the mass of the particle, v = |v| is its speed and the function G(v) is to be determined. Since the kinetic energy of a stationary object vanishes, G(0) = 0. Consider an elastic collision between two particles of mass m and M m, with speeds in Scm of v and V respectively, in which the outgoing and incoming velocity vectors are perpendicular. (See ?gure 2(a).) Assume that the mass M is so large that in frame Scm its speed V 1, and hence we can use the non-relativistic expressions for the momentum and kinetic energy of mass M. The magnitudes of the momenta of m and M are equal in Scm , implying mγ (v)v = MV .
See footnote 1.
B Y-K Hu
Figure 2. Collision between particles of mass m and M
m in (a) centre of momentum and (b) laboratory frames of reference. Dashed and solid lines indicate before and after the collision, respectively.
The Scm frame pre- and post-collision velocities of mass m are vcm = (v, 0) and vcm = (0, v), respectively, and of mass M are Vcm = (?V , 0) and Vcm = (0, ?V ), respectively. Transforming these to the Slab frame which moves at velocity (?V , 0) with respect to Scm (see ?gure 2(b)) using (1a) and (1b) gives vlab = ((v + V )/(1 + vV ), 0), vlab = √ √ (V , v 1 ? V 2 ), Vlab = (0, 0) and Vlab = (V , ?V 1 ? V 2 ). By conservation of kinetic energy in an elastic collision in the Slab frame and (8), mG(vlab ) = mG(vlab ) + Expanding vlab , vlab and Vlab M 2 V . 2 lab to the ?rst order in V gives (10)
vlab ≈ v + V (1 ? v 2 ), vlab ≈ v, √ Vlab ≈ 2V .
(11a) (11b) (11c)
Substituting these into (10) and Taylor expanding the G(vlab ) term on the left-hand side about v give, to the ?rst order in V ,3 m G(v) + dG(u) du V (1 ? v 2 ) = mG(v) + MV 2 .
Substituting MV 2 = mγ (v)vV (from (9)) into (12) leads to dG du =
γ (v)v v = , 2 1?v (1 ? v 2 )3/2 1 (1 ? u2 )1/2
which upon integration yields G(v) ? G(0) = = γ (v) ? 1.
Since G(0) = 0, this yields (from (8) and reintroducing c) T = m(γ (v) ? 1)c2 .
2.4. E = mc2
Consider the initial situation as in section 2.3, except that the speed V of mass M can be relativistic, and after collision the two particles merge into one composite particle. In Scm , Mγ (V )V = mγ (v)v, and after the collision the composite particle is stationary. In Slab
The term 2MV 2 in (12) is actually the ?rst order in V , because M is of order V ?1 (see (9)).
Relativistic momentum and kinetic energy, and E = mc2
which moves with velocity (?V , 0) with respect to Scm , before the collision particle M is v+V stationary and particle m moves with velocity vlab = 1+vV , 0 , and after the collision the composite particle moves with velocity (V , 0). The magnitude of the total momentum in Slab before the collision is Plab = mγ (vlab )vlab = mγ (v)γ (V )(v +V ). If the mass of the composite particle does not change, then the momentum of the composite particle after the collision in Slab would be (M + m)γ (V )V = Plab in general, violating conservation of momentum. Therefore, the mass of the composite particle must change by m such that momentum is conserved in Slab , i.e. mγ (v)γ (V )(v + V ) = (M + m + m)γ (V )V . (15) Using mγ (v)v = Mγ (V )V to eliminate v on the left-hand side of (15) and cancelling γ (V )V on both sides give m = m(γ (v) ? 1) + M(γ (V ) ? 1). (16) From section 2.3, the right-hand side of (16) is equal to ? T , the total change in kinetic energy in Scm (since particles m and M start with speeds v and V , respectively, and both are stationary at the end). By conservation of total energy, E + T = 0, where E is the change of energy associated with the change in mass. Hence, E = ? T = m or (reintroducing c and making the plausible assumption that a zero mass object with zero velocity has zero energy4 ) E = mc2 . Finally, combining the results of sections 2.3 and 2.4 gives the total energy of a particle of mass m moving with speed v, E + T = Etotal = mγ (v)c2 .
3. Concluding remarks
It should be noted that these derivations do not guarantee that the momentum and total energy are conserved in all inertial reference frames or in all collisions. They only show the forms that the momentum, kinetic energy and energy–mass relation must have, given momentum and energy conservation. Once these expressions are known, when the 4-momentum is introduced its components will be recognized as the total energy and momentum. The covariance of the momentum 4-vector can then be used to demonstrate the momentum and total energy conservation in all inertial frames. The conservation of momentum can be shown to be a consequence of conservation of energy5 , and, as be?tting an experimental science, the conservation of energy ultimately depends on experimental observations.
 See, e.g., Grif?ths D J 1999 Introduction to Electrodynamics 3rd edn (Upper Saddle River, NJ: Prentice-Hall) Taylor J R 2005 Classical Mechanics (Sausalito, CA: University Science Books)  See, e.g., Lewis G L and Tolman R C 1909 The principle of relativity, and non-Newtonian mechanics Phil. Mag. 18 510–23 Penrose R and Rindler W 1965 Energy conservation as a basis of relativistic mechanics Am. J. Phys. 33 55–9 Ehlers J, Rindler W and Penrose R 1965 Energy conservation as a basis of relativistic mechanics: II Am. J. Phys. 33 995–7 Baird L C 1980 Relativistic mass Am. J. Phys. 48 779 Gupta P D 1981 Relativistic mass Am. J. Phys. 49 890 Peters P C 1986 An alternate derivation of relativistic momentum Am. J. Phys. 54 804–8 Fegenbaum M J and Mermin N D 1988 E = mc2 Am. J. Phys. 56 18–21
4 Special relativity by itself does not specify that zero mass gives zero energy. See Feigenbaum M J and Mermin N D in . To identify zero energy with zero mass, one needs general relativity. See, e.g., . 5 See Penrose R and Rindler W, Ehlers J, Rindler W and Penrose R, Siman Y and Husson N, and Feigenbaum M J and Mermin N D in . Note that these derivations explicitly or implicitly assume isotropy of space and translational invariance.
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