Hidden Momentum in a Spinning Sphere

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1 Hidden Momentum in a Spinning Sphere 1 Problem Kirk T. MDonald Joseph Henry Laboratories, Prineton University, Prineton, NJ 8544 (August 16, 212; updated June 3, 217 A spinning sphere at rest has zero total momentum. Dedue the mass/energy and the momentum of the spinning sphere as observed in an inertial frame where the sphere has veloity v perpendiular to its axis of rotation, taking relativisti effets into aount. Give results aurate to order 1/ 2 where is the speed of light in vauum. Also dedue the loation of the enter of mass/energy of the moving, spinning sphere. Does the moving sphere have hidden momentum [1]? f P hidden P Mv m (x x m (p ρv b darea = (x x m dvol, (1 where P is the total momentum of the subsystem, M = U/ 2 is its total mass, U is its total energy, is the speed of light in vauum, x m is its enter of mass/energy, v m = dx m /dt, p is its momentum density, ρ = u/ 2 is its mass density, u is its energy density, v b is the veloity (field of its, and f μ = Tμν x, (2 ν is the 4-fore density exerted on the subsystem by the rest of the system, with T μν being the stress-energy-momentum 4-tensor of the subsystem. 1 2 Solution 2.1 Mass and Momentum In its rest frame the sphere has mass M and is not spinning. In the lab frame of the observer the geometri enter (entroid of the sphere has veloity v = v x. Taking the sphere to be entered on the origin in the frame in whih its enter is at rest and the sphere is spinning (the frame 2, a volume element (with rest mass dm about r =(x,y,z hasveloity u = ω r =( ωy,ωx,, (3 1 As disussed in se. 3 of [1], we do not advoate replaing v m by v entroid = v in definition (1, where the entroid is the geometri enter of the sphere in the lab frame. 2 The frame is obtained from the lab frame by a Lorentz transformation with veloity v. The rest frame of the sphere differs from the frame by a rotation about the ω-axis. 1

2 defining ω = ω ẑ to be the angular veloity of the sphere in the frame, where ωa is small ompared to the speed of sound inside the sphere. In the lab frame this element has veloity by u = v + u + u /γ 1+v u / 2 = (v ωy ˆx + ωx ŷ/γ 1 ωv y / 2, where γ = 1 1 v 2 / 2. (4 The relativisti mass dm (= energy/ 2 of the volume element in the lab frame is given dm = γ u dm = ( dm 1 u2 / dm 2 (1+ u2 dm v2 2ωv y + ω 2 (x 2 + y 2, (5 2 2 keeps terms only to order 1/ 2. Hene, ( M = dm M 1+ v2 2 + kω2 a 2, ( realling that the moment of inertia of the sphere in its rest frame is I = (x 2 +y 2 dm = km 2a2 with k =2/5. Note that the relativisti mass M in the lab frame does not equal γ M M (1 + v 2/22 as holds for a nonspinning sphere. 3 The x-omponent of the momentum of the sphere in the lab frame is P x = u x dm (v ωy (1+ ωv ( y dm 2 1+ v2 2ωv y + ω 2 (x 2 + y 2 2 ( 2 M v 1+ v2 2 + kω2 a 2 = Mv (7 That is, the lab-frame momentum is related by P = Mv = Mv m, (8 noting that the enter of mass/energy of the sphere has veloity v in the lab frame (as an be onfirmed using eq. (11 below. As suh, the momentum in the lab frame is not hidden. This last statement is onsistent with the definition (1 in that the of the (sub system an be taken outside the sphere, so that the integral vanishes and P hidden =. Also, for a system in isolation, suh as the present example, the 4-divergene of the stress tensor is zero, so that f = in partiular. Thus, P hidden = aording to the seond form of eq. (1 as well. 4 3 The mass M of the spinning sphere whose enter is at rest in the frame follows from eq. (6 with v =. 4 A basi onsequene of the definition (1 is that an isolated system an have no hidden momentum, so it should not be a surprise that P hidden = in the present example. 2

3 2.2 Loation of the Center of Mass/Energy (First Analysis The position x m of the sphere in the lab frame is given by ( Mx m = x dm (x ˆx + y ŷ + z ẑ dm 1+ v2 2ωv y + ω 2 (x 2 + y [ ( ] ( (x + v t 1 v2 ˆx + y ŷ + z ẑ dm v2 2ωv y + ω 2 (x 2 + y Mv t ˆx v kma 2 ω ŷ Mx 2 2 S v, (9 2 2 where x = v t = v t ˆx is the geometri enter of the moving sphere in the lab frame, S = I ω = km a 2 ω is the (spin angular momentum of the sphere about its geometri enter, and the moment of inertia I is I = [(x v t 2 + y 2 ] dm ( ] ( [x 2 1 v2 + y 2 dm 2 1+ v2 2ωv y + ω 2 (x 2 + y M a ( ω2 a 2 kma 2. ( To order 1/ 2, x m x ka2 ω v 2 2 x S v 2M 2. (11 The enter of mass/energy of the moving sphere is not at its geometri enter x, but is shifted to the side of the sphere that has the higher speed in the lab frame, as the relativisti mass is higher there. The spinning, translating sphere (or better, Lorentz-ontrated spheroid annot be desribed as a rigid body, in that the enter of mass/energy does not rotate with the matter of the spheroid. The above analysis did not take into aount that the rolling hoop has internal fores/stresses, In the lab frame these stresses are largest near the top of the hoop, where the speed is the greatest. Sine there is mass/energy assoiated with the internal stresses, it seems likely that the position of the enter of mass/energy is even higher above the entroid, x = vt, than indiated in eq. (11. We pursue this additional upward shift in the following setion. 3

4 2.3 A More General Argument This setion follows [2]. See also [3, 4, 5], se. 64 of [6] 5 and [7, 8, 9]. This topi has an extensive history in onsiderations of the quantum position operator. For a review, see [1]. The mehanial behavior of a marosopi subsystem an be desribed with the aid of its (symmetri stress-energy-momentum 4-tensor T μν.thequantity T P μ =(U/, P i μ =(U/, P = dvol. (12 desribes the total energy and momentum of the subsystem, although P μ is not truly a 4-vetor unless the subsystem is isolated. 6 The total mass/energy of the subsystem is U = T dvol, (13 and we define the effetive mass of the subsystem as M = U T = 2 dvol = 2 ρdvol, (14 where we define the effetive mass density of the subsystem to be ρ = T / 2. The enter of mass/energy of the subsystem is at position x μ m = 1 T x μ dvol, x m = 1 T x dvol. (15 U M 2 where x μ =(t, x, as haraterizing the oordinates of the enter of mass/energy of the subsystem. In general, the lab-frame quantity x μ m is not a 4-vetor, and it is not the Lorentz transformation x μ of the quantity x μ x μ m = 1 T x μ dvol, (16 U where the frame is the (instantaneous frame of the subsystem in whih its total 3- momentum is zero (but where the angular veloity is still ω, T =P i i = dvol. (17 We denote the lab-frame transform of the quantity x μ as x μ, whih we will all the entroid. 7 Note that x = t = x m. The veloity of the boost from the frame to the lab frame is 5 In thermodynamis a losed subsystem an have exhange of energy, but not matter, with other subsystem, whereas an isolated subsystem has no exhange of mass/energy. The term losed system in [2, 6] orresponds to the term isolated system of thermodynamis. 6 In ase of a nonisolated system, P μ of eq. (12 has been alled a false 4-vetor [11]. 7 The oordinates x μ are alled those of the proper enter of mass in [6]. 4

5 v m of eq. (35. Hene, the veloity v of the entroid is the same as the veloity v m of the enter of mass/energy, even though the position x is not neessarily that same as x m. As seen in se. 2.2 the differene between x m and x in the lab frame is related to the presene of angular momentum in the subsystem, so we introdue the quantity, x L μν μ T ν x ν T μ = dvol, (18 as the (antisymmetri angular momentum 4-tensor of the subsystem. Further, we introdue the spin angular momentum tensors, defined by S μν = L μν (x μ P ν x ν P μ, (19 and Sm μν = L μν (x μ mp ν x ν mp μ, (2 whih subtrat away the angular momentum assoiated with the energy/momentum of the entroid, and of the enter of mass/energy, respetively. For an antisymmetri 4-tensor A μν we onstrut two 3-vetors a and ã and aording to a =(a 23,a 31,a 12 and ã =(a 1,a 2,a 3. (21 Then, for either of the spin 4-tensors (19-(2 we an write S = L x P, S = L Mx + tp, and so x = 1 M where from eqs. (18, L = ( L S + tp, (22 x p dvol, and L = Mxm tp. (23 In partiular, from eqs. (19 and (22-(23 we obtain an expression for the lab-frame 3- position of the entroid, 8 x = xm S M. (24 This result was dedued in the lab frame, but it also holds in the frame, where x = x m, so it must be that S =. Then, sine P =,wehavethat S μν Pν =. (25 IF S μν and P μ are a 4-tensor and a 4-vetor, respetively, with respet to Lorentz transformations, then in the lab frame we have that 9 =S μν P ν, S P = forμ =, M S = S P, for μ =1, 2, 3, (27 8 Using eq. (2 rather than (19 leads only to S m =, whih also follows diretly from eq. (2. 9 In the frame, x μ = x μ m, so we also have that the spin tensors are the same in this frame, S μν = Sm μν, and in partiular S = S m =. SineP =,wehavethatsm μνp ν =. IfSμν m were a 4-tensor with respet to Lorentz transformations, then in the lab frame we would have that =SmP μν ν, S m P = forµ =, M S m = S m P, for µ =1, 2, 3. (26 This ontradits the fat that S m =,soweinferthatsm μν is not a tensor under Lorentz transformations. 5

6 and x = x m S P = x M 2 2 m S v. (28 M 2 as in eq. (11, but now the term in S is twie as large. 1, How General is the General Argument? The above argument may not hold for a subsystem that is not isolated from the rest of the Universe, as was taitly assumed above in eq. (27, and in writing P = Mv. As disussed in se. 3 of [1], we an dedue a relation (35 between P and Mv for a nonisolated subsystem. The reader may wish to skip ahead to that point. In detail, from eq. (14 we find and dm dt T = dvol + d dt (Mx m = dm dt x m + M dx m dt T = x dvol + T 2 (v b darea, (29 T 2 x (v b darea, (3 where x μ =(t, x, μ = / x μ =( / t,, and v b is the veloity (field of the. Hene, Mv m = M dx m T T = (x x m dvol + dt (x x m(v 2 b darea T = (x x m dvol + (x x m (ρv b darea. (31 While the stress-energy-momentum tensor for an isolated system has zero 4-divergene, this is not neessarily the ase for the subsystem under onsideration. Rather, the possibly nonzero 4-vetor, f μ = ν T νμ = ν T μν, (32 desribes the 4-fore density exerted by the subsystem on the rest of the system. If the stress-energy-momentum tensor is nonzero just inside the, the 4-fore density an have delta-funtion ontributions on the (as T μν is defined to be zero outside the, whih must be onsidered arefully in the following. Of ourse, f μ is zero outside the of the subsystem. Using eq. (32 we an write T = f j T j, (33 1 In [5] it is shown that S P = S m P = L P and that x = x m + S m P/M 2 2 (with the + sign miswritten as a. 11 The result (28 appears in eq. (8 of [12], with v taken to be the veloity of the observer relative to the sphere, i.e., v of this note. This result was also disussed around eq. (7 of [13], with the laim that x rather than x m is the true enter of mass. 6

7 where both f and j T j an have delta funtions on the. Then, noting that the momentum density p has omponents p j = T j /, the volume integral in eq. (31 an be written as T f (x x m dvol = (x x j T j m dvol (x x m dvol f = (x x j [T j (x x m ] T j j x m dvol dvol + dvol f = (x x T j (x x m m dvol darea j + p dvol f = (x x m dvol (x x m (p darea+p. (34 Combining eqs. (31 and (34, we have v m = P M 1 x m (p ρv b darea + M (x 1 f M (x x m dvol. (35 The integrals vanish for an isolated subsystem, in whih ase v m = P/M, as expeted. The integrals in eq. (35 an be zero even for subsystems that are not isolated. For example, if the subsystems oupy disjoint spatial volumes the integrals are zero. This holds for any reasonable hoie of subsystems of an all-mehanial system, as onsidered in [14]. The volume integral an be nonzero only if the subsystems are both fields, or matter and fields, whih oupy that same spatial volume. We now restrit our disussion to examples in whih the integrals are zero, so v m = P/M, and in addition the total mass and energy, M and U, of the subsystem are onstant in time. In this ase, the magnitude of the momentum P, andp μ P μ = M 2, are is also onstant in time. Furthermore we suppose that the angular momentum tensor L μν is onstant in time, as holds for eah of two subsystems that interat via a entral fore. We still fae the issue of whether relation (27 holds in the lab frame of a nonisolated subsystem. Referenes [1] K.T. MDonald, On the Definition of Hidden Momentum (July 9, 212, [2] C. Møller, On the Definition of the Centre of Gravity of an Arbitrary Closed System in the Theory of Relativity, Comm. Dublin Inst. Adv. Study A 5, 3 (1949, [3] A.D. Fokker, Relativiteitstheorie (Noordhoof, Groningen, 1929, Time and Spae, Weight and Inertia (Pergamon, 1965, 7

8 [4] A. Papapetrou, Drehimpuls- und Shwerpunktsatz in der relativistishen Mehanik, Praktika Aad. Athénes 14, 54 (1939, Drehimpuls- und Shwerpunktsatz in der Dirashen Theorie, PraktikaAad.Athénes 15, 54 (194, [5] M.H.L. Prye, The mass-entre in the restrited theory of relativity and its onnexion with the quantum theory of elementary partiles, Pro. Roy. So. London A 195, 62 (1948, [6] C. Møller, The Theory of Relativity, 1st ed. (Clarendon Press, Oxford, 1952, [7] H.C. Corben, Classial and Quantum Theories of Spinning Partiles (Holden-Day, San Franiso, 1968, Chap. II, [8] D. Alba, L. Lusanna and M. Pauri, Centers of mass and rotational kinematis for the relativisti N-body problem in the rest-frame instant form, J. Math. Phys. 43, 1677 (22, [9] D. Alba, H.W. Crater and L. Lusanna, Hamiltonian relativisti two-body problem: enter of mass and orbit reonstrution, J. Phys. A 4, 9585 (27, [1] M. Lorente and P. Roman, General expressions for the position and spin operators of relativisti systems, J. Math. Phys. 15, 7 (1974, [11] E. Comay, Lorentz transformation of a system arrying Hidden Momentum, Am. J. Phys. 68, 17 (2, [12] L. Filipe Costa et al., Mathisson s helial motions for a spinning partile: Are they unphysial? Phys. Rev. D 85, 241 (212, Seealsose.3of[2]. [13] S.E. Gralla, A.I. Harte, and R.M. Wald, Bobbing and kiks in eletromagnetism and gravity, Phys. Rev. D 81, 1412 (21, [14] K.T. MDonald, Hidden Momentum in a Linked Pair of Gyrostats (Aug. 17, 212, 8