Relativistic rotation dynamics formalism and examples

Transcription

1 epl draft Relativistic rotation dynamics formalism and examples J Güémez 1 (a), M iolhais 2 (b) and Luis A ernández 3 (c) 1 Department of Applied Physics, University of Cantabria, E Santander, Spain 2 Department of Physics and CisUC, University of Coimbra, P Coimbra, Portugal 3 Department of Mathematics, Statistics and Computation, University of Cantabria, E Santander, Spain PACS PACS PACS p Special relativity Bb Rotational kinematics dc Rotational dynamics Abstract We obtain the relativistic rotation dynamical equations by using the Hamiltonian- Lagrangian formalism. After establishing the dynamical equation for the relativistic rotation, we derive the relation between the rotational kinetic energy variation and the angular momentum variation. Then, we discuss three examples that illustrate how this special relativity formalism deals with rotational processes of systems acted upon by external torques, including cases when dissipative forces are present. Introduction. The relativistic rotation [1], including the inertia and the moment of inertia of rotating systems [2], and also the spin equation [3], has deserved some interest, certainly due to the qualitative differences in comparison with the classical situation [4]. When a solid body rotates with angular velocity ω, its internal energy in such pure rotational state, E 0 (ω), is larger than the energy of the same body, E 0, in the absence of the rotation. Hence, the rotational kinetic energy, i.e. the kinetic energy with respect to the frame in which the system s linear momentum vanishes, is internal to the system, like in classical mechanics [5]. According to Einstein s principle of the inertia of the energy [6], the inertia of a rotating body M(ω) = E 0 (ω)c 2 > E 0 c 2 depends on its angular velocity ω and, surely, also depends on other magnitudes, such as its temperature, its chemical composition, etc. The body s inertia dependence on its angular velocity is a pure relativistic effect: the inertia of a rotating object (a compact star, a black-hole, a galaxy, etc.) is higher than the inertia of the very same non-rotating object. In a similar way, the moment of inertia of a relativistic rotating body, I(ω), is angular velocity dependent (dislike in classical mechanics). This article is about relativistic rotations around a principal axis and it is organized as follows. In sec. we work out the formalism to deal with the relativistic rotation and (a) guemezj@unican.es (b) tmanuel@teor.fis.uc.pt (c) luisal.fdez@unican.es derive a remarkable relation between the variation of the rotational kinetic energy and the variation of the angular momentum. The development of the formalism includes considering the Hamiltonian-Lagrangian formalism for rotating systems. In sec. we present three examples to show how the formalism works: we study a rotating ring submitted to conservative forces, then we discuss the effect of frictional forces in a rotating disc and, finally, we study a fireworks wheel, a system in which there is a mechanical energy increase due to chemical reactions. In sec. the main conclusions are drawn. Relativistic rotation. In this paper we extend to rotations a previously developed formalism for relativistic translations [7] based, in part, on the relation d[γ(v)c 2 ] = v x d[γ(v)v x ] + v y d[γ(v)v y ] + v z d[γ(v)v z ], (1) between the variation of the kinetic energy and the variation of the linear momentum of a system with constant inertia, where γ(v) = ( 1 v 2 /c 2) 1/2, with c the speed of light. Here v = (v x, v y, v z ) denotes the velocity of the system moving as a whole in a certain reference frame. This is the velocity of the reference frame in which the linear momentum of the system vanishes. The classical counterpart of expression (1), d[γ(v)c 2 ] = v d[γ(v) v], is simply 1 2 dv2 = v d v. Characteristic functions and classical limit. We consider a solid body (a ring, a disc, a bar, etc.) robust enough so that its plastic distortion and strain rep-1

2 J Güémez, M iolhais, Luis A ernández sponse to stress, caused by the stationary revolving state, is negligible. Let us consider such a rigid solid body of volume, and constant density, ρ, rotating with angular velocity ω around a principal axis. Differently from classical mechanics, now the inertia of the body is angular velocity dependent, M(ω), and it is defined by [2] M(ω) = ρ γ (ω r(x)) dx, (2) where r(x) is the distance from the point x to the rotation axis. Similarly, the moment of inertia is also angular velocity dependent, I(ω), and it is given by I(ω) = ρ r 2 (x)γ (ωr(x)) dx. (3) or a non-rotating body, ω = 0, and the classical inertia, M(0), and the moment of inertia, I(0), are given by M(0) = ρdx, I(0) = ρ r 2 (x)dx. (4) Here we assume that M(0) and I(0) are fixed quantities. Taking into account that the relativistic factor, γ(v), and its derivative, γ (v) dγ(v)/dv, are strictly increasing functions, we can differentiate the angular velocity dependent functions (2) and (3) under the integral sign, leading to and d [ M(ω)c 2] = c 2 d [I(ω)ω] Now we use the following identity, γ (v) = to obtain: from (5), d [ M(ω)c 2] and, from (7), d [I(ω)ω] ργ (ωr(x)) r(x)dx, (5) = ω ργ (ωr(x)) r 3 (x)dx + (6) + ρr 2 (x)γ (ωr(x)) dx. (7) v c 2 γ(v), (0 v < c) (8) v2 c 2 r 2 (x) = ω ργ (ωr(x)) c 2 ω 2 r 2 dx ; (9) (x) = + = ργ (ωr(x)) ρr 2 (x)γ (ωr(x)) dx ργ (ωr(x)) ω 2 r 4 (x) c 2 ω 2 r 2 dx + (10) (x) c 2 r 2 (x) c 2 ω 2 r 2 dx. (11) (x) inally, by termwise dividing equation (9) and equation (11) one arrives at d [ M(ω)c 2] = ω. (12) d[i(ω)ω] This mathematical proof, based on the identity (8), is general for M(ω) and I(ω) as defined by (2) and (3), respectively, with fixed M(0) and I(0) see equation (4). Equation (12) expresses the relation between the energy variation and the angular momentum variation in the course of a rotation. It is the rotational counterpart of equation (1) that applies to translations. Equation (12) can also be expressed in a physically more meaningful way as dk rot dj = ω, (13) (just as in non-relativistic physics) where K rot = [M(ω) M(0)] c 2 is the (relativistic) rotational kinetic energy of the body and J = I(ω) ω is its (relativistic) angular momentum. The classical equivalent of equation (12) is simply = ω. The coherence of the formalism presented so far can be checked by studying its classical limit, as c +. or a given body, it turns out to be convenient to define two characteristic functions: ζ(ω), for the inertia, and χ(ω), for the moment of inertia; they are defined by ζ(ω) = M(ω) M(0) and χ(ω) = I(ω) I(0), (14) respectively. It is imperative that, in the limit c +, the relativistic rotational kinetic energy, K rot = E 0 (ω) E 0, with K rot = M(0) [ζ(ω) 1] c 2, transforms into the corresponding classical rotational kinetic energy. Let us check this point by taking the limit lim K rot = c + lim c + 1 (1/c 2 ) [ ργ (ωr(x)) dx ρdx (15) The L Hôpital rule can be applied and, after differentiation with respect to c, one obtains lim K 1 rot = lim c + c + 2 ω 2 r 2 (x)ρdx [1 ω2 r 2 (x)/c 2 ] 3. (16) The denominator reduces to 1 and, using the second definition in (4), one obtains lim K rot = 1 c + 2 I(0)ω2. (17) Together with lim c + χ(ω) = 1, this is indeed the correct classical limit with I(0) = I. Relativistic Lagrangian formalism for rotations. or a body (a ring, a disc, a bar, etc.) that rotates with angular velocity ω and moves as a whole with linear velocity v, its total (relativistic) energy is given by ]. p-2

3 Relativistic rotation dynamics formalism and examples E(ω, v) = γ(v)m(ω)c 2, according to Einstein s principle of the inertia of the energy [6]. or a process in which the total linear momentum of the body is zero, v = 0, and the resulting net force of all external forces applied on it is also zero. The body s total energy, i.e., its rest energy since the linear momentum p = γ(v)m(ω)v = 0, is E 0 (ω) = M(0)ζ(ω)c 2. This energy does include the internal rotational kinetic energy, which depends on ω. or a body that rotates submitted to a set of external conservative forces { i ext } forces exerted by a work reservoir [8], each of which with arm r i one has the following relativistic Hamiltonian H(ω, θ) = M(0)ζ(ω)c 2 + i V i (θ), (18) where each potential energy is such that (dx i = r i dθ) V i(θ) θ = V i(θ) x i x i θ = i ext r i = τ ext i. (19) The external forces, whose resultant must be zero, exert a zero net linear impulse on the body, but they may perform work on it (as well as angular impulse). rom the first Hamilton equation d θ = J ω = M(0)ζ(ω)c2, (20) J and from equation (12), one has, for the angular momentum J, J = I(0)χ(ω)ω. (21) rom the second Hamilton equation d J = θ, (22) one obtains the relativistic Euler-Poinsot equation [4] for the body s rotation, namely ( ) d[i(0)χ(ω)ω] = i ext r i = Γ ext. (23) i rom Hamiltonian (18), one obtains the relativistic Lagrangian, L(ω, θ) = ωj H(ω, θ), i.e. L(ω, θ) = I(0)χ(ω)ω 2 M(0)ζ(ω)c 2 V (θ), (24) where V (θ) represents the sum of all potential energies. It is worth noting that, from the Euler-Lagrange equation d L(ω, θ) L(ω, θ) = 0. (25) ω θ applied to Lagrangian (24), one can also obtain the relativistic Euler-Poinsot equation (23). On the other hand, from equation (23) and equation (12) one derives the pseudo-work kinematic rotational energy variation [9], namely ( ) d[m(0)ζ(ω)c 2 ] = i r i dθ. (26) i Because this equation can also be expressed as d[m(0)ζ(ω)c 2 ] + V i i θ dθ = 0, from Hamiltonian (18) one concludes that = 0 and one has = 0. (27) Mechanical energy is conserved for a process when conservative forces act on the system. Equations for a rotational process. Angular impulse angular momentum variation relativistic equation According to the previous Hamiltonian-Lagrangian description, one has the relativistic Euler-Poinsot equation for relativistic rotation dynamics d [I(0)χ(ω)ω] = Γ ext, (28) where Γ ext = j τ j ext is the total resulting torque for the external forces (conservative or non-conservative). After integration in a time interval [0, t], assuming constant torque, one gets the finite equation ] I(0) [χ(ω f )ω f χ(ω i )ω i = Γ ext t. (29) This equation is the relativistic counterpart of the corresponding non-relativistic equation I (ω f ω i ) = Γ ext t, where the left hand side is the angular momentum variation and the right hand side is the net angular impulse. Pseudo-work rotational kinetic variation relativistic equation Multiplying both sides of equation (28) by ω = dθ/, and taking into account equation (12), one obtains the equation for the relativistic pseudo-work rotational kinetic energy variation: d [ M(0)ζ(ω)c 2] = Γ ext dθ. (30) The right hand side includes the contribution of both conservative or non-conservative forces. After integration, using equation (12), between an initial and a final state, and for a constant torque, one gets the corresponding finite equation [ M(0) ζ(ω f ) c 2 ζ(ω i ) c 2] = Γ ext (θ f θ i ). (31) This equation is the relativistic counterpart of the corresponding non-relativistic equation 1 2 I ( ωf 2 ) ω2 i = Γ ext (θ f θ i ), where the left hand side is the rotational kinetic energy variation and the right hand side is, in general, pseudo-work [9]. Note that, in the absence of equation (12), it would not be possible, from the Euler-Poinsot equation, to obtain the pseudo-work rotational kinetic energy variation equation and its relation with the net external torque and the rotation angle. p-3

4 J Güémez, M iolhais, Luis A ernández Examples. In this section we present three examples of relativistic rotations, namely a rotating ring, with only conservative forces acting on it; a rotating disc when dissipative forces are also present; and the relativistic analogue of a fireworks wheel, where the kinetic rotational energy increase results from chemical reactions.! i r 1 r 2 (i) R ig. 1: Rotating ring acted upon only by conservative forces: (i) initial state and (f) final state, after a time interval [0, t]. Rotating ring (conservative forces). or a ring of radius R and moment of inertia I ring (0) = M(0)R 2, rotating with angular velocity ω around its axis, its characteristic functions (14) ζ ring (ω) and χ ring (ω) are [2] ζ ring (ω) = γ (ωr), χ ring (ω) = γ (ωr), (32) respectively. We study the rotating process depicted in ig. 1, representing a rotating ring of radius R mounted in a rigid massless structure. Two conservative forces, 1 = (, 0, 0) and 2 = (, 0, 0), are applied to the system, with arms r 1 and r 2, respectively. Since the resultant force vanishes, ext = 0, the ring does not move as a whole in the reference frame where its initial total linear momentum is zero [7]. or the description of the rotating ring we integrate equation (28) in the time interval [0, t], obtaining [I ring (0) is supposed to be fixed] I ring (0) [χ ring (ω f )ω f χ ring (ω i )ω i ] = Γ ext t, (33) where the external net torque is Γ ext = (r 1 + r 2 ). The corresponding non-relativistic equation for this process is the well known expression I ring (ω f ω i ) = Γ ext t. Assuming, in (33), that the initial angular velocity is zero, the angular velocity, simply denoted by ω, as a function of time, is obtained, as well as the angular dependence θ(t), yielding [11] ω(t) = θ(t) =! f (f) αt 1 + (αrt) 2 /c 2, (34) c 2 αr (αrt)2 c 2 1, (35) where α = Γ ext /I ring (0). It is straightforward to check the following limits: [ ] [ ] ω(t) θ(t) lim = α, lim t 0 t t 0 t 2 = 1 2 α ; (36) lim ω(t) = ω L = c [ ] θ(t) t + R, lim = ω L, (37) t + t where ω L is the limiting angular velocity for the ring and the limit t 0 corresponds to the classical limit. The maximum angular velocity may also be derived from the requirement that the ring characteristic functions (32) are real functions. As the above limit indicates, that angular velocity is attained only for t + and the energy transferred to the system is lim ω ωl M(0)[ζ ring (ω) 1] c 2 = +. Inserting the characteristic mass function for a ring, equation (32), into equation (31), for zero initial angle and zero initial angular velocity, one obtains the explicit dependence of the angular velocity with the rotation angle (we are setting ω f = ω and θ f = θ): ω(θ) = c R { 1 [ 1 + ( )] } 2 1/2 aθ c 2, (38) where a = Γ ext /M(0) = αr 2. We can also obtain the limits of the angular velocity for large and small angles: [ ] ω(θ) lim ω(θ) = ω L, lim = 2α. (39) θ + θ 0 θ 1/2 Rotating disc with friction. Now, we analyse the process described in ig. 2, that includes frictional forces f 1 = (0, f, 0) and f 2 = (0, f, 0), applied on a disc. f!i r 2 (i) r 1 R f ig. 2: Rotating disc subjected to dissipative forces: (i) initial state and (f) final state, after a time interval [0, t]. or a disc of radius R, rotating around its axis, the moment of inertia is I disc (0) = 1 2 M(0)R2 and [2] [ ( c ) 2 ( ) ] 2 1/2 ωr ζ disc (ω) = 2 ωr 1 1 c,(40) { [ χ disc (ω) = 8 ( c ) ( ) ] 2 ωr 3 ωr 2 c [ ( ) ] 2 1/2 } ωr 1. (41) c The inertia of the disc varies with time, since M(ω) = M(0)ζ disc (ω), but M(0) is assumed to be fixed.! f (f) p-4

5 Relativistic rotation dynamics formalism and examples In order to describe the linear displacement of the body as a whole, one should use the general impulse-linear momentum variation equation d [γ(v)m(ω)v] = ext [7]. Since = 0 and f 1 + f 2 = 0, the resultant external force vanishes, ext = 0, and therefore M(ω f )γ(v f )v f M(ω i )γ(v i )v i = 0. (42) During this process, the disc, as a whole system, does not move (v i = v f = 0). In order to consider the (phenomenological) frictional forces acting on the rim of the disc, we introduce Rayleigh s dissipation function [12] R(θ) = 2f Rθ. The corresponding Euler-Lagrange equation with Rayleigh s function is then given by d L(ω, θ) L(ω, θ) = R(θ). (43) ω θ θ rom this, one obtains the corresponding relativistic Euler-Poinsot equation d[i disc χ disc (ω)ω] = Γ ext (44) for this process, where Γ ext = (r 1 +r 2 ) 2fR is the total resultant torque and the right hand side of equation (44) is the net angular impulse applied on the disc. rom this equation and (12) one obtains [9] d[m(0)ζ disc (ω)c 2 ] = (r 1 + r 2 )dθ 2fRdθ, (45) Because equation (45) can be expressed as d[m(0)ζ disc (ω)c 2 (r 1 + r 2 )θ] = 2fRdθ, and, from Hamiltonian (18), one obtains Interesting enough, thermal effects for this process can be obtained in the classical limit [4] with δq = 2fRω, (49) where Q is the energy dissipated as heat. Thus, = δq < 0, (50) is the mechanical-thermodynamical equation characterizing this kind of mechanical energy dissipation process with frictional forces. ireworks wheel (rotational energy production). A fireworks wheel consists of a rigid bar, of length L = 2R, with two (massless) gun powder filled cartridges attached at its ends, as represented in ig. 3. We shall consider the relativistic analogue of such an apparatus and assume that the inertia of the bar, M(0), and its moment of inertia, I(0), are fixed quantities. or a rotating bar of length L and moment of inertia I bar (0) = 1 12 M(0)L2, when rotating around the axis perpendicular to its centre, it is not difficult to obtain that bar s characteristic functions, ζ bar (ω) and χ bar (ω), are ζ bar (ω) = 2c ( ) ωl ωl arcsin, (51) 2c [ ( ) χ bar (ω) = 12c3 ωl ω 3 L 3 arcsin ωl ] 4 ω2 L 2 2c 4c c 2,(52) respectively. = 2fRω. (46) Therefore, part of the work provided by forces 1 and 2 is dissipated. or the rotational description of the process, equations (28) and (30) apply, and the following relations hold:! i! f I disc (0) [χ disc (ω f )ω f χ disc (ω i )ω i ] = Γ ext t, (47) M(0) [ζ disc (ω f ) ζ disc (ω i )] c 2 = Γ ext (θ f θ i )(48), Even without working out the explicit function ω(t), and since the characteristic functions (40) and (41) should be real, one may readily conclude that the disc s maximum angular velocity is ω L = c/r. When this maximum angular velocity is achieved, the outermost infinitesimal layer of the disc attains the linear velocity c = ω L R and the total energy of the system, obtained from (40) and (48) is 2M(0)c 2 [2]. Therefore, assuming absence of friction forces, the maximum work performed by the binary acting on the ropes is M(0)c 2. When friction forces are acting on the system, there is mechanical energy dissipation. In order to reach the maximum angular velocity, the forces on the ropes should then perform an additional work equal to the heat dissipated in the surrounding. ig. 3: ireworks wheel: (i) initial state and (f) final state, just when no more powder is left the cartridges. Due to the gun powder combustion, a pair of two opposite forces, 1 (t) = ( cos θ 1 (t), sin θ 1 (t), 0) and 2 (t) = ( cos (π + θ 1 (t)), sin(π + θ 1 (t)), 0), of magnitude, assumed to be constant, are applied at the ends of the bar. The resultant external force is zero, ext = 0, the bar as a whole remains at rest, and the system rotates without friction around the centre of the bar due to the external torque Γ ext = 2R. p-5

6 J Güémez, M iolhais, Luis A ernández The Hamiltonian is simply given by Its corresponding Lagrangian is H(ω) = M(0)ζ bar (ω)c 2. (53) L(ω) = I bar (0)χ bar (ω)ω 2 M(0)ζ bar (ω)c 2. (54) Similarly to example 2, we can introduce, in the present case, an energy production function, P ξ (θ) = 2 ξ Rθ, where ξ is a force applied on the bar resulting from the chemical reactions. The corresponding Euler-Lagrange equation is then given by d L(ω, θ) L(ω, θ) = P ξ(θ). (55) ω θ θ rom here, one obtains the corresponding relativistic Euler-Poinsot equation d[i bar (0)χ bar (ω)ω] = 2 ξ R, (56) for the process. rom this equation, by using equation (12) one obtains d[m(0)ζ bar (ω)c 2 ] = 2 ξ Rdθ, (57) and, from the Hamiltonian (53), and from (57) one obtains = 2 ξrω. (58) The integration of equations (56) and (57) in the time interval [0, t] leads to I bar (0) [χ bar (ω f )ω f χ bar (ω i )ω i ] = 2 ξ R t, (59) M(0) [ζ bar (ω f ) ζ bar (ω i )] c 2 = 2 ξ R θ. (60) These equations allow one to obtain the dependence of the angular velocity with time and with the rotation angle, though the explicit form of such equations is hard to work out. Nevertheless, the maximum angular velocity can be obtained from (51) or (52) which must be real positive functions. Hence, that limiting angular velocity is ω L = c/r and this corresponds to a linear speed of the bar ends equal to c. The maximum energy increase of the bar, if it is initially at rest, is obtained from the left hand side of (60) and it is given by lim ω ωl M(0)[ζ bar (ω) 1] c 2 = ( π 2 1) M(0)c 2. The decrease of the Gibbs free energy associated with the chemical reactions goes into work and from the classical description it is possible to write for this process [4] G ξ = 2 ξ Rω. (61) This process is reversible: the total produced mechanical energy can be completely, at least in principle, transformed into the corresponding G ξ increment for a chemical reaction [13]. One has = G ξ > 0, (62) as the mechanical-thermodynamical equation that characterizes this kind of mechanical energy production process from chemical reactions. Conclusions. In this paper we addressed the relativistic rotation by developing an appropriate formalism, based on relativistic Lagrangians and applying it to three examples. The formalism includes the remarkable relation (12) which links the vector description of the rotation process, involving the angular impulse and the variation of the angular momentum, to the scalar approach that relies on the rotational kinetic energy and pseudowork. The examples have been chosen to emphasize this relationship between the angular momentum-angular momentum equation and the pseudo-work-variation equation of the rotational kinetic energy. through that relativistic rotation theorem. A process has been analyzed in which only conservative forces (related to a work reservoir) intervene, a process that evolves with conservation of mechanical energy. It has also been analyzed a process in which dissipative forces intervene, now with dissipation of mechanical energy (and heat production). inally, a process of production of kinetic energy of rotation, due to forces that come from chemical reactions has been analyzed, to show how the Lagrangian formalism treats this type of processes with a Gibbs free energy decrease. REERENCES [1] Ø Grøn, Energy considerations in connection with a relativistic rotating ring, Am. J. Phys (1982) [2] C M L Leonard, The relativistic classical increase of energy and angular momentum with rotation, Am. J. Phys (1984) [3] S Pahor and J Strnad, Statics in Special Relativity, Nuovo Cimento 20 B, (1974) [4] J Güémez, M. iolhais, Thermodynamics of rotating systems selected examples, Eur. J. Phys. 35 (2014) (14pp). [5] A J Mallinckro, H. S. Leff, All about work, Am. J. Phys (1992) [6] J Güémez, An undergraduate exercise in the first law of relativistic thermodynamics, Eur. J. Phys (2010) [7] J Güémez, M. iolhais, L A ernández, Relativistic mechanical-thermodynamical formalism description of inelastic collisions, Eur. J. Phys. 37 (2015) (21pp) [8] J Anacleto, Work reservoirs in thermodynamics, Eur. J. Phys (2010) [9] B A Sherwood, Pseudo-work and real work, Am. J. Phys (1983) [10] Ø Grøn The asynchronous formulation of relativistic statics and thermodynamics, Nuovo Cimento (1973) [11] D E ahnline, Parallels between relativistic and classical dynamics for introductory courses Am. J. Phys (1975) [12] T W B Kibble, H Berkshire, Classical Mechanics, 4th Ed., Addison Wesley Longman, Essex p. 189 [13] J Güémez, M. iolhais, rom mechanics to thermodynamics: analysis of selected examples, Eur. J. Phys (2013) p-6