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Dr Dev Raj Mishra. International Journal of Engineering Research and Applications www.ijera.com ISSN: 2248-9622, Vol. 10, Issue 7, (Series-IV) July 2020, pp. 11-15

RESEARCH ARTICLE OPEN ACCESS

A Brief Study on Covariance of Newtonian Mechanics

Dr Dev Raj Mishra Department of Physics, R.H.Government Post Graduate College, Kashipur, U.S.Nagar, Uttarakhand -244713 INDIA.

ABSTRACT The possibility of a covariant formulation of Newton’s second law of motion using Energy-Momentum four- vector was explored. This covariant formulation explains the relationships between the electromagnetic potentials and the electromagnetic fields. It successfully explains the observation of pseudo-forces in nature for example the observed up-thrust in a falling or a falling lift and the observed centrifugal force in circular motion. Keywords– Covariance,Electromagnetic Field ,Four- vectors, time, Relativity

------Date of Submission: 06-07-2020 Date of Acceptance: 21-07-2020 ------

I. INTRODUCTION the applied force and the momentum. The applied The invariance under is also the gradient of potential energy. transformation expresses the proposition that the Therefore we have two quantities – the momentum laws of physics are same for the observers in and the energy, which are also part of a four-vector different frame of reference in relative motion to called four-momentum or Energy-Momentum four- each other [1]. The laws of Physics should be vector. Therefore the covariant formulation of the expressible in what is known as a covariant form if it second law, as we shall discuss below, can be has to be independent of the choice of the frame of accomplished using Energy-Momentum four-vector reference of the observer.Principle of Covariance requires the formulation of physical laws using only II. RESULTS AND DISCUSSIONS those variables the measurement of which the Let an event E of force 퐅 = Fx퐢 + Fy퐣 + observer in different frames of reference can Fz퐤where Fx , Fy , Fz are components of the force, unambiguously correlate [2]. These are the being applied to a particle P with spatial co- quantities which have spatial as well as temporal ordinates (x, y, z) occursat time, say t. Therefore the components such as four-co-ordinates,four- four co-ordinates of event E aret, x, y and z, denoted vectorsand [3]. in terms of four-coordinate as xμ = The non-covariant form of Newton’s law of x0, x1 , x2, x3 choosing system of units such that the Gravitation led to the formulation of Einstein’s Field velocity of light, c=1. Equations in General theory of Relativity. These The separation of two events in Minkowski Field Equations use Einstein tensor, a derived form space – time using Einstein summation convention, of Ricci’s Tensor,is a tensor and therefore confirms is given by to the laws of covariance. The laws of ds = g dxμ dxν (1) Electromagnetism were also brought to the format μν The tensorg = gμν = diag 1, −1, −1, −1 is prescribed by the requirement of μν called metric of the space-time frame. Covariance.Maxwell Equations are expressed in μ compact covariant forms using electromagnetic field Alsodx denote the separation of two events along μ tensor Fμν and current density four-vector jμ . four-coordinate x . The other laws of physics of pre-relativistic Newton’s Second law of motion, as stated eraneed to be modified so as to confirm to the above, tells us that rate of change of momentumpis principle of covariance. Newton’s second law of proportional to the applied force Fand takes place in Motion is one of the suitable candidates for this its direction. Mathematically we can write it as d퐩 modification. As we know Newton’s Second law of 퐅 = (2) motion can be stated as “The rate of change of dt momentum is proportional to the applied force and Here퐩 = px 퐢 + py 퐣 + pz퐤is the momentum and takes place in its direction”. It has two quantities – px , py and pzare its spatial components.As we

www.ijera.com DOI: 10.9790/9622-1007041115 11 | P a g e

Dr Dev Raj Mishra. International Journal of Engineering Research and Applications www.ijera.com ISSN: 2248-9622, Vol. 10, Issue 7, (Series-IV) July 2020, pp. 11-15 know, the units of force are so defined to make the 훛 횲 퐩 = ퟎ (13) constant of proportionality in above equationas The tensor 흏 횲 퐩 has six components given unity. Then x-component of the force is given by by 흏 횲 퐩 흁흂 = ∂μpν − ∂νpμ for 휇, 휈 = 0,1,2,3. ∂p x (3) Therefore assigning different values Fx = 0 ∂t to흁 and 흂and substituting p0 = g00p = g00E = Of course this force can be a fieldand therefore 1 E andp1 = g11 p = g11 px = −px and so on for possessing a unique value at each point. other four-momentum components, we obtain the Further, we know that force is a gradient of the following six relations between four components of potential energy V of the particle. Mathematically Energy-Momentum four-vector. 퐅 = −∇V (4) ∂E ∂px ∂ ∂ ∂ − = (i) where∇= 퐢 + 퐣 + 퐤 is the gradient operator. ∂x ∂y ∂z ∂x ∂t ∂E ∂py The x-component of the force from Eq. (4)is − = (ii) given by ∂y ∂t ∂V ∂E ∂pz F = − (5) − = (iii) x ∂x ∂z ∂t ∂py ∂p Comparing the two equations, we get = x (iv) ∂V ∂px ∂x ∂y − = (6) ∂p ∂p ∂x ∂t z = x (v) This result, however is not covariant in the ∂x ∂z sense 퐩 is a three vector. The corresponding four - ∂p ∂py z = (vi) vector isp = E, px , py , pz , E being the total ∂y ∂z energy of the particle. The total energy E is the sum (14) of potential and kinetic energies and also kinetic The first three relations are between spatial variation energy is not a function of co-ordinates. Therefore of energy and the temporal rate of change of three- we can safely replace V by Ein above equation momentum. These relations prove the fact that ∂E ∂p whenever there is a spatial variation in Energy; a − = x (7) ∂x ∂t force appears in the direction of the decrease in Replacing coordinates x = t, x , y, z byxμ = energy. This force we usually call as pseudo- force. 0 1 2 3 μ x , x , x , x and p = E, px , py , pz by p = The examples of these forces are upward force in a p0, p1 , p2, p3 , we can write above equation as falling lift or frame of reference, centrifugal force on ∂p0 ∂p1 a particle doing circular motion etc. The last three − = (8) relations just correlate the spatial variations of the ∂x1 ∂x0 ∂ components of the three- momentum. Further using notation∂ for , Eq. (8) becomes μ ∂xμ Following are the illustrative examples of these ν μ ∂μ p + ∂νp = 0 (9) results. forμ = 0 and ν = 1. Lowering indices of four-momentum pμ using III. ILLUSTRATIVE EXAMPLES metric tensorgμν , Eq. (9) can be written as The illustrative examples chosen here have να μβ ∂μ g pα + ∂ν g pβ = 0 (10) kinetic energy variations which are the prominent But all off-diagonal elements of the metric gμν are energy variations in these cases but tensor Eq. (13) να νν μβ is equally and actually valid for the total energy zero; thereforeg pα = g pν and g pβ = gμμ p .we get fromEq. (10) variations only. μ gνν ∂ p + gμμ ∂ p = 0 (11) μ ν ν μ 3.1. Relation between Electromagnetic forμ = 0 and ν = 1. μν potentials But gμν = g = diag 1, −1, −1, −1 , therefore Let an electron having charge –e is placed in an gμμ = 1 푎푛푑 gνν = −1forμ = 0 and ν = 1. We get electromagnetic field. The electric field 퐄 = Ex퐢 + from Eq. (11) E 퐣 + E 퐤 and magnetic field 퐁 = B 퐢 + B 퐣 + ∂ p − ∂ p = 0 (12) y z x y μ ν ν μ B 퐤 are expressed in terms of the electric potential φ forμ = 0 and ν = 1. z and the magnetic potential Aby the following We see that left hand side of above equationis equation μν component of an anti-symmetric tensor ∂퐀 ∂ Λ pforμ = 0 and ν = 1. Let us generalize this 퐄 = −∇φ − and 퐁 = ∇ × 퐀 (15) ∂t result assuming it true for allμ and ν,we can write Further Energy E and Momentum p of the the covariant form substitute for Newton’s Second electron in presence of electromagnetic fields Law of Motion as modifies to

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Dr Dev Raj Mishra. International Journal of Engineering Research and Applications www.ijera.com ISSN: 2248-9622, Vol. 10, Issue 7, (Series-IV) July 2020, pp. 11-15

E → E − eφ, 퐩 → 퐩 − e퐀 (16) Therefore the first three equationsof the We have from Eq. (14(i)) formulationexpressed by Eq.(14) successfully ∂E ∂p correlate the force experience by a charged particle − = x ∂x ∂t with the corresponding electric field components of Let us assume the variation in Energy is only the electromagnetic field tensor Fμν . Therefore if the due to the electromagnetic fields, then, from Eq. (16) formulation is complete, the last three equations in x - direction should correlate the motion of the particle with the ∂ −eφ ∂ p − eA magnetic field components of the tensor.Let us − = x x ∂x ∂t consider Eq. (14)(iv) Rearranging terms, we get ∂py ∂px 1 ∂p ∂φ ∂A = x = + x (17) ∂x ∂y e ∂t ∂x ∂t Howeverthe components of the momentum modifyin But from Eq.(15), x-component of electric field Ex presence of electromagnetic potentials in accordance is given by with theEq. (16). Therefore we get ∂φ ∂Ax E = − − (18) ∂ py − eAy ∂ px − eAx x ∂x ∂t = Combining Eqs.(17) and (18), we can write ∂x ∂y Rearranging terms, we get 1 ∂px ∂φ ∂Ax = + = −E ∂py ∂p ∂Ay ∂A e ∂t ∂x ∂t x − x = e − x (23) Therefore we conclude that the Electric field in ∂x ∂y ∂x ∂y x-direction is given by According to Eq. (15), magnetic field 퐁 is related to 1 ∂p the vector potential 퐀by the equationB= ∇ × 퐀, E = − x x e ∂t therefore It is therefore, the rate of change of momentum ∂Ay ∂A − x = B i. e. z component of 퐁 (24) classically speaking the force experienced by a unit ∂x ∂y z positive charge. Note here the charge is –e. This is Also the left hand side of Eq.(23)can be written as also the known definition of E . x ∂py ∂px ∂vy ∂vx Reordering above equation − = m − (25) ∂x ∂y ∂x ∂y 1 ∂px ∂φ ∂Ax −Ex = = + (19) Here m is the mass of the electron and 퐯 = vx퐢 + e ∂t ∂x ∂t Writing in terms of the four-potentialAμ = vy 퐣 + vz퐤is its velocity.But we know from calculus A0, A1, A2, A3 = φ, A , A , A and the four co- ∂vy ∂vx x y z − = ∇ × v z = 2ωz ordinates, we get ∂x ∂y 0 1 Here ω is the z-component of the angular velocity i ∂px ∂A ∂A z −E = = + of the particle. Therefore from Eq. (25) x e ∂t ∂x1 ∂x0 ∂ ∂py ∂p (26) Further using notation∂ for on right hand side, − x = 2mω μ ∂xμ ∂x ∂y z we get Substituting results from Eqs.(24)and (26), in Eq. i ∂p x 0 1 (23), we get −Ex = = ∂1A + ∂0A e ∂t (20) ∂A 0 1 0α 1α y ∂Ax But ∂1A + ∂0A = ∂1 g Aα + ∂0 g Aα 2mωz = e − = eBz All off-diagonal elements of the metric gμν are zero, ∂x ∂y 0α 00 1α 11 Dividing throughout by charge e and rearranging, thereforeg Aα = g A0 and g Aα = g A1. We get from the above equation we get ∂A 0 1 00 11 2m y ∂Ax ∂1A + ∂0A = ∂1 g A0 + ∂0 g A1 Bz = ωz = − (27) 00 11 e ∂x ∂y But as we knowg = 1 and g = −1, we get μ 0 1 2 3 0 1 As A = A , A , A , A = φ, Ax, Ay, Az we can ∂1A + ∂0A = ∂1A0 − ∂0A1 = 퐹10 and also 퐹 = g g 퐹10 = −퐹10 (21) write 10 11 00 2 1 10 ∂Ay ∂A ∂A ∂A Here F is the x-t component of the − x = − Electromagnetic Field TensorF휇휈 = ∂휇 Aν − ∂νA휇 ∂x ∂y ∂x1 ∂x2 Therefore from Eq.(20), we get ∂ Further using notation∂μ for on right hand side, i ∂p ∂xμ −E = x = −퐹10 (22) we get x e ∂t ∂Ay ∂A Similarly it can be shown that F20 = E and − x = ∂ A2− ∂ A1 y ∂x ∂y 1 2 F30 = E using equations (14)(ii) and (14)(iii) z Therefore from Eq. (27), we get respectively.

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Dr Dev Raj Mishra. International Journal of Engineering Research and Applications www.ijera.com ISSN: 2248-9622, Vol. 10, Issue 7, (Series-IV) July 2020, pp. 11-15

2m (28) acceleration is along negative z axis) as shown in B = ω = ∂ A2− ∂ A1 z e z 1 2 Fig.1. 2 1 2α 1α But ∂1A − ∂2A = ∂1 g Aα − ∂2 g Aα Let the frame falls from an initial height h. The But all off-diagonal elements of the metric change in energy at a height z is equal to the gain in μν 1 tensorg are zero, therefore the kinetic energy mz 2 of the object. As we know 2α 22 1α 11 2 g Aα = g A2and g Aα = g A1. Therefore we from elementary kinematics have 2 0 − z = −2az(h − z). (33) 2 1 22 11 ∂1A − ∂2A = ∂1 g A2 − ∂2 g A1 Therefore theincrease in kinetic energy of the 11 22 1 But as we knowg = g = −1, we get objectis given by E = mz 2 = ma (h − z). ∂ A2− ∂ A1 = ∂ A − ∂ A = 퐹 2 z 1 2 2 1 1 2 21 Therefore Also 퐹 = g g 퐹21 = 퐹21 = −퐹12 (29) 21 22 11 ∂E ∂ ma (h − z) (34) Finally,from Eqs.(28) and (29), we get = z = −ma ∂z ∂z z 2m 21 12 Bz = ωz = 퐹 = −퐹 (30) We get from Eq.(14(iii)) e ∂E ∂pz Therefore Bz is associated with the angular + = 0 (35) motion of the particle and is (2, 1) component of the ∂z ∂t Electro-magnetic Field TensorFμν . Similarly it can Therefore 2m ∂p ∂E be shown that F13 = −퐹31 = B = ω and z = − = ma (36) y e y z 2m ∂t ∂z F32 = −퐹23 = B = ω using Eqs.(14(v)) This shows that the object experiences a force x e x and(14(vi)) respectively. maz along positive z direction which is the actually Let us consider an electron moving with observed pseudo force in a frame moving downward velocity 풗enters a perpendicular magnetic field B. It experiences a Lorentz force which provides for the necessary centripetal force. The electron therefore, undergoes a circular motion with an angular velocity 휔, given as h-z 푚푣2 푚 퐵푒푣 = or 퐵 = 휔 (31) 푟 푒 +푚풂풛 Therefore we have arrived at the angular P frequency (see Eq. (30)which is half of the expected z- value in Eq. (31). This requires an appropriate h explanation which we shall reserve for our future axis communication. Our present discussion, however, successfully illustrates the fact that the magnetic z -풂풛 field causes rotation and the angular velocity is proportional to the applied magnetic field. Thus various components forElectromagnetic μν Field Tensor using Minkowskimetric gμν = g = diag 1, −1, −1, −1 can be arranged in the matrix x-axis form as below Fig.1. Frame at particle P in a container moving 0 −Ex −Ey −Ez downwards with acceleration 풂풛. The particle Ex 0 −Bz By Fμν = (32) experiences a force m풂 upwards. E B 0 −B 풛 y z x Ez −By Bx 0 with accelerationaz. This is exactly the matrix predicted from other methods.Therefore we have obtained all the 3.3. Centrifugal Force components of starting from Let a particle of mass m is moving in a circle of ourbasic covariant Eq. (14). radius r with an angular velocityω, as shown in Fig.2. The magnitude of the velocity of the particle 3.2. Pseudo force in a freely falling frame at any instant isv = ωr. The kinetic energy, which is Consider an object of mass m in a frame of the only variable part of the total energy of the reference falling vertically downward along z-axis particle under non-relativistic conditions, is given by 1 with acceleration – 퐚퐳 (negative sign to show that the E = mv2 2

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Dr Dev Raj Mishra. International Journal of Engineering Research and Applications www.ijera.com ISSN: 2248-9622, Vol. 10, Issue 7, (Series-IV) July 2020, pp. 11-15

∂E ∂v vector components indicate that the variation in one Therefore = mv (37) ∂r ∂r component is related to that in the other component. This relationship can lead to new results in v 푚푣2 theoretical physics. v v+ v 푟 ACKNOWLEDGEMENT Author will like to thank Head, Department of v Physics, Kumaun University, Nainital, Uttarakhand, India and Principal, R.H. Govt PG College, Kashipur (U.S. Nagar), India for their generous support in pursuing these studies.

REFERENCES [1]. A. Einstein A Brief Outline of the Fig. 2: Particle rotating in circular path with Development of the Theory of Relativity velocity v experiences an outward force Nature 106, 1921, 782-784. 푚푣2 of . [2]. R.Adler,M.Bazin, , &M. Schiffer, 푟 Introduction to (Second The change in velocity is ∆v = ω∆r and is Edition) (McGraw-Hill, New York, 1975). along −∆r i.e. points towards the center of the circle. [3]. Bernard F. SchutzA First Course in General Therefore ∆v = −ω∆r. For small variations, this Relativity (Second Edition) (Cambridge becomes University Press, New York, 2009). ∂v ∆v [4]. Weblink: = lim = −ω www.cambridge.org/9780521887052 ∂r ∆r→0 ∆r Therefore from Eq.(37), we have ∂E DrDev Raj Mishra Gold = −mvω (38) ∂r Medallist at Graduation, did Writing again Eq. (14(i)) M.Sc. from M.D. University, Rohtak (Haryana) India. Did his ∂E ∂px + = 0 (39) Ph.D. in High Temperature ∂x ∂t Let the particle is at an infinitely small Superconductivity from displacement from the x-axis,we can safely replace x Cryogenics in National Physical Laboratory by r, to get and University of Delhi, Delhi, India. Joined as Lecturer in the Department of Higher Education, ∂E ∂pr + = 0 Government of Uttarakhand in Jan. 2000. Presently ∂r ∂t Then from Eq. (38), we get working as Professor at Department of Physics, ∂p ∂E R.H. P.G. College, Kashipur, U.S.Nagar, India. r = − = mvω (40) ∂t ∂r Therefore ∂p mv2 v r = as ω = (41) ∂t r r This is obviously outward force acting along positive r direction that is alongp푟 and is known as centrifugal force. These were few examples to illustrate the validity of 훛 횲 퐩 = ퟎ as covariant form substitute for Newton’s Second law.

IV. CONCLUSION Thecovariant form of Newton’s Second Law of motionsuccessfully explain the various pseudo-forces observed in nature and also confirmthe known relations between the electromagnetic potentials,fields and the field tensor components.The relationship between spatial temporal variations of Energy-Momentum four-

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