Charged particle in a electromagnetic field 2
Seminar 4: CHARGED PARTICLE IN ELECTROMAGNETIC FIELD Introduction Let take Lagrange’s equations in the form that follows from D’Alembert’s principle, d dt ∂T ∂ ˙q j − ∂T ∂q j = Q j , (1) and suppose that the generalized force is derivable not from the potential V (q j ) but from a more general function U(q j , ˙q j ), by the prescription Q j = − ∂U ∂q j + d dt ∂U ∂ ˙q j . (2) Substituting this expression for the f orce into Eqs. (1) yields d dt ∂T ∂ ˙q j − ∂T ∂q j = − ∂U ∂q j + d dt ∂U ∂ ˙q j , (3) or d dt ∂(T − U) ∂ ˙q j − ∂(T − U) ∂q j = 0. (4) It follows that these equations can be written in the form of Lagrange’s equations, d dt ∂L ∂ ˙q j − ∂L ∂q j = 0, (5) if we use as the Lagrangian the function L = T − U. (6) The function U is called usually the velocity-dependent potential (sometimes the term generalized potential is also used). It can be thought that the possibility of using such a strange potential is purely academic but this is not the case! On the contrary, it appears that all the fundamental forces in physics can be expressed in the form (2), for a suitably chosen potential function U. Its near practical importance relates to the theory of an electric charge in an electromagnetic field. As you know, a charge q moving with the velocity v in an electromagnetic field, containing both an electric, E, and magnetic, B, fields, experiences a force (Note: 1/c appears in Gauss system of units, in SI system it will be absent) F = q E + 1 c (v × B) , (7) which is called the Lorentz force. Both vectors E(r, t) and B(r, t) are continuous functions of time t and position r = (x, y, z) derivable from the scalar and vector potentials ϕ(r, t) and B(r, t) by E = −∇ϕ − 1 c ∂A ∂t (8) and B = ∇ × A. (9) Here, ∇ is the differential operator defined in Cartesian coordinates by ∇ ≡ ∂ ∂x , ∂ ∂y , ∂ ∂z , (10) so that ∇ϕ = ∂ϕ ∂x ˆ x + ∂ϕ ∂y ˆ y + ∂ϕ ∂z ˆ z (11) and ∇ × A = ˆ x ˆ y ˆ z ∂ ∂x ∂ ∂y ∂ ∂z A x A y A z , (12) where ˆ x, ˆ y and ˆ z are the unit vectors along x-, y- and z axes, respectively. Notice that the electromagnetic field defined by Eqs. (8-9) don’t change when the potentials are transformed according to: ϕ = ϕ − ∂ψ ∂t , A = A + ∇ψ, (13) where ψ is an arbitrary function of the coordinates and time. These transformations are known as the gauge transformations. Problem 14. Lagrangian of Charged Particle in Electromagnetic Field Show that the Lagrangian of a particle with the charge q moving with the velocity v in an electromagnetic field given by the scalar and vector potentials ϕ and A is L = 1 2 mv 2 − qϕ + q c A · v. (14) Solution: Taking the vectors of the electric and magnetic fields E and B represented in terms of the scalar and vector potentials in accordance with Eqs. (8) and (9), we have for the Lorentz force F = q −∇ϕ − 1 c ∂A ∂t + 1 c v × (∇ × A) . (15) Let calculate, say, the x-component of this vector force, F x = q −(∇ϕ) x − 1 c ∂A x ∂t + 1 c v × (∇ × A) x . (16) Using the definitions (11) and (12), we obtain (∇ϕ) x = ∂ϕ ∂x (17) and v × (∇ × A) x = v y ∂A y ∂x − ∂A x ∂y − v z ∂A x ∂z − ∂A z ∂x = v y ∂A y ∂x + v z ∂A z ∂x + v x ∂A x ∂x −v y ∂A x ∂y − v z ∂A x ∂z − v x ∂A x ∂x = ∂ ∂x (v · A) − dA x dt + ∂A x ∂t , (18) since ∂ ∂x (v · A) = ∂ ∂x (v x A x + v y A y + v z A z ) = v x ∂A x ∂x + v y ∂A y ∂x + v z ∂A z ∂x (19) and dA x dt = ∂A x ∂t + v x ∂A x ∂x + v y ∂A x ∂y + v z ∂A x ∂z . (20) Substituting Eqs. (17) and (18) into Eq. (16), we finally obtain F x = q − ∂ϕ ∂x − 1 c ∂A x ∂t + 1 c ∂ ∂x (v · A) − 1 c dA x dt + 1 c ∂A x ∂t = q − ∂ ∂x ϕ − 1 c v · A − 1 c dA x dt = q − ∂ ∂x ϕ − 1 c v · A + d dt ∂ ∂v x ϕ − 1 c v · A , (21) which can be written as F x = − ∂U ∂x + d dt ∂U ∂v x , (22) where we introduce the function U = qϕ − q c v · A = qϕ − q c A · v. (23) Note that the term in the right-hand side of this equation coincides with the potential V defined by Eq. (2.119) from our Lecture notes in a particular case of a single particle of charge q. Comparing the expression for the Lorentz force in the form (22) with the definition (2) of the generalized force in terms of the velocity-dependent potential, we see that in our case this potential is defined just by the equation (23). This observation immediately yields the Lagrangian in the desired form given by Eq. (6), L = T − U = 1 2 mv 2 − qϕ + q c A · v. (24) Problem 15. Calculate the conjugate momentum p and the energy function h for a particle of the mass m and charge q in an electromagnetic field given by the scalar and vector potentials ϕ and A. Solution: The x-comp onent of the conjugate momentum is p x = ∂L ∂ ˙x = ∂ ∂v x 1 2 m[v 2 x +v 2 y +v 2 z ]− qϕ+ q c [A x v x +A y v y +A z v z ] = mv x + q c A x . (25) The same for y- and z-components p y = mv y + q c A y (26) and p z = mv z + q c A z . (27) The second terms in these expressions play the role of a potential momentum. The energy function is h = j ˙q j ∂L ∂ ˙q j − L = j ˙q j p j − L = v x p x + v y p y + v z p z − 1 2 mv 2 − qϕ + q c A · v = v x mv x + q c A x + v y mv y + q c A y + v z mv z + q c A z − 1 2 mv 2 − qϕ + q c A · v = mv 2 + q c A · v − 1 2 mv 2 − qϕ + q c A · v = 1 2 mv 2 + qϕ = T + qϕ. (28) If A and ϕ are independent of t, then L does not depend on t explicitly and the energy function is constant, that is T + qϕ = const. (29) Since the second term in this equation is nothing but the potential energy of a charge particle, we see that in this case the total energy of the system is conserved as it might be. Problem 16. A particle of the mass m and charge q moves in a constant magnetic field B = (0, 0, B). (30) Show that the orbit of a particle is a helix. Solution: Let specify the scalar and vector potentials for the case when the electric field is absent, E = 0, (31) and magnetic field has only non-zero component B z = B. Keeping in mind that this component is expressed in terms of the vector potential A as B z = ∂A y ∂x − ∂A x ∂y , (32) we can choose the potentials in the form A = (0, Bx, 0), ϕ = 0. (33) With this choice, the Lagrangian is L = m 2 ( ˙x 2 + ˙y 2 + ˙z 2 ) + q c Bx ˙y. (34) The corresp onding Lagrange’s equation are written as d dt (m ˙x) − q c B ˙y = 0 d dt (m ˙y + q c Bx) = 0 d dt (m ˙z) = 0. (35) From the second and third equations it follows ˙y = C − ωx (36) and z = z 0 + Dt, (37) where C, D, z 0 are constants and the substitution ω = qB mc (38) has been made. With the value of the ˙y given by (36), the first equation from (35) takes the form ¨x − ω(C − ωx) = 0, (39) or ¨x + ω 2 (x − x 0 ) = 0, (40) where x 0 = C ω . (41) The general solution of Eq. (40) can b e written as x − x 0 = a cos(ωt + δ). (42) Then ˙y = C − ωx = ω(x 0 − x) = −aω cos(ωt + δ). (43) Integrating this equation yields y = −a sin(ωt + δ) + y 0 . (44) Finally, combining Eqs. (42) and (44) we obtain (x − x 0 ) 2 + (y − y 0 ) 2 = a 2 . (45) Together with Eq. (37) for z-component, this defines a helix as a trajectory of a particle. Problem 17. Find the eigenfrequencies for an isotropic three-dimensional har- monic oscillator realized as a particle of charge q placed in a uniform magnetic, B, and electric, E, fields which are mutually perpendicular and take their directions along z- and x-axes, respectively. Solution: As the particle is an isotropic harmonic oscillator and has the charge q, its potential energy may be written as V = 1 2 kr 2 + qϕ − q c A · v ≡ 1 2 mω 2 0 r 2 + qϕ − q c A · v, (46) where we introduced the natural angular frequency of an isotropic oscillator ω 0 = k m , (47) and r = (x, y, z) is the displacement of the particle from the origin. We can easily check also that the configuration of electric and magnetic field given in the problem can be realized by the choice of the scalar and vector potentials in the form ϕ = −Ex, A = − 1 2 By, 1 2 Bx, 0 . (48) Indeed, with this choice we have E = −∇ϕ = ˆxE (49) and B z = ∂A y ∂x − ∂A x ∂y = ∂ ∂x 1 2 Bx − ∂ ∂y − 1 2 By = 1 2 B + 1 2 B = B, (50) as it should be. It is instructive to notice that in previous problem to obtain the same result for B we used another choice of A given by Eq. (33). Now we are able to write the total Lagrangian as L = 1 2 m( ˙x 2 + ˙y 2 + ˙z 2 ) − 1 2 mω 2 0 (x 2 + y 2 + z 2 ) + qEx + 1 2c qB(− ˙xy + x ˙y). (51) This Lagrangian create Lagrange’s equations ¨x + ω 2 0 x − qB mc ˙y − qE m = 0 ¨y + ω 2 0 y + qB mc ˙x = 0 ¨z + ω 2 0 z = 0. (52) The general solution of the last equation is z = z 0 cos(ω 0 t + δ), (53) that is the oscillation in z-direction takes place with the natural angular frequency ω 0 . By the change of variables x = x + qE mω 2 0 , (54) the first two equations can be represented in a more symmetric form ¨x + ω 2 0 x − ω L ˙y = 0 ¨y + ω 2 0 y + ω L ˙x = 0, (55) where ω L = qB mc . (56) This is the system of the coupled linear differential equations of the second order, and hence we can try a solution of type x = Ae iωt , y = Be iωt . (57) Then the system (55) changes to the system of the algebraic equations which is written in matrix form as ω 2 0 − ω 2 −iω L ω iω L ω ω 2 0 − ω 2 A B = 0. (58) So the secular equation is ω 2 0 − ω 2 −iω L ω iω L ω ω 2 0 − ω 2 = (ω 2 0 − ω 2 ) 2 − (ω L ω) 2 = 0. (59) This equation is equivalent two the pair of equations, ω 2 + ω L ω − ω 2 0 = 0; ω 2 − ω L ω − ω 2 0 = 0, (60) which has two positive roots ω + = 1 2 [ω L + ω 2 L + 4ω 2 0 ], ω − = 1 2 [−ω L + ω 2 L + 4ω 2 0 ]. (61) Hence the oscillator in a combined electric and magnetic field exhibits the three eigenfrequencies ω 0 , ω + and ω − . Note that first mode does not depend on the applied fields at all, and the last two modes are caused by the magnetic field alone, whereas the electric field only causes a displacement qE mω 2 0 along its direction. For a weak magnetic field, ω L << ω 0 , (62) the frequencies ω ± are approximated to the form ω + = ω 0 + ω L 2 , ω − = ω 0 − ω L 2 , (63) while in an opposite case of a strong magnetic field, ω L >> ω 0 , (64) we have ω + ≈ 1 2 ω L + 1 + 2ω 2 0 ω 2 L = ω L + ω 2 0 ω L , ω − ≈ 1 2 −ω L 1 + 2ω 2 0 ω 2 L = ω 2 0 ω L . (65) . − ∂ψ ∂t , A = A + ∇ψ, (13) where ψ is an arbitrary function of the coordinates and time. These transformations are known as the gauge transformations. Problem 14. Lagrangian of Charged Particle in Electromagnetic. Electromagnetic Field Show that the Lagrangian of a particle with the charge q moving with the velocity v in an electromagnetic field given by the scalar and vector potentials ϕ and A is L = 1 2 mv 2 −. Seminar 4: CHARGED PARTICLE IN ELECTROMAGNETIC FIELD Introduction Let take Lagrange’s equations in the form that follows from D’Alembert’s principle, d dt ∂T ∂ ˙q j − ∂T ∂q j = Q j , (1) and