# Seminar 4: CHARGED PARTICLE IN ELECTROMAGNETIC FIELD. q j

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1 Seminar 4: CHARGED PARTICLE IN ELECTROMAGNETIC FIELD Introduction Let take Lagrange s equations in the form that follows from D Alembert s principle, ) d T T = Q j, 1) dt q j q j suppose that the generalized force is derivable not from the potential V q j ) but from a more general function Uq j, q j ), by the prescription Q j = U + d ) U. ) q j dt q j Substituting this expression for the force into Eqs. 1) yields ) d T T = U + d ) U, 3) dt q j q j q j dt q j or ) d T U) dt q j T U) q j = 0. 4) It follows that these equations can be written in the form of Lagrange s equations, ) d L L = 0, 5) dt q j q j 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 ), 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, magnetic, B, fields, experiences a force Note: 1/c appears in Gauss system of units, in SI system it will be absent) which is called the Lorentz force. F = q [ E + v B) ], 7) Both vectors Er, t) Br, t) are continuous functions of time t position r = x, y, z) derivable from the scalar vector

2 potentials ϕr, t) Br, t) by E = ϕ A t 8) B = A. 9) Here, is the differential operator defined in Cartesian coordinates by, y, ), 10) z so that ϕ = ϕ ϕ ˆx + y ŷ + ϕ z ẑ 11) ˆx ŷ ẑ A = y z, A x A y A z 1) where ˆx, ŷ ẑ are the unit vectors along x-, y- 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 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 vector potentials ϕ A is L = 1 mv qϕ + qa v. 14) Solution: Taking the vectors of the electric magnetic fields E B represented in terms of the scalar vector potentials in accordance with Eqs. 8) 9), we have for the Lorentz force F = q [ ϕ A t + v A) ) ]. 15)

3 Let calculate, say, the x-component of this vector force, [ F x = q ϕ) x A x + v A) ) ]. 16) t x Using the definitions 11) 1), we obtain since v A) ) x = v y = v y A y v y A x y ϕ) x = ϕ Ay + v z Az v z Ax z ) Ax y Ax vz z + v x Ax v x Ax ) Az 17) = dax v A) + Ax, 18) dt t v A) = v A x xa x + v y A y + v z A z ) = v x + v A y y + v A z z da x = A x A x + v x dt t + v A x y y + v A x z z. 0) Substituting Eqs. 17) 18) into Eq. 16), we finally obtain F x = q [ ϕ Ax + dax v A) t dt q [ ] [ + Ax t = q ) ] ϕ v A da x dt = 19) ) )] ϕ v A + d dt v x ϕ v A, 1) which can be written as F x = U + d U, ) dt v x where we introduce the function U = qϕ qv A = qϕ qa v. 3) Note that the term in the right-h side of this equation coincides with the potential V defined by Eq..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 ) with the definition ) 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 3). immediately yields the Lagrangian in the desired form given by Eq. 6), This observation L = T U = 1 mv qϕ + qa v. 4)

4 Problem 15. Calculate the conjugate momentum p the energy function h for a particle of the mass m charge q in an electromagnetic field given by the scalar vector potentials ϕ A. Solution: The x-component of the conjugate momentum is p x = L ẋ = v x { } 1 m[v x + vy + vz] qϕ + q[a x v x + A y v y + A z v z ] The same for y- z-components = mv x + qa x. 5) p y = mv y + qa y 6) p z = mv z + qa z. 7) 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 mv qϕ + qa v ) = v x mvx + qa x ) + vy mvy + qa y ) + vz mvz + qa z ) 1 mv qϕ + qa v ) = mv + qa v 1 mv qϕ + qa v ) = 1 mv + qϕ = T + qϕ. 8) If A ϕ are independent of t, then L does not depend on t explicitly the energy function is constant, that is T + qϕ = const. 9) 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 charge q moves in a constant magnetic field B = 0, 0, B). 30) Show that the orbit of a particle is a helix.

5 Solution: Let specify the scalar vector potentials for the case when the electric field is absent, E = 0, 31) 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 we can choose the potentials in the form With this choice, the Lagrangian is B z = A y A x y, 3) A = 0, Bx, 0), ϕ = 0. 33) L = m ẋ + ẏ + ż ) + qbxẏ. 34) The corresponding Lagrange s equation are written as d mẋ) qbẏ = 0 dt d mẏ + qbx) = 0 35) dt d mż) = 0. dt From the second third equations it follows ẏ = C ωx 36) z = z 0 + Dt, 37) where C, D, z 0 are constants the substitution ω = qb m 38) has been made. With the value of the ẏ given by 36), the first equation from 35) takes the form ẍ ωc ωx) = 0, 39) or where ẍ + ω x x 0 ) = 0, 40) x 0 = C ω. 41)

6 The general solution of Eq. 40) can be written as x x 0 = a cosωt + δ). 4) Then ẏ = C ωx = ωx 0 x) = aω cosωt + δ). 43) Integrating this equation yields y = a sinωt + δ) + y 0. 44) Finally, combining Eqs. 4) 44) we obtain x x 0 ) + y y 0 ) = a. 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 harmonic oscillator realized as a particle of charge q placed in a uniform magnetic, B, electric, E, fields which are mutually perpendicular take their directions along z- x-axes, respectively. Solution: As the particle is an isotropic harmonic oscillator has the charge q, its potential energy may be written as V = 1 kr + qϕ qa v 1 mω 0r + qϕ qa v, 46) where we introduced the natural angular frequency of an isotropic oscillator k ω 0 = m, 47) r = x, y, z) is the displacement of the particle from the origin. We can easily check also that the configuration of electric magnetic field given in the problem can be realized by the choice of the scalar vector potentials in the form Indeed, with this choice we have ϕ = Ex, A = 1 By, 1 Bx, 0). 48) E = ϕ = ˆxE 49)

7 B z = A y A x y = [ 1 Bx] [ 1 y By] = 1 B + 1 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 mẋ + ẏ + ż ) 1 mω 0x + y + z ) + qex + 1 qb ẋy + xẏ). 51) This Lagrangian create Lagrange s equations ẍ + ω0x qb ẏ qe = 0 m m ÿ + ω 0y + qb ẋ = 0 m z + ω0z = 0. The general solution of the last equation is 5) z = z 0 cosω 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, 54) mω0 the first two equations can be represented in a more symmetric form { ẍ + ω 0x ω L ẏ = 0 ÿ + ω 0y + ω L ẋ = 0, 55) where ω L = qb m. 56) This is the system of the coupled linear differential equations of the second order, 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 0 ω ) ) iω L ω A iω L ω ω0 ω = 0. 58) B So the secular equation is ω0 ω iω L ω iω L ω ω0 ω = ω 0 ω ) ω L ω) = 0. 59)

8 This equation is equivalent two the pair of equations, { ω + ω L ω ω 0 = 0; ω ω L ω ω 0 = 0, 60) which has two positive roots ω + = 1[ω L + ωl + 4ω0], ω = 1[ ω L + ωl + 4ω0]. 61) Hence the oscillator in a combined electric magnetic field exhibits the three eigenfrequencies ω 0, ω + ω. Note that first mode does not depend on the applied fields at all, the last two modes are caused by the magnetic field alone, whereas the electric field only causes a displacement magnetic field, qe mω 0 the frequencies ω ± are approximated to the form along its direction. For a weak ω L << ω 0, 6) ω + = ω 0 + ω L, while in an opposite case of a strong magnetic field, we have ω = ω 0 ω L, 63) ω L >> ω 0, 64) ω + [ 1 ωl + )] 1 + ω 0 ω = L ωl + ω 0 ω 1 [ ωl 1 + ω 0 ω L ω L, )] = ω 0 ω L. 65)

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