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Angular Momentum, Electromagnetic Waves Lecture33: Electromagnetic Theory Professor D. K. Ghosh, Physics Department, I.I.T., Bombay As before, we keep in view the four Maxwellβs equations for all our discussions. π» β πΈβ = π β = πππππ βπ»β π· π0 β =0 π»β π΅ β ππ΅ ππ‘ β ππ· β =π½+ π»×π» ππ‘ π» × πΈβ = β In the last lecture, we have seen that the electromagnetic field carries both energy and momentum and any discussion on conservation of these two quantities must keep this into account. In this lecture, we will first talk about the possibility of angular momentum being associated with the electromagnetic field and in the second half, introduce the electromagnetic waves. Angular Momentum Electromagnetic field, in addition to storing energy and momentum, also has angular momentum and this can be exchanged with the charged particles of the system. π 1 β . Based We have seen that the momentum density of the electromagnetic field is given by π 2 = π 2 πΈβ × π» on this, we define the angular momentum density of the electromagnetic field about an arbitrary origin as πππ = 1 β) π × (πΈβ × π» π2 so that the angular momentum of the field is given by πΏβππ = 1 β )ππ β« π × (πΈβ × π» π2 Let us start with a collection of charges and currents. The expression for the force density (i.e. the rate of change of momentum density) for the particles is given by the Lorentz force expression π β) π = π × (ππΈβ + π½ × π΅ ππ‘ πππβ We will do some aklgebraic manipulation on the right hand side by using Maxellβs equations, We take the medium to be the free space. 1 ππΈβ 1 β β π0 . With this First, we substitute π0 (β β πΈβ ) for ο²ο and replace for the current density, π½ = π π» × π΅ ππ‘ 0 we have, π 1 ππΈβ β β π0 ) × π΅ β] ππππβ = π × [π0 πΈβ(β β πΈβ ) + ( π» × π΅ ππ‘ π0 ππ‘ We next add some terms to makes this expression look symmetrical in electric and magnetic fields. To 1 β (β β π΅ β ) , which is identically zero. To make the second term match the first term we add a term π΅ π0 symmetric, we add and subract a term π0 ( π» × πΈβ ) × πΈβ . With these we get, π 1 ππΈβ β (β β π΅ β )βπ΅ β × (π» × π΅ β )] β π0 π × ( × π΅ β) ππππβ = π0 π × [πΈβ (β β πΈβ ) β πΈβ × ( π» × πΈβ )] + π × [π΅ ππ‘ π0 ππ‘ + π0 π × πΈβ × ( π» × πΈβ ) where we have interchanged the order of cross product in a couple of places by changing sign. β ππ΅ The last two terms on right can be simplified as follows. Using π» × πΈβ = β ππ‘ , we can write, βπ0 π × ( β ππΈβ ππΈβ ππ΅ β ) + π0 π × πΈβ × ( π» × πΈβ ) = βπ0 π × ( × π΅ β ) β π0 π × πΈβ × (β ) ×π΅ ππ‘ ππ‘ ππ‘ β) π(πΈβ × π΅ ππ = βπ0 π × = βπ0 π0 π × ππ‘ ππ‘ π = β πππ ππ‘ Thus we have, π 1 β (β β π΅ β )βπ΅ β × (π» × π΅ β )] (ππππβ + πππ ) = π0 π × [πΈβ (β β πΈβ ) β πΈβ × ( π» × πΈβ )] + π × [π΅ ππ‘ π0 We recall that the elements of the stress tensor was defined as 1 1 1 β πΌ,π½ = π0 [πΈπΌ πΈπ½ β |πΈ|2 πΏπΌ,π½ ] + [π΅πΌ π΅π½ β |π΅|2 πΏπΌ,π½ ] π 2 π0 2 β‘), so that It can be shown (see Tutorial assignment 1) that the right hand side simplifies to π × (β β π π β‘) + πππ ) = π × (β β π (π ππ‘ πππβ On integrating over the volume one gets a statement of conservation of angular momentum. π β‘) β ππ + β« πππ π 3 π) = β« π × (π (πΏβ ππ‘ πππβ π£ππ 2 which sates that the total rate of change of momentum (of particles and field) is equal to the flux of the torque through the surface. Example : The Feynman Paradox β a variant of Feynman disk We discuss a variant of the famous Feynman disk problem, which is left as an exercise. There is an infinite line charge of charge density βπ is surrounded by an insulating cylindrical surface of radius a having a surface charge density π = + π , 2ππ so that the net charge of the system is zero. The cylinder can freely rotate about the z axis, which coincides with the line charge. Because of Gaussβs law, the electric field exists only within the cylinder. The system is β = π΅0 π§Μ along the z axis. The system is initially at rest. If immersed in a uniform magnetic field π΅ the magnetic field is now reduced to zero, the cylinder will be found to rotate. βπ The explanation of rotation lies in conservation of angular momentum. As the system was initially at rest, the initial mechanical angular momentum is zero. The field angular momentum can be calculated as follows. The electric field is given by π πΈβ = β πΜ . The magnetic field is circumferential 2ππ0 π π β = π΅0 π§Μ . and is given by π΅ Thus, the initial field angular momentum per unit length is given by, (taking the angular momentum about the axis) πΏβ = β« (π × =β where we have used π0 π0 = π ) 2ππ ππ π2 π ππ΅0 ππ΅0 π2 2π β« [πΜ × (πΜ × π§Μ )]πππ = π§Μ 2π 2 0 1 . π2 This is the net angular momentum of the field plus the mechanical system because the latter is at rest. If the magnetic field is reduced to zero, because of a changing flux through any surface parallel to the xy plane, there is an azimuthal current generated. This is caused by the rotating cylinder which rotates with an angular speed π, giving rise to an angular momentum πΌπ, where I is the moment of inertia per unit length of the cylinder about the z axis. If the time period of rotation is π, the current per unit length is given by π½π = π π = 2πππ π 2π = πππ = ππ . 2π Since the current is circumferential, the magnetic field (like in a solenoid) is along the z direction and is β πππππ = π0 π½π§Μ = π0 ππ π§Μ . The final angular momentum is thus given by given by π΅ 2π ππ΅πππππ π2 π0 π2 π2 π§Μ = π 2 4π 3 . Thus we must have, πΌπ + π0 π2 π2 ππ΅0 π2 π= 4π 2 which allows us to determine π π= ππ΅0 π2 1 π 2 πΌ + 0 π2 π 2 4π Example : Radiation Pressure An experimentally verifiable example of the fact that the electromagnetic field stores momentum is provided by the pressure exerted by radiation confined inside a cavity. Consider an enclosure in the shape of a rectangular parallelepiped and let us consider the right most wall of the enclosure. The normal direction to the wall is to the let and the fieldexerts a force on this wall along the x direction. The force exerted on an area dS of the wall is β‘ β ππ given by ππΉ = π Since the force is exerted in x direction we only need πΜ 1 1 1 ππ₯π₯ = π0 (πΈπ₯2 β |πΈ|2 ) + (π΅π₯2 β |π΅|2 ) 2 π0 2 If the radiation is isotropic, we can write, 1 3 1 3 πΈπ₯2 = |πΈ|2 and π΅π₯2 = |π΅|2 so that, 1 π΅2 1 ππΉπ₯ = (π0 πΈ 2 + )= π’ 6 2π0 3 where u is the energy density. This, incidentally, was the starting point for proving Stefan Boltzmannβs law. Plane Wave solutions to Maxwellβs Equations In the following we obtain a special solution to the Maxwellβs equations for a linear , isotropic medium which is free of sources of charges and currents. We have then the following equations to solve: π» β πΈβ = 0 β =0 π»β π΅ π» × πΈβ = β 4 β ππ΅ ππ‘ β = ππ π»×π΅ ππΈβ ππ‘ Taking the curl of the third equation, and substituting the first equation therein, we have, π» × (π» × πΈβ ) = π»(π» β πΈβ ) β π» 2 πΈβ = βπ» 2 πΈβ = β β) π(π» × π΅ ππ‘ Substituting last equation in this, we get, π» 2 πΈβ = ππ π 2 πΈβ ππ‘ 2 In a similar way, we get an identical looking equation for the magnetic field, β = ππ π»2π΅ β π2π΅ 2 ππ‘ These represent wave equations with the velocity of the wave being 1/βππ. It may be noted that, in a general curvilinear coordinate system, we do not have, (π» 2 πΈβ)π₯ = π» 2 πΈβπ₯ . However, it would be true in a Cartesian coordinates where the unit vectors are fixed. Let us look at βplane waveβ solutions to these equations. A plane wave is one for which the surfaces of constant phases, viz. wavefronts, are planes. Time harmonic solutions are of the form sin(πβ β π β ππ‘) or cos(πβ β π β ππ‘). However, mathematically it turns out to be simple to consider an exponential form and take, at the end of calculations, the real or the imaginary part. We take the solutions to be of the form, β πΈβ = πΈβ0 π π(πβ πβππ‘) β =π΅ β 0 π π(πββ πβππ‘) π΅ Note that the surfaces of constant phase are given by πβ β π β ππ‘ = constant At any time t, since ππ‘ = constant, we have, the surface given by π = πβ β π = constant Represent the wave-fronts. This is obviously an equation to a plane.πβ is known as the βpropagation vectorβ. As time increases these wave fronts move forward, I.e the distance from source increases. One β could also look at the backward moving waves which would be of the form πΈβ = πΈβ0 π π(πβ π+ππ‘) . 5 As time progresses, the surfaces of constant phase satisfy the equation |π|π β ππ‘ = constant where π = πΜ β π. Thus the angular frequency ο·ο ο ο ο©s given by π=π ππ ππ‘ = π π ππ ππ‘ = π£ is known as the βphase velocityβ. Substituting our solutions into the two divergence equations, we get, πβ β πΈβ = 0 β =0 πβ β π΅ Thus the direction of both electric and magnetic fields are perpendicular to the propagation vector. If we restrict ourselves to non-conducting media, we would, in addition, have, from the curl equation, β = ππ π»×π΅ ππΈβ ππ‘ β = βπππππΈβ ππβ × π΅ This also shows that the electric field is perpendicular to the magnetic field vector. Thus the electric field, the magnetic field and the direction of propagation form a right handed triad. If the propagation vector is along the z direction, the electric and magnetic field vectors will be in the xy plane being perpendicular each other. Note that using the exponential form has the advantage that the action of operator π» is equivalent to replacing it by ππβ and the time derivative is equivalent to a multiplication by β ππ. β = βπππππΈβ with πβ to get, We can take the cross product of the equation ππβ × π΅ β ) = βππππβ × πΈβ πβ × (πβ × π΅ expanding the scalar triple product, β ) β π2π΅ β = βππππβ × πΈβ πβ (πβ β π΅ β = 0, we get, substituting πβ β π΅ β = π΅ πππ πππ πβ × πΈβ = πΜ × πΈβ 2 π π = The ratio of the strength of the magnetic field to that of the electric field is given by 6 πππ π = 1 π£ Where we have used the expressions for the velocity of the wave, π£ = π π = 1 . βππ In free space this velocity is the same as the speed of light which explains why the magnetic field associate with the electromagnetic field is difficult to observe. E βπ B In the above figure the electric field variation is shown by the green curve and that of magnetic field by the orange curve. We have seen that electromagnetic field stores energy and momentum. A propagating electromagnetic wave carries energy and momentum in its field. The energy density of the electromagnetic wave is given by |π΅|2 1 π’ = π|πΈ|2 + 2 2π |π΅|2 1 = π (|πΈ|2 + ) = π|πΈ|2 2 ππ where we have used the relation |πΈ| = |π΅| βππ . The Poynting vector is given by β = π = πΈβ × π» β πΈβ × π΅ π Recall that the difference between H and B is due to bound currents which cannot transport energy. Replacing the real form of the electromagnetic field, we get, taking the electric field along the x direction, the magnetic field along the y direction and the propagation vector along the z direction, 7 π = ππ0 πΈ 2 cos 2(ππ§ β ππ‘)πΜ where we have assumed the electromagnetic wave to be travelling in free space. The intensity of the wave is defined to be the time average of the Poynting vector, 1 πΌ = ππ0 πΈ 2 πΜ 2 where the factor of ½ comes from the average of the cos 2 (ππ§ β ππ‘) over a period. In general, given the propagation direction, the electric and the magnetic fields are contained in a perpendicular to it. The fields can point in arbitrary direction in the plane as long as they are perpendicular to each other. If the direction of the electric field is random with time, the wave will be known as an βunpolarized waveβ. Specifying the direction of the electric field (or equivalently of the orthogonal magnetic field) is known as a statement on the statement of polarization of the wave. Consider the expression for the electric field at a point at a given time. Assuming that it lies in the xy plane, we can write the following expression for the electric field πΈβ (π§, π‘) = π π[(πΈ0π₯ πΜ + πΈ0π¦ πΜ)π π(ππ§βππ‘) ] In general, the quantities inside the bracket are complex. However, we can specify some specific relations between them. 1. Suppose πΈ0π₯ and πΈ0π¦ are in phase, i.e., suppose, πΈ0π₯ = |πΈ0π₯ |π ππ πΈ0π¦ = |πΈ0π¦ |π ππ then we can express the field as πΈβ (π§, π‘) = (|πΈ0π₯ |πΜ + |πΈ0π¦ |πΜ) cos(ππ§ β ππ‘ + π) 2 The magnitude of the electric vector changes from 0 to β|πΈ0π₯ |2 + |πΈ0π¦ | but its direction remains constant as the wave propagates along the z direction. Such a wave is called βlinearly polarizedβ wave. The wave is also called plane βplane polarizedβ as at a given time the electric vectors at various locations are contained in a plane. 2. Instead of the phase between the x and y components of the electric field being the same, suppose the two components maintain a constant phase difference as the wave moves along, we have πΈ0π₯ = |πΈ0π₯ | πΈ0π¦ = |πΈ0π¦ |π ππ π If the phase difference happens to be 2 , we can write the expression for the electric field as πΈβ (π§, π‘) = π π[(πΈ0π₯ πΜ + πΈ0π¦ πΜ)π π(ππ§βππ‘) ] 8 = |πΈ0π₯ | cos(ππ§ β ππ‘)πΜ + |πΈ0π¦ | sin(ππ§ β ππ‘)πΜ Consider the time variation of electric field at a particular point in space, say at z=0. As t π increases from 0 to 2π , πΈπ₯ decreases from its value |πΈ0π₯ | to zero while πΈπ¦ increases from zero to |πΈ0π¦ | the electric vector rotating counterclockwise describing an ellipse. This is known as βelliptic polarizationβ. (In the general case of an arbitrary but constant phase difference, the state of polarization is elliptic with axes making an angle with x and y axes. Alternatively, a state of arbitrary polarization can be expressed as a linear combination of circular or linear polarizations.) 3. A special case of elliptic polarization is βcircular polarizationβ where the amplitudes of the two components are the same, viz., |πΈ0π₯ | = |πΈ0π¦ |, when the electric vector at a point describes a circle with time. Angular Momentum, Electromagnetic Waves Lecture33: Electromagnetic Theory Professor D. K. Ghosh, Physics Department, I.I.T., Bombay 9 Tutorial Assignment 1. Prove the relation π0 π × [πΈβ (β β πΈβ ) β πΈβ × ( π» × πΈβ )] + 1 β (β β π΅ β)βπ΅ β × (π» × π΅ β )] = π × (β β π β‘) π × [π΅ π0 2. Two concentric shells of radii a and b carry charges ±π. At the centre of the shells a dipole of magnetic moment ππΜ is located. Find the angular momentum in the electromagnetic field for this system. Solutions to Tutorial Assignments 1. We will simplify only the electrical field term, the magnetic field term follows identically. We have, πΈβ (β β πΈβ ) β πΈβ × ( π» × πΈβ ) ππΈπ₯ ππΈπ¦ ππΈπ§ + + ) β (πΜπΈπ₯ + πΜπΈπ¦ + πΜ πΈπ§ ) ππ₯ ππ¦ ππ§ ππΈπ¦ ππΈπ₯ ππΈπ§ ππΈπ¦ ππΈπ₯ ππΈπ§ × [πΜ ( β β ) + πΜ ( β ) + πΜ ( )] ππ¦ ππ§ ππ§ ππ₯ ππ₯ ππ¦ = (πΜπΈπ₯ + πΜπΈπ¦ + πΜ πΈπ§ ) ( Let us consider the x component of both sides, ππΈπ¦ ππΈπ₯ ππΈπ₯ ππΈπ¦ ππΈπ§ ππΈπ₯ ππΈπ§ + + β β ) [πΈβ (β β πΈβ ) β πΈβ × ( π» × πΈβ )]π₯ = πΈπ₯ ( ) β πΈπ¦ ( ) + πΈπ§ ( ππ₯ ππ¦ ππ§ ππ₯ ππ¦ ππ§ ππ₯ 1 π π π = (πΈπ₯2 β πΈπ¦2 β πΈπ§2 ) + (πΈπ₯ πΈπ¦ ) + (πΈπ₯ πΈπ§ ) 2 ππ₯ ππ¦ ππ§ π 2 π π π = πΈ β |πΈ|2 + (πΈ πΈ ) + (πΈπ₯ πΈπ§ ) ππ₯ π₯ ππ₯ ππ¦ π₯ π¦ ππ§ The y and z components follow by symmetry. The x component of the divergence of the stress tensor can be seen to be the same, as, ππ ππ ππ β‘) = π₯π₯ + π₯π¦ + π₯π§ (β β π π₯ ππ₯ ππ¦ ππ§ 2 π πΈ π π = (πΈπ₯2 β ) + (πΈπ₯ πΈπ¦ ) + (πΈπ₯ πΈπ§ ) ππ₯ 2 ππ¦ ππ§ 10 β = 2. In a spherical coordinate system, the magnetic field due to the dipole is given by π΅ π0 π [2 cos ππΜ + sin ππΜ]. The electric field is confined only within the shells and is given by πΈβ = 4π π 3 π πΜ . 4ππ0 π 2 The Poynting vector is given by β πΈβ × π΅ ππ = sin ππΜ π0 16π 2 π0 π 5 The angular momentum density about the centre of the shells is then given by π= π ππ =β sin ππΜ 2 π 16π 2 π0 π 4 The total angular momentum of the electromagnetic field can be obtained by integrating this over the volume. However, since πΜ is not a constant unit vector, we first convert this to Cartesian and write the total angular momentum as ππ 1 πΏβ = β β« 4 sin π [cos π cos π πΜ + cos π sin ππΜ β sin ππΜ ]π3 π 2 16π π0 π π× The first two terms, when integrated gives zero because the integral over πvanishes. We are left with ππ 1 πΏβ = πΜ β« 4 sin2 ππ3 π 2 16π π0 π π π ππ 1 Μ 2π β« = π ππ β« sin3 πππ 2 16π 2 π0 π π 0 ππ 1 1 4 ππ 1 1 = ( β ) × πΜ = ( β ) πΜ 8ππ0 π π 3 6ππ0 π π 11 Angular Momentum, Electromagnetic Waves Lecture33: Electromagnetic Theory Professor D. K. Ghosh, Physics Department, I.I.T., Bombay Self Assessment Questions 1. A metal sphere of radius R has a charge Q and is uniformly magnetized with a ββ . Calculate the angular momentum of the field about the centre of the magnetization π sphere. 2. In Problem 1, if the magnetization of the sphere is gradually and uniformly reduced to zero (probably by heating the sphere through the Curie temperature), calculate the torque exerted by the induced electric field on the sphere and show that the angular momentum is conserved in the process. Solutions to Self Assessment Questions 1. The charge being uniformly distributed over its surface, the electric field exists only outside π the sphere and is given by πΈβ = 2 πΜ . The magnetic field inside is constant and is given β = outside the sphere by π΅ 4ππ0 π π0 π [2 cos ππΜ 4π π 3 + sin ππΜ]. One can calculate the Poynting vector and angular momentum density closely following Problem 2 of the tutorial. The total angular momentum is obtained by integrating over all space outside the sphere. Following this we get πππ0 1 πΏβ = πΜ β« 4 sin2 ππ3 π πΜ 2 16π π β π ππ 1 = π 2π β« ππ β« sin3 πππ πΜ 2 16π 2 0 π π 0 πππ0 1 4 = πΜ 8π π 3 Substituting π = 4ππ 3 π, 3 we get 12 πΏβ = 2πππ0 2 π πΜ 9 2. When the sphere is slowly demagnetized from M to zero, there is a changing flux which induces an electric field. The induced electric field is circumferential and can be calculated from Faradayβs law. The magnetic field due to the magnetized sphere inside the sphere is 2 constant and is given by 3 π0 π. Consider a circumferential loop on the surface between polar angles π and π + ππ. The radius of the circular loop is π sin π and the area of the loop 2 3 is π(π sin π) 2 . The changing flux through this area is π0 ππ π(π sin π) 2 ππ‘ and this changing flux is equal to the emf induced in the loop of radius π sin π, so that we have, the electric field magnitude to be given by 2 ππ πΈ (2ππ sin π) = π0 π(π sin π) 2 3 ππ‘ which gives, 1 ππ πΈβ = π0 π sin π πΜ 3 ππ‘ If we consider an area element ππ = 2ππ 2 sin πππ of the surface, the charge on the surface π ππ = π2ππ 2 sin πππ = 2 sin πππ experiences a force 1 ππ π 1 ππ 2 ππΉ = π0 π π ππ π sin πππ = π0 ππ sin πππ 6 ππ‘ π 6 ππ‘ The torque about the z axis is obtained by multiplying this with the distance π sin π of this element about the z axis. π ππ 3 ππ = π 2 π0 sin πππ 6 ππ‘ Total torque is obtained by integrating over the angle π ππ π 3 2π 2 ππ π = π 2 π0 β« sin πππ = π π0 6 ππ‘ 0 9 ππ‘ The change in angular momentum is β« πππ‘ = 13 2π 2 π π0 π, 9 as was obtained in Problem 1.