Physics 4183 Electricity and Magnetism II. Covariant Formulation of Electrodynamics-1

Transcription

1 Physics 4183 Electricity and Magnetism II Covariant Formulation of Electrodynamics 1 Introduction Having briefly discussed the origins of relativity, the Lorentz transformations, 4-vectors and tensors, and invariants, we are now prepared to discuss the relativity as it applies to electromagnetism. We will start with the formal aspects, that is we will discuss the covariant forms of the continuity equation, and the Maxwell equations. From this point we will move on to discussing how the fields transform. 1.1 The Continuity Equation We will use the continuity equation charge conservation) as the starting point in our development of a covariant form of electrodynamics. The continuity equation has the simple form of time derivative of a scalar function plus the divergence of a vector function equal to a constant ρ + J = 0 1) The derivatives can be written in the compact form µ µ = ct, x, y, ) z 2) If we assume that a current density 4-vector can be define as J µ = cρ, J), then the continuity equation takes on the very compact form µ J µ = 0 3) which is an invariant. This states that charge is conserved locally in all Lorentz frames; note the derivatives and the components of the J µ are not the same in all frames, but the form of the equation remains the same, the continuity equation is covariant. Based on the invariance of the continuity equation, the 4-vector J µ transforms like J ν = Λ ν µj µ µ J µ = µ Λ µ ν Λ ν µ J µ = ν J ν 4) An alternate way of show that the charge and current densities form a 4-vector, is to recall that in a given frame where the charge density has a relative velocity to the observer the current density is given by J = ρ v 5) where ρ corresponds to the charge density as observed in the observers frame. The charge and current densities written in terms of the charge density in its rest frame is ρ = γρ 0 J x = γv x ρ 0 6) Covariant Formulation of Electrodynamics-1

2 where for simplicity we have oriented our axis parallel to the direction of motion. Again, recall that the boost factor can be written in terms of derivatives γ = dt dτ γv x = dt dx dτ dt = dx dτ but this implies that γ is the zeroth component of the 4-velocity and v x γ is the first component of the 4-velocity. Therefore, the charge and current densities can be written as 7) J µ = ρ 0 u µ 8) The first question that we must ask ourselves, does this make sense? We start by looking at the charge density. Transforming from its rest frame, the charge density is note in the rest frame of the charge density the current density is zero) ρ = γρ while the current density is J 1 = cβ x γρ 9) To see whether this makes sense or not, let s consider a charge distribution place in a cube of length l in its rest frame. The cube is assumed to be traveling along the x-axis at a velocity β x, with one of the sides parallel to the x-axis see Fig. 1). The length of the cube is therefore contracted along the x-axis l = l 10) γ Next we assume that the container has impenetrable walls, therefore the total charge remains within the cube in both frame. The charge density as observed relative to the unprimed coordinates is ρ = Q l 3 /γ = γρ 11) as we had already deduced from the invariance of the continuity equation. The current density is the charge density multiplied by the velocity J x = cβ x γρ. The continuity equation implies charge conservation. The total charge is the volume integral over the charge density. In the rest frame, the total charge is Q = ρ dx dy dz 12) V In the unprimed frame, the total charge is Q = ρdx dy dz = V V γρ dx γ dy dz = Q 13) therefore both observers see the same total charge, and our original assumption of treating the charge and current densities as components of a 4-vector is correct. Covariant Formulation of Electrodynamics-2

3 t Frame S β x PSfrag replacements x Figure 1: Box in uniform motion relative to the frame S. 1.2 The Maxwell Equations As was previously shown, the Maxwell equations are not covariant. The example we used was Faraday s law. In the rest frame of a loop, the induced EMF is equal to minus the rate of change of the magnetic flux, while in a moving frame, the induce EMF has an added term that depends on the relative velocity of the two frames E d l = Φ B E v B) d l = Φ B Even though this is correct and consistent with what relativity tells us, it is not in the spirit of Einstein s first postulate, which states the the laws of physics are independent of what inertial frame they are viewed from. In this instance, we expect that Maxwell s equations retain the same form in all frames; the equations must be covariant form invariant), the values of the fields can change, but the form of the equation cannot. So how do we show the equations to be covariant? To begin with, let s write the Maxwell equations in free space 14) E = ρ ɛ 0 B = 0 These equation can be expressed in terms of potential E = B B = µ0 J + µ0 ɛ 0 E 15) B = 0 B = A 16) E = B E = Φ + A ) Let s assume that these equations can be rewritten in a covariant form by defining the 4-potential A µ = Φ/c, A ) 17) Covariant Formulation of Electrodynamics-3

4 where we justify this by noting that using the 4-current density leads back to the equations for the potentials in terms of there sources 1 A µ = µ 0 J µ 4π R d3 r 18) It is clear that the current density is a 4-vector, based on our previous arguments, but for the 4-potential to be a 4-vector the ratio d 3 r/ R must be an invariant. We start with an observer in the frame S, which is stationary with respect to the field point. The field point is located at 0, 0, 0, 0). The source point, in this frame, is located at ct, x, y, z) where ct = x 2 + y 2 + z 2 corresponds to the propagation time between the two points, in other words, we have to take into account that any change that occurs at the source takes some amount of time to be felt elsewhere. The component of R along its direction is R = ct. Let s now ask what an observer sees in the S frame that is traveling at a velocity β to the right relative to S. We will also assume that the time origins coincide t = t = 0. The value of the component of R along its direction is given by R = ct = γct βx) = Rγ 1 + x ) R β = Rγ1 + β cos θ) 19) see Fig. 2) The quantity d 3 r must take into account that changes that occur in the back end of the volume element takes a time cdt longer to propagate to the field point. The differential element is d 3 r = dx dy dz, where dy and dz are invariant. The other element is given by dx = γdx βc dt) 20) but as we stated, the time difference between front and back has to be account for dr = c dt dx = γdx + βdr) = γdx 1 + β dr ) = γdx1 + β cos θ) 21) dx Therefore d 3 r R = d3 r 22) R The 4-potential as defined above it a 4-vector, since it is composed of a 4-vector J µ and an invariant d 3 r/r. This 4-vector is given by A µ = φ/c, A x, A y, A z ) 23) where the factor of 1/c comes from the definition of the zeroth component of the 4-current cρ) and the units used in Eq. 18. Using the 4-potential, the relation between the electric field and the potential is written as follows E j c = j A A j) = j A 0 0 A j) = 0 A j j A 0) = F 0j 24) where we use µ = g µν ν µ = 0, 1, 2, 3 ) 25) 1 In order to prove that the 4-potential as defined has the correct transformation properties, requires the introduction of the retarded potential. That is the propagation speed of the field lines has to be taken into account. We will return to this subject later. Covariant Formulation of Electrodynamics-4

5 PSfrag replacements S S R dx θ dr θ x Figure 2: This shows the geometry used to show the invariance of d 3 r/r. The magnetic field becomes B k = i A j j A i) = i A j j A i) = F ij 26) Combining the two sets together leads to 0 E x /c E y /c E z /c F µν = µ A ν ν A µ = E x /c 0 B z B y E y /c B z 0 B x 27) E z /c B y B x 0 where µ corresponds to the row and ν to the column, and this tensor is called the Maxwell field strength tensor. Note that we started out with electric and magnetic fields, and ended up with a tensor that combines the two fields together. To get back to the Maxwell equations, we need to find mathematical operations that give back the curl and divergence of the fields. Notice that if we apply µ on F 0µ we get the divergence of the electric field E = ρ ɛ 0 c µ F 0µ = x E x + y E y + z E z = µ 0 J 0 = µ 0 cɛ 0 c 2 ρ ɛ 0 = ρ ɛ 0 28) while j acting on F ij we get the curl of the magnetic field, as an example take i = 1 [ B ] x = µ 0J x + ɛ 0 µ 0 E x ν F 1ν = 0 F F F F 13 = y B z z B y 1 c 2 te x = µ 0 J x 29) Therefore, taking the derivative of the field strength tensor gives ν F µν = µ 0 J µ 30) To arrive at the homogeneous set of Maxwell equations, expand them out using the field strength tensor B = 0 1 F F F 21 = 0 31) and [ ] E + B = 0 x 2 F F F 23 = 0 32) Covariant Formulation of Electrodynamics-5

6 The two equations can be combined to give α F βγ + γ F αβ + β F γα = ɛαβγδ F γδ = 0 33) where the totally antisymmetric tensor ɛ αβγδ is defined as + 1 if α = 0, β = 1, γ = 2, δ = 3 and even permutations of the indices ɛ αβγδ = 1 for odd permutations of the indices 0 if any indices are the same 34) This equation can be put into a simpler form by using a duality transformation as discussed earlier in the semester. We make the replacements E B and B E; note that since there is no magnetic charge it transform into not magentic charge after the transformation. The new new tensor, G µν, leads to the following form for the homogeneous Maxwell equation 0 B x B y B z ν G µν = 0 G µν = B x 0 E z /c E y /c B y E z /c 0 E y /c 35) B z E y /c E x /c 0 Note that the equation for G µν is the same equation as Eq. 33 for F µν using G αβ = 1 2 ɛαβγδ F γδ. 1.3 The Lorentz Force To complete the discussion of the equations of electrodynamics, we examine the Lorentz force, and put it into a covariant form. The standard form of the Lorentz force is d p ) dt = q E + v B Clearly this is not in a form that allows for simple transformations. Therefore, we rewrite the Lorentz force in replacing the time derivatives with the proper time derivatives, and add in the proper time derivative of p 0 d p ) dτ = q u 0 E/c + u B and where the term dp 0 /dτ is the rate of change of energy of the particle 36) dp 0 dτ = q u E 37) W = F l dw dt = F v 38) note there is no magnetic field term since it doesn t change the energy of a electrically charged particle. In the rest frame of the observer, this reduces to the standard form of the Lorentz force. Combining the Lorentz force with the components of the field strength tensor, we get which is in a covariant form the product of tensor and 4-vector). dp α dτ = qu βf αβ 39) Covariant Formulation of Electrodynamics-6

7 1.4 Field Transformations The final item that needs to calculated is the transformation equations of the electric and magnetic fields. We will determine these by examining how the Maxwell field strength tensor transforms. If we blindly apply the rules that we have discussed, we find that the field strength tensor transforms like F α β = Λ α µλ β νf µν 40) one transformation tensor per index. tensor in terms of the potential and transform each of the 4-vectors To verify that this is correct, we write the field strength F µν = µ A ν ν A µ 41) F α β = α A β β A α = Λ α µ µ Λ β νa ν Λ β ν ν Λ α µa µ = Λ β µλ α ν [ µ A ν ν A µ ] = Λ β µλ α νf µν 42) where we have used the known 4-vector transformations. Based on the transformation of the field strength tensor, we can now derive the transformations for the electric and magnetic fields. As usual, we will assume that the primed frame travels with velocity β along the x-axis of the unprimed frame. The y and y axis are assumed to be parallel as are the z and z axis. In matrix notation, the Lorentz transformation is γ βγ 0 0 [Λ µν ] = βγ γ ) We can start by calculating E x where we only write down the terms that are non-zero) E x = F 0 1 = Λ 0 µλ 1 νf µν = Λ 0 1Λ 1 0F 10 + Λ 0 0Λ 1 1F 01 = 1 β 2 )γ 2 E x = E x 44) therefore the component of the electric field along direction of relative motion stays the same. Next we E y the transformation for E z will have a similar form) E y = F 0 2 = Λ 0 µλ 2 νf µν = Λ 0 0Λ 2 2F 02 + Λ 0 1Λ 2 2F 12 = γe y βγb z = γ E y cβb z ) 45) while the E z transforms like E z = F 0 3 = Λ 0 µλ 3 νf µν = Λ 0 0Λ 3 3F 03 + Λ 0 1Λ 3 3F 13 = γ E z + cβb y ) 46) To calculate the transformations for the magnetic field, we proceed in the same manner as for calculating the electric fields. Start with B x B x = F 2 3 = Λ 2 µλ 3 νf µν = F 23 = B x 47) as was the case for E x. The B y component transforms as B y = F 31 = Λ 3 µλ 1 νf µν = Λ 3 3 Λ 1 0F 30 + Λ 1 1F 31) = γ B y + βe z /c) 48) Covariant Formulation of Electrodynamics-7

8 and finally the transformation for B z B z = F 1 2 = Λ 1 µλ 2 νf µν = Λ 2 2 Combining all the transformation equations together we have Λ 1 0F 02 + Λ 1 1F 12) = γ B z βe y /c) 49) E x = E x E y = γe y βcb z ) E z = γe z + βcb y ) 50) B x = B x B y = γb y + β/ce z ) B z = γb z β/ce y ) Covariant Formulation of Electrodynamics-8