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UIUC Physics 436 EM Fields & Sources II Fall Semester, 2015 Lect. Notes 18.5 Prof. Steven Errede LECTURE NOTES 18.5 The Lorentz Transformation of E and B Fields: We have seen that one observer’s E -field is another’s B -field (or a mixture of the two), as viewed from different inertial reference frames (IRF’s). What are the mathematical rules / physical laws of {special} relativity that govern the transformations of E B in going from one IRF(S) to another IRF(S') ??? In the immediately preceding lecture notes, the reader may have noticed some tacit / implicit assumptions were made, which we now make explicit: 1) Electric charge q (like c, the speed of light) is a Lorentz invariant scalar quantity. No matter how fast/slow an electrically-charged particle is moving, the strength of its electric charge is always the same, viewed from any/all IRF’s: e 1.602 1019 Coulombs. {n.b. electric charge is also a conserved quantity, valid in any/all IRF’s.} Since the speed of light c is a Lorentz invariant quantity, then since c 1 o o then so is c 1 o o ; the scalar EM properties of the vacuum - o and o are separately Lorentz invariant scalar quantities, i.e. 2 o 8.85 1012 Farads/meter o 4 107 Henrys/meter same in any/all IRF’s 2) The Lorentz transformation rules for E and B are the same, no matter how the E and B fields are produced - e.g. from sources: q (charges) and/or currents I, or from fields: e.g. E B t , etc. The Relativistic Parallel-Plate Capacitor: The simplest possible electric field: Consider a large -plate capacitor at rest in IRF(S0). It has surface charge density 0 (Coul/m2) on the top/bottom plates respectively and has plate dimensions 0 and w0 {in IRF(S0)!} separated by a small distance d 0 , w0 . In IRF(S0): Electric field as seen in IRF(S0): E0 E0 yˆ 0 yˆ E0 0 o o No B -field is present in IRF(S0): B0 0 no currents present! n.b. E0 is non-zero only in the gap region between -plates {i.e. neglect fringe field} © Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois 2005-2015. All Rights Reserved. 1 UIUC Physics 436 EM Fields & Sources II Fall Semester, 2015 Lect. Notes 18.5 Prof. Steven Errede Now consider examining this same capacitor setup from a different IRF(S), one which is moving to the right at speed v0 (as viewed from the rest frame IRF(S0) of the -plate capacitor) i.e. v v0 xˆ is the velocity of IRF(S) relative to IRF(S0). Viewed from the moving frame IRF(S), the -plate capacitor is moving to the left (i.e. along the x̂ -axis) with speed –v0. The plates along the direction of motion have Lorentz-contracted by a factor of 0 1 1 v0 c , 2 i.e. the length of plates as seen by an observer in IRF(S) is: 0 0 1 v0 c 0 . 2 {n.b. the plate separation d and plate width w are unchanged in IRF(S), since both d and w are to direction of motion!!} Since: And: 0 Qtot Q tot Area w but: Qtot = Lorentz invariant quantity Qtot Qtot A0 0 w0 but: w = w0 and d = d0 since both to direction of motion. Thus: Qtot w 0 0 w0 But: 0 0 0 0 0 0 0 0 0 0 0 Thus: 0 0 but since: 0 1 0 The surface charge density on the plates of capacitor in IRF(S) is higher than in IRF(S0) by factor of 0 1 1 v0 c To an observer in IRF(S), the plates of the -plate capacitor are moving in the v0 xˆ direction. Thus the electric field E in IRF(S) is: E yˆ 0 0 yˆ 0 E0 where: 0 0 o o E 0 E0 , E0 0 yˆ o The superscript is to explicitly remind us that E 0 E0 is for E -fields to the direction of motion. Here, v v0 xˆ between IRF’s. 2 © Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois 2005-2015. All Rights Reserved. 2 UIUC Physics 436 EM Fields & Sources II Fall Semester, 2015 Lect. Notes 18.5 Prof. Steven Errede Now consider what happens when we rotate the {isolated} -plate capacitor by 90o in IRF(S0), then E0 0 xˆ in rest frame IRF(S0), but in the moving frame IRF(S): 0 The electric field in IRF(S) is: E xˆ E o but: Qtot Q Q tot tot 0 Area w 0 w0 E xˆ 0 xˆ E0 E E0 o o 0 w w0 E -field in IRF(S0) The plate separation distance d is Lorentz-contracted in IRF(S): d d 0 0 but has no effect on E {in IRF(S)}, because E does not depend on d! Why??? Because here, the -plates were first charged up (e.g. from a battery) and then disconnected from the battery! Potential difference: V IRF S V0 IRF S0 n.b. IRF choice has intimate connection to gauge invariance!!! V0 V V0 V V V0 Since: E but: d d 0 0 E0 d d0 d0 0 d0 d d0 V IRF S 0 V0 IRF S0 The -plate capacitor is deliberately not connected to an external battery (which would keep V = constant, but then we would have 0 in the case and 0 in the case. Currents would then flow (transitorially) in both situations… Note that we also want to hang on to/utilize the Lorentz-invariant nature of Qtot, which is another reason why the battery is disconnected… © Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois 2005-2015. All Rights Reserved. 3 UIUC Physics 436 EM Fields & Sources II Fall Semester, 2015 Lect. Notes 18.5 Prof. Steven Errede But wait !!! This isn’t the entire story for the EM fields in IRF(S) !!! In the first example of the “horizontal” -plate capacitor, which was moving with relative velocity v v0 xˆ as seen by an observer in IRF(S), shown in the figure below: The moving surface-charged plates of the -plate capacitor as viewed by an observer in IRF(S) constitute surface currents: K v0 xˆ (Amps/m) with: 0 0 (Coul/m2). The two surface currents create a magnetic field in IRF(S) between the plates of the -plate capacitor !!! Side / edge-on view: B-fields produced (use right-hand rule): Ampere’s Circuital Law: C w Bd o I encl o K dw 0 2 Bw o K w K v K v B y 0 o zˆ o 0 zˆ y 0 B y d o zˆ o 0 zˆ 2 2 2 2 o K o v0 o K o v0 B y d zˆ zˆ B y 0 zˆ zˆ y 0 2 2 2 2 Add B and B to get the total B -field: v v Btot y d o 0 zˆ o 0 zˆ 0 2 2 v v Btot 0 y d o 0 zˆ o 0 zˆ o v0 zˆ In IRF(S): 2 2 v v o 0 zˆ o 0 zˆ 0 Btot 0 y 2 2 4 © Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois 2005-2015. All Rights Reserved. UIUC Physics 436 EM Fields & Sources II Fall Semester, 2015 Lect. Notes 18.5 Prof. Steven Errede Thus, in IRF(S) {which moves with velocity v v0 xˆ relative to IRF(S0)} we have for the horizontal -plate capacitor: 0 yˆ 0 E0 yˆ where: 0 0 and: 0 E exists only in region 0 y d : E yˆ 0 o o B exists only in region 0 y d : B o v0 zˆ 0 o 0 v0 zˆ 1c 0 0 E0 zˆ 1 1 02 0 v0 c In IRF(S): The fact that B exists / is non-zero only where E exists / is non-zero is not an accident / not a “mere” coincidence! Note also that {here}: B 1c E where: 0 x̂ and: E Eyˆ 0 E0 yˆ 0 0 yˆ . o The space-time properties associated with rest frame IRF(S0) are rotated (Lorentz-transformed) in going to IRF(S). E0 in IRF(S0) only exists between plates in rest frame IRF(S0) {neglecting fringe field} Gets space-time rotated (Lorentz-transformed) in going to moving frame IRF(S) → E and B in IRF(S) exist only between the capacitor plates in IRF(S). E and B between plates in IRF(S) comes from / is associated with E0 between plates in IRF(S0) Point is: EM field energy density uEM (x,y,z,t) must be non-zero in a given IRF in order to have EM fields present at space-time point (x,y,z,t)! In IRF(S0): u0 r0 , t0 12 o E02 21 o 02 E0 only EM field energy densities u0 and u ≠ 0 only in region between plates of u r , t 12 o E 2 r , t 2 1o B 2 r , t ||-plate capacitor. In IRF(S): 12 o 02 E02 12 o 02 02 E02 12 o 02 1 02 E02 n.b. If EM energy density u0 = 0 in one IRF(S0) → u = 0 in another IRF(S). Poyntings Vector: In IRF(S0): S0 r0 , t0 In IRF(S): S r , t 1 E0 r0 , t0 B0 r0 , t0 0 1 E r,t B r,t o o 1 o 0 0 2 0 o 0 v0 yˆ zˆ 02 0 v0 xˆ 02 o E02 v0 xˆ o o xˆ © Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois 2005-2015. All Rights Reserved. 5 UIUC Physics 436 EM Fields & Sources II Fall Semester, 2015 Lect. Notes 18.5 Prof. Steven Errede Momentary Aside: Hidden Momentum Associated with the Relativistic Parallel Plate Capacitor. Between plates of -plate capacitor in IRF(S): E yˆ 0 0 yˆ where: 0 0 and: 0 0 0 B o v0 zˆ 0 o 0 v0 zˆ 1 1 02 0 v0 c xˆ yˆ zˆ xˆ 2 1 02 02 v0 2 0 v0 E B 0 xˆ yˆ zˆ xˆ Between plates of -plate capacitor (only!): S yˆ zˆ 0 0 0 zˆ xˆ yˆ Poynting’s vector points in the direction of motion of the -plate capacitor, as seen by an observer in IRF(S), and only exists/is non-zero in the region between plates of the -plate capacitor. EM field linear momentum density in IRF(S): EM o o S 02 o 02 v0 xˆ . - Points in the direction of motion of the -plate capacitor, as seen by an observer in IRF(S), - Only exists/is non-zero in the region between plates of the -plate capacitor. - n.b. In rest frame IRF(S0): B0 = 0 Poynting’s vector S0 0 and thus 0EM 0 ! EM field linear momentum in IRF(S): pEM EM Volume(IRF(S )) EM VS Volume (in IRF(S)): VS wd 0 0 w0 d 0 , where: 0 0 , w w0 and: d d 0 {here}. Volume (in IRF(S0): V0 0 w0 d 0 . Thus: VS V0 0 . wd pEM 02 o 02 v0VS xˆ 02 o 02 v0 0 0 0 xˆ 0 0 02 v0 0 w0 d 0 xˆ 0 0 02 v0V0 xˆ 0 V0 1 phidden 2 m E where m = magnetic dipole moment c where: m IA zˆ with: I K d Kw where: K v0 . Note that: m B . The “hidden momentum” in IRF(S) is: The cross-sectional area is: A d 0 d 0 since: 0 0 and: d d 0 , w w0 . 0 Side view (in IRF(S)): 0 0 m IA zˆ Kwd v0 w0 0 d 0 zˆ 0 0 0 v0 w0 0 0 d 0 zˆ 0 v0 0 w0 d 0 zˆ E 0 0 yˆ o 6 © Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois 2005-2015. All Rights Reserved. UIUC Physics 436 EM Fields & Sources II Fall Semester, 2015 Lect. Notes 18.5 Prof. Steven Errede In IRF(S): xˆ 0 0 02v0 1 1 ˆ ˆ phidden 2 0 v0 0 w0 d 0 0 z y w0 d 0 xˆ 0 o 02 v0V0 xˆ using 2 o o 0 2 c c o o c V0 1 Thus in IRF(S): phidden 2 m B 0 o 02 v0V0 xˆ But: pEM o E B VS 0 0 02 v0V0 xˆ c Thus, we (again) see that: phidden pEM for the relativistic -plate capacitor !!! 1 E0 B0 0 pEM 0 . Note that in IRF(S0): B0 0 S0 o 1 But note that: phidden 2 m0 B0 0 , thus phidden pEM is also valid in IRF(S0). c The numerical value of hidden momentum is reference frame dependent, i.e. it is not a Lorentz-invariant quantity, just as relativistic momentum p {in general} is reference frame dependent/is not a Lorentz-invariant quantity. Now let us return to task of determining the Lorentz transformation rules for E and B : For the case of the relativistic horizontal -plate capacitor, let us consider a third IRF(S′) that travels to the right (i.e. in the x̂ -direction) with velocity v vxˆ relative to IRF(S). yˆ and: B o vzˆ In IRF(S′), the EM fields are: E o From use of Einstein’s velocity addition rule: v is the velocity of IRF(S′) relative to IRF(S0): v v0 1 with: and: 0 2 2 1 vv0 c 1 v c We need to express E and B {defined in IRF(S′)} in terms of E and B {defined in IRF(S)}. v 0 E yˆ yˆ where: o E yˆ 0 o o 1 1 v c 2 but: E yˆ {in IRF(S)} o B o vzˆ o 0 vzˆ but: 0 0 and: 0 1 and: 0 2 0 1 v0 c E E yˆ 0 0 o B o vzˆ but: B 0 o 0 v0 zˆ o v0 zˆ {in IRF(S)} 0 © Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois 2005-2015. All Rights Reserved. 7 UIUC Physics 436 EM Fields & Sources II Fall Semester, 2015 Lect. Notes 18.5 Prof. Steven Errede 1 v0 c 1 vv0 c 2 1 vv 1 20 where: Now (after some algebra): 2 2 2 c 0 1 v c 1 v c 1 v c 2 But: v v v0 1 vv 0 In IRF(S′): c2 1 and: 1 vv0 c 2 with: 2 0 1 v c E E yˆ 1 vv0 c 2 yˆ o 0 0 o vv 2 0 ˆ B o v z 1 vv0 c o 1 vv0 c 2 0 zˆ o v v0 zˆ Compare these to the E and B fields in IRF(S): E yˆ 0 0 yˆ where: 0 0 and: 0 In IRF(S): 0 0 B o v0 zˆ 0 o 0 v0 zˆ 1 1 02 0 v0 c Using: o 1 c 2 o we can rewrite E in IRF(S′) as: E yˆ o vv0 yˆ yˆ v o v0 yˆ o o Bz in IRF(S ) Ey in IRF(S ) Very Useful Table(s): xˆ yˆ zˆ yˆ xˆ zˆ yˆ zˆ xˆ zˆ yˆ xˆ zˆ xˆ yˆ xˆ zˆ yˆ where: v vxˆ and: B Bz zˆ v B vBz xˆ zˆ vBz yˆ yˆ E E y yˆ E y vBz yˆ E y vBz yˆ Or simply: E y E y vBz Likewise, again using: o 1 c 2 o we can rewrite B in IRF(S′) as: v B o v v0 zˆ o v0 zˆ o vzˆ o v0 zˆ 2 zˆ c o B Bz zˆ in IRF(S ) Ey in IRF(S ) v v v B Bz zˆ Bz zˆ 2 E y zˆ Bz 2 E y zˆ Or simply: Bz Bz 2 E y c c c Thus, we now know how the Ey and Bz fields transform. 8 © Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois 2005-2015. All Rights Reserved. UIUC Physics 436 EM Fields & Sources II Fall Semester, 2015 Lect. Notes 18.5 Prof. Steven Errede Next, in order to obtain the Lorentz transformation rules for Ez and By, we align the capacitor plates parallel to x-y plane instead of x-z plane as shown in the figure below: In IRF(S) the {now} rotated fields are: E rot zˆ Ezrot zˆ o B rot o v0 yˆ Byrot yˆ Use the right-hand rule to get correct sign !!! The corresponding E and B fields in IRF(S') are (repeating the above methodology): Ez Ez vBy and: v By By 2 Ez c As we have already seen by orienting the plates of capacitor parallel to y-z plane: Ex Ex ← n.b. there was no accompanying B -field in this case! Thus we are not able to deduce the Lorentz transformation for Bx ( to direction of motion) from the -plate capacitor problem… So alternatively, let us consider the long solenoid problem, with solenoid {and current flow} oriented as shown in the figure below: In IRF(S): In IRF(S): E 0 B Bx xˆ o nIxˆ We want to view this from IRF(S'), which is moving with velocity v vxˆ relative to IRF(S). In IRF(S): n N L = # turns/unit length, N = total # of turns, L = length of solenoid in IRF(S). Viewed by an observer in IRF(S'), the solenoid length contracts: L L in IRF(S') In IRF(S'): n N N 1 1 n = # turns/unit length in IRF(S'), where: 2 2 L L 1 1 v c © Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois 2005-2015. All Rights Reserved. 9 UIUC Physics 436 EM Fields & Sources II Fall Semester, 2015 Lect. Notes 18.5 Prof. Steven Errede However, time also dilates in IRF(S') relative to IRF(S) – affects currents: I dQ dt dQ dt dQ 1 dt 1 I in IRF(S'). But: in IRF(S) I I I dt dt dt dt dt dt 1 B Bx xˆ o nI xˆ o n I xˆ o nIxˆ Bx xˆ B Bx Bx Longitudinal / parallel-to-boost direction B-field does not change !!! Thus, we now have a complete set of Lorentz transformation rules for E and B , for a Lorentz transformation from IRF(S) to IRF(S'), where IRF(S') is moving with velocity v vxˆ relative to IRF(S): Ex Ex Bx Bx 1 1 2 E y E y vBz v By By 2 Ez c Ez Ez vBy v Bz Bz 2 E y c v c Just stare at/ponder these relations for a while – take your (proper) {space-}time… Do you possibly see a wee bit of Maxwell’s equations afoot here ??? ;) Two limiting cases warrant special attention: v v 1.) If B 0 in lab IRF(S), then in IRF(S') we have: B 2 Ez yˆ E y zˆ 2 Ez yˆ E y zˆ c c 1 But: v vxˆ B 2 v E in IRF(S') !!! c 2.) If E 0 in lab IRF(S), then in IRF(S') we have: E v Bz yˆ By zˆ v Bz yˆ By zˆ But: v vxˆ E v B in IRF(S') ← i.e. the magnetic part of Lorentz force law !!! Griffiths Example 12.13: The Electric Field of a Point Charge in Uniform Motion A point charge q is at rest in IRF(S0). An observer is in IRF(S), which moves to the right (i.e. in the x̂ direction) at speed v0 relative to IRF(S0). What is the E -field of the electric charge q, as viewed from the moving frame IRF(S)? 10 © Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois 2005-2015. All Rights Reserved. UIUC Physics 436 EM Fields & Sources II Fall Semester, 2015 Lect. Notes 18.5 Prof. Steven Errede In the rest frame IRF(S0) of the point charge q, the {static} electric field of the point charge q is: r0 1 q 2 2 2 2 E0 rˆ0 where: r0 x0 y0 z0 and: rˆ0 r0 4 o r02 x0 q E0x and: r0 x0 xˆ y0 yˆ z0 zˆ 3 4 o x 2 y 2 z 2 2 E0 y E0z q 0 0 y0 4 o x 2 y 2 z 2 3 2 0 0 0 q z0 4 o x 2 y 2 z 2 3 2 0 0 0 0 1 1 u0 c But: E 0 E0 and: E E0 0 2 Then in IRF(S), which is moving to right (i.e. in the x̂ direction) at speed v0 relative to IRF(S): Ex E0x E y 0 E0 y Ez 0 E0z q x0 4 o x 2 y 2 z 2 3 2 0 0 0 q 0 y0 4 o x 2 y 2 z 2 0 0 0 q 3 0 z0 2 n.b. These relations are expressed in terms of the IRF(S0) coordinates (x0, y0, z0) of the field point P. 4 o x 2 y 2 z 2 3 2 0 0 0 However we want/need the IRF(S) E expressed in terms of the IRF(S) coordinates (x, y, z) of the field point P. Use the inverse Lorentz transformation on the coordinates: In IRF(S) at the IRF(S) present time t: Observation/field point in IRF(S) at present time t: Inverse Lorentz Transformation: x0 0 x v0t 0 Rx R Rx xˆ Ry yˆ Rz zˆ y0 y R y 0 1 1 v0 c 2 z0 z Rz Then: x02 y02 z02 r02 02 Rx2 Ry2 Rz2 = 02 R 2 cos 2 R 2 sin 2 © Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois 2005-2015. All Rights Reserved. 11 UIUC Physics 436 EM Fields & Sources II Fall Semester, 2015 Lect. Notes 18.5 Prof. Steven Errede From the above figure: Rx R cos Rx2 R 2 cos 2 . Then since: R 2 Rx2 Ry2 Rz2 R 2 y Rz2 R 2 sin 2 Since: R 2 R 2 cos 2 sin 2 . In the moving frame IRF(S): Ex Ey Ez 0 Rx q Picked up Lorentz factor 4 o 2 R 2 cos 2 R 2 sin 2 3 2 0 0 Ry q 4 o 2 R 2 cos 2 R 2 sin 2 3 2 0 Picked up Lorentz factor 0 Rz q Or: E q 0 0 4 o 2 cos 2 sin 2 3 2 0 q 4 o R Rx xˆ Ry yˆ Rz zˆ R R Rˆ R ˆ ˆ 3 where: 3 2 since: R RR or: R R R R R 0 03 2 cos 2 1 2 sin 2 0 3 2 Rˆ R2 with: v0 2 1 02 c 1 q Rˆ q Thus: E 3 2 4 o 4 o v0 2 2 2 R 2 1 sin 1 sin c Thus: from Lorentz transformation of field v0 2 1 c E 0 3 E from Lorentz transformation of coordinate 4 o 2 R 2 cos 2 R 2 sin 2 2 0 0R q E with: 4 o 2 R 2 cos 2 R 2 sin 2 3 2 0 q v0 2 1 c 3 2 2 4 o v0 2 1 sin c v0 2 1 c 2 v0 2 2 2 1 sin sin sin c 2 Oliver Heaviside’s 1888 expression for the retarded Rˆ electric field Eret r , t expressed in terms of the IRF(S) R2 present time t !!! See/compare with Physics 436 Lect. Notes 12, p. 21-24 / Griffiths Ch. 10, Example 10.4, p. 439-440. In the moving frame IRF(S), the unit vector R̂ points along the line from the present position of the charged particle at time t !!! E R because the 0 factor is present in the numerator of each of the xˆ , yˆ , zˆ components !!! 12 3 Rˆ R2 © Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois 2005-2015. All Rights Reserved. UIUC Physics 436 EM Fields & Sources II Fall Semester, 2015 Lect. Notes 18.5 Prof. Steven Errede Griffiths Example 12.14: Magnetic Field of a Point Charge in Uniform Motion A point electric charge q moves with constant velocity v vzˆ in the lab frame IRF(S). Find the magnetic field B in IRF(S) associated with this moving point electric charge. Note that in the rest frame of the electric charge {IRF(S0)} that B0 0 everywhere. 1 In IRF(S) the electric charge q is moving with velocity v vzˆ and: B 2 v E . c 1 v c q But the electric field of moving point charge q in IRF(S) is: E 4 o (See pages 3-6 above/Griffiths Example 12.13) 1 v c 2 sin 2 2 3 2 Rˆ R2 E Rˆ sin qv 1 v c o 1 1 ˆ where: cos 1 vˆ Rˆ , v Rˆ v sin ˆ . B 2 vE 3 2 2 c 4 2R 2 1 v c sin 2 If the point electric charge q is heading directly towards you (i.e. along the ẑ direction) then ̂ points counter-clockwise: (out of page) o NOTE: In the non-relativistic limit ( v c ): B 4 v Rˆ q 2 R The Biot-Savart Law for a moving point charge !!! NOTE ALSO: H.C. Øersted discovered the link between electricity and magnetism in 1820. It wasn’t until 1905, with Einstein’s special relativity paper that a handful of humans on this planet finally understood the profound nature of this relationship – a time span of 85 years – approximately a human lifetime passed! © Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois 2005-2015. All Rights Reserved. 13 UIUC Physics 436 EM Fields & Sources II 14 Fall Semester, 2015 Lect. Notes 18.5 Prof. Steven Errede © Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois 2005-2015. All Rights Reserved. UIUC Physics 436 EM Fields & Sources II Fall Semester, 2015 Lect. Notes 18.5 Prof. Steven Errede © Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois 2005-2015. All Rights Reserved. 15 UIUC Physics 436 EM Fields & Sources II 16 Fall Semester, 2015 Lect. Notes 18.5 Prof. Steven Errede © Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois 2005-2015. All Rights Reserved. UIUC Physics 436 EM Fields & Sources II Fall Semester, 2015 Lect. Notes 18.5 Prof. Steven Errede The Electromagnetic Field Tensor F v The relations for the Lorentz transformation of E and B in the lab frame IRF(S) to E and B in IRF(S'), where IRF(S') is moving with velocity v vxˆ relative to IRF(S) are: components Ex Ex B components Bx Bx E y E y cBz By By Ez c 1 1 2 Ez Ez cBy Bz Bz E y c vc components B components It is readily apparent that the six components of E and B field certainly do not transform like the spatial / 3-D vector parts of e.g. two separate contravariant 4-vectors ( E and B ), because the (orthogonal) components of E and B to the direction of the Lorentz transformation are mixed together {as seen in the case for B 0 in IRF(S) resulting in B c12 v E in IRF(S') and the case for E 0 in IRF(S) resulting in E v B in IRF(S')}. © Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois 2005-2015. All Rights Reserved. 17 UIUC Physics 436 EM Fields & Sources II Fall Semester, 2015 Lect. Notes 18.5 Prof. Steven Errede Note also that the form of the Lorentz transformation for vs. components is “switched” {here} for the EM fields vs. 4-vectors!!! Recall that true relativistic 4-vectors transform by the rule: a v a v , e.g. for a Lorentz boost from IRF(S) → IRF(S') along v vxˆ . Specifically, recall that x v x v is explicitly: ct ct x Parallel components: x x ct Perpendicular y y Components: z z Ex Ex Bx Bx Ez Ez cBy Bz Bz E y c E y E y cBz By By Ez c For a Lorentz transformation x v x v from IRF(S) → IRF(S') along v vxˆ , the v tensor has the form: Row index v 0 0 0 0 0 0 1 0 0 0 n.b. v is a symmetric matrix 0 1 Column index Thus, there is no way that the six components of the E and B fields can be construed as being the spatial / 3-D vector components of contravariant 4-vectors, e.g. E and B . (What would be their temporal components/ scalar counterparts: E 0 and B 0 = ???) It turns out that the 3-D spatial vectors E and B are the six unique components of a 4 4 rank-two EM field strength tensor, F v !!! A 4 4 rank two tensor t Lorentz transforms via two -factors (one for each index): t v t v In matrix form: t t T t because is symmetric matrix Where t is: Column # 0 t 00 10 t t 20 t Column t 30 Row 18 1 2 t 01 t 02 t11 t12 21 22 t t t 31 t 32 3 t 03 t13 t 23 t 33 Row # t -t sx - t 1 sy - t 2 3 sz - t 0 t - sx t - sy sx - sx sx - s y s y - sx sy - sy sz - sx sz - s y t - sz sx - sz s y - sz sz - sz n.b. Each element of tensor t has temporalspatial meaning! © Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois 2005-2015. All Rights Reserved. UIUC Physics 436 EM Fields & Sources II Fall Semester, 2015 Lect. Notes 18.5 Prof. Steven Errede The 16 components of the 4 4 second rank tensor t need not all be different (e.g. v ) – the 16 components of t may have symmetry / anti-symmetry properties: Symmetric 4 4 rank two tensor: Anti-symmetric 4 4 rank two tensor: t t t t For the case of a symmetric 4 4 rank two tensor t t , of the 16 total components, 10 are unique, but 6 are repeats: t01 = t10, t02 = t20, t03 = t30 t12 = t21, t13 = t31 t32 = t23 tsym . t 00 t 01 t 02 01 11 12 t t t 02 12 22 t t t 03 13 23 t t t t 03 t13 t 23 t 33 For the case of an anti-symmetric 4 4 rank two tensor t t , of the 16 total components, only 6 are unique, 6 are repeats (but with a minus sign!) and the 4 diagonal elements must all be 0 (i.e. t00 = t11 = t22 = t33 0): t01 = t10, t02 = t20, t03 = t30 t12 = t21, t13 = t31 t32 = t23 tanti sym. 0 01 t 02 t 03 t t 01 t 02 0 t12 t12 0 t t 23 13 t 03 t13 t 23 0 So it would seem that the six components of the spatial / 3-D E and B vectors can be represented by an anti-symmetric 4 4 rank two contravariant tensor !!! v v Let’s investigate how the Lorentz transformation rule t t for a 4 4 rank two anti-symmetric tensor tanti works (6 unique non-zero components). Starting with t 01 0 t 1 : sym. Column # 0 v or 0 0 1 2 0 0 3 0 0 1 0 Row # 0 0 0 1 0 1 2 3 tanti sym. 0 01 t 02 t 03 t t 01 t 02 0 t12 t12 0 t t 23 13 t 03 t13 t 23 0 © Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois 2005-2015. All Rights Reserved. 19 UIUC Physics 436 EM Fields & Sources II Fall Semester, 2015 00 Row # We see that 0 0 , unless 0 or 1: Column # Row # We also see that 1 0 , unless 0 or 1: Column # There are only 4 non-zero terms in the sum: Lect. Notes 18.5 Prof. Steven Errede 01 10 11 t 01 0 t 1 0 0t 00 10 0 0t 0111 01t10 10 01t1111 But: t00 = 0 and t11 = 0, whereas: t01 = t10. t 01 0 0 11 0110 t 01 t 01 2 2 2 t 01 2 1 2 t 01 01 01 But: 2 1 1 2 t t v v One (e.g. you!!!) can work through the remaining 5 of the 6 distinct cases for t t (or for completeness’ sake, why not work out all 16 cases explicitly!!!) . . . The complete set of six rules for the Lorentz transformation of a 4 4 rank two contravariant v v anti-symmetric tensor t t are: t 01 t 01 ↔ Ex Ex t 02 t 02 t12 ↔ E y E y cBz t 03 t 03 t 31 ↔ Ez Ez cBy t 23 t 23 ↔ Bx Bx t 31 t 31 t 03 ↔ By By Ez c t 12 t12 t 02 ↔ Bz Bz E y c As can be seen, the above Lorentz transformation rules for this tensor are indeed precisely those we derived on physical grounds for the EM fields! Thus, we can now explicitly construct the so-called electromagnetic field tensor F v – a 4 4 rank two contravariant anti-symmetric tensor: F v 20 F 00 0 10 F F 01 20 F F 02 30 03 F F F 01 F 02 F 11 0 F 12 F 21 F 12 F 22 0 F 31 F 32 F 23 0 F 13 F 31 Ex Ey F 23 F 33 0 Ez F 03 Ex Ey 0 cBz cBz 0 cBy cBx Ez cBy cBx 0 © Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois 2005-2015. All Rights Reserved. UIUC Physics 436 EM Fields & Sources II F v 0 Ex Ey Ez Fall Semester, 2015 Ex Ey 0 cBz cBz 0 cBy cBx Ez cBy cBx 0 Lect. Notes 18.5 Prof. Steven Errede E-field representation of F v (SI units: Volts/m). Note that there exist alternate/equivalent forms of the field tensor F v . If we divide our F v by c, we get Griffith’s definition of F v : i.e. 1 FGriffiths FOurs c Ex c E y c Ez c 0 0 Bz Ex c By E y c Bz 0 Bx 0 By Bx Ez c B-field representation of F v (SI units: Tesla) Note that there also exists an equivalent, but different way of embedding E and B in an anti-symmetric 4 4 rank two tensor: Instead of comparing: a) Ex Ex , with: a') t 01 t 01 , E y E y cBz , t 02 t 02 t12 , Ez Ez cBy t 03 t 03 t 31 and also comparing: b) cBx cBx , cBy cBy z , cBz cBz y with: b') t 23 t 23 , t 31 t 31 t 03 , t 12 t12 t 02 We could instead compare {a) with b')} and {b) with a')}, and thereby obtain the so-called anti-symmetric rank-two dual tensor G v : G v 0 cBx cBy cBz cBx cBy 0 Ez Ez 0 Ey Ex cBz Ey Ex 0 E-field representation of G v (SI units: Volts/m) Note further that the G v elements can be obtained directly from F v by carrying out a duality transformation (!!!): E cB, cB E Additionally note that a duality transformation leaves the Lorentz transformation of E and B unchanged !!! (i.e. duality = 90o). © Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois 2005-2015. All Rights Reserved. 21 UIUC Physics 436 EM Fields & Sources II Fall Semester, 2015 Lect. Notes 18.5 Prof. Steven Errede Again, note that our representation of the dual tensor G v is simply related to Griffiths by dividing ours by c, i.e.: n.b. For the sake of compatibility with Griffiths book, we will use his definitions of the field strength tensors. Please be aware that these tensor definitions do vary in different books! 1 v v GGriffiths Gours c 0 Bx By Bz 0 Ez c E y c 0 Ez c Ex c 0 E y c Ex c Bx By Bz B-field representation of G v (SI units: Tesla) With Einstein’s 1905 publication of his paper on special relativity, physicists of that era soon realized that the E and B -fields were indeed intimately related to each other, because of the unique structure of space-time that is associated with the universe in which we live. Prior to Øersted’s 1820 discovery that there was indeed a relation between electric vs. magnetic phenomena, physicists thought that electricity and magnetism were separate / distinct entities. Maxwell’s theory of electromagnetism “unified” these two phenomena as one, but it was not until Einstein’s 1905 paper, that physicists truly understood why they were so-related! It is indeed amazing that the electromagnetic field is a 4 4 rank two anti-symmetric contravariant tensor, F v (or equivalently, G v ) !!! (six components of which are unique.) The contravariant and covariant forms of the field strength tensor are: Ex c E y c Ez c Ex c E y c Ez c 0 0 0 Bz 0 Bz Ex c By Ex c By and: F v E y c Bz 0 0 Bx Bx E y c Bz 0 0 By By Bx Bx Ez c Ez c F v The contravariant and covariant forms of the dual field strength tensor are: G v 0 Bx By Bz 0 0 Ez c E y c Bx G v and: By 0 Ez c Ex c 0 E y c Ex c Bz Bx By Bz Bx By Bz 0 Ez c E y c 0 Ez c Ex c 0 E y c Ex c An arbitrary relativistic contravariant 4-vector x is related to its covariant version x by: x g v x v and: x g v xv , where the “flat” space-time metric g v g v {here} is defined: g v gv 22 1 0 0 0 0 1 0 0 0 0 1 0 0 0 with: g g 0 1 is the 4-D space-time version of the 3-D Kroenecker -function ij , i.e. 0 if and 1 for 0,1, 2,3 . © Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois 2005-2015. All Rights Reserved. UIUC Physics 436 EM Fields & Sources II Fall Semester, 2015 Lect. Notes 18.5 Prof. Steven Errede Likewise, the field strength tensors F v and F v , G v and G v are formally related to each other by: F g F v g v or: F g F v g v and: G g G v g v or: G g G v g v . The dual field strength tensor G v is formally related to the field strength tensor F v by: G 12 F v where: is the 4×4×4×4 = 256 element totally anti-symmetric tensor of rank four, the 4-D generalization of the 2-D Levi-Cività symbol ( ij ) : 1 (1) for even (odd) permutations of 0-1-2-3, respectively, 0 if any two indices are equal/same. We have learned that the 4-vector product of any two (bona-fide) relativistic 4-vectors, e.g. a b a b = Lorentz invariant quantity (i.e. a b a b = same value in all/any IRF’s) Likewise, the tensor product A v B v A v B v of any two relativistic rank-2 tensors is also a Lorentz invariant quantity. Thus, the tensor products of EM field tensors F v and G v , namely: F v F v , G v G v , and: F v G v F v G v are all Lorentz invariant quantities. Specifically, it can be shown that (see Griffiths Problem 12.50, p. 537): 1 F v F v G v G v 2 B 2 2 E 2 uM uE ← Lorentz invariant quantity !!! c And: F v G v F v G v 4 E B ← Lorentz invariant quantity !!! c Obviously, F v F v F v F v , G v G v G v G v and F vG v F v G v are the only Lorentz invariants that can be formed / constructed using F v and G v !!! Note that it is by no means obvious/clear how to form such Lorentz invariant EM field quantities starting directly from the E and B fields themselves !!! n.b. Since EM energy density uEM 12 o E 2 2 1o B 2 , Poynting’s vector S 1o E B , EM field linear momentum density EM o o S and EM field angular momentum density EM r EM cannot be formed from any linear combinations of F v F v , G v G v or F v G v , then uEM , S , EM and EM cannot be Lorentz invariant quantities! © Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois 2005-2015. All Rights Reserved. 23