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Liquid Filled Capacitor Submitted by: I.D. 065628786 The problem: Parallel-plate capacitor built of two round conducting plates with the area A √ is filled with dielectric liquid with a dielectric constant . The distance between the plates is d d A. Due to a leaking, the surface of the liquid lowers at a constant rate u. 1. Find the capacitance as a function of t. 2. Find the electric field inside the capacitor. 3. Find the magnetic field between the capacitor’s plates. The solution: 1. Each part of the capacitor (the part filled with the liquid and the part without the liquid) have a different dielectric constant, therefore they act as two different capacitors connected in series. C1 = C2 = Cef f = 0 A ut A d − ut C1 C2 = C1 + C2 (1) (2) 0 A A ut · d−ut 0 A A ut + d−ut = 0 A ut( − 0 ) + 0 d (3) 2. Let the electrical field at the top part of the capacitor be E0 and the electrical field in the bottom part of the capacitor be E. The we find the relation between E, E0 from the Gauss’ law (in which we substitute 0 by ): E0 A = EA = E = q 0 q 0 E0 (4) (5) (6) 1 Then Z V = E0 = E = ~ · d~l = E0 ut + 0 E0 (d − ut) E V ut( − 0 ) + 0 d V 0 ut( − 0 ) + 0 d (7) (8) (9) 3) Magnetic field: ~ ·B ~ =0⇒B ~r = 0 ∇ (10) For the top part of the capacitor: I Z Z Z d d ~ ~ J · d~a + µ0 0 E0 · d~a = µo 0 E~0 · d~a dt dt d V 0 πr2 u( − 0 )πr2 B0φ · 2πr = µo = −µo 0 dt ut( − 0 ) + 0 d (ut( − 0 ) + 0 d)2 µ0 u( − 0 )0 r B0φ = − 2 (ut( − 0 ) + 0 d)2 ~ · d~l = µ0 B For the bottom part of the capacitor: I Z ~ · d~l = µo d ~ · d~a B E dt d V 0 πr2 u( − 0 )0 πr2 Bφ · 2πr = µo = −µo dt ut( − 0 ) + 0 d (ut( − 0 ) + 0 d)2 µ0 u( − 0 )0 r Bφ = − = Boφ 2 (ut( − 0 ) + 0 d)2 Pay attention that the magnetic field is continuous along the capacitor. 2 (11) (12) (13) (14) (15) (16) Induced Electromagnetic Fields Submitted by: I.D. 309026722 The problem: ~ r, t) = B0 r sin ωtẑ in some region of space. There is a magnetic field B(~ 1. Determine in which coordinates the magnetic field. 2. Find the induced electric field at the point: ~r0 = r0 x̂. 3. Find the current density ~j. What can you conclude about the region at which there are these magnetic and electric fields? The solution: ~ = 0. 1.Magnetic field must obey the second Maxwell’s equation ∇ · B Cylindrical coordinates X ∂ h1 h2 h3 Bi 1 ~ ·B ~ = ∇ h 1 h2 h3 ∂xi hi h1 = 1, h2 = r, h3 = 1, h1 h2 h3 = r ~ = (0, 0, B0 r sin ωt) B 1 ∂ B0 r2 sin ωt = 0 r ∂z Spherical coordinates. ~ ·B ~ = ∇ (1) (2) (3) (4) h1 = 1, h2 = r, h3 = r sin θ, h1 h2 h3 = r2 sin θ 1 ~ ~ ~ ·B ~ = ∇ ∇ · B = 5B0 sin ωt cos θ 2 r sin θ Therefore, the coordinates are cylindrical. (5) (6) 2. The Faraday’s law is ~ ~ ×E ~ = − ∂B ∇ ∂t (7) ~ is only in the z-direction, therefore, we only look at the component of the rotor that is in the B same direction: ∂Eφ 1 ∂Er ∂(B0 r sin φ) − =− (8) ∂r r ∂φ ∂t The electrical field is induced by the magnetic field and is not formed by the presence of electrical charges, therefore, the radial component of the electrical field equals zero: ~ · E~r = 4πρ = 0 ∇ (9) and Z Eφ = − ∂B dr = ∂t Z r ωB0 r cos(ωt)dr = 0 ωB0 r2 cos(ωt) 2 (10) Thus at r~0 = ro x̂ 2 ~ = ωB0 r0 cos(ωt) φ̂ E 2 (11) 1 3. We use the Maxwell’s equation: ~ ~ ×B ~ = µ0 J~ + µ0 0 ∂ E ∇ ∂t (12) Then J~ = 2 2 ~ 1 ~ ~ − 1 ∂ E = − B0 sin(ωt)φ̂ + 1 ω Bo r sin(ωt) φ̂ ∇×B µ0 0 ∂t µ0 0 2 (13) (14) From the expression above we can see that J~ grows to infinity like r2 . Physically, this is impossible, therefore, we conclude that the region in which the magnetic and electrical field are present is finite. 2 Fields inside a rolling charged cylinder Submitted by: I.D. 310159025 Problem A cylinder of a radius R, a length L, charded with a surface density charge σ, is rolling with a linear velocity u c without sliding. Find the electric and magnetic fields inside the cylinder. Solution Let S be the rest system of the cylinder. The laboratory system is S 0 . The rolling direction is −x̂. the current in the S system is: σ2πRLω q = T 2π u ω = R I = σuL I = (1) (2) (3) where ω is the angular velocity. From the Ampere’s law we can find out the magnetic field in the S system: I ~ · d~r = 4πKI B B = 4πKσuẑ (4) (5) The electric field in the S system is zero: ~ = 4πkρ ⇒ ρ = 0 ⇒ E = 0 divE (6) Since u c, then γ ' 1 and σ ' σ 0 . Using the Lorentz transformations we can find the fields in the lab system. The S 0 system moves with the velocity u to the x direction Electric field E 0 Ek0 = Ek = 0 (7) 0 E⊥ = γ0 E⊥ + u × B⊥ = γ0 uB ⊥ ' 4πKσu2 ŷ (8) Magnetic field B 0 Bk0 = Bk = 0 0 B⊥ (9) = γ0 B⊥ + u × E⊥ = γ0 B⊥ ' 4πKσuẑ = B⊥ 1 (10) 1 F~ ′ ⊥ = F~⊥ γ :ihhqexhwl` gk q lr lirtn `ed jkitl ,darnd zkxrna dgepna `vnp −q orhnd () 2 q F~2 = − 2 x̂ 6= −F~lab d ihhqexhwl`d dydn wx raep q −q −q lr gkd okl ,darnd zkxrna ihpbn dy xvei epi` .o`k miiwzn epi` oeheip ly iyilyd wegd ik al eniy .( :ze`ad qegid zekxrn .darnd q1 q1 q2 .(ef zkxrna dgepna ) . ) ly z` xibp S S1 S2 zkxrn - ly dgepnd zkxrn - ly dgepnd zkxrn - :xibp ok enk . . :xerya β21 = S1 S2 - l - l epgzity S S - n xarna upxel ixhnxt - n xarna upxel ixhnxt - γ1 , β1 γ2 , β2 zeiexidn xeaigl ze`gqepa mb ynzyp β2 − β1 1 − β1 β2 γ21 = γ1 γ2 (1 − β1 β2 ) q2 ” znpy itk x(2) xear l S2 ly zexidnd lynl) pke S1 , zkxrna npy - e itk S1 x x zekxrnd oia xarna dpzynd eyexit x(1) x21 S1 S2 dpzynd eyexit lebd ok enk .( . xy`k zkxrna zkxrnde :mixibn ep` :dxrd ~1 = v1 x̂ ⇒ β1 = v1 β c c ~2 = − v2 x̂ ⇒ β2 = − v2 β c c :darnd zkxrn - :lawpe rp orhn ly ilnygd dyl r̂ = ŷ, r = d, θ = dgqepa π 2 S • :aivp (`) v1 2 ~ 1 = q1 (1 − ( c ) ) ŷ = γ1 q1 ŷ E v1 2 3/2 2 d d2 ((1 − ( c ) ) xvei q2 iaeig orhny meyn riten ( − ) - d oniq) r̂ = −ŷ, r = d, θ = ŷ ( π 2 aivp q2 : xear ” l pk (a) xiv lr dhn oeeika dy v2 2 ~ 2 = − q2 (1 − ( c ) ) ŷ = −γ2 q2 ŷ E d2 ((1 − ( vc2 )2 )3/2 d2 zkxrna dyd o`k q2 z` xveiy aygl ,`idylk ihpbnd dligz qegi dydn yi zkxrna okl mb .ef qt` j` ,epayigy zkxrna `ed ~ B v1 m`) ~2 E ilnygd q1 ~×E ~ ′ = −β ~′ B zexidn 4 yi - dydn le raep xg`n :xywa q1 z`f lr gkd ,darnd ynzyp ,ihpbnd (b) .2 S2 :`idy , S - l ziqgi ’β’ ~×E ~ 2 k ẑ −β ly zexidnd `ed o`k :lawpe - d ik al miype ,( okl .( β2 q2 ly dgepnd zkxrna v2 c x̂ ly dxbdl m`zda) = −β2 x̂ v2 2 ~ 2 = v2 q2 (1 − ( c ) ) ẑ = −β2 γ2 q2 ẑ B v2 2 3/2 2 c ((1 − ( c ) ) d d2 :`ed F~1 da dewpa) lr upxel gke ~ 2 + v1 B2 (x̂ × ẑ)) = q1 (−γ2 q2 ŷ + β1 γ2 β2 q2 ŷ) q1 (E c d2 d2 q1 q2 γ2 − (1 − β1 β2 )ŷ d2 = = `vnp q1 `ed q1 xveiy ihpbnd q1 dyd . lr gkl ddf dxeva ayegn q2 lr gkd () q2 :( v2 2 ~ 1 = v1 q1 (1 − ( c ) ) ẑ = β1 γ1 q1 ẑ B v2 2 3/2 2 c ((1 − ( c ) ) d d2 q2 :`ed F~2 lr gkde ~ 1 + v2 B1 (−x̂ × ẑ)) = q2 (γ1 q1 ŷ − β2 γ1 β1 q1 ŷ) q2 (E c d2 d2 q1 q2 γ1 (1 − β1 β2 )ŷ d2 = = q1 : ly dgepnd zkxrn - :ihhqexhwl` dy `ed S1 • ~ (1) E 1 (`) ~ (1) = q1 ŷ = q1 ŷ E 1 d21 d2 dpi` drepzd oeeikl zkpe`nd dhpixe`ewy xg`n z`f , d1 = d y jkn raep oexg`d oeieeyd .upxel zivnxetqpxh zgz ilnyg dyl dgqepd zervn`a ayegn zexidnd xnelk) mini`znd γ - e q1 β znerl q2 mixhnxtd :lawp .dn`zda γ21 - e q1 β21 z` dgqepa epneqe q2 ly dgepnd zkxrna ly ziqgid zexidna ynzydl aivdle dl`yd S1 ( - a ly ~ (1) E 2 dpzyn ilnygd dyd (a) yi mrtd xy`k ,rp orhn ly dgepna zligza eayeg dl` dtevl qgia q2 ly mixhnxt .ef zexidnl 2 ~ (1) = q2 (1 − β21 ) (−ŷ) = − q2 γ21 ŷ = − q2 γ1 γ2 (1 − β1 β2 )ŷ E 2 2 2 d1 (1 − β21 )3/2 d2 d2 zkxrna dgepna q1 - e xg`n q1 lr gk lirtn epi` df dy : (1) ~ (1) F~1 = q1 E 2 zkxrnl S ~ (1) E 2 j` , ilnygd dydn S1 - a ihpbn dy xvei wx raep q1 lr gkd q2 okl .ef q1 q2 = − 2 γ1 γ2 (1 − β1 β2 )ŷ d zkxrndn gkd ly divnxetqpxhd zxfra zexiyi df iehia lawl mb ozip :dxrd) .(dgepna wiwlgd da :heyt `ed q2 lr gkdy jk ,ala ihhqexhwl` dy xvei q1 orhnd S1 S1 zkxrna (1) ~ (1) = q1 q2 ŷ F~2 = q2 E 1 d2 izya rp (b) q2 - e xg`n ,gkl (dheytd) divnxetqpxhd S2 .( zkxrna 5 zgqepa q1 ynzydl lr gkd ozip aeyiga l ” `l o`k) pk) zekxrnd () q2 : ly dgepnd q1 :zexidna rp β12 = zkxrn - S2 • ef zkxrna (`) β1 − β2 1 − β1 β2 γ :`ed ely - d xhnxt okle γ12 = γ1 γ2 (1 − β1 β2 ) :`ed xvei `edy ilnygd dyde 2 ~ (2) = q1 (1 − β12 ) (ŷ) = q1 γ12 ŷ = q1 γ1 γ2 (1 − β1 β2 )ŷ E 1 2 d2 (1 − β12 )3/2 d2 d2 :ef zkxrna ihhqexhwl` dy wx xvei q2 (a) ~ (2) = − q2 ŷ = − q2 ŷ E 2 d22 d2 :ala ihhqexhwl` gk `ed q1 lr gkd (b) (2) ~ (2) = − q1 q2 ŷ F~1 = q1 E 2 d2 okl S2 . zkxrna dgepna q2 - e xg`n q2 lr gk lirtn epi` df dy j` :ala ihhqexhwl` gk `ed ,ihpbn dy q2 lr q1 xvei q1 () lirtny gkd (2) ~ (2) = q1 q2 γ1 γ2 (1 − β1 β2 )ŷ F~2 = q2 E 1 d2 darnd zkxrna epayigy gkd ly dxiyi divnxetqpxh zxfra gkd z` lawl ozip o`k mb S2 .( zkxrna dgepna :ze`ad q2 - e xg`n) qegid zekxrn .darnd q . (λ− ) λ+ . .( β+ = ly dgepnd zkxrn - ) miiaeigd miprhnd ly dgepnd zkxrn - oia zeivnxetqpxhd S S′ S− S+ - miililyd miprhnd ly dgepnd zkxrn :zekxrnd β= z` xibp zkxrn - ly mixhnxtd z`e v x̂ , γ trans. S → S ′ c v0 x̂ , γ+ trans. S → S+ c β− = − v0 x̂ , γ− trans. S → S− c :lawp zeiexidn xeaigl zizeqgid dgqepd ′ β+ = ′ β− = β+ − β 1 − β+ β β− − β 1 − β− β 6 zervn`a .3 Cavity within a cylinder Submitted by: I.D. 066370016 The problem: There is a cylindrical cavity of a radius b inside an infinite cylinder. The cylinder’s radius is R and it is charged with a uniform charge density ρ. The distance between the axis of the cavity and the axis of the cylinder is a. Find the electric and the magnetic field as follows: 1. The electric field within the cavity. 2. By using the electric field, find the magnetic field inside the cavity. Consider the cylinder moving with a velocity v in it’s axis direction. 3. Compare your result to direct method of finding the magnetic field (q.e 45 4 102). The solution: 1. If we look carefully, and ignoring the cavity, then we notice that we have cylindrical symmetry. Therefore, we can use the Gauss’ law in order to calculate the electric field which is produced by the electrical charges in the cylinder and in the cavity (we will consider them negative). According to the formula ~ = 4πkρ Div E (1) For the entire cylinder we will use a cylindrical Gaussian surface with a radius r and a height h. Because then the electric field will be perpendicular to the surface and equal at every point on it. So by using the Gauss’ law we get I Z Z ~ ~ Ed~s = Div Edv = 4πk ρdv (2) v s v We know that the charge density is uniform and the surface area is 2πrh we get: I I ~ E · d~s = Er,cylinder ds = Er,cylinder 2πrh s (3) s Z 4πk Z ρdv = 4πkρ v dv = 4πkρπr2 h (4) v Er,cylinder = 2πkρr (5) where a − b ≤ r ≤ a + b. And with a vector notation we get that the electric field produced by the cylinder alone a distance r from the cylinder axis: ~ r,cylinder = 2πkρ~r E (6) 1 From the same considerations we can calculate in the same manner the electric field which is produced by the cavity, because we assign every infinitesimal volume element a charge density −ρ and say that it is uniform, in order to ”create” the cavity in the first place. Hence the electric field from the cavity is: I I ~ Ed~s = Er,cavity ds = Er,cavity 2π |~r − ~a| h (7) Zs Zs −ρdv = −4πkρ 4πk v dv = −4πkρπ |~r − ~a|2 h (8) v ~ r,cavity = −2πkρ(~r − ~a) E (9) By using the superposition idea we can find the total electric field inside the cavity. ~ T OT E ~ r,cylinder + E ~ r,cavity = E (10) = 2πkρ~r − 2πkρ(~r − ~a) (11) = 2πkρ~a (12) We can see that the total electric field inside the cavity is constant everywhere inside the cavity and pointed along the vector ~a. 2. It is given that the entire cylinder is moving along it’s axis with a velocity v. Let us choose a cylindrical coordinates so that ~v = vẑ. We shall use two system frame that are moving relative to each other. The Lab system is not moving and we shall call it the S system. The cylinder is in the rest in another system which is moving with the velocity v, and we shall call it S 0 . In S 0 the parallel electric field is zero, hence the parallel electric field in S is also zero, according to the equation: Ez = Ez0 = 0 (13) In addition, in S 0 there is no magnetic field at all, i.e B 0 = 0. So in S the parallel magnetic field is also zero, i.e Bz = 0. According to the transformation (from S to S 0 ): ~ × E) ~ ~ 0 = γ(B ~⊥ − 1 V B ⊥ c2 (14) And the opposite transformation (from S 0 to S): ~ ×E ~ 0) ~ ⊥ = γ(B ~0 + 1 V B ⊥ c2 ~⊥ = γ 1 V ~ ×E ~0 B c2 Calculating the cross r̂ 0 ~ ~ V × E = 0 Er (15) (16) ~ ×E ~0 product of V ϕ̂ ẑ 0 v = Er v ϕ̂ 0 0 (17) ~ ⊥ is along the ϕ̂ direction, i.e Which mean that the direction of B ~⊥ = γ 1 V ~ ×E ~ 0 = γ 1 Er v ϕ̂ = γ 1 2πkρav ϕ̂ B 2 c c2 c2 2 (18) By using the following relations we can arrive to a simpler solution 1 µ0 ε0 1 k = 4πε0 1 1 γρ ~⊥ = γ B 2π( )ρav ϕ̂ = µ0 av ϕ̂ 1 2 √ 4πε0 2 ( µ0 ε0 ) c = √ (19) (20) (21) 3. In order to calculate directly the magnetic field we first have to understand that we have an infinite cylinder which is uniformly charged and the charge is moving. So we can analog this to an infinite wire, with a radius R, and a cavity parallel to it’s axis at a distance a and with a radius b, which has a current. The current density is J~cylinder = J ẑ. Due to the cylindrical symmetry we can use Ampere’ law to calculate the magnetic field. First, let us conclude, by using the right hand rule, the direction of the magnetic field. If we shall put our thumb along the z axis (parallel to the current), we can see that the direction of the magnetic field is pointed along ϕ̂. In addition we shall indicate the existence of the cavity by a current which is J~cavity = −J~cylinder = −J ẑ. According to the Maxwell’s equation ~ = 4πκJ~ + rotB ~ κ ∂E k ∂t (22) ~ We know that ∂∂tE = 0 because there is no change in the electric field and we have no EMF. So the Maxwell equation reduces to ~ = 4πκJ~ rotB (23) By using Stokes’s theorem we get: I Z Z Z ~ ~ ~ ~ B · dl = rotBd~s = 4πκ Jd~s = µ0 J~ · d~s s L s (24) s If we choose a closed circular curved the magnetic field which is created by the cylinder or the cavity will be tanget to the curve and equal at every point. Also the current density is uniform so we can consider it as a constant, i.e we can take it out of the integral. I I ~ ~l = Bϕ dl = Bϕ 2πr Bd (25) L L Z µ0 s ~ s = µ0 J Jd~ Z ds = µ0 Jπr2 (26) s Bϕ 2πr = µ0 Jπr2 µ0 Jr Bϕ = 2 (27) (28) And in vector notation Bϕ will be: ~ cylinder = µ0 Jr ϕ̂ = µ0 J ~r × ẑ B 2 2 (29) 3 By using the same consideration we can find the magnetic field from the cavity: I I ~ ~ Bdl = Bϕ dl = Bϕ 2π |~r − ~a| L (30) L Z ~ s = µ0 J Jd~ µ0 Z s ds = −µ0 Jπ |~r − ~a|2 (31) s Bϕ 2π |~r − ~a| = −µ0 Jπ |~r − ~a|2 µ0 J |~r − ~a| Bϕ = − 2 (32) (33) And in vector notation Bϕ will be: ~ cavity = − µ0 J |~r − ~a| ϕ̂ = − µ0 J (~r − ~a) × ẑ B 2 2 (34) ~ cylinder and B ~ cavity to find the total magnetic field in Using the superposition principle for both B the cavity, we get: ~ T OT B ~ cylinder + B ~ cavity = B µ0 J µ0 J = ~r × ẑ − (~r − ~a) × ẑ 2 2 µ0 Ja µ0 J ~a × ẑ = ϕ̂ = 2 2 (35) (36) (37) In the previus question we have found the magnetic field respect to the lab system. ~ ⊥ = µ0 γρ av ϕ̂ B 2 (38) In order to get the right answer for the magnetic field we have to ”move” to the cylinder system, 0 0 to the S 0 system. We can do this by presenting γρ as ρ because ρ = γρ. And if we look carefully 0 and the dimensions of the product ρ v we will get J 0 . Eventually the magnetic field will be also ~ 0 = µ0 Ja ϕ̂ B ⊥ 2 (39) So both answers are correct. 4 Lorentz transformations of electromagnetic fields Submitted by: I.D. 36760742 The problem: ~ = B0 ẑ. It’s velocity is V ~ = V0 x̂. A neutral conducting ball is moving in a constant magnetic field B 1. Find the electric and magnetic fields in the ball’s rest frame. 2. Find the electric and magnetic fields inside the ball in the lab frame. 3. Is the magnetic field inside the ball equal to the one outside of it? Explain. 4. What is the induced electric dipole? The solution: Through out the solution i’ll use: S 0 as the ball’s f.o.r. (frame of reference), and S as the lab’s. 1. Lab f.o.r.: Ex = Ey = Ez = 0 (1) Bx = By = 0 (2) Bz = B0 (3) In the ball’s rest frame: The electric field is zero since the ball is a neutral conductor. Then, using the Lorentz transformation, we find: Ex 0 = Ek 0 = 0 E⊥ 0 (4) = γ0 (E⊥ + V0 × B) = γ0 (V0 × B) = −γ0 V0 B0 ŷ 0 Bx = Bx = 0 B⊥ 0 (5) (6) 2 = γ0 (B⊥ − V0 × E⊥ /c ) = γ0 B⊥ = γ0 B0 ẑ (7) 2. Inside the ball, ball’s f.o.r.: E0 = 0 B 0 (8) = γ0 B0 ẑ (9) The magnetic field remains the same outside and inside the ball. Again, using the Lorentz transformations to find the fields inside the ball in the lab’s f.o.r.: Ex = Ek = Ek 0 = 0 (10) 0 0 E⊥ = γ0 (E⊥ − V0 × B⊥ ) = γ0 (−V0 × B⊥ ) = 0 Bx = Bx = 0 0 γ02 V0 B0 ŷ (11) (12) 0 2 B⊥ = γ0 (B⊥ − V0 × E⊥ /c ) = γ02 B0 ẑ (13) 3. As we have seen, the magnetic field inside and outside of the ball (lab’s f.o.r.) are not the same: Bin = γ02 B0 ẑ (14) Bout = B0 ẑ (15) 1 The magnetic field inside the ball is increased due to the electrical currents flowing in it. Also the electrical field inside the conducting ball is non-zero due to the induced charge/dipole. 4. In the ball’s rest frame the electric dipole is induced (look at e 36 1 028) p~0 = R3 ~ 0 R3 E = − γ0 V0 B0 ŷ k k (16) The surface charge density is then σ0 = − 3 γ0 V0 B0 cos θ0 4πk (17) where θ0 = r̂ · ẑ in the ball’s rest frame. ~ Hence in the lab frame the dipole will not change, We can think of the dipole moment as p~ = q d. ~ since the charge is invariant and d is perpendicular to the direction of the movement. But pay attention that σ 0 6= σ and cos θ 6= cos θ0 . 2 Lorentz Invariance Submitted by: I.D. 036769412 The problem: ~ · B, ~ 1 B 2 − 0 E 2 are Lorentz invariance,i.e. prove that the next Prove that the expressions E µ0 identities: ~ ·B ~ =E ~0 · B ~0 1. E 2. 1 2 µ0 B − 0 E 2 = 1 02 µ0 B − 0 E 02 are true for any two reference frame. The solution: We will solve this problem in MKS. In case we have tow reference frames S and S’ which travels at ~ = V0 x̂ the Lorentz transformations for the electric and magnetic field are: a speed V B~|| = B~||0 (1) 0 0 E~⊥ ~ ~ ~ B⊥ = γ0 (B⊥ + V0 × 2 ) c E~ = E~0 (3) 0 0 E~⊥ = γ0 (E~⊥ − V~0 × E~⊥ ) (4) || (2) || ~ ·B ~ =E ~0 · B ~ 0 . first we can divide any one of those four vectors to two parts: 1. let us show that E ~ = B~⊥ + B~|| B ~0 (5) 0 0 0 0 B = B~⊥ + B~|| ~ = E~⊥ + E~|| E ~ 0 = E~⊥ + E~|| E (6) (7) (8) and now we: ~ ·B ~ = (E~⊥ + E~|| ) · (B~⊥ + B~|| ) = E~⊥ · B~⊥ + E~⊥ · B~|| + B~⊥ · E~|| + E~|| · B~|| E (9) Since E~⊥ ⊥B~|| and B~⊥ ⊥E~|| we get: ~ ·B ~ = E~⊥ · B~⊥ + E~|| · B~|| E (10) and in the same manner we get: ~0 · B ~ 0 = E~⊥ 0 · B~⊥ 0 + E~|| 0 · B~|| 0 E (11) using the lorentz transformation (formulas 1-4)we have: 0 0 0 0 0 0 0 E~⊥ 2 ~ 0 ~ ~ ~ ~ ~ ~ ~ ~ (12) E · B = E|| · B|| + γ0 (E⊥ − V0 × E⊥ ) · (B⊥ + V0 × 2 ) = E~|| · B~|| + c " ! ! # ~⊥ 0 ~⊥ 0 0 0 0 0 0 0 E E + γ02 E~⊥ · B~⊥ + E~⊥ V~0 × 2 − (V~0 × E~⊥ ) · B~⊥ − V~0 × 2 · (V~0 × E~⊥ ) c c 1 and then we have: 0 0 ~ ·B ~ = E~⊥ · B~⊥ + E ! 1 1− V02 c2 0 0 E~⊥ · B~⊥ − V02 c2 0 0 E~⊥ · B~⊥ (13) and finally we have: ~ ·B ~ = E~|| 0 · B~|| 0 + E~⊥ 0 · B~⊥ 0 = E ~0 · B ~0 E (14) 2. let us now show that µ10 B 2 − 0 E 2 = µ10 B 02 − 0 E 02 , multiplying the both sides of the equality by µ0 gives us B 2 − 0 µ0 E 2 = B 02 − 0 µ0 E 02 and since µ010 = c2 we have that we need to proof B 2 − c12 E 2 = B 02 − c12 E 02 . proving that we will use some of the calculations we already did. from 5 and 7 we get: B2 − 1 1 2 E = (B~⊥ + B~|| ) · (B~⊥ + B~|| ) − 2 (E~⊥ + E~|| ) · (E~⊥ + E~|| ) = 2 c c 1 1 = B~⊥ · B~⊥ + B~|| · B~|| − 2 E~⊥ · E~⊥ − 2 E~|| · E~|| c c (15) and in the same manner: B 02 − 0 0 0 0 0 0 0 0 1 02 1 1 E = B~⊥ · B~⊥ + B~|| · B~|| − 2 E~⊥ · E~⊥ − 2 E~|| · E~|| 2 c c c (16) using lorentz transformation we get: B2 − ~ 0 ~ 0 1 2 ~|| 0 · B~|| 0 − 1 E~|| 0 · E~|| 0 + γ02 (B~⊥ 0 + V~0 × E⊥ ) · (B~⊥ 0 + V~0 × E⊥ ) − E = B c2 c2 c2 c2 0 0 0 0 1 − 2 γ02 (E~⊥ − V~0 × B~⊥ ) · (E~⊥ − V~0 × B~⊥ ) c (17) and then we will get: # " ~⊥ 0 ~⊥ 0 0 0 0 02 0 0 1 E 1 E B 2 − 2 E 2 = B~|| · B~|| − 2 E~|| · E~|| + γ02 B~⊥ + 2B~⊥ · (V~0 × 2 ) + (V~0 × 2 )2 −(18) c c c c h i 02 0 0 0 1 − 2 γ02 E~⊥ − 2E~⊥ · (V~0 × B~⊥ ) + (V~0 × B~⊥ )2 c 0 0 since V~0 ||E~⊥ ||B~⊥ we get: B2 − 0 0 0 02 0 1 1 2 E = B~|| · B~|| − 2 E~|| · E~|| + γ02 B~⊥ + 2 c c 0 E~⊥ ~ V0 × 2 c !2 h i − 1 γ02 E~⊥ 02 + (V~0 × B~⊥ 0 )2 (19) c2 and then we will have: 0 0 02 1 2 1 ~ 0 ~ 0 V02 ~ 02 1 2 ~ 02 V02 ~ 02 2 ~ ~ ~ B − 2 E = B|| · B|| − 2 E|| · E|| + γ0 B⊥ − 2 B⊥ − 2 γ0 E⊥ − 2 E⊥ c c c c c 2 (20) and finally we have: B2 − 0 0 0 0 0 0 0 0 1 2 1 1 1 E = B~⊥ · B~⊥ + B~|| · B~|| − 2 E~⊥ · E~⊥ − 2 E~|| · E~|| = B 02 − 2 E 02 2 c c c c 2 (21)
