� � Lecture 3 8.251 Spring 2007 Lecture 3 - Topics • Relativistic electrodynamics. Gauss’ law • • Gravitation and Planck’s length Reading: Zwiebach, Sections: 3.1 - 3.6 Electromagnetism and Relativity Maxwell’s Equations Source-Free Equations: 1 ∂B�� × E�= − c ∂t (1) � · B�= 0 (2) With Sources (Charge, Current): � · E�= ρ (3) 1 ∂E�1 � × B�− c ∂t = cJ�(4) Notes: 1. E and B have same units. 3. ρ is charge density [charge/volume]. Here no 0 or 4π - those constants would get messy in higher dimensions. 4. J�is current density [current/area] E�, B�are dynamical variables. d�p �1 �dt = q E + c�v × B 1� � � � ( � �� � Lecture 3 8.251 Spring 2007 Solve the source free equations � · B�= 0 solved by B�= � × A�. (Used to have � × E�= 0, E = −�Φ) True equation: 1 ∂ 1 ∂A�E + E +� × �c ∂t (� × A�) = � × �c � × ∂t 1 ∂A�= � × E�+ c ∂t = 0 So: 1 ∂A�E�+ c ∂t = −�Φ (Φ scalar) Thus: 1 ∂A�E�= −�Φ − c ∂t E, B�) encoded as (Φ, A) Φ, A are the fundamental quantities we’ll use Gauge Transformations A�→ A�� = A�+ � B�� = � × A� = � × (A + �) = B�function of �x,t. � function = vector. 1 ∂Φ → Φ� = Φ − c ∂t 1 ∂ 1 ∂ E�� = −�(Φ�) = −� Φ − c ∂t − c ∂t (A + �) = E�So under gauge transformations, E�and B�fields unchanged! �↔�(Φ, A) g.t. (Φ�, A�) (Physically equivalent) Suppose 2 sets of potentials give the same E�’s and B�’s. Not guaranteed to be gauge-related. Suppose we have 4-vector Aµ = (Φ, A�) then Aµ = (−Φ, A�) 2Lecture 3 8.251 Spring 2007 Take ∂x∂ µ . Have indices from ∂x∂ µ and from Aµ so will get a 4x4 matrix. Have two important quantities (E and B) with 3 components each 6 important quantities. Hint that we should get a symmetric matrix. ⇒ Fµν = ∂Aν ∂Aµ = ∂µAν − ∂ν Aµ∂xµ − ∂xν Fµν = −Fνµ 1 ∂Ai ∂ Foi = c ∂t − ∂xi (−Φ) = −Ei F12 = ∂xAy − ∂yAx = Bz ⎞⎛ 0 −Ex −Ey −Ez Ex 0 Bz −By Bx Fµν = ⎜⎜⎝ ⎟⎟⎠Ey −Bz 0 By Ez −Bx 0 What happens under gauge transformation? Aµ → Aµ�= Aµ + ∂µ Then get: F �= ∂µA�µν ν − ∂ν A�µ = ∂µ(Aν + ∂ν ) − ∂ν (Aµ + ∂µ) = Fµν + ∂µ∂ν − ∂ν ∂µ = Fµν Define: Tλµν = ∂xFµν + ∂µFνλ + ∂ν Fλµ Note indices are cyclic. Some interesting symmetries: Tλµν = −Tµλν Tλµν = −Tλνµ So Tλµν is totally antisymmetric. A totally symmetric object in 4D has only 4 nontrivial components so Tλµν = 0 gives you 4 equations. 3Lecture 3 8.251 Spring 2007 Tλµν = 0 = ∂λ(−∂ν Aµ) + ∂µ(∂ν Aλ) + ∂ν (∂λAµ − ∂µAλ) Charge Q is a Lorentz invar. Not everything that is conserved is a Lorentz invar. eg. energy. Since Q is both conserved and a Lorentz invar, (cρ, J�) form a 4-vector Jµ Now let’s do what a typical theoretical physicist does for a living: guess the equation! F µν ≈ Jµ No, derivatives not right. ∂F µν /∂xν ≈ Jµ No, constants not right. 1 F µν /∂xµ = Jµ Correct, amazingly! (even sign) c µ = 0: ∂F 0ν /∂xν = ρ ∂F 0i/∂xi = ρ F0i = −Ei F 0i = Ei So � · E�= ρ verified! Electromagnetism in a nutshell: Fµν = ∂µAν − ∂ν Aµ ∂F µν Jµ = ∂xν c Consider electromagnetism in 2D xy plane. Get rid of Ez component: ⎛ ⎞ 0 −Ex −Ey Fµν = ⎝ Ex 0 Bz ⎠ Ey −Bz 0 But what about Bz? Doesn’t push particle out of the plane (v × Bz with v in the xy plane remains in xy plane) but rename Bz as B, a scalar. 4Lecture 3 8.251 Spring 2007 How about in 4D spatial dimensions? ⎞⎛ F = ⎜⎜⎜⎜⎝ −Ex −Ey −Ez −EN 0 ∗ ∗ ∗ ∗ ∗ 0 ∗ 0 0 ⎟⎟⎟⎟⎠ So get tensor B! It’s a coincidence that in our 3D spacial world E and B are both vectors. Let’s look at � · E�= ρ in all dimensions. Notation: Circle S� is a 1D manifold, the boundary of a ball B2 Sphere S2(R) : x212223 = R2+ x + xBall B3(R) : x212223≤ R2+ x + x5� � Lecture 3 8.251 Spring 2007 When talking about S2(R), call it S2 (R = 1 implied) Vol(S1) = 2π Vol(S2) = 4π Vol(S3) = 2π2 22π d Vol(Sd−1) = Γ( d 2 ) All you need to know about the Gamma function: Γ(1/2) = √π Γ(1) = 1 Γ(x + 1) = xΓ(x) Γ(n) = (n − 1)! for n ∈ Z Γ(x) = ∞ dte−ttx−1 for x > 0 0 Calculating � · E�in d = 3 and general d dimensions. d = 3: Ed(vol) = ρd(vol) = q� · �This represents the flux of E�through S2(r) E(r) vol(S2(r)) = q B3(r) � B3 (r) · 6� � Lecture 3 8.251 Spring 2007 E(r) 4πr2 = q· 1 qE(r) = 4π r2 This falls off much faster at large r and increases much faster as small r. General d: Ed(vol) = ρd(vol) = q� · �This represents the flux of E�through Sd−1(r) E(r) vol(Sd−1(r)) = q Bd(r) Bd(r) · Γ(d/2) qE(r) = 2πd/2 rd−1 Electric field of a point charge in d dimensions. If there are extra dimensions, then would see larger E at very small distances.