String Theory- Electrodynamics

� � 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 = qE + 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�) 2 Lecture 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 0ByEz −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= FµνDefine: 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. 3 Lecture 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. 4 Lecture 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+ x 5 � � 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 q E(r) = 4πr2 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.7