Homework # 3 Solution Key

Transcription

1 PHYSICS 631: Geneal Relativity Homewok # 3 Solution Key 1. You e on you hono not to do this one by hand. I ealize you can use a compute o simply look it up. Please don t. In a flat space, the metic in spheical coodinates,,θ,φ is: g µν = sin 2 θ (a) Please compute all non-zeo Chistoffel symbols fo this system. Fist, you need to ealize that the only non-zeo deivatives ae: g θθ, = 2 g φφ, = 2 sin 2 θ g φφ,θ = 2 2 sin θ cos θ Since the metic is diagonal, we e only to get tems with those thee combinations of paametes. If the non-zeo deivative is with espect to the coodinate and appeas upstais in the Chistoffel symbol, we e going to get a minus sign. Γ θθ = Γ θ θ = 1 Γ φφ = sin 2 θ (b) Compute the divegence The geneal expansion is: Γ φ φ = 1 Γ θ φφ = sin θ cos θ Γ φ θφ = cot θ ;α ;α =,α + Γ α αβv β So we only get to include those Chistoffel symbols that have the same upstais and downstais components. This means: If you like, this can be simplified to: ;α = V, + V θ,θ + V φ,φ + 2 V + cot θv θ V = V + V θ θ + cot θv θ + V φ φ Quick note: if you e tying to make this look the same as the spheical divegance that you might find in the inside cove of Jackson, fo example, you e going to be disappointed. The components of a vecto descibed in Jackson ae nomalized, but ous ae not.

2 2. Conside a vecto in 2-d flat space: at the position v = î i = î (a) Expess the vecto and position in pola coodinates. This is petty staightfowad. Both ae: v = = ˆ (b) Suppose you moved the vecto to a position (in Catesian coodinates): f = î + 0.1ĵ without changing the vecto itself. What would the components of the vecto, v be in pola coodinates at the final position? Method 1: Use a staight geometic appoach. Tun the Catesian coodinates into pola coodinates. The adial component is simply computed via: V = V ˆ = 1 and the θ component via: o so V µ V µ = 1 = 1 + (V θ ) 2 V θ = ( ) 1/ V f e 0.1 e θ (c) Method 2: Use the elation: ;β = 0 along the cuve. I m going to be moe pecise than is necessay hee, and will save my appoximations until the end. Fist, we need to wite ou diffeential equations: V, = 0 V,θ = Γ θθv θ = V θ V, θ = Γ θ θv θ = V θ V,θ θ = Γ θ θv = V So, conside moving along a cuve in the -diection, a geneic vecto tansfoms as: V2 = V1

3 and V θ 2 = V θ (It s a petty staightfowad diffeential equation). Now, suppose we go along a cuve of constant but vay θ? diffeential equation with the solution: Geneically, we get a coupled V 3 = V 2 cos(δθ) + B sin(δθ) And: V θ = 1 dv dθ = V 2 sin(δθ) + B cos(δθ) Since fo δθ = 0, this must yield the oiginal, we get: and thus: B = V θ 2 V3 = V2 cos(δθ) + V2 θ sin(δθ) = V1 cos(δθ) + 1 V1 θ sin(δθ) V3 θ = V2 θ cos(δθ) V 2 sin(δθ) 2 = 1 V1 θ cos(δθ) V 1 sin(δθ) 2 1 This is exact. In ou case, we e moving fom 1 = 1 to 2 =, and fom θ = 0 to sin θ = cos θ = Plugging and chugging, we get: and V = 1 V θ = both of which ae exactly what I got in the pevious section. We have the one-fom with components: (a) p α,β p µ = ( x 2 + 3y y 2 + 3x ) p α,β = ( 2x 3 3 2y )

4 (b) Deivatives and then tansfom. The tansfomation matix is: Λ α α = xα ( x α cos θ sin θ = sin θ cos θ ) But in ode to solve fo this, we fist need to expess the deivatives in tems of pola coodinates: ( ) 2 cos θ 3 p α,β = 3 2 sin θ So, multiplying it out, we get: p ; = 6 sin θ cos θ + 2 sin 3 θ + 2 cos 3 θ p ;θ = p θ; = (2 sin θ cos 2 θ 6 cos 2 θ cos θ + 2 cos 3 θ and p θ;θ = 2 2 sin θ cos θ( sin θ + cos θ 3) Hadly an elegant expession! (c) p α;β This is a staightfowad application, which fist equies us to compute p α = ( (6 sin θ cos θ + cos 3 θ + sin 3 θ) 2 [ sin θ(cos θ sin θ) cos 2 θ ] ) Fom thee, confiming the esult fom the pevious pat is staightfowad. I could e-wite it if you like, but I m not going to. Fistly, the non-zeo deivatives of the tenso A µν ae: A, = 2 A θ, = sin θ A θ,θ = cos θ A θ, = cos θ A θ,θ = sin θ A θθ,θ = sec 2 θ Combined with the non-zeo Chistoffel symbols: Γ θθ = ; Γ θ θ = Γ θ θ = 1

5 gives us the following in tems of Eq. 5.65: A = 2 A θ = 2 sin θ A θ = 2 cos θ A θθ = 2 tan θ θ A = 2 (sin θ cos θ) θ A θ = (1 + cos θ tan θ) θ A θ = (1 sin θ tan θ) θ A θθ = sec 2 θ + cos θ + sin θ The weak field metic: g µν = (1 + 2φ) φ φ φ Since any deivatives ae going to be 1st ode in φ, the 1st ode expansion of the Chistoffel symbols can be appoximated as: g µν = η µν Likewise, the only non-zeo tems ae: g µν,λ = 2δ µν φ,λ In othe wods, we e going to get non-zeo Chistoffel symbols if two of the indices ae the same. These will be of the following fom: Γ 0 ii = φ,0 Γ i 0i = φ,0 Γ 0 00 = φ,0 Γ i 00 = φ,i Γ 0 0i = φ,i Γ j ii = φ,j Γ i ij = φ,j 6. A cosmic sting is a theoetical constuct (neve obseved in natue) which is infinitely long, and has a mass density pe unit length of λ. The coodinates descibing the spacetime suounding a cosmic sting ae: and which has a metic: g µν = x µ = t R φ z R 2 (1 4λ)

6 (a) Compute the volume element, dv, nea the cosmic sting. Fist, take the negative squae oot of the deteminant: g = R 2 (1 4λ) so dv = gdx 0 dx 1 dx 2 dx 3 = 1 4λRdRdφdzdt (b) Compute all non-zeo Chistoffel symbols. Please be smat, and stat by noting that a) the metic is diagonal (fo easy invesion), and b) most deivatives cancel. The only non-zeo deivatives ae: g 22,1 = 2R(1 4λ) and thus, the only non-zeo Chistoffel symbols ae: Γ 2 12 = Γ 2 21 = 1 R (c) Compute the distance between two points sepaated by dx µ = dr, and all othe coodinates equal to zeo. Fom that, compute the distance fom the sting itself out to a distance of R = 1. Vey staightfowad: ds 2 = dr 2 ds = dr and thus: s = 1 0 dr = 1 (d) Compute the distance between two points, each R = 1 fom the sting sepaated by an angle dφ (with all othe dx µ = 0). Using that, what is the total distance tavesed by a paticle coveing a cicula obit of R = 1 aound the cosmic sting? Hee, the distance is: ds = 1 4λdφ and thus, the distance aound a cicula obit is: s = 2π 1 4λ (e) Compae pats c) and d) in the context of the nomal elationship between adius and cicumfeence. That is, does C = 2π? If not, what should it be eplaced with? This is odd. It is not 2π times the adius (unless λ = 0).