MATH423 String Theory Solutions 4. = 0 τ = f(s). (1) dτ ds = dxµ dτ f (s) (2) dτ 2 [f (s)] 2 + dxµ. dτ f (s) (3)

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1 1. MATH43 String Theory Solutions 4 x = 0 τ = fs). 1) = = f s) ) x = x [f s)] + f s) 3) equation of motion is x = 0 if an only if f s) = 0 i.e. fs) = As + B with A, B constants. i.e. allowe reparametrisations are a shift of origin an a rescaling We have to take the variation x τ) x τ) + δx τ) in the action τf S = mc η ν. The variation gives τf x S + δs = mc η + δx ) x ν + δx ν ) ) ) τf = mc η + δx ν + δxν τf = mc η ν + δx ν + δx ν + δx ) τf = mc η ν + ν δx + δx δx ν ) δx ν We have to Taylor expan the square root. Recall that Taylor expansion for an arbitrary function to first orer is fx 0 + ) = fx 0 ) + f x 0 ). 1

2 Here, fx 0 ) = η ν an = η δx ν + δx Hence we get ropping terms of orer δ ) τf S + δs = mc η ν mc τf 1 η δx η α αβ τf 1 = δs = mc η δx η αβ α ν β = mc Integrating by parts V U = V U) UV ) with, τf δs = mc V = η α β αβ η α β αβ τf, U = δx we have δx τf δx δx ν ) ν β η α β αβ mc η α β αβ δx τ). The first term vanishes by the conitions δx ) = δx τ f ) = 0. Since δx τ) is arbitrary in the omain of integration, The secon term vanishes iff the integran is ienticaly zero i.e. mc η α β αβ = 0 4) This equation is in manifestly reparameterization invariant form. Inee, the object between the brackets is clearly reparameterization invariant: mc η α β αβ = mc η α β αβ This follows from the chain rule =. The erivative in front of the square bracket oes not spoil the reparameterization invariance since [ ] = [ ] = 0 [ ] = 0. When we choose τ = s η ν = η ν = ) ) = 1

3 Equation 4) then becomes [ mc ] = p = The relativistic version of Newton s secon law is p t = ) m v = F. t 5) 1 v or in covariant form with p = ) m = f 6) f = v F, F. 1 v we want to show that 6) is the same as 5). Note Similarly, or an v p t = v ) m v = t 1 v = p0 = p0 t m v v t 1 v ) 3 = v F t = v F or p0 = p t = p t = F p = F 1 v p = f = ) m t 1 v v F 1 v = p0 t Here s is the proper time an = 1 v t. Note that we fixe c = 1. 3

4 a. 1 S = mv t + q c A x) t t where A = Φ, A). 1 S = mv t + q A vt q Φt = Lt. c where L = 1 mv + q c A v qφ b. c. p = L v = m v + q c A H = p v L = mv + q c A v 1 mv q c A v + qφ = 1 mv + qφ We have to replace v by p. From part b) v = 1 m p q A c ) hence H = 1 p q A) m c + qφ We are intereste in variation of the action S = mc + q c I, I = A xτ)) τ). 7) when we let x τ) x τ) + δx τ). We note that δa xτ)) A xτ) + δxτ)) A xτ)) = A x ν δxν τ), 4

5 where A is calculate at x = xτ). The variation of the firstpart wss one x ν in problem. The variation of the secon part is obtaine from I + δi = = = A xτ) + δxτ)) x + δx ) A xτ)) + A x ν δxν ) + δx ) A xτ)) + A ) δxν xν + A xτ)) δx where we roppe the term of orer δ in the last line. We therefore have δi = A δxν xν + A xτ)) δx. We exchange ν in the first term an rewrite the secon using a total erivative: δi = δx A ν ν [ ] x + A δx ) δx A. We assume that δx vanishes at the en of, so the total erivative first term in square brackets) vanishes. Using the chain rule for the secon term in square brackets we fin δi = δx Aν x A ) ν x ν = δx ν F This conclues the variation of I. The variation of the first term in 7) is obtaine by varying the path x τ) x τ)+δx τ), as was one in problem, or alternatively as we i in the lectures, δs = mc τf = mc δ) = mc η δx η δx τ)) ν ν ) τf δx τ)) The first term is a bounary term an vanishes by imposing δx )) = δx τ f )) = 0. 5 mcη )) ν

6 τf δs = δx τ)) mcη )) ν The momentum four vector is given by p ν = mu ν = mc ν, where u ν is the velocity four vector. Hence, τf δs = δx p ν τ))η = τf δx τ)) p Combining the variation of the two terms in the action for a charge particle in an electromagnetic fiel we get p = q c F ν 6. Our starting point is the variation δs = mc δ). 8) With the path parametrise by an arbitrary τ we have ) = g xτ)) ν ). 9) Uner the variation xτ) xτ) + δxτ) the variation of the metric is The variation of 9 gives δg xτ)) = g x + δx) g x) = g x α δxα τ). δ) = g ν δx ν x α δxα ) g ) Diviing by an cancelling one factor of, we have δ) = 1 g ν x α δxα g δx α α. 6

7 inserting this result into 8 we have, δs = mc 1 g ν x α δxα g α δx α ). Integrating by parts an ropping total erivatives that vanish since δx = 0 at the bounaries of the worl line 1 δs = mc δx α g ν x α [ ]) g α. Since δx α is arbitrary an requiring that the variation vanishes fernishes the equation of motion which is given by [ ] g α = 1 g ν x α. This is the geoesic equation. Since s an arbitrary parameter we can choose τ = s. We then have [ ] g α = 1 g ν x α. 10) Expaning the erivatives in 10 gives x g α + g α ν x ν 1 g ν x α = 0. The last two terms on the left han sie can be combine to give x g α + 1 g α x g ) ν ν x α = 0. Since the expression in parenthesis multiplies ν, an is symmetric in the summation inices an ν, it can be rewritten in a form which is also symmetric in an ν as x g α + 1 gα x + g να ν x g ) ν x α = 0. multiplying throughout by g λα an using g λα g α = δα λ gives x λ + 1 gα gλα x + g να ν x g ) ν x α = 0. 7

8 The equation takes the form with x λ + ν Γλ = 0. Γ λ = 1 gα gλα x + g να ν x g ). x α 8