PS2-solutions

PS2-solutions - Phys 253a 1 Problem Set 2 Solutions...

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Phys 253a 1 Problem Set 2 Solutions September 30, 2010 1. (a) Remember that d D x has mass dimension - D , while derivatives have mass di- mension 1. We’ll denote the mass dimension of a field φ by [ φ ]. Then since the action has mass dimension 0, 0 = [ d D x F 2 μν ] = 2[ F μν ] - D = 2[ A μ ] + 2 - D 0 = [ d D x φ * ± φ ] = 2[ φ ] + 2 - D In other words, [ φ ] = [ A μ ] = D - 2 2 . This is a general result for any bosonic field in a relativistic theory, since the kinetic term will always have two derivatives and be quadratic in the fields. Like the kinetic terms, the interaction terms should have dimension D , so D = [ gA μ φ * μ φ ] = [ g ] + 1 + 2[ φ ] + [ A μ ] = [ g ] + 1 + 3 D 2 - 3 So we’ve found [ g ] = 4 - D 2 . We also have D = [ λφ 3 ] = [ λ ] + 3[ φ ] = [ λ ] + 3 D - 6 2 so that [ λ ] = 6 - D 2 . (b) The electromagnetic interaction is renormalizable exactly when D = 4. The φ 3 interaction is renormalizable when D = 6 (which is precisely why Srednicki works in 6 dimensions for the first part of his text). 2. The transformation Y : ( t,x,y,z ) ( t,x, - y,z ), a reflection in the y direction, is indeed a Lorentz transformation, since it clearly preserves the Lorentz-invariant inner product. The reason we don’t talk about Y very much is that it’s equivalent to the product of P and an already familiar Lorentz transformation. Recall that a rotation by an angle θ in 2-dimensions is represented by a matrix cos θ - sin θ sin θ cos θ !
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Phys 253a 2 When θ = π , this transformation is just minus the identity. Thus, we can flip the sign of any two coordinates with a 180 rotation in the plane spanned by those coordinates. If we apply P , and then rotate by π in the x - z plane, we get Y . So while Y isn’t continuously connected to the identity, it is continuously connected to P . 3. (a) We know from class the equations of motion for a massless vector in the presence of a source, so all we need to do is add the variation of the mass term: 0 = μ F μν + m 2 A ν + J ν Taking the divergence gives m 2 · A + · J = 0 (the μ ν F μν term drops out because it is the contraction of a symmetric tensor with an antisymmetric one). We are assuming that the current is divergenceless, so we have · A = 0 and the equation of motion becomes 0 = ± A ν + m 2 A ν + J ν (b) Since the source is static, we expect the potential A 0 to be independent of time. Thus, we’d like to solve 0 = ( ± + m 2 ) A 0 + J 0 = ( -∇ 2 + m 2 ) A 0 ( x ) + J 0 This is easiest in Fourier space: ( k 2 + m 2 ) e A 0 ( k ) = - e J 0 ( k ) = - e A 0 ( x ) = - e Z d 3 k (2 π ) 3 e i x · k k 2 + m 2 This had better be invariant under rotations of x (and it’s not hard to prove that it is, using a change of integration variables), so we’re free to choose x = (0 , 0 ,r ). Moving to spherical coordinates ( k,θ,φ ), we get A 0 ( r ) = e (2 π ) 3 Z k 2 dk dφd (cos θ ) e ikr cos θ k 2 + m 2 = e (2 π ) 2 Z 0 k 2 dk 1 ikr e ikr - e - ikr k 2 + m 2 = e (2 π ) 2 ir Z -∞ k dk e ikr k 2 + m 2 In the last line, the region of integration where k > 0 gives the e ikr term in the second line, while the region where k < 0 gives the - e - ikr term.
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Phys 253a 3 (c) The integral in (c) is along the real axis in the complex- k plane. Note that, because of the exponential e ikr
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PS2-solutions - Phys 253a 1 Problem Set 2 Solutions...

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