PS2-solutions

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

This preview shows pages 1–4. Sign up to view the full content.

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 !

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
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.
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

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 8

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

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online