peskin2 - Solutions to Problems in Peskin and Schroeder An...

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Solutions to Problems in Peskin and Schroeder, An Introduction To Quantum Field Theory Homer Reid November 16, 2002 Chapter 2 Problem 2.1 (a) The Lagrangian density is L = - 1 4 F μν F μν . Since F ij = - F ji there are really only 6 terms in the sum: = - 1 2 bracketleftBig F 01 F 01 + F 02 F 02 + F 03 F 03 + F 12 F 12 + F 13 F 13 + F 23 F 23 bracketrightBig . Also, F ij = - F ij if one of i or j is zero, and F ij = + F ij if both i, j are nonzero, so we may write = + 1 2 bracketleftbig F 2 01 + F 2 02 + F 2 03 - F 2 12 - F 2 13 - F 2 23 bracketrightbig = 1 2 bracketleftbig ( d 0 A 1 - d 1 A 0 ) 2 + ( d 0 A 2 - d 1 A 2 ) 2 + ( d 0 A 3 - d 1 A 3 ) 2 - ( d 1 A 2 - d 2 A 1 ) 2 - ( d 1 A 3 - d 3 A 1 ) 2 - ( d 1 A 2 - d 2 A 1 ) . 2 bracketrightbig . Evidently we have L ∂A ν = 0 L ( μ A ν ) = braceleftBigg ( μ A ν - ν A μ ) , μν = 0 - ( μ A ν - ν A μ ) , μν negationslash = 0 . = - F μν = F νμ . 1
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Homer Reid’s Solutions to Peskin and Schroeder Problems: Chapter 2 2 Hence the Euler-Lagrange equation is 0 = μ L ( μ A ν ) = μ F νμ = μ ν A μ - μ μ A ν . Identifying E i = - F 0 i and B k = - epsilon1 ijk F ij , the Euler-Lagrange equations be- come that 1 E 1 + 2 E 2 + 3 E 3 = 0 ( ∇ · E = 0) epsilon1 ijk i B j = 0 E k parenleftbigg ∇ × B = E ∂t parenrightbigg . which are two of Maxwell’s equations. (b) First let’s review this whole Noether theorem business. Suppose we make a change, parameterized by a small parameter α, under which the fields in the Lagrangian undergo the transformations φ i ( x ) φ prime i ( x ) = φ i ( x ) + α Δ φ i ( x ) (1) where the Δ φ i ( x ) are fields themselves. This transformation does not affect the equations of motion if the Lagrangian of the system is only modified by the addition of the divergence of a four-vector, i.e. if L → L prime = L + α∂ μ J μ ( x ) . (2) On the other hand, computing directly from (1) we can also write the trans- formed Lagrange density as L → L prime = L + summationdisplay i braceleftbigg L ∂φ i ( α Δ φ i ) + L ( μ φ i ) ( α∂ μ Δ φ i ) bracerightbigg = L + α summationdisplay i braceleftbigg L ∂φ i φ i ) + μ parenleftbigg L ( μ φ i ) Δ φ i parenrightbigg - Δ φ i μ L ( μ φ i ) bracerightbigg (where we used the chain rule, ( fg ) = f∂g + g∂f ) = L + α summationdisplay i braceleftbigg μ parenleftbigg L ( μ φ i ) Δ φ i parenrightbigg + Δ φ i parenleftbigg L ∂φ i - μ L ( μ φ i ) parenrightbiggbracerightbigg = L + α∂ μ summationdisplay i parenleftbigg L ( μ φ i ) Δ φ i parenrightbigg where we used the Euler-Lagrange equation in going to the last line. Comparing with (2) we see that μ summationdisplay i L ( μ φ i ) Δ φ i = μ J μ
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Homer Reid’s Solutions to Peskin and Schroeder Problems: Chapter 2 3 or μ j μ = 0 , j μ = J μ - summationdisplay i Δ φ i L ( μ φ i ) .
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