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

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Unformatted text preview: 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 F . 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 A 1- d 1 A ) 2 + ( d A 2- d 1 A 2 ) 2 + ( d 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 = 0 L ( A ) = braceleftBigg ( A - A ) , = 0- ( A - A ) , negationslash = 0 . =- F = F . 1 Homer Reids Solutions to Peskin and Schroeder Problems: Chapter 2 2 Hence the Euler-Lagrange equation is 0 = L ( A ) = F = A - A . Identifying E i =- F 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 = E k parenleftbigg B = E t parenrightbigg . which are two of Maxwells equations. (b) First lets review this whole Noether theorem business. Suppose we make a change, parameterized by a small parameter , under which the fields in the Lagrangian undergo the transformations i ( x ) prime i ( x ) = i ( x ) + i ( x ) (1) where the 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 + J ( 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 i ( i ) + L ( i ) ( i ) bracerightbigg = L + summationdisplay i braceleftbigg L i ( i ) + parenleftbigg L ( i ) i parenrightbigg- i L ( i ) bracerightbigg (where we used the chain rule, ( fg ) = fg + gf ) = L + summationdisplay i braceleftbigg parenleftbigg L ( i ) i parenrightbigg + i parenleftbigg L i- L ( i ) parenrightbiggbracerightbigg = L + summationdisplay i parenleftbigg L ( i ) i parenrightbigg where we used the Euler-Lagrange equation in going to the last line. Comparing with (2) we see that summationdisplay i L ( i ) i = J Homer Reids Solutions to Peskin and Schroeder Problems: Chapter 2 3 or...
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This note was uploaded on 02/07/2011 for the course PHYS 1002 taught by Professor Imanpour during the Spring '11 term at Alabama.

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

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