soln11

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.��Solution for Chapter 11(compiled by Guodong Wang)January 21, 20031 Problem A.(BT-11.1)[by Alexander Putilin/01]We have the wave equation (BT-11.12):"2� 1� = (K + �)"(" � �) + �"2� (1)"t2 3Writing � as� = "� + " � A (2)and substituting in (1) we get"2 4 "2" � � - (K + �)"2� + " � � A - �"2A = 0 (3)"t2 3 "t2Since � and A are independent each of the two terms in (3) vanishes.Thuswe get separate equations for scalar and vector potentials.1.1"2�- c2 "2� = �(t) (4)"t2 L4K+ �3where cL = and �(t) is arbitrary function of time.�Notice that � is not defined uniquely by (2).If we make the transformation� �! � +�(t) with arbitrary �(t), "� will not change.Using this gauge freedomwe can set �(t) = 0.To see that it is possible substitute � �! � + �(t) into (4).Then"2�- c2 "2� = �(t) - � (t) = 0 (5)"t2 Ltif we choose �(t) = dt t dt �(t ).1.2"2A- c2 "2A = "�(t, x) (6)"t2 T�where cT = and � is arbitrary.As in the previous case we can remove�"� term using gauge transformation A �! A + "�.Under this transformation(6) becomes1"2A "2�- c2 "2A = " �(t, x) - + c2 "2� = 0 (7)"t2 T "t2 TWe should choose � such that"2- c2 "2 � = � (8)"t2 T(8) is just a wave equation with source term.It always has solution.2 Problem A.(BT-11.2)[by Guodong Wang/03]Including the effect of gravity into equation (BT-11.12),"2� 1� = (K + �)"(" � �) + �"2� + �g (9)"t2 3Consider small pertubations to the displacement and density, Substituing � �!�0 + ��, � �! �0 + � into Eq.(9), Keeping the small pertubations to the firstorder, we obtain the static equation10 = (K + �)"(" � �0) + �"2�0 + �0g (10)3which describe the static displacement �0 produced by gravity g, and the waveequation"2� 1�0 = (K + �)"(" � �) + �"2� + ��g (11)"t2 3Recalling Eq.(BT-11.10), �� -�0" � �, Eq.(11) can be written as"2� 1�0 = (K + �)"(" � �) + �"2� - �0" � �g.(12)"t2 3The first term and the second term on the right side of Eq.(12) are of the sameorder of magnitude.The ratio of the third to the second term is of the order of�0g� [ Pobierz całość w formacie PDF ]

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