MOTION OF IXDIVIDUAT. DOMAIN WALLS 101") 



we sec at once that the solution for (/[ as a function of / as the domain 

 wall passes the position of our small x'olume is: 



g'lit) =\e-"' j g'Ut)e"'dL (19) 



We have ignored the solution of the homogeneous equation, as it con- 

 tains a factor c~^'^ . 



In order to evaluate (15) for our small volume, we need a more ex- 

 plicit value for gi . From the Fourier transformation we may write: 



g'Ut) = r Gi=o(co)e'"' dc,. (20) 



J— 00 



Consequently: 



g[{t) = i e-'" j[ G'Uo:)e'"''-"''' dco dt. (21) 



This can be integrated over t. If we do this, then bring the factor e'"" 

 out from under the integral sign, we find: 



g[{t) = f ^^ 6^""' Jco. (22) 



J-00 i + 1(J}T 



Now the frequencies for which ^^•e get a contribution to the integral in 

 (22) have an upper bound determined by the velocity and the thickness 

 of the domain wall, of course. The faster the domain wall moves, the 

 higher the upper bound on the frequency. We have been careful to make 

 our measurements near zero wall velocity, and the data clearly extra- 

 polates Hnearly to zero. This was done in order to minimize the possibility 

 of wall distortion, but it also is consistent with our assumption of al- 

 most isothermal conditions. We are therefore justified in assuming that 

 cor <K 1 for all the frequencies in Gi«(w), and we can expand the factor 

 1/(1 -f icor) in (22). We obtain: 



f/i(0 = f " [1 - (icor) + (iccrf + • • •]Gl«(a;)e''^' do:, (23) 



and if we form the derivatives of (20) and compare them with the terms 

 in (23), we see that: 



,;(0 = ,;.(0-.!%^ + r= '%«+■•. (24) 



