L. 1-D Soliton Solutions

Transcription

1 L. -D Soliton Solutions Consider a -D soliton described by H = f + U(ϕ) ( Ezawa: f = ħ c, U = ħ c U Ezawa ) with E-L eq. d ϕ - d U = 0 d ϕ d = d - d U = 0 = ± - U + u 0 = 0 ( u 0 = const ) = ± (U - u 0) (U - u 0 ) ϕ() or - 0 = ± ϕ( 0 ) d U = d U ( c ± = constant ) (U - u 0 ) & H = f [ U(ϕ) -u 0 ] Kinks for Real K-G ϕ 4 Field For the real K-G ϕ 4 field, (see 7.3._SolitaryWavesKinksAndSolitons.pdf ) H = f ( ϕ) + ϕ - v U(ϕ) = 4 ϕ - v For kinks, boundary conditions are ϕ(± ) = ±v. U (± ) = 0 ± = ± (-u 0 ) = 0 u 0 = 0 Let η = ϕ - v ϕ = ± η + v = ± U(η) = 4 η d η η + v To avoid the proliferation of ± sins, we ll work only with the + sin & consider the - case only at the end.

2 L._-DSolitonSolutions.nb = U + c = d η η η + v + c + a = - a tanh - + a a = - v tanh- η + v v + c η + v = -v tanh v ( - c) ϕ = ± η + v ϕ = v tanh v ( - c) tanh(),-tanh() - Since tanh 0 = 0, we can write ϕ() -ϕ ( 0 ) = ±v tanh v ( - 0) Usin tanh(± ) ±, we see that the +(-) sin denotes kink (anti-kink). Settin ξ = v we have ϕ() - ϕ ( 0 ) = ±v tanh - 0 ξ so that ξ is a measure of the width of the transition reion where ϕ chanes sinificantly. Incidentally, settin η = ϕ - v leads to a solution of the form ϕ() = ±v coth -c ξ However, since lim 0 ± coth = ±, it can t be used for our purposes. Without loss of enerality, we can set 0 = 0 so that ϕ() = ±v tanh ξ

3 L._-DSolitonSolutions.nb 3 ϕ - v = v tanh ϕ = ± v ξ sech ξ ξ - = -v sech ξ H = f ( ϕ) + ϕ - v = f v ξ = f v4 sech 4 + v4 sech 4 ξ Kinks for Sine-Gordon Field See 7.4._Sine-GordonSolitons.pdf. ξ H = f ( ϕ) + ( - cosϕ) ( Ezawa: f = ħ c, U = ħ c U Ezawa ) U(ϕ) = ( - cosϕ) = sin ϕ = ± (U - u 0 ) = ± sin ϕ - u 0 = ± - u 0 - cos ϕ Let κ = - u 0 = ± κ - κ cos ϕ Note that the interal is periodic with period π, which means that if ϕ() is a solution, so are ϕ() + π n for n = 0, ±, ±,... Since is real, we must have κ if ϕ is unrestricted. Another way to write the solution is - 0 = ± ϕ() κ ϕ( 0 ) - κ cos ϕ

4 4 L._-DSolitonSolutions.nb S-G Kinks lim ϕ() = 0 = 0 κ ϕ( 0 ) - κ cos ϕ which is possible only if the interand has a pole at ϕ = 0, i.e., κ = u 0 = 0 so that = ± sin ϕ = ± sin ϕ where we ve set κ = without lost of enerality. Note that the interand now has poles at ϕ = n π, for n = 0, ±,... Usin sin ϕ = ln tan ϕ we have = ± ln tan ϕ +c ± ϕ ± () = tan - ep ± - c ± Usin tan - = π 4, tan - 0 = 0 & tan - ( ) = π we can write where ϕ ± () = ϕ( 0 ) 4 π tan- ep ± - 0 ϕ + (± ) = ϕ( 0) 0 Thus, ϕ +/- denotes a kink / anti-kink. tan - e, tan - e - 0 ϕ - (± ) = ϕ ( 0 ) Kink & anti-kink As mentioned before, ϕ +/- + π n with n = 0, ±, ±... is also a leitimate kink / anti-kink. On the other hand, the heiht of a kink / anti-kink is ϕ( 0 ). Thus, in order to have multiple kinks / anti-kinks co-eistin in the system, the heiht must equal to

5 L._-DSolitonSolutions.nb 5 the period, i.e., ϕ( 0 ) = π By definition, only such kinks / anti-kinks can be called solitons. A wave that vanishes at both ± would be ϕ() = ϕ( 0 ) 4 π tan- ep Unfortunately, since ϕ is not defined at = 0, it is not admissible as a solution. κ 0 For κ 0, = ± κ +c ± - κ cos ϕ ± κ ϕ() ϕ() = ± κ ( c ±) or ϕ() -ϕ ( 0 ) = ± κ ( - 0) H For a kink centered at the oriin with ϕ(0) = ( n + ) π & rises from n π to (n + ) π : ϕ() = 4 tan - ep + n π d tan- = + ϕ = 4 tan ϕ 4 = = + e sech e e + tann π - e tann π = - cosϕ = sin ϕ = tan ϕ + tan ϕ e ( ϕ) = 6 for n = even -e - for n = odd e = + e 4 sech Usin tan A = tan A - tan A we see that tan ϕ = e - e for n = even - e- - e - for n = odd = e = -csch - e

6 6 L._-DSolitonSolutions.nb + tan ϕ = + e = - e coth 8 e - cosϕ = = + e sech Hence H = f ( ϕ) + ( - cosϕ) = 6 f e = + e e E = 6 f - + e = 8 f d y 0 = 8 f - = 8 f + y 0 ( + y) 4 f sech E-L eq., Static Case E-L eq. for the S-G field is c t t ϕ - ϕ + sin ϕ = 0 Consider the static kink ( see section H ) ϕ() = 4 tan - ep + π n d sech = -sech tanh ϕ = sech ϕ = - sech tanh - cosϕ = sech sinϕ = - 4 sech tanh sinϕ = - sech tanh Toether with t ϕ = 0, we see that ϕ() does satisfy the E-L eq. E-L eq. The time-dependent solution is obtained by a Lorentz boost of the static solution in the rest frame to a frame movin with velocity u. Hence

7 L._-DSolitonSolutions.nb 7 With X = ϕ(t, ) = 4 tan - - u t ep + π n - u c - u t - u c, we have ϕ(t, ) = 4 tan - e X + π n ϕ = sech X - u c ϕ = - sech X tanh X - u c sinϕ = - sech X tanh X u & t ϕ = - sech X - u c t t ϕ = - u sech X tanh X - u c Thus, ϕ(t, ) does satisfy the E-L eq. c t t ϕ - ϕ + sin ϕ = 0 Similarly, with - cosϕ = sech X we have H = f ( ϕ) + ( - cosϕ) = f 4 sech X u c sech X = f E = f = f - u c sech X - u c - u c - u c sech X - - u c - u c - d X sech X sech = tanh tanh(± ) = ± E = 4 f - u c - u c