where ⃗ I ± ≡ ( I 1 $_{±}$( k ) , I 2 $_{±}$( k ) , I 3 $_{±}$( k )). The method to calculate I ± is described in [7, 20]. In the limit ω → ∞ , I ± exponentially reduces to 0, unless at least one of the following conditions on the integrand is met. One is the existence of discontinuities of a ′ $^{i}$( u ) (or b ′ $^{i}$( v )). The contribution of a discontinuity of a ′ to I$_{+}$ ( k ) is where u ∗ is the position of the discontinuity, a ∗ = a ( u ∗ ) and we assume a ′ jumps from a ′ − to a ′ + at u = u ∗ . The region of length ∼ ω − 1 around u = u ∗ contributes to this value. We find I i $_{+}$( k ) ∼ ψ/ω , where ψ denotes sharpness of the kink. The other condition is the existence of stationary points of the phase of the integrand, i.e. ω ( u − ˆ k · a ( u )) / 2 (or ω ( − v − ˆ k · b ( v )) / 2). This condition is expressed as One can find the energy of the GW burst from ONE kink by picking up the contribution from the discontinuity for one of I ± (say, I$_{+}$ ) and from the stationary point for the other (say, I − ). The energy emitted when the kink is located at the world sheet coordinate ( u, v ) = ( u ∗ , v$_{s}$ ) is evaluated by substituting (26) into I$_{+}$ and (28) into I − in (25). Then we obtain where the subscript s represents the value at v = v$_{s}$ . The energy is radiated at every moment toward the direction of − b ′ from the kink. This is the formula which represents energy emitted in a short period in a small solid angle. The total energy emitted in the I i $_{+}$( k ) ≃ − 2 iω ( a ′ i + 1 − ˆ k · a ′ + − a ′ i − 1 − ˆ k · a ′ − ) e iω ( u $_{$_{∗}$−}$ˆ k · a$_{∗}$ ) / 2 (26) 1 − ˆ k · a $^{′}$( u$_{s}$ ) = 0 (or − 1 − ˆ k · b $^{′}$( v$_{s}$ ) = 0) (27) dE d Ω dω ( ω, − ω b ′ $_{s}$) = Γ $^{2}$( 2 3 ) 3 Gµ $^{2}$ω − 4 3 ( 12 | b ′ s · b ′′′ s $_{|}$) 4 3 × ( 1 (1 + b ′ s · a $^{′}$$_{+}$) 2 − 2 a ′ + · a ′ − (1 + b ′ s · a $^{′}$$_{+}$)(1 + b ′ s · a ′ − ) + 1 (1 + b ′ s · a ′ − ) 2 ) b ′′ 2 s , (29) 10 at the point u = u$_{s}$ ( v = v$_{s}$ ). The contribution of the stationary point of the phase to I$_{+}$ ( k ) is For any value of a ′ ( b $^{′}$), there is one direction ˆ k that satisfies (27), i.e. ˆ k = a ′ ( − b $^{′}$). Therefore, every point of u can contribute to I i $_{+}$( k ) for one direction ˆ k . I i $_{+}$( k ) ≃ 1 ω 2 / $_{3}$a ′′ $_{i}$( u$_{s}$ ) e iω ( u$_{s}$ $_{−}$ˆ k · a ( u$_{s}$ )) / 2 ( 12 | ˆ k · a $^{′′′}$( u$_{s}$ ) $_{|}$) 2 / 3 i √$_{3}$ Γ(2 / 3) . (28)