5.4 BIons

Perhaps one of the most pedagogical examples of brane solitons are BIons. These were first described in [128Jump To The Next Citation Point, 234Jump To The Next Citation Point] and correspond to on-shell supersymmetric D-brane configurations representing a fundamental string ending on the D-brane, i.e., the defining property of the D-brane itself. They correspond to the array of branes
Dp : 1 . . .p x x x x (288 ) F : x x x x x p + 1 x x x .
Working in the static gauge describes the vacuum infinite Dp-brane. The static soliton excites a transverse scalar field (y = y (σa )) and the electric field (V = V (σa )) 0 0, while setting the magnetic components of the gauge field (Va) to zero
μ μ i i X = σ , X = c, Xp+1 (σa) = y(σa ) , V0 = V0(σa ). (289 )
The gauge invariant character of the scalar ensures its physical observability as a deformation of the flat world volume geometry described by the global static gauge, whereas the electric field can be understood as associated to the end of the open string, which is seen as a charged particle from the world volume perspective. A second way of arguing the necessity for such electric charge is to remember that fundamental strings are electrically charged under the NS-NS two form. The latter appears in the effective action through the gauge invariant form β„±. Thus, turning on V0 is equivalent to turning such charge33.

Supersymmetry analysis:
Let me analyse the amount of supersymmetry preserved by configurations (289View Equation) in type IIA and type IIB, separately. In both cases, the matrix 𝒒 μν + β„± μν equals

( ) − 1 F0b 𝒒μν + β„± μν = − F0a δab + ∂ay∂by = ⇒ − det(𝒒μν + β„± μν) = det(δab + ∂ay∂by − F0aF0b)(290 )
while the induced gamma matrices are decomposed as
γ0 = Γ 0, γa = Γ a + ∂ay Γ y, (291 )
where a stands for world space indices. Due to the electric ansatz for the gauge field, the kappa symmetry matrix Γ κ has only two contributions. In particular, for type IIA (p = 2k )
( ( ) ) ---------1------------1---- μ0...μp k+1 p + 1 k Γ κ = ∘-----------------(p + 1)!πœ€ γμ0...μpΓβ™― + 2 F μ0μ1γμ3...μpΓ β™― . (292 ) − det(𝒒 μν + β„± μν)
Summing over world volume time, one obtains
1 1 ( p ) Γ κ = ∘----------------- -πœ€a1...ap Γ 0γa1...apΓ kβ™―+1+ -F0a1γa2...apΓ kβ™― . (293 ) − det(𝒒 μν + β„±μν) p! 2
Using the duality relation
√ -------- πœ€i1...ikjk+1...jp+1γjk+1...jp+1 = (− 1)k(k−1)βˆ•2(p + 1 − k)!γi1...ik − det𝒒 Γ 0...p, (294 )
one can write the first term on the right-hand side of Eq. (293View Equation) as
Γ Γ k+1 − Γ b∂ y Γ Γ Γ k+1. (295 ) 0...p β™― b y 0...p β™―
Using the same duality relation and proceeding in an analogous way, the second term equals
F0aΓ aΓ 1...pΓ k + F0a∂byΓ yΓ abΓ 1...pΓ k. (296 ) β™― β™―
Inserting Eqs. (295View Equation) and (296View Equation), the kappa symmetry preserving condition can be expressed as
∘ -------------- [ ] − det(𝒒 + β„± )πœ– = 1 + Γ aΓ 0Γ β™―(F0a − ∂ay Γ 0yΓ β™―) − Γ abF0a ∂byΓ 0yΓ β™― Γ 0...pΓ kβ™―+1πœ– . (297 )
Given the physical interpretation of the sought soliton, one imposes the following two supersymmetry projection conditions
Γ Γ k+1πœ– = πœ– (298 ) 0...p β™― Γ 0yΓ β™―πœ– = πœ– (299 )
corresponding to having a type IIA Dp-brane along directions 1,...,p and a fundamental string along the transverse direction y. Since both Clifford valued matrices commute, the dimensionality of the subspace of solutions is eight, as corresponds to preserving ν = 1βˆ•4 of the bulk supersymmetry. Plugging these projections into Eq. (297View Equation), the kappa symmetry preserving condition reduces to
∘ -------------- ( a ab ) − det(𝒒 + β„±)πœ– = 1 + Γ Γ 0Γ β™―(F0a − ∂ay) − Γ F0a∂by πœ–. (300 )
It is clear that the BPS condition
F0a = ∂ay, (301 )
derived from requiring the coefficient of Γ aΓ Γ 0 β™― to vanish, solves Eq. (300View Equation). Indeed, the last term in Eq. (300View Equation) vanishes due to antisymmetry, whereas the square root of the determinant equals one, whenever Eq. (301View Equation) holds. The analysis for type IIB Dp-branes (p = 2k + 1) works analogously by appropriately dealing with the different bulk fermion chiralities, i.e., one should replace Γ k β™― by τk iτ2 3. Thus, the supersymmetry projection conditions (298View Equation) and (299View Equation) are replaced by
k+1 Γ 0...pτ3 iτ2πœ–, = πœ– (302 ) Γ 0yτ3πœ– = πœ–, (303 )
corresponding to having a type IIB Dp-brane along the directions 1,...,p and a fundamental string along the transverse direction y.

Satisfying the BPS equation (301View Equation) does not guarantee the on-shell nature of the configuration. Given the non-triviality of the gauge field, Gauss’ law ∂ Ea = 0 a must be imposed, where Ea is the conjugate momentum to the electric field, which reduces to

Ea = -∂β„’- = δabF0b, (304 ) ∂ Λ™Va
when Eq. (301View Equation) is satisfied. Thus, the transverse scalar y must be a harmonic function on the p-dimensional D-brane world space
∂a∂ay = 0 . (305 )

Hamiltonian analysis:
Using the phase space formulation of the D-brane Lagrangian in Eqs. (222View Equation) and (223View Equation), I will reproduce the BPS bound (301View Equation) and interpret the charges carried by BIons. Working in static gauge, the world space diffeomorphism constraints are trivially solved for static configurations, i.e., Pi = 0, and in the absence of magnetic gauge field excitations, i.e., Fab = 0. The Hamiltonian constraint can be solved for the energy density [225Jump To The Next Citation Point]

β„°2 a b T-2- = E E 𝒒ab + det 𝒒ab. (306 ) Dp
Since det 𝒒ab = 1 + (∂y )2, Eq. (306View Equation) is equivalent to [225Jump To The Next Citation Point]
β„°2 --2-= (1 ± Ea ∂ay )2 + (E βˆ“ ∂y)2. (307 ) TDp
There exists an energy bound
-β„°-- a T ≥ 1 + |E ∂ay|, (308 ) Dp
being saturated if and only if
Ea = ± ∂ay. (309 )
This is precisely the relation (301View Equation) derived from the solution to the kappa symmetry preserving condition (214View Equation) (the sign is related to the sign of the fundamental string charge). Thus, the total energy integrated over the D-brane world space ω satisfies
E ≥ E0 + |Zel|, (310 )
where Zel is the charge
∫ Z = Ea ∂ y. (311 ) el ω a
To interpret this charge as the charge carried by a string, consider the most symmetric solution to Eq. (305View Equation), for Dp-branes with p ≥ 3, depending on the radial coordinate in world space r, i.e., r2 = σa σbδab,
q y(σa) = ---------, (312 ) Ωp −1rp−2
where Ω p stands for the volume of the unit p-sphere. This describes a charge q at the origin. Gauss’s law allows us to express the energy as an integral over a (hyper)sphere of radius δ surrounding the charge. Since y = y(δ) is constant over this (hyper)sphere, one has
∫ E = lδim→0 |y (δ ) dβƒ—S ⋅Eβƒ—| r=δ = q lδi→m0 y(δ). (313 )
Thus, the energy is infinite since y → ∞ as δ → 0, but this divergence has its physical origin on the infinite length of a string of finite and constant tension q [128, 234]. See [164] for a discussion of the D-string case, corresponding to string junctions.
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