= Solution
A supersymmetric type-II <D-brane> carries a <Ramond–Ramond potential> charge and saturates the corresponding generalized <BPS bound in supersymmetry>: its <brane tension> equals the charge-dependent central/tensorial-charge bound in the relevant normalization. It is therefore a <BPS state>, preserving sixteen of the thirty-two spacetime <supercharges>. Identical parallel branes with the same orientation impose the same supersymmetry projector. This statement does not apply to arbitrary unstable non-BPS branes or to the bosonic branes of the preceding parts.
One direct demonstration of the <parallel D-brane no-force identity> is the annulus vacuum amplitude. For separation $r$, its oscillator-dependent part is proportional to
$$
\mathcal A(r)\propto V_{p+1}\int_0^\infty\frac{dt}{2t}(8\pi^2\alpha't)^{-(p+1)/2}
e^{-r^2t/(2\pi\alpha')}
\frac{f_3(q)^8-f_4(q)^8-f_2(q)^8}{f_1(q)^8},\qquad q=e^{-\pi t}.
$$
Here
$$
\begin{aligned}
f_1(q)&=q^{1/12}\prod_{n\ge1}(1-q^{2n}),&
f_2(q)&=\sqrt2\,q^{1/12}\prod_{n\ge1}(1+q^{2n}),\\
f_3(q)&=q^{-1/24}\prod_{n\ge1}(1+q^{2n-1}),&
f_4(q)&=q^{-1/24}\prod_{n\ge1}(1-q^{2n-1}).
\end{aligned}
$$
The first two terms describe the projected <Neveu–Schwarz sector>, and the last is the oppositely signed <Ramond sector> contribution. The <Jacobi abstruse identity> gives $f_3^8-f_4^8-f_2^8=0$ for every $t$. Therefore
$$
\boxed{\mathcal A(r)=0,\qquad -\frac{dV(r)}{dr}=0.}
$$
In the closed-string channel, attractive <graviton> and <dilaton> exchange cancel repulsive <Ramond–Ramond potential> exchange. The classical probe calculation in the <D-brane supergravity solution> gives the same cancellation. The common supersymmetry allows separation to remain a flat modulus; relative angles, reversed RR charge or supersymmetry-breaking backgrounds can remove this protection. For a brane–antibrane pair the RR interaction reverses and does not cancel the NS-NS attraction. The charge and half-supersymmetry interpretation is established in https://arxiv.org/abs/hep-th/9510017[Dirichlet-Branes and Ramond-Ramond Charges].
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