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In "String Theory and M-Theory: a modern introduction" by K.Becker, M. Becker and J.H.Schwarz, they say that BPS D-brane is stable as it preserves half of the Supersymmetry. I really want to understand more about this statement and see detail calculations. What is the mechanism of D-brane stability? Is there any derivation for the instability of space-filling D-brane (so that open string tachyon will be eliminated from the theory)?
Thank you.
BPS objects are stable because they're the lightest objects with given values of certain conserved charges. So there exists no potential final state that would be lighter and that the BPS state could decay into, by conserving the energy. The excess energy may be invested to the kinetic energy of the final energy but a deficit energy means that the decay is prohibited.
As an analogy, note that the electron has to be stable because there exists no lighter $Q=-e$ object than the electron (and positron).
BPS objects are either those preserving some (enough) supersymmetry; or objects saturating the BPS bound $M=Q$, schematically speaking (for branes, it's the tension equal to the charge density; coefficients should be inserted everywhere). These conditions are equivalent because $$ \{Q,Q\} = M-Z $$ schematically, for some conserved supercharge $Q$. So the expectation value of $\{Q,Q\}$ in the BPS state is zero – because $Q$ annihilates the state – but it's also equal to $M-Z$ which means that the mass is equal to the charge. For non-BPS states, we have the strict $M\gt Z$. Here, $Z$ is the conserved central charge.
