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D0-branes and counting

  1. Nov 9, 2011 #1

    haushofer

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    Hi,

    I'm reading Becker2Schwarz, chapter 5.1, about D0 branes in the GS formalism. They introduce kappa-symmetry, and end the section with "without this symmetry there would be the wrong number of propagating degrees of freedom".

    I'm trying to understand that. The fermions [tex]\Theta^a[/tex] have, for D=10, [tex]2^5=32[/tex] complex components. But they are Majorana-Weyl, so this brings this number back to [tex]\frac{32}{4}=8[/tex] complex components. Kappa-symmetry implies that half of these fermions are gauge degrees of freedom, giving us 8 real components.

    However, for the D0-brane, which is a particle, we start with 10 real components [tex]X^{\mu}[/tex]. Choosing e.g. the static gauge brings this back to 9 components. Obviously, to have as many bosonic degrees of freedom as fermionic (8 real), I need to get rid of another bosonic degree of freedom. How do I do that?

    Perhaps I'm also confused by worldsheet SUSY versus target space SUSY; to realize target space SUSY on [tex]\{\Theta^a,X^{\mu}\}[/tex] one doesn't need this kappa symmetry, right?

    Any help is appreciated :)
     
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  3. Nov 9, 2011 #2

    haushofer

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    Let me add something: for the massless case m=0, one can write down the superparticle in both RNS and GS formalism, and also realize both worldline- and targetspace SUSY. (this is also treated in chapter 4 of BBS). In this case the counting works: a massless particle has 8 bosonic degrees of freedom, and so has the spinor.

    For the massive case one realizes target space SUSY via

    [tex]
    \delta \Theta^A = \epsilon^A, \ \ \ \ \delta X^{\mu} = \bar{\epsilon}^A\Gamma^{\mu}\Theta^A
    [/tex]
    where A=1,...,N labels the amount of SUSY. This algebra implies that a translation P on Theta is zero, so Theta is a zero eigenvector of P, and hence that P is not invertible. Then one doesn't need to have the same amount of X-components and Theta-components, right? I wouldn't know how to realize now wordline supersymmetry.

    Maybe this section of BBS is a bit unclear; it now seems to me that this kappa symmetry is not per se about obtaining the same amount of X- and Theta components, but just about obtaining the right amount of SUSY. Does that make sense?
     
  4. Nov 9, 2011 #3

    The number of propagating degrees of freedom are determined by the equation of motion (5.18).
     
  5. Nov 9, 2011 #4

    haushofer

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    And (5.15) for the X, of course. But I can't fully understand what is "wrong" if kappa symmetry is not there for the massive D0-brane. Is it because the D0-brane breaks N=2 SUSY to N=1 SUSY, and that kappa symmetry takes care of this in the GS formalism?
     
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