Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

What does ##\delta F=0## mean?

  1. Dec 29, 2015 #1
    Links for [1] and [2] are below.

    Please have a look here section 12.6 [1]. It says here that

    Given the action of a supergravity theory, it is generally useful to search for solutions of the classical equations of motion. It is most useful to obtain solutions that can be interpreted as backgrounds or vacua. Fluctuations above the background are then treated quantum mechanically. The backgrounds that are considered have vanishing values of fermions, and are thus determined by a value of the metric, the vector fields (or higher forms) and scalar fields. One common background is Minkowski space, but there are others such as anti-de Sitter space, certain black holes, cosmic strings, branes or pp-waves, which are all supersymmetric, i.e. they ‘preserve some supersymmetry’. This means that the background is invariant under a subset of the local supersymmetries of the supergravity theory. For a preserved supersymmetry, the local SUSY variations of all fields must vanish when the background solution is substituted. This leads to conditions of the generic form $$δ(\epsilon) \text{boson} = \text{fermion} = 0, \hspace{.5cm} δ(\epsilon)\text{fermion} = \text{boson} = 0. \tag{12.15}$$

    I discussed this a little with ACuriousMind in this thread [2]. My question was if supersymmetric background means that it preserves some supersymmetry, then as the text above says, the SUSY variations (12.15) must vanish. This results in for example that $$\delta \text{fermion}=0.$$ So, what does this equation imply? Does it imply that for a background to be supersymmetric, then fermions must not transform into bosons but rather stay fermions? I must be confused about this because this doesn't make much sense now does it?

    Note: I would like to note ACuriousMind's answer there on where he said

    "You're completely misunderstanding what a "preserved supersymmetry" is. That's not a requirement on the theory, it is just a requirement on the classical solutions, that they be mapped onto themselves under the preserved supersymmetric transformation (i.e. δX=0 for all fields), and not onto another solution (this is just a weird way of stating that the solution doesn't break the supersymmetry)."

    This got me confused, what does he mean by this answer?

    [1]: http://books.google.com.lb/books?id=KFUhAwAAQBAJ&pg=PA249&lpg=PA249&dq=van proeyen freedman supergravity It is most useful to obtain solutions that can be interpreted as backgrounds or vacua.&source=bl&ots=vh-QrPO7je&sig=ZDggO4dPDDVcNeiqaC8ojY8clSQ&hl=ar&sa=X&ved=0ahUKEwjUzdDYj__JAhXG1BoKHaqyBX0Q6AEIGjAA#v=onepage&q=van proeyen freedman supergravity It is most useful to obtain solutions that can be interpreted as backgrounds or vacua.&f=false
    [2]: http://physics.stackexchange.com/qu...etry/226342?noredirect=1#comment489467_226342
     
  2. jcsd
  3. Dec 30, 2015 #2

    haushofer

    User Avatar
    Science Advisor

    I think you can compare this with GR. In GR, one has gauge invariance under gct's,

    [tex]
    \delta g_{\mu\nu} = 2 \nabla_{(\mu} \xi_{\nu)}
    [/tex]

    The Einstein Field Equations are invariant under these transformations. However, particular solutions (e.g. Schwarzschild) are NOT. This corresponds to the fact that coordinates can only be interpreted with a metri and a solution is given by an equivalence class ('gauge orbit', as the sophisticated say). Now, given a solution and a particular coordinate system, the solution is not invariant under the beforementioned gct's. Its symmetries are given by the Killing equations

    [tex]
    \delta g_{\mu\nu} = 2 \nabla_{(\mu} \xi_{\nu)} = 0
    [/tex]
    Now xi is not a gauge parameter anymore, but a vector field which generates your symmetry! As such the background 'breaks' the gct's to a subgroup generated by the Killing vectors.

    Now apply this reasoning to your SUSY-case :)
     
  4. Dec 30, 2015 #3
    @haushofer thanks very much for your answers. I really like the fact that you put an analogy but I did not get the whole picture of the analogy and how it plays a role in answering my first question which says: "Does it imply that for a background to be supersymmetric, then fermions must not transform into bosons but rather stay fermions?" Also, when the most confusing of all was ACuriousMind's answer that I quoted above and his mentioning of classical solutions (What does he mean mean by classical solutions anyway and what brought this up?:frown:). I would very much appreciate if you can elaborate on these points as I have been thinking about those confusions for two days.
     
  5. Dec 30, 2015 #4

    haushofer

    User Avatar
    Science Advisor

    Classical solutions are solutions of the classical eom. Deviations (degrees of freedom) from these solutions, ( backgrounds), can then be quantized. This is what you can do in the standard model (th Higgs vev could be seen as a background of the Higgsfield; deviations from this background are called Higgs bosons), or e.g. quantizing Fierz Pauli theory on Minkowski spacetime. But classical fermionic backgrounds must vanish, see e.g.

    https://www.physicsforums.com/threads/why-a-bosonic-background-in-sugra.840522/

    Fermions can still go to bosons, but the parameter epsilon is constrained, see eqn.12.15 of Van Proeyen.

    So backgrounds which preserve some susy are analogous to backgrounds (like Schwarzschild) in GR which preserve 'some coordinate transformations'. These transfo's are generated by Killin vectors. Analogously, backgrounds which preserve some susy have Killing spinors.
     
  6. Dec 30, 2015 #5
    even though ##\delta F=0##? Doesn't this quote contradict with the ##\delta F=0##?
     
  7. Dec 30, 2015 #6
    @haushofer from what I read the answer is not quite precise and does not answer the questions samuelphysics is asking about. What he's asking is in short "What does it mean to have a preserved supersymmetry". He's practically asking what does it mean that local supersymmetry variations must vanish in that preserved supersmmetry. The confusion arises (also my confusion after reading all this) from the fact that he does not understand the physical meaning of $\delta F=0$ where he asked you several times, does it mean that in a preserved supersymmetry fermions do not undergo transformation into bosons? (which I would like to read our answer on this too).

    Please, would be great if you can explain this without the analogy (if this is not too much to ask for). Thanks!:oldsmile:
     
  8. Dec 31, 2015 #7

    haushofer

    User Avatar
    Science Advisor

    I'm sorry if I'm not being clear. Yes, as I stated in post #4, I'd say bosons still go to fermions and vice versa. From the OP: "t is just a requirement on the classical solutions, that they be mapped onto themselves under the preserved supersymmetric transformation (i.e. δX=0 for all fields), and not onto another solution". "Onto themselves" means that the bosonic background is mapped to the fermionic one and vice versa. To take the GR- analogy: under the flow of a Killing vector, you can still shift points of e.g. the Schwarzschild solution (it's not like you suddenly have to stay at the same point), but the solution itself (numerical value of the components) doesn't change. Analogously, preserving a SUSY background doesn't mean that suddenly bosons don't go to fermions and vice versa anymore. (I'm sorry, again the analogy, but that's how I understand it; SUSY is merely an extension of this to superspace :P )

    It can be confusing (the GR-case can be confusing already, let alone SUSY :P ). So let's take a look at the simplest example I can think of: N=1, D=4 SUGRA. I will not care about numerical factors.

    For this theory we take the multiplet consisting of the metric and the gravitino. The symmetries are local Lorentz transformations (let's ignore these), general coordinate transformations (gct's) and local SUSY. The transformation rules for the latter are

    [tex]
    \delta g_{\mu\nu} = \bar{\epsilon}\gamma_{(\mu}\psi_{\nu)}, \ \ \ \ \ \ \delta \psi_{\mu} = D_{\mu} \epsilon = \partial_{\mu} \epsilon - \omega_{\mu}{}^{ab}\gamma_{ab}\epsilon
    [/tex]

    Now we look at a classical solution: the Minkowski-background and a vanishing gravitino background,

    [tex]
    g_{\mu\nu} \equiv \eta_{\mu\nu}, \ \ \ \ \ \ \psi_{\mu} \equiv 0
    [/tex]

    To preserve this solution (with this I mean the Minkowski AND vanishing gravitino background!) we need that the SUSY transformations of these solutions vanish. As such these solutions are mapped onto each other. The vanishing of SUSY-transfo of the metric,

    [tex]
    \delta g_{\mu\nu} = \bar{\epsilon}\gamma_{(\mu}\psi_{\nu)} \equiv 0
    [/tex]

    is trivial, because the gravitino solution is taken to be zero. That's easy. Physically, under this transformation the Minkowski solution is mapped onto the gravitino solution (which is zero). The gravitino variation becomes (the spin connection vanishes for the Minkowski background)

    [tex]
    \delta \psi_{\mu} = D_{\mu} \epsilon \equiv \partial_{\mu} \epsilon \equiv 0
    [/tex]

    I'd say the gravitino is still transformed to the metric (or Vielbein) and vice versa: their numerical values are mapped onto each other! (I think this is the confusing part; I must say I'm also a bit confused now :P ) E.g, on the right of the last equation there is still a spin connection; the gravitino is still transformed into the spin connection, but both have the classical solution zero.This is accomplished by taking epsilon to be constant, making SUSY global. The gct's finally are broken to the Killing vectors which generate the Poincare group (also global). You can check that the {Q,Q}~P commutator is then realized on this solution.

    Does this help?
     
    Last edited: Dec 31, 2015
  9. Dec 31, 2015 #8

    haushofer

    User Avatar
    Science Advisor

    Btw, if bosons transform to bosons and fermions to fermions, it wouldn't be a susytransfo, right?

    Anyway, happy 2016 and if something is still not clear, let me know :)
     
  10. Dec 31, 2015 #9
    This sounds more like "fermion" is still transformed into the "boson" so is it that spin connection is a boson? or else what is the point of mentioning this?

    Do you mean that the metric is a boson?

    happy 2016 @haushofer
     
  11. Dec 31, 2015 #10
    Thanks a lot for your explanation @haushofer ! I have read it but still I have to reread it with more attention, and will then get back to you next year if I have any questions :p. However, I wanted to wish you a happy new year.:oldbiggrin:
     
  12. Jan 1, 2016 #11

    haushofer

    User Avatar
    Science Advisor

    The metric (or Vielbein) is a bosonic field, so the spin connection, which is constructed out of the metric (or vielbein) and its derivatives is also bosonic. You can find the expression in every SUGRA book, e.g. Van Proeyen. It is similar to the fact that the connection in GR in curved indices depends on the metric.

    To be more precise: the boson we are talking about is the graviton, having spin 2. It is defined as the metric perturbation of a chosen background, like Minkowski or AdS. This is analogous to QFT, where you define a particle to be an excitation of a quantum field with respect to a vacuum expectation value (vev). The vev of the metric field is here the classical solution "Minkowski-space". The fact that the graviton has spin 2 comes from representation theory.

    So yes, fermions are still transformed to bosons, but only under a subgroup of the local SUSY algebra we started out with. Just like the Killing vectors of e.g. the Schwarzschild solution are a subgroup of gct's and still move points around (but in such a way to keep the metric components invariant). If this would not be the case, I wouldn't know why we would still call it SUSY.

    I will also take a look at a less trivial vacuum solution, like AdS, of N=1, D=4; it is treated in Ortin's Gravity and Strings, if you want to check it out for yourself.

    I wish you and Emilie a very SUSY 2016 :P
     
  13. Jan 10, 2016 #12
    Hello again, I have a question please @haushofer

    So, when you say
    , you mean the metric (boson) transforms into "0" which is the same thing as ##\psi_{\mu}## (fermion) because the solution above was ##\psi_{\mu}=0##.

    BUT, when you say
    , the fermion (##\psi_{\mu}##) transforms into what?Should it transform into a boson? Which in outr case is the metric ##\eta_{\mu\nu}##?
     
  14. Jan 11, 2016 #13

    haushofer

    User Avatar
    Science Advisor

    The covariant derivative D on the susy-parameter epsilon contains the spin connection, which depends on the vielbein and its derivatives. So the gravitino transforms into the vielbein and its derivatives (or, equivalently, the metric and its derivatives).
     
  15. Jan 11, 2016 #14

    haushofer

    User Avatar
    Science Advisor

    I was alerted that there was another question, but can't find it here.
     
  16. Jan 23, 2016 #15
  17. Jan 23, 2016 #16

    haushofer

    User Avatar
    Science Advisor

    You can regard susy/sugra as a classical field theory and then quantize it. Quantum fluctuations are then considered with respect to a classical solution of the field equations. Would you know of any other way of quantizing? :)
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook