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Understanding AdS and quantization in AdS

  1. Jun 9, 2012 #1
    Hello everyone,

    For a project that may involve some work with Anti de Sitter space, I want to understand

    1. what the AdS metric looks like
    2. how to set up and solve boundary value problems in AdS, e.g. scalar field KG equation in AdS
    3. how to quantize scalar fields in AdS

    For 1, I am reading the authoritative review by Aharony, Gubser, Maldacena, Ooguri and Oz, entitled 'Large N field thoeries, string theory and gravity'. I also tried reading Witten's 1998 paper on AdS and Holography, but there were lots of things I could not relate to (which are becoming less fuzzy with the Aharony paper).

    What would be good references for 2 and 3? By good, I mean pedagogically oriented, with some examples.

    Thanks in advance!
     
  2. jcsd
  3. Jun 9, 2012 #2

    samalkhaiat

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    hep-th/9804035 and hep-th/9805145
     
  4. Jun 10, 2012 #3
    Thank you Sam, I'll have a look at these papers and reply on this thread in case I need something else, in which case you can probably point me in the right direction again.
     
  5. Jun 11, 2012 #4
    Why does equation (2) in hep-th/9804035 bear a [itex]\sqrt{g}[/itex] and not a [itex]\sqrt{-g}[/itex]? What does Riemannian signature mean? Also, why is there only a spatial derivative? Shouldn't the Lagrangian density look something like

    [tex]\mathscr{L} = \frac{1}{2}\left[\left(\partial_{0}\phi\right)^2 + |\nabla \phi|^2 \right] + \sum_{n \geq 3}\frac{\lambda_n}{n!}\phi^n[/tex]
     
  6. Jun 11, 2012 #5
    It just means signature (++...+) (d+1 values), sometimes called Euclidean signature. Since there's no singled out time coordinate with opposite signature, there's just one gradient operator covering all the coordinates.
     
  7. Jun 11, 2012 #6
    Thanks sheaf. This probably has to do more with QFT in curved space, but how does equation (4) come about? I have seen this before, but I have not studied QFT in curved space in detail, so I'm not absolutely sure. And is equation (5) a general result? That is, can the surface term always be written in this manner?
     
  8. Jun 11, 2012 #7

    samalkhaiat

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    Don’t get put off by this minor (unnecessary) complication; set [itex]\sqrt{g} = 1[/itex] if you like. He uses the Euclidean signature with [itex]x_{0}=it[/itex]. Eq(4) is the Green’s function equation associated with Eq(3). He uses the curved Euclidean Laplacian
    [tex]\nabla^{2}= \frac{1}{\sqrt{g}} \partial_{i}\left(g^{ij} \sqrt{g}\partial_{j}\right).[/tex]
    To find Eq(5) multiply Eq(3) by [itex]G[/itex] and Eq(4) by [itex]\phi[/itex] subtract then integrate using Dirichlet boundary condition after applying Green’s theorem to transform volume integrals into surface integrals.

    Sam
     
  9. Jun 13, 2012 #8
    Thanks for your reply. I think I need to back up a bit and look at the curved space formulation of QFT quickly. I tried looking at the book by Birrell and Davies, but I got a little lost there probably because of the notation (I have an old dilapidated version with a poor print). Could you suggest a reference I could work through?

    As for what you suggested, this is what I have

    [tex]\int_{\Omega}d^{d+1}y\sqrt{g}\left[ G(x,y)(\nabla_{x}^2-m^2)\phi(x)-\phi(x)(\nabla_x^2-m^2)G(x,y)\right] = \int_{\Omega}d^{d+1}y\sqrt{g} \left[G(x,y)\sum_{n\geq 3}\frac{\lambda_n}{(n-1)!}\phi(x)^{n-1}-\frac{\phi(x)\delta(x-y)}{\sqrt{g(x)}}\right][/tex]

    I get the logic about transforming the left hand side using Green's theorem, but can you explain how h enters equation (5)? Also, what form of the Green's theorem is being used here anyway?

    I'm sorry these questions may be silly, but I'm just trying to work out every detail of this paper and discovering in the process several small topics which I might have glossed over before. So please bear with me.

    EDIT -- What is the expression for the gradient here?
     
    Last edited: Jun 13, 2012
  10. Jun 13, 2012 #9
    Also, isn't there a misprint in equation (8)? The derivative with respect to time is only of first order..
     
  11. Jun 13, 2012 #10

    samalkhaiat

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    No, it is first order. You need to take a step back and consult some lecture notes on the Ads space and its parametrizations. Try J. L. Peterson; hep-th/9902131.

    Sam
     
  12. Jun 13, 2012 #11

    samalkhaiat

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    Birrell & Davies is good. You can also try Hawking & Ellis, “The large scal structure of spacetime” or Isham, (1978), “QFT in curved spacetime – a general mathematical framework”. But you are missing the point; your miss-understanding has nothing to do with the curved-space formulation. The induced metric shows up even in flat space. This happen when you change the coordinates in the surface integrals from [itex]y \in D[/itex] to [itex]y \in \partial D[/itex]. To see this, consider the flat space equations
    [tex]\partial_{\mu}\partial^{\mu}\phi = J[/tex]
    [tex]\partial_{\mu}\partial^{\mu} G = \delta[/tex]
    Green’s theorem let you write [itex]\phi[/itex] in terms of [itex]G, \partial G[/itex] and [itex]\partial \phi[/itex];
    [tex]
    \phi (x) = \int_{D\subset \mathbb{R}^{n+1}} d^{n + 1} y \ J(y) \ G(x - y) + \int_{\partial D \subset \mathbb{R}^{n}} d^{n}z \ \sqrt{h}\ n^{i}\ \{ \phi(z) \partial_{i}G(x-z) – G(x-z) \partial_{i}\phi(z) \},
    [/tex]
    where [itex]z^{i}, i = 1, 2,…n[/itex] are coordinates on [itex]\partial D[/itex] and
    [tex]h = \det |\frac{\partial y}{\partial z}|,[/tex]
    is the determinate of the induced metric on [itex]\partial D[/itex];
    [tex]h_{ij} = \frac{\partial y^{\mu}}{\partial z^{i}}\frac{\partial y_{\mu}}{\partial z^{j}}.[/tex]

    Sam
     
  13. Jun 14, 2012 #12
    Thanks for your replies Sam. I am looking at the reference you suggested.

    Meanwhile, I worked out some GR related quantities from the AdS_5 metric, for practice and familiarity and tried to generalize them to AdS_(d+1). Can you see if these look okay?

    I consulted A Practical Guide to AdS/CFT by Dosch for the preliminary construction of the metric.

    So for AdS_5, I can write the metric as

    [tex]ds_{AdS}^2 = \frac{L^2}{z^2}(-(dx^1)^2 - (dx^2)^2 - (dx^3)^2 + (dx^4)^2 - (dz)^2)[/tex]

    Define a 5D pseudo Minkowski metric tensor [itex]\eta_{\mu\nu}[/itex] to have the following matrix representation.

    [tex]\{\eta_{\mu\nu}\} = \left(\begin{array}{ccccc} -1 & 0 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 & 0 \\ 0 & 0 & -1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & -1\end{array}\right)[/tex]

    So the metric tensor can be written as

    [tex]\{g_{\mu\nu}\} = \frac{L^2}{z^2}\{\eta_{\mu\nu}\}[/tex]

    The Christoffel symbol is

    [tex]\Gamma^{\rho}_{\mu\nu} = -\frac{1}{z}({\delta_{\mu}}^{5}{\delta_{\nu}}^{\rho}+{\delta_{\nu}}^{5}{\delta_{\mu}}^{\rho}-\eta^{\rho 5}\eta_{\mu\nu})[/tex]

    so the Riemann curvature tensor turns out to be

    [tex]{R^{\lambda}}_{\mu\nu\kappa} = -\frac{\eta^{55}}{z^2}(\eta_{\mu\nu}{\delta_{\kappa}}^{\lambda} - \eta_{\mu\kappa}{\delta_{\nu}}^{\lambda})[/tex]

    The Ricci tensor is

    [tex]R_{\mu\nu} = \frac{4 \eta^{55}}{z^2}\eta_{\mu\nu}[/tex]

    and the Ricci scalar is

    [tex]R = -\frac{20}{L^2}[/tex]

    These results seem okay, but when I plug them into Einstein's equation, I get a positive cosmological constant. Shouldn't I get a negative cosmological constant? That's the first thing I remember reading about AdS. In fact, if my calculations are correct then for AdS_(d+1) I should have

    [tex]R_{\mu\nu} = -\frac{d}{L^2}g_{\mu\nu}[/tex]

    and

    [tex]R = -\frac{d(d+1)}{L^2}[/tex]

    From these, I get

    [tex]\Lambda = \frac{d(d-1)}{2L^2}[/tex]

    which is positive. What's going wrong?
     
    Last edited: Jun 14, 2012
  14. Jun 14, 2012 #13
    I think I understand some part of it at least. The idea is to identify the metric tensor [itex]g_{\mu\nu}[/itex], and then construct the derivative

    [tex]\nabla^2 = \frac{1}{\sqrt{|g|}}\partial_{i}(\sqrt{|g|}g^{ij} \partial_{j})[/tex]

    which should match equation (8) (of hep-th/9804045). But for me it does not:

    [tex]ds^2 = \frac{1}{x_0^2}\sum_{i=0}^{d}dx_{i}^{2} = g_{\mu\nu}dx^{\mu}dx^{\nu}[/tex]

    which implies

    [tex]g_{\mu\nu} = \frac{1}{x_{0}^2}\mathbb{I}_{d+1}[/tex]

    so that

    [tex]g^{\mu\nu} = x_0^2\mathbb{I}_{d+1}[/tex][/tex]

    Now,

    [tex]g = \text{det}(g_{\mu\nu}) = \left(\frac{1}{x_{0}^2}\right)^{d+1}[/tex]

    which gives [itex]\sqrt{|g|} = \frac{1}{x_{0}^{d+1}}[/itex]. Substituting all this into the formula for the laplacian, one gets

    [tex]\nabla^2 = x_{0}^{d+1}\partial_{i}(x_{0}^{d+1}x_{0}^2\delta^{ij}\partial_j)[/tex]

    which does not give the form of the Laplacian given in equation (8).


    Secondly, as I understand, eqn (7) is the Euclidean representation of the AdS metric. This is the same as the equation derived in D'Hoker and Freedman's review (hep-th/0201253). There, the Euclidean metric is

    [tex]ds^2 = -dY_{-1}^2 + dY_{0}^2 + dY_{1}^2 + \cdots + dY_{d}^2[/tex]

    Now, they say the following, which I do not quite understand:

    I don't get this: I can see that

    [tex]\sum_{i=1}^{d}dY_{i}^2 = \frac{d\textbf{z}^2}{z_0^2}[/tex]

    But I do not see how [itex]Y_{-1}[/itex] and [itex]Y_{0}[/itex] can be clubbed quite this way:

    [tex]dY_{-1} + dY_{0} = -\frac{1}{z_0^2}dz_0[/tex]

    whereas

    [tex]dY_{0}^2 - dY_{-1}^2 = -(dY_{-1}-dY_{0})(dY_{-1}+dY_{0}) = \frac{1}{z_{0}^2}(dY_{-1}-dY_{0})dz_0[/tex]

    What happens to the difference term? Is there some approximation being used here?
     
    Last edited: Jun 14, 2012
  15. Jun 14, 2012 #14

    samalkhaiat

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    If you are new to the subject, you should stick with one notation. This subject suffers from notational nightmare. So, comparing results from different papers is not an easy task even for experts. You also seem to make few algebraic mistakes, for example, you thought that
    [tex]\sum_{i=1}^{d}(dY_{i})^{2} = \frac{d\mathbf{z}^{2}}{z_{0}^{2}}[/tex]
    which is wrong! The right hand side is actually
    [tex]\left( \frac{z_{0}dz_{i} – z_{i}dz_{0}}{z_{0}^{2}}\right)^{2}.[/tex]
    You also forget that you have the Ads equation
    [tex]-1 = - Y_{-1}^{2} + Y_{0}^{2} + Y_{i}^{2},[/tex]
    which you can write as
    [tex]z_{0} + \frac{z_{i}^{2}}{z_{0}} = - (Y_{0}-Y_{-1}).[/tex]
    Differentiate this and use it in the metric and you should find correct result.

    Sam
     
  16. Jun 14, 2012 #15
    Thanks Sam. The paper by D'Hoker and Freedman is what I have to get through, but since you suggested the paper by Muck and Viswanathan, I started looking into it as well.

    Referring first to D'Hoker and Freedman, from equation 5.13 we have

    [tex]-Y_{-1}^{2} + Y_{0}^{2} + Y_{1}^{2} + \cdots + Y_{d}^{2} = -1[/tex]

    with induced metric

    [tex]ds^2 = -dY_{-1}^{2} + dY_{0}^{2} + dY_{1}^{2} + \cdots + dY_{d}^{2}[/tex]

    Define

    [tex]Y_{-1} + Y_{0} = \frac{1}{z_0}[/tex]
    and
    [tex]z_i = z_0 Y_i[/tex] for [itex]i = 1, \ldots, d[/itex]

    So,

    [tex]ds^2 = -dY_{-1}^2 + dY_{0}^2 + \frac{1}{z_0^2} d\vec{z}^2[/tex]

    where

    [tex]d\vec{z}^2 = dY_{1}^{2} + \cdots + dY_{d}^{2}[/tex]

    The AdS equation is actually

    [tex]-Y_{-1}^{2} + Y_{0}^{2} + \sum_{i=1}^{d}Y_{i}^2 = -1[/tex]

    that is,

    [tex](Y_{0}-Y_{-1})\frac{1}{z_0} + \frac{1}{z_{0}^2}\sum_{i=1}^{d}z_{i}^2 = -1[/tex]

    which can be written as,

    [tex]Y_{0}-Y_{-1} = -z_0 - \frac{1}{z_0}\sum_{i=1}^{d}z_{i}^2[/tex]

    so that

    [tex]dY_{0} - dY_{-1} = -dz_0 + \frac{dz_{0}}{z_0^2}\sum_{i=1}^{d}z_{i}^2 - \frac{2}{z_0}\sum_{i=1}^{d}z_{i} dz_{i}[/tex]

    Therefore

    [tex]dY_{0}^2-dY_{-1}^2 = -\frac{dz_{0}}{z_{0}^2}\left[-dz_0 + \frac{dz_{0}}{z_0^2}\sum_{i=1}^{d}z_{i}^2 - \frac{2}{z_0}\sum_{i=1}^{d}z_{i} dz_{i}\right][/tex]
     
    Last edited: Jun 14, 2012
  17. Jun 15, 2012 #16

    samalkhaiat

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    So,
    [tex]
    \sum (dY_{i})^{2} = \sum \left( d ( \frac{z_{i}}{z_{0}}) \right)^{2} = \sum \left( \frac{z_{0}dz_{i} - z_{i}dz_{0}}{z_{0}^{2}}\right)^{2}.
    [/tex]
    Now add this (after expanding the square) to
    you find the resuls.I am not going to correct algebraic mistake anymore, you have got to do these things yourself.
    Good Luck
    Sam
     
  18. Jun 15, 2012 #17
     
  19. Jun 20, 2012 #18
    In section 6.3 of D'Hoker and Freedman's paper, they give the following result for the geodesic distance in AdS:

    [tex]d(z,w) = \int_{w}^{z}ds = \ln\left(\frac{1+\sqrt{1-\zeta^2}}{\zeta}\right)[/tex]

    with

    [tex]\zeta \equiv \frac{2 z_0 w_0}{z_0^2 + w_0^2 + (\vec{z}-\vec{w})^2}[/tex]

    How does one derive this?
     
  20. Jun 20, 2012 #19

    Ben Niehoff

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    First you have to solve the geodesic equation. It's not hard if you use some tricks. AdS is a homogeneous space. In general, homogeneous spaces include spheres, hyperbolic spaces, dS and AdS spaces. Every homogeneous space embeds isometrically as a quadric surface in a flat space of one dimension higher; this flat space may turn out to have mixed signature (++..+ for spheres, -+..+ for hyperbolic spaces and dS, --+..+ for AdS).

    So think about the 2-sphere embedded in R^3. Can you see a very simple way to determine the 2-sphere's geodesics from this embedding? There should be a fairly obvious way to use symmetry to your advantage.

    The same technique applies in AdS. Then, once you have the geodesics of AdS, you simply integrate the line element along them.

    It would also help to read about hyperbolic spaces, particularly the Poincare upper-half-plane model, because it applies very well to AdS spaces, and the UHP model corresponds to the coordinates you are using mostly in this thread.
     
  21. Jun 21, 2012 #20
    Thanks Ben, but could you pleas esuggest references where this might be worked out or something? I need to figure this out reasonably quickly, in order to move on to the solution of the free wave equation in AdS. and I am hard pressed for time.
     
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