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Worldsheet ghost stress-energy tensor

  1. Feb 12, 2010 #1
    Hey guys,

    I haven't posted on here for quite a while, so hello to everybody.

    I've been trying to derive the stress-energy tensor for the ghost LaGrangian:

    [tex] \int d^2 \sigma \sqrt{g} \left( b_{\alpha\beta} \nabla^\alpha c^{\beta} + \omega b^{\alpha\beta} g_{\alpha\beta} \right)[/tex]

    for quite a while now. It is a conformal field theory and to word out the central charge the most efficient way is through the T(z)T(w) OPE. Both fields, b and c, are anticommuting. b is a symmetric tensor. [tex]\omega[/tex] is a Lagrange multiplier to enforce the tracelessness of b. I've got tantalisingly close, but no cigar. So really need some help here ;).

    Firstly, I gauge fixed the WS metric to be a flat one, so that [tex]\nabla^\alpha c^{\beta} = \partial^{\alpha} c^{\beta} [/tex].

    Now you can proceed two ways.

    1) Couple the theory to gravity (which involves reinstating the covariant derivative), and vary the metric. This gives a symmetric SE tensor. This I have done, and it works out fine.

    2) Peform an infinitesimal change in the coordinates [tex]x^\mu \rightarrow x^\mu + \epsilon^\mu (x)[/tex]. The change in the action will be of the form
    [tex]\int d^2 \sigma T_{\alpha\beta} \partial^\alpha \epsilon^\beta[/tex],
    so you can read off the SE tensor.

    The method I want to make work is the second (because I'm stubborn and want to do it both ways).

    In complex coordinates, I get:

    [tex] T_{zz} \equiv T = b\partial c + c \partial b [/tex] .

    Compare this with the correct result, as quoted in Polchinski, BBS, Steve Tong's DAMTP notes:

    [tex] T_{zz} \equiv T = 2 \partial c b + c \partial b [/tex] .

    Now the tensor obtained this method is not symmetric, so there are things you can add to it to make it so. But I don't think I've found the 'right thing' to bring it into the standard symmetric form. Nor do I know how much this matters the context of the CFT. By adding various things conserved quantities to the the SE tensor I've found quite a few different forms. Can they all be used in the CFT OPEs, to find the central charge, etc? I'm a bit worried that I can find quite so many different forms for a single object.

    Thanks for listening to my rambling. I know it's reasonably technical, but I've been stuck on it for weeks, so any help is appreciated. I've check my calculation as best I can, although there are quite a few terms (e.g. from the tensor transformations of b and c),so it's open to error.

    Ideas anybody? Some of these mentor bods floating around? ;)
  2. jcsd
  3. Feb 13, 2010 #2

    Physics Monkey

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    Hi AntideSitter,

    Welcome! You have one of my favorite spacetimes for a name.

    I think your question is a bit complex. Perhaps it would be helpful to get the basics down first, just so we're all on the same page. I assume you've gotten the OPE of b with c. If b and c are to be Weyl invariant, then their conformal dimension is fixed by their tensor structure. Do you understand how this determines the coefficients of the terms in the stress tensor? I don't mean how to derive them, just what the coefficients need to be to get the answer you want.

    I think if you've gotten that far, and it seems like you have, then you must just be making some technical error when doing the variation. Can you write out your expressions for the variation of b and c under coordinate transformations? I'm assuming your notation is [tex] b = b_{z z} [/tex] and [tex] c = c^z [/tex] in your action?

  4. Feb 14, 2010 #3
    Hi Physics Monkey, thanks for the reply to my long-winded question! And yes, hopefully I'm starting a PhD in holography in September, so the named seemed appropriate ;).

    OK so your reply got me thinking. I see how if we want b and c to be Weyl invariant, then their tensor weights imply their conformal weights. I'm not sure about how this in in turn constrains the coefficients in the SE tensor. Could you explain this?

    Here is my setup. I do an active coordinate transformation:

    [tex] x_\mu \rightarrow x'^\mu = x^\mu + \epsilon^\mu (x) [/tex]

    Under this we have, to linear order in [tex]\epsilon[/tex]:

    [tex] c^\alpha \rightarrow \frac{\partial{x'^\alpha}}{\partial{x^\beta}} c^\beta (x') = (\delta^\alpha _{\beta} + \partial_\beta \epsilon^\alpha ) (c^\beta + \partial_\gamma c^\beta \epsilon^\gamma) [/tex]

    [tex]b \rightarrow \frac{\partial{x^\gamma}}{\partial{x'^\alpha}} \frac{\partial{x^\delta}}{\partial{x'^\beta}} b_{\gamma\delta} (x') = (\delta^\gamma _{\alpha} - \partial_\alpha \epsilon^\gamma ) (\delta^\delta _{\beta} - \partial_\beta \epsilon^\delta ) (b_{\gamma\delta} + \partial_\mu b_{\gamma\delta} \epsilon^\mu) [/tex]

    In carried out the calculation in index notation because it seemed cleaner that way. In complex coordinates these transformations become:

    [tex] c \rightarrow c + \partial c \epsilon + c \partial \epsilon + \bar{\partial} c \bar{\epsilon}[/tex]

    [tex] b \rightarrow b + \partial b \epsilon - 2 b \partial \epsilon + \bar{\partial} b \bar{\epsilon}[/tex]

    The derivative [tex]\partial_{\alpha} \rightarrow \frac{\partial x^{\beta}}{\partial x'^{\alpha}} \partial_{\beta}[/tex] .

    My method was simply to stick all of these transformations into the action, then integrate by parts (throwing away the surface terms) until I got something proportional to [tex] \partial^{\alpha} \epsilon^{\beta}[/tex], and called that thing [tex] T_{\alpha \beta}[/tex].

    The measure also changes, but this doesn't affect my calculation because it only contributes to [tex]T_{z \bar{z}}[/tex].

  5. Feb 16, 2010 #4

    Physics Monkey

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    That's great, I hope you have a great time. Any idea what sort of systems you'd like to study?

    What I meant is that if you guess the two terms of the stress tensor to be [tex] \partial c b [/tex] and [tex] c \partial b [/tex] on dimensional grounds, then can you fix the coefficients? For example, by requiring that b and c be primary fields of weight 2 and -1 i.e. fields with weights determined by the index structure alone. This is sort of a third "guessing" method of determing the right stress tensor. As you say, the stress tensor is partially arbitary in any field theory, but you should be able to check that whatever extra terms you add, they don't affect these assignments.

    Your basic setup looks ok to me, but I'll be back to offer some more detailed comments later.
  6. Feb 19, 2010 #5
    Hello there again,

    Assuming the terms in the stress tensor T which you suggested, I did a holomorphic coordinate transformation on it.

    That is, I assumed the tensor to have the form:
    [tex]\alpha\, \partial c b + \beta\, c \partial b[/tex]
    and then transformed according to the rules for b, c and [tex]\partial[/tex] which I gave above. Using the equations of motion, I ended up with:

    [tex]T' = T +(\partial T) \epsilon + (\alpha -2\beta)cb\partial^2 \epsilon - (\alpha \beta)T\partial \epsilon[/tex]

    Now we expect T to transform under conformal tranformations like a weight 2 field, plus an inhomogeneous bit [tex] \propto\epsilon'''[/tex]coming from the Weyl transformation. Since I didn't do the Weyl transformation above, I expect there to be no [tex]\epsilon'''[/tex] bit. Thus I demand the expression above to match that for a weight 2 tensor. This means:

    [tex]\alpha -2\beta = 0 \quad \alpha + \beta = 2 [/tex]

    so[tex] \alpha = \frac{4}{3}, \quad \beta = \frac{2}{3}[/tex].

    This agrees with the standard result up to a factor. Does anybody know how that factor would change things in the CFT algebras? For one you'd calculate the wrong central charge, if you assumed the usual
    [tex]T(z)T(w) = \frac{c/2}{(z-w)^4} + ...[/tex].

    What is the criterion for T to be the "right" stress tensor to appear in these formulas?

    I wouldn't care so much about this is if so much of modern physics were not built on CFTs. Until I get this sorted, I still have this niggling feeling in the back of my mind, you know?

    Thanks for reading.
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