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Spin 2 graviton

  1. Oct 1, 2005 #1
    How do we know that a graviton has to be spin 2?
     
  2. jcsd
  3. Oct 2, 2005 #2

    samalkhaiat

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    I WILL ASSUME THAT YOU KNOW THE REPRESENTATION THEORY OF GROUPS IN PARTICULAR,THE LORENTZ GROUP SO(1,3).
    NOW,ANY TENSOR REPRESENTATION CAN BE REGARDED AS A DIRECT PRODUCT OF VECTOR REPRESENTATIONS, SO IT CONSISTS ONLY OF IRREDUCIBLE COMPONENTS WITH j(1)+j(2) AN INTEGER [j(1),j(2) ARE ANGULAR MOMENTUM FOR THE TWO SU(2) GROUP, REMEMBER SO(1,3) = SU(2) X SU(2)].
    NOW TAKE TWO VECTOR REPRES. (1/2,1/2) OF SO(1,3) AND FORM THE PRODUCT;
    (1/2,1/2) x (1/2,1/2) = (1,1) + {(1,0)+(0,1)} + (0,0),
    YOU KNOW(I ASSUME) THAT
    (0,0) = SCALAR(spin-0) SATISFAYING KLEIN-GORDON EQUATION
    {(1,0)+(0,1)} = ANTISYMMETRIC TENSOR or ~VECTOR (spin-1)SATISFAYING;
    1) PROCA'S EQUATION FORTHE MASSIVE CASE. 2)MAXWELL'S EQUTIONS FOR THE MASSLESS CASE.
    so what is (1,1)? of course it is spin-2 component. so let us try a symmetric second rank tensor;
    THE MASSIVE CASE: AS YOU MIGHT KNOW, MASSIVE SPIN-2 FIELD HAS 2S+1=5 INDEPENDENT COMPONENTS, THUS WE NEED TO IMPOSE FIVE CONDITIONS ON THE SYMMETRIC TENSOR IN ORDER TO DESCRIBE THE SPIN-2 FIELD (C^ab say). NOTE THAT THE SCALAR (0,0) CORRESPONDS TO THE TRACE OF C^ab AND CANBE SUBTRACTED BY THE CONDITION;
    Tr(C^ab)=0 (CONDITION No1)
    NEXT WE FORBID THE EXISTENCE OF ANY OTHER INDEPENDENT SEMMETRIC TENSOR (EXCEPT THE SOURCE TENSOR T^ab) AS WELL AS OF ANY VECTOR MADE OUT OF C^ab AND T^ab, i.e
    T AND C ARE DIVERGENLESS TRNSORS. THIS GIVES US THE REMAINING 4 CONDITIONS ON THE TENSOR C WHICH IS NOW SATISFAYING THE FIELD EQUATION;
    (box + m^2) C^ab = kT^ab,
    where (box + m^2) is the KLEIN-GORDON OPERATOR, AND k IS A COUPLING CONSTANT.
    THE MASSLESS CASE ; PUTING m=0 IN THE ABOVE EQUATION, WE GET;
    (box) C^ab = kT^ab.
    BUT WE HAVE 5 INDEPENDENT COMPONENTS IN THE TENSOR C AND THIS IS NOT RIGHT BECAUSE MASSLSS FIELDS HAVE ONLY TWO INDEPENDENT COMPONENTS.
    BUT IF WE START WITH A TEN COMPONENTS FIELD
    h^ab = h^ba, and write the most general linear second order equation for the symmetric, massless field h^ab in presence of the conserved source T^ab, then we write the equation in term of the tensor;
    G^ab = h^ab - 1/2*g^ab*h, where g is Lorentz metric and
    h = Tr(h^ab),
    WE FIND THAT OUR EQUATION IS INVARIANT UNDER THE GAUGE TRANSFORMATION;
    h^ab--->h^ab - d^af^b - d^bf^a
    THE GAUGE FUNCTIONS, f^a, reduces the number of independent compo. in h^ab from 10 to 6. IF WE NOW IMPOSE GAUGE FIXING CONDITIONS;
    d¬aG^ab = 0, WE LEFT WITH ONLY TWO PHYSICALLY SIGNIFICANT COMPONENTS, AND THE FIELD EQUATION BECOMES;
    box (G^ab) = k* T^ab.

    NOW I AM READY TO ANSWER YOUR QUETION:

    THE ABOVE EQUATION IS EXACTLY THE LINEAR FORM OF EINSTEIN'S EQUTION WITH
    h^ab is a small deviation from the flat metric AND THEY CALL IT GRAVITON.
     
  4. Oct 3, 2005 #3

    dextercioby

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    The linearized gravity field (aka "Pauli-Fierz field") is a symmetric second rank tensor. Using Young tableaux formalism, such a tensor corresponds to a line made up of two boxes and therefore the field carries the spin 2. Because the P-F field has two degrees of freedom (you can convince of that by doing the Hamiltonian analysis), at quantum levels they correspond to the two possible eigenvalues of the helicity operator:[itex] \pm 2 [/itex].

    Daniel.
     
  5. Oct 3, 2005 #4

    samalkhaiat

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    This is to dextercioby, Howmany box would the spiner representation of SO(1,3) HAVE?
    You are talking about Lorentz (or Poincare) group, aren't you?
    I am interested to read about YOUNG TABLEAUX treatment of LORENTZ & POICARE' GROUPS. Would you please point out a reference for me?
    In doing young's tableaux for two vector [(box) x (box)], you get:
    4 x 4 = 10 + 6
    of course the 10 (two horizontal boxes) represents our symmetric 2nd rank tensor (1,1). the 6 (2 boxes on top of each other) is the antisymmetric tensor which,in post #2, I wrote [(0,1)+(1,0)]. where is the trivial (spin-0) representation (0,0)? and what kind of box would you give it?

    REGARDS

    SAM
     
  6. Oct 4, 2005 #5

    dextercioby

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    Yeah, it's quite delicate. First of all, Young tableaux should be used for the [itex] \mbox{SL(2,\mathbb{C})} [/itex] group seen as a subgroup of [itex] \mbox{GL(2,\mathbb{C})} [/itex]. You then (should) know the connection between the restricted Lorentz group (as a subgroup of [itex] \mbox{SO(1,3)} [/itex]) and [itex] \mbox{SL(2,\mathbb{C})} [/itex] and just then ascribe boxes to vectors (or covectors). The [itex] (0,0) [/itex] irrep of [itex] \mbox{SL(2,\mathbb{C})} [/itex] has no box, it's a scalar.

    BTW, one selects the [itex] (1,1) [/itex] irrep of [itex] \mbox{SL(2,\mathbb{C})} [/itex] from the [itex] h_{\mu\nu} [/itex] Pauli-Fierz field in a covariant & elegant manner, redefining the field

    [tex] \bar{h}_{\mu\nu}=:h_{\mu\nu}-\frac{1}{4}\eta_{\mu\nu} h^{\lambda}{}_{\lambda} [/tex]

    to a traceless one which carries 9 independent components, as it should.

    Daniel.
     
  7. Oct 4, 2005 #6
    Hey, thanks so much for the responses.

    I haven't had a chance to absorb this all yet - I have some QFT HMWK (!) on flipped SU(5) so might actually be closer to understanding some of the responses you've been good enough to write for me.

    I'll have some questions soon - but thanks beforehand!
     
  8. Oct 5, 2005 #7

    samalkhaiat

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    This is again for dextercioby.
    I would like to clarify few points;
    1) saying it is delicate, does not answer my questions.
    SL(2,C), the universal covering group of the proper orthochronous Lorentz group SO(1,3), deals with linear mappings in the 2-dimensional complex, symplectic space or the spinor space.It has on direct "physically intuitive" meaning (you could call this delicate!). This is exactly why I did not mention SL(2,C) in post #2, even though it is a very convenient starting point to constuct all relevant grometrical objects out of spinors:
    By considering the two fundamental 2-spinor of SL(2,C) (dotted,undotted) or ( left L, write R) i.e (1/2,0) and (0,1/2), now :
    Dirac spinor is L+R,
    4-vector (spin-1 field) is (0,1/2) X (1/2,0) = (1/2,1/2)
    spinorvector (spin-3/2 field) is (1,1/2) in
    (1/2,1/2) X (1/2,0) = (1,1/2) + (0,1/2),
    and of course our symmetric tensor(spin-2) part of:
    (1/2,1/2) X (1/2,1/2) = [(0,0) + (1,1)] + [(0,1) + (1,0)]
    So,to make life easy, I was usin SL(2,C) without naming it.
    2) I never asked about the Pauli-Feirz method. If you read post#2 carefully, you find it, more or less, is the Pauli-feirz method for constructing "relativistically invariant wave equation for massive and massless spin-2 field".
    3) My questions for you were about the use of young tableaux in Lorentz group.
    As you might know Young assigns a box for the fundamental REP of the group in question.But you assigned a box for 4-vector which is not FUND. REP of SL(2,C).
    Now if you have used so(1,3) instead, you could have been right, because one can regard 4-vec as some sort of FUND.REP of so(1,3). However my questions remain unanswered even in this case: howmany boxes would one assign to the spinor rep. of so(1,3), or if you want to complicate things: is the FUND.REP of SL(2,C) (2-SPINOR) represented by a box in young tableaux?
    As for (0,0) field, young assigns a stack of vertical boxes for the trivial REP = Identity = scalar = singlet = et cetera. so why (0,0) has no boxex?
    FINALLY I WANT YOU TO BE KIND AND NAME A REFERENCE WHICH USES YOUNG TABLEAUX FOR SO(1,3) AND/OR SL(2,C).

    REGARDS


    SAM
     
  9. Oct 6, 2005 #8

    dextercioby

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    The standard reference for [itex] \mbox{SL(2,\mathbb{C})} [/itex] finite dimensional irreps through Young tableaux are
    -S. Weinberg "Lectures on Particles and Field Theory", Brandeis Summer Institute on Theoretical Physics, Volume II, Prentice Hall, 1965 and
    -P. Moussa, R. Stora: "Methods in Subnuclear Physics", Volume II, Gordon & Breach, NY, 1968.

    Daniel.
     
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