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Hypothetical graviton : Why a spin of 2 ?

  1. Jul 6, 2011 #1

    Why have the hypothetical gauge boson of gravitation, the so-called graviton, a spin of 2 ?

    What about his chirality ? (projection of spin on his propagation direction)

    I mean, does it mean that the graviton can have a polarisation like a photon, but that this polarisation is two times faster ? What is the difference and why a spin of 2 ? If it's a boson, a spin of 1 could be 'enough' :tongue:

  2. jcsd
  3. Jul 6, 2011 #2


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    The free limit of general relativity is described by a symmetric tensor of rank 2 in a flat Minkownski spacetime which under restricted Lorentz transformations transforms like a spin 2 field. But spin is a quantum concept, so a quantum theory of gravitation (free limit or GR) describes spin 2 fields (free or interacting).
  4. Jul 6, 2011 #3
    Spin 1 or vector bosons have two charge types so they can attract or repel. Spin 2 charge carriers have only one charge type and can only attract. I'll have to think about the polarization question.
  5. Jul 8, 2011 #4


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    You mean like the photon?
  6. Jul 10, 2011 #5
  7. Jul 10, 2011 #6


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    A gravitational wave has two polarization states, similar to an electromagnetic wave, but different. :smile: Its polarization is transverse to the propagation vector, but instead of a being a transverse vector the field is a transverse tensor. Which means: the passage of a gravitational wave through a swarm of test particles has a shearing effect - an oscillatory stretching and squeezing. Picture a circle being squeezed into an ellipse, then back again and squeezed the other way.

    For an electromagnetic wave the two polarization states are at 90 degrees from each other. For a gravitational wave the two possible (linear) polarizations are tilted at 45 degrees. Equivalently, you can take combinations of the polarization states that are circularly polarized, a shear that rotates either to the left or to the right.
  8. Jul 10, 2011 #7


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    I must have misinterpreted your remark. The photon has spin 1, but does not have two charge types.
  9. Jul 17, 2011 #8
    It doesn't have two charge types, it carries two charge types.
  10. Jul 21, 2011 #9
    a good source of information is the feynman lectures on gravitation. he shows that a spin 2 field reproduces experiment: ie perihelion shift predicted by gravity, and the deflection of light by a massive object, whereas a spin 0 and 1 field do not. I think Misner, Thorne, and Wheelers book goes through this as well. Also he pictorially describes the graviton polarization states that Bill K was describing.
  11. Jul 29, 2011 #10
    Does this mean you can't have a rank two tensor field that is described by a spin-1 boson?
  12. Jul 30, 2011 #11
    No one has said anything so let me be more specific.
    Here Sam says that spin-1 satisfies Maxwell's equations. But it also seems that rank-2 tensors correspond to spin-2 gauge bosons. But could you have a rank-2 symmetric tensor field that satisfies Maxwell's equations and have a boson of spin-1 be the quanta of the field?
    \Box B_{\alpha\beta}=0
  13. Jul 30, 2011 #12


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    Rank-2 antisymmetric tensor field describes spin-1 (boson). In the massive representation, this is same as vector field.
    Rank-2 symmetric tensor is the appropriate object for spin-2 (boson) field. The field equation of this spin-2 tensor can be written in a Maxwell-like form, but it fails to describe electromagnetism correctly.
    The Maxwell equations contain the Bianchi identity
    If you assume that F is symmetric, then contraction with [itex]\eta^{ab}[/itex] gives you
    [tex]\partial^{c}F_{ca} + \frac{1}{2}\partial_{a}F^{c}_{c}(x) = 0[/tex]
    This causes a lot of complications;
    If you write
    [tex]F_{ab} = \partial_{a}A_{b} + \partial_{b}A_{a}[/tex]
    you then get
    [tex]\partial^{2}A_{a} + 2 \partial_{a} ( \partial . A) = 0[/tex]
    This does not look anything like the Maxwell equations and of course it is not gauge invariant. So, you are stuck with the ten original independent components of the symmetric tensor F. One can be little bit smarter and write down the most general linear, second order differential equation for free massless symmetric tensor [itex]h^{ab}[/itex] as
    [tex]\partial^{2}h^{ab} + A \partial_{c} \partial^{(a}h^{b)c} + B \partial^{a}\partial^{b}h^{c}_{c} + C \eta^{ab}\partial^{2}h^{c}_{c} + D \eta^{ab}\partial_{c}\partial_{e}h^{ce}=0[/tex]
    Differentiation leads to
    [tex]A + 1=0, \ \ A + D =0, \ \mbox{and} \ \ B + C =0[/tex]
    Setting B = 1, we find
    [tex]\partial^{2}G^{ab} + \eta^{ab}\partial_{c}\partial_{e}G^{ce}= \partial_{c}\partial^{(a}G^{b)c} \ \ (1)[/tex]
    [tex]G^{ab}= h^{ab} - \frac{1}{2}\eta^{ab}h^{c}_{c}[/tex]
    Equation (1) is invariant under the gauge transformation
    [tex]h^{ab}\rightarrow h^{ab} + \partial^{a}f^{b} +\partial^{b}f^{a}[/tex]
    This gauge freedom reduces the number of independent components in h from ten to six; this seems promising because six is the number of components of the physical fields [itex](\vec{E},\vec{B})[/itex]. However, unfortunately no physical electromagnetic quantity (including E and B themselves) can be obtained from the symmetric tensor G. Notice also that the tensor G can only couple to a symmetric tensor current, something that does not exist in electromagnetism.
    If we now impose the so called gauge fixing condition
    we left with only 2 physically significant components, and the (coupled) field equation becomes
    [tex]\partial^{2}G_{ab}=J^{ab} \ \ (2)[/tex]
    Eq(2) is nothing but the linear version of Einstein’s equation.
    Ok, now ignore the above and ask how one can obtain the Maxwell equations from the Poincare algebra? With a bit of work one can show that only ANTISYMMETRIC spin matrix can produce the full set of Maxwell equations.


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