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Coupling of the graviton

  1. Mar 21, 2005 #1
    I was wondering if anyone has any insights or can guide me to some decent literature of the coupling of the graviton to mass. I became intrigued about this when thinking about presenting the gravitational interaction in a Feynman diagram. If anyone has any ideas or knows I would greatly appreciate it.
    Josh
     
  2. jcsd
  3. Mar 21, 2005 #2
    This an idea coming from QFT. In QFT you have a quantum field that you must imagine as being a mattress built out of a gazillion springs (harmonic oscillators). Now suppose we but two objects on that mattress, it will start to vibrate. This vibration of the quantumfield (the mattress) corresponds to a particle called the gauge boson or force carrier which really brings over the interaction between the two particles.

    Now, a mass is represented by the stress energy tensor from general relativity, which is a symmetric lorentz tensor. Such a tensor has 5 components. In order to put a mass on the mattress is QFT, we have to multiply this tensor with the field (which is the actual mattress, remember ?)

    The specific symmetry of this tensor leads to the fact that our field must have 5 degrees of freedom, otherwise this multiplication is not 'valid' (not Lorentz covariant). A field with 5 degrees of freedom expresses a particle of spin 2 because spin 2 really means 2*2+1 = 5 degrees of freedom (think over the degenerate energylevels in QM)

    regards
    marlon

    ps : read A. Zee's QFT in a Nutshell

    or check out my journal : the info on the web entry
     
  4. Mar 22, 2005 #3
    Thanks for the help. I had never really asked why the graviton had to have spin 2. I will look some more into QFT. It will probably be a couple of years before I am able to take a class in it.
    Josh
     
  5. Mar 22, 2005 #4

    dextercioby

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    1.The stress-energy tensor of matter fields [itex] \Theta_{\mu\nu} [/itex] which appears in the RHS of Einstein equation is a second rank SYMMETRIC 4TENSOR which has 10 (sic) independent components (outta 16 altogether).
    2.The gravitational field has 2 (sic) degrees of freedom.You can show this very elegantly in the weak field limit,where one not only can prove this fact,but can also quantize it (the gravity field).

    The graviton is the quanta of the gravitational field.A spin 2 particle with 2 degrees of freedom/polarizations/eigenvalues of the helicity operator:+2 & -2...

    Daniel.
     
  6. Mar 22, 2005 #5

    dextercioby

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    It's David Hilbert's fault.He was the first to come up with the Lagrangian or the gravitational field

    [tex] S_{L}^{H}[g_{\mu\nu}]=\int \left(d^{4}x\sqrt{-g}\right) \ R [/tex]

    from which it can be shown that this classical field has 2 degrees of freedom & at quantum level describes massless particles of spin 2.

    Daniel.
     
  7. Mar 22, 2005 #6
    dextercioby, you are again wrong...here we go again...pitty cause i really liked the mutual ignorance.

    A massive spin 2 field has 5 degrees of freedom, caracterized by the five polarization tensors which are each symmetric in their indices. You will also have one tracelessness condition when you multiply the g-tensor with these polarization tensors.

    The stress energy tensor has ten degrees of freedom, i never argued that. You misinterpreted (again) my words. In order for a field, to couple to this tensor, this field will have to be symmetric in its two indices. This field will have ten degrees of freedom.

    But minus 5 degrees of frredom coming from the polarization tensors (well, their product with the k-four vector that is) and minus 1 coming from the traceleness relation. 5 degrees of freedom remain.

    To the OP : disregard dextercioby's post, since it is partially incorrect and it contains incorrect corrections,...hmmm

    marlon
     
  8. Mar 22, 2005 #7
    The photon has spin 1 because the field (A-field) describing it will couple to the four vector J.

    At first you might think that there is one degree of freedom to much, but this 4th variable is eliminated by the equation [tex]\partial_{\mu}A^{\mu} = 0[/tex]

    Thus, three degrees of freedom remain : this is a spin one particle...

    regards
    marlon
     
  9. Mar 22, 2005 #8

    dextercioby

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    Marlon,what is a degree of liberty...?From your post,you have no idea.

    Daniel.
     
  10. Mar 22, 2005 #9
    First of all it's a 'degree of freedom'

    Basically in this case this concept denotes the number of components in a tensor. They can be reduced whenever some kind of equation, expressing some linear connection between those components, is given.

    BTW this is a nice tactic dexter, asking something you don't know yourself but making it look like other people's incompetence...you should have become a sollicitor :rofl:

    marlon
     
  11. Mar 22, 2005 #10
    The components don't always have to be independent. Just look at the influence of gauges in field theory...In the general case of the stress energy tensor, the components are independent, but not all tensors respect that property.

    regards
    marlon
     
  12. Mar 22, 2005 #11

    dextercioby

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    I wasn't mentioning gauge wrt to matter fields,because in QFT (SM),there are no I-st class constrained systems acting as matter fields.

    As for degrees of freedom,you just proved you have no idea whatsoever.

    Daniel.
     
  13. Mar 22, 2005 #12
    Well, the graviton picture arises from the principles of QFT. You just take as the perturbation of the field (normally denoted by the J-tensor in the path integral formalism) the stress energy tensor.

    Don't sell this blablabla...

    :rofl:
    How is that ?

    Please, prove me wrong on the matter as to why the graviton has spin two and thus five degrees of freedom (and the other way around).

    Before you start your charade, i suggest you read QFT in a Nutshell (page 32 in the chapter called 'bypassing einstein'), you will find the exact same definitions and explanaitions there. Or are you gonna deny the content of this book :rofl: ?

    Should be interesting, though in respect of our 'mutual ingorance'-promise (which you broke) i won't respond any more to your posts...

    regards from your best friend

    marlon
     
  14. Mar 22, 2005 #13

    pervect

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    If you really must, Feynmann has some notes about Gravity approached from a QFT perspective. "Freyman's lectures on Gravitation".

    But if you don't already know QFT I wouldn't recommend it.

    These are published as a collection of lecture notes and not as a complete theory. The actual theory runs into some difficulties down the road (IIRC it won't renormalize). So it's not a complete theory, just a collection of lecture notes for an approach to gravity that Feynman himself eventually wound up abandoning.

    It's noted in the book (which I actually looked at, for some reason which I find unfathomable our local library actually has a copy of it!) that for tutorial purposes, there are much better sources.

    I belive one of them is even online

    http://xxx.lanl.gov/abs/astro-ph/0006423

    From the abstract

    I can't say I really follow the paper - I think the classical approach to gravity is a lot easier. If someone is in the position of already knowing QFT really well, while also not knowing differential geoemtry, the approach in this paper might be of special interest. Otherwise, it's probably a lot easier to learn some differential geometry than it is to learn QFT at the required level.
    .
     
  15. Mar 22, 2005 #14

    pervect

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    I think a lot of people on this thread are talking apples and oranges. Take the electromagnetic field - does it have two degrees of freedom, because a photon has a spin of +1 or -1, or does it have six degrees of freedom, because there are 3 components of the electric field, and three of the magnetic which are all independent?

    I seem to recall that gravity has 5 degrees of freedom in empty space, and 10 in general.

    http://www.physlink.com/Education/AskExperts/ae98.cfm

    has a lot more detail, but I have to reread it.
     
  16. Mar 22, 2005 #15
    The biggest problem with QFT AND gravity is the cosmological constant (let's call it M). In the absence of gravity, the Lagrangian L is invariant under L ---> L-M, where M is a constant. In classical physics this is true because only the variation of the L is important in the Euler-Lagr equantions. In QFT, the M yields a multiplicative factor in the Feynmann path integral, but that is no problem since we can omit that.

    However if gravity is 'turned on' (i mean, included in the theory, which is NOT the case in the SM) the invariant under L-M is broken. This means that gravity 'knows' about M...The associated shift in the Hamiltonian H--> H + int(d³xM) is not invariant under coordinate transformations x-->x'(x)...so that is a big problem.

    This M represents a constant energy or mass per unit volume permeating the universe.What is this mass and how big is it ? A possible candidate is the vacuum fluctuations...check out my journal for some references to peer reviewed articles.

    In particle physics this M is supposed to be 0 but that cannot be the case because of cosmological measurements

    regards
    marlon
     
  17. Mar 22, 2005 #16
    The EM-field has THREE degrees of freedom because of 2j+1 and j=1 for this field (and thus for photons)...Otherwise there cannot be a coupling with this field and the vectorcurrent J (the charged particles)

    In classical physics there are indeed two degrees of freedom. The spin 0 state is not possible because we can never bring the massless photon to a restframe and then look at how the rotationgroup works on the photon-spin states. One of the polarization vectors (the longitudinal mode) will decouple from the EM-interactions...Thus two degrees of freedom remain...

    The three values for s initially arise from the fact that the field theory is applied for a massive spin 1 particle.

    However virtual photons DO have three spin values...
    marlon
     
    Last edited: Mar 22, 2005
  18. Mar 22, 2005 #17
    Pervect,

    The degrees of freedom are just the number of linear independent variables that you need in order to describe some physical quantity in a certain representation. They can be denoted by the quantumnumbers...For example you will need three polarization tensors in order to represent a massive spin one particle in the SO(3) representation. If you take away mass, SO(3) will become SO(2)...The polarization tensors can be seen as a fundamental representation, just like the three colours in the case of the SU(3) color symmetry of QCD.

    A massive spin 1 particle is just described in terms of these three (in the SO(3)-case) polarizations or in terms of the two polarizations in the photon case (SO(2))

    marlon
     
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