# Graviton vs gluon ball

1. May 19, 2008

### glueball8

Can graviton possible act like gluon ball? Why not?

2. May 19, 2008

### mathman

Gravity (mediated by graviton - yet to be discovered) obeys an inverse square law. Strong force (mediated by gluons) is completely different. I can't see any connection. What did you have in mind?

3. May 19, 2008

### arivero

An spin 2 glueball? It should be massive (this short distance).

4. May 19, 2008

### humanino

I don't have Zee's book right here, but I think somewhere he mentions the "missing" symmetric (color) gluon as a candidate for graviton, since it would not be confined, carry an inverse distance square coupling basically to the amount of mass. This of course is just presented as an example of how one can do things wrong. Maybe that is what confuses the OP ?

5. May 19, 2008

### masudr

I imagine Zee is talking about the

$$| \bar{r}r \rangle + | \bar{g}g \rangle + | \bar{b}b \rangle$$

combination of gluons (which is the trivial representation of SU(3))? This reason this cannot be the graviton is because it would couple to all baryons with the same strength (since all baryons are colourless), and yet baryons have different mass and should couple differently to the graviton.

Since there is an allowed 3 gluon vertex and 4 gluon vertex in QCD, we can in principle have a bound state of gluons.

I have yet to see the Lagrangian for a QFT of gravity, so I can't comment about the allowed vertices, and indeed the possibility of a bound state.

Last edited: May 19, 2008
6. May 19, 2008

### humanino

This is not obvious. One can argue that "there are more gluons in a more massive hadrons". Anyway, the idea has yet other caveats.
This also is not obvious at all ! Despites extensive searches (to which I participate, I am not saying this is worthless), we have only very indirects mixing hints in only a few mesons (not even to mention pure glueballs). Besides, it is sometimes dangerous to jump to the conclusion "because something is not forbidden, it must be realized in Nature". Not all models of confinements agree on the possibility of free glueballs. The presence of light quarks, either in the valence or only in the (virtual) sea, is sometimes believed to be essential.

7. May 20, 2008

### Orion1

Affirmative, by setting their fine structure constants to equivalence:

Gravitational fine structure constant equivalent to strong fine structure constant:
$$\boxed{\alpha_g = \alpha_s}$$

8. May 20, 2008

### humanino

No, as indicated by arivero the graviton has spin-2 (couples to the EMT). The gluon has spin-1 (vector).

9. May 20, 2008

### arivero

On other hand, if you only want to mimic attractiveness of like charges, then a spin 0 glueball could do the trick (again ,were it massless). Probably it would fail to predict the orbit of Mercury.

10. May 20, 2008

### Orion1

EMT = Energy Momentum Tensor

Pure massless gluons and pure massless graviton 'balls' probably cannot exist within the framework of QCD and QGD, both are required to couple between masses as bosonic force carriers.

However, I would still defend gluon-gravitational Grand Unification for a degenerate QCD QGP plasma (ball), which was my original presumption for a 'gluon ball'.

Gravitational fine structure constant equivalent to strong fine structure constant:
$$\boxed{\alpha_g = \alpha_s}$$

Degenerate Grand Unification QCD-QGD QGP mass:
$$m_g = \sqrt{\frac{\hbar c \alpha_s(m_Z)}{G}}$$

$$\boxed{m_g = 7.46426018425756 \cdot 10^{- 9} \; \text{kg}}$$
$$\boxed{E_g = 4.18714467091867 \cdot 10^{18} \; \text{GeV}}$$

Reference:
Quark-gluon plasma - Wikipedia

Last edited: May 20, 2008
11. May 20, 2008

### humanino

Ah but that is very interesting ! It was not clear from the beginning of the discussion anybody was interested into the AdS/CFT correspondence and the so-called "RHIC fireball". As far as I am aware, the first serious (courageous) attempt in this direction was The RHIC fireball as a dual black hole. This line has been very productive. A recent review can be found in Introduction to String Theory and Gauge/Gravity duality for students in QCD and QGP phenomenology.

12. May 20, 2008

### humanino

I don't want to hack this thread but... I'll do it anyway :tongue2:
How would you write down a lagrangian involving only a scalar field, supposed to be the most relevant component of the graviton field in a certain low energy/long distance limit, while keeping non-abelian effects ?
I was thinking to try someything like
$$S=\int d^4x \left[\partial h\partial h+\sqrt{G}h\partial h\partial h+Gh^2\partial h\partial h+\cdots + \sqrt{G}hT \right]$$
by directly replacing in the usual expansion the graviton with a scalar field. Do I make sens ? :uhh:

13. May 20, 2008

### Orion1

The lagrangian density for a massless spin-two field:
$$\mathcal{L} = - \frac{1}{2} \partial_{\lambda} h_{\mu \nu} \partial^{\lambda} h^{\mu \nu} + \partial_{\lambda} h_{\mu}^{\lambda} \partial_{\nu} h^{\mu \nu} - \partial_{\mu} h^{\mu \nu} \partial_{\nu} h + \frac{1}{2} \partial_{\lambda} h \partial^{\lambda} h$$

Einstein gravity weak field limit:
$$h \equiv h_{\mu}^{\mu}$$

Scalar field theory:
$$\mathcal{S}=\int \mathrm{d}^{D-1}x \, \mathrm{d}t \mathcal{L}$$

Solution for a scalar field lagrangian density massless spin-two field:
$$\boxed{\mathcal{S}=\int \mathrm{d}^{D-1}x \, \mathrm{d}t \left[ - \frac{1}{2} \partial_{\lambda} h_{\mu \nu} \partial^{\lambda} h^{\mu \nu} + \partial_{\lambda} h_{\mu}^{\lambda} \partial_{\nu} h^{\mu \nu} - \partial_{\mu} h^{\mu \nu} \partial_{\nu} h + \frac{1}{2} \partial_{\lambda} h \partial^{\lambda} h \right]}$$

Tensor scalar theory solution: ???
$$\boxed{T^{\mu\nu}_g = \frac{c^4}{8 \pi G} \left[ - \frac{1}{2} \partial_{\lambda} h_{\mu \nu} \partial^{\lambda} h^{\mu \nu} + \partial_{\lambda} h_{\mu}^{\lambda} \partial_{\nu} h^{\mu \nu} - \partial_{\mu} h^{\mu \nu} \partial_{\nu} h + \frac{1}{2} \partial_{\lambda} h \partial^{\lambda} h \right]}}$$

Reference:
Scalar field theory - Wikipedia
Multi-Graviton Theory
Scalar theories of gravitation - Wikipedia

14. May 20, 2008

### blechman

Unfortunately, I don't think this works - I remember playing with this idea in grad school, and we found that we can always kill such "nonabelian" terms with a (nonlinear) field redefinition. VERY sketchily:

$$S = \int d^4x Z(\phi)(\partial\phi)^2$$

Let $\phi' = \phi/\sqrt{Z(\phi)}$. So at the end of the day, we end up with a free scalar field!

You have to be a little careful with this since the Path Integral measure can also change, but I think the result is still rigorous. Check out Tomas Ortin's textbook "Gravity and Strings" - it talks about these kinds of theories.

We can have 4-derivative terms, but these are a problem - they introduce ghosts and other nasties. And besides, they wouldn't correspond to the usual "scalar-Einstein-Hilbert action".

Last edited: May 20, 2008
15. May 20, 2008

### glueball8

Hmm I'm in high school. So I don't get the math. Can anyone explain the concepts?

16. May 21, 2008

### humanino

I will keep playing with it and try to redefine my field to "gauge" away my non-abelian terms. Intuitively, it does seem odd because any non-abelian group acting on my scalar could only do so trivially. I expected this to be fishy actually.
Arg, I just bought Kiefer's books and barely began it
So little time...

17. May 21, 2008

### blechman

it's NOT a "gauge" issue - it's a field redefinition. quite different. there's no symmetry argument, it's just that these nonlinear terms can all go into a sort-of wave-function renormalization constant, and are therefore irrelevant; that is: they don't lead to any physical observables. the "Z" I mention is NOT a transformation.

A similar argument fails for the usual graviton due to the tensor nature of the interactions - you cannot "reabsorb" the nonlinear terms in that case. Scalar fields are a little bit TOO trivial.

yeah, there's a lot to read out there. Have fun!

Ortin does explain scalar gravity theories very well, including why they fail (wrong precession of mercury; no bending of light,...). This is, of course, not to be confused with Brans-Dicke-Jordan theory, where there is BOTH scalar AND tensor.

18. May 23, 2008

### Haelfix

Just glancing at it briefly, I think there is a topological term in one of those nonlinear bits which is why it doesn't work and can be redefined away.

I could be wrong though.

19. May 23, 2008

### blechman

Hi Bright Wang.
I'm sorry to say this thread has been hijacked! I don't think I can explain the concepts without a LOT of time and several blackboards, but the quick no-math summary is like this:

you asked about gluballs and gravitons. From my point of view: glueballs cannot be gravitons for a number of reasons - they're massive, first of all, and the graviton is supposed to be massless; they interact with the STRONG nuclear force (note emphasis on strong!) while gravity is weak - in fact, it's super-super-super-SUPER-SSSUUUPPPEEERRR!!!!!!! weak!

Then people started talking about the "singlet gluon", which is a possible particle that could come out of the theory of strong nuclear force (called "Quantum ChromoDynamics", or QCD). This particle does not seem to exist in nature (and there are plenty of satisfying explanations out there for why this is so), but if it did, it would interact weakly and be massless. HOWEVER, it would have the wrong spin (a quantum mechanical angular momentum) - in particular, it would have spin angular momentum $\hbar$ while the graviton should have spin angular momentum $2\hbar$ - that's a very important factor of 2! In case you don't know $\hbar\sim 10^{-34}$ Joules-seconds is Planck's constant - it has units of angular momentum.

Then the thread started to go awry from your point of view! people started (wisely!) asking about ways to modify Einstein's theory of general relativity to get some interesting results. This is where the friendly explanations must stop, I'm afraid, as things start to get more technical.

I hope that quick and sloppy summary helps you understand what happened.

Last edited: May 23, 2008
20. May 24, 2008

### humanino

Yes, that's why I put "gauge" and not gauge
Thanks for the correction though.