Hypothetical graviton : Why a spin of 2 ?

In summary, a spin-2 field described by a spin-1 boson corresponds to a quanta of the field that is not correctly described by the Maxwell equations.
  • #1
A_theory_2012
1
0
Hi,

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' :-p

Thanks
 
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  • #2
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).
 
  • #3
A_theory_2012 said:
Hi,

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' :-p

Thanks

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.
 
  • #4
cosmik debris said:
Spin 1 or vector bosons have two charge types so they can attract or repel.
You mean like the photon?
 
  • #5
clem said:
You mean like the photon?

Yes.
 
  • #6
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.
 
  • #7
cosmik debris said:
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.
I must have misinterpreted your remark. The photon has spin 1, but does not have two charge types.
 
  • #8
clem said:
I must have misinterpreted your remark. The photon has spin 1, but does not have two charge types.

It doesn't have two charge types, it carries two charge types.
 
  • #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.
 
  • #10
Does this mean you can't have a rank two tensor field that is described by a spin-1 boson?
 
  • #11
No one has said anything so let me be more specific.
samalkhaiat said:
I can show you that the two postulates of SR ( plus, of course, causality) is all what you need to conclude that:
spin-0 representation satisfies the K-G equation.
spin-1/2 representation satisfies the Dirac equation.
spin-1 rep. satisfies the Maxwell equations.
spin-3/2 rep. satisfies the R-S equation. and
spin-2 rep. satisfies the linearized Einstein equation.

regards


sam
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?
[tex]
\Box B_{\alpha\beta}=0
[/tex]
 
  • #12
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
[tex]\partial_{a}F_{bc}+\partial_{c}F_{ab}+\partial_{b}F_{ca}=0[/tex]
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]
where
[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
[tex]\partial_{a}G^{ab}=0[/tex]
we left with only 2 physically significant components, and the (coupled) field equation becomes
[tex]\partial^{2}G_{ab}=J^{ab} \ \ (2)[/tex]
where
[tex]\partial_{a}J^{ab}=0[/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.

regards

sam
 

Related to Hypothetical graviton : Why a spin of 2 ?

1. What is a hypothetical graviton?

A hypothetical graviton is a theoretical particle that is believed to be the carrier of the force of gravity in quantum physics. It has not yet been observed or proven to exist, but scientists have proposed its existence based on mathematical models.

2. How does a graviton have a spin of 2?

In quantum mechanics, spin is a fundamental property of particles that describes their angular momentum. The graviton is predicted to have a spin of 2 because it is a massless particle, similar to the photon, which also has a spin of 2. This is consistent with the mathematical equations that describe how gravitons interact with matter.

3. Why is a spin of 2 significant for a graviton?

A spin of 2 is significant for a graviton because it plays a crucial role in the theory of general relativity. In this theory, gravity is described as the curvature of spacetime, and a spin-2 particle is necessary to mediate this curvature. Without a spin of 2, the theory of general relativity would not be consistent with the observed behavior of gravity.

4. Can the spin of a graviton be measured?

Currently, there is no experimental evidence for the existence of gravitons, so their spin cannot be directly measured. However, some researchers are working on ways to indirectly detect the presence of gravitons by observing their effects on other particles.

5. What are some potential implications of a graviton having a spin of 2?

If the existence of gravitons is confirmed and their spin is confirmed to be 2, it would provide strong support for the theory of general relativity and our understanding of gravity. It could also help bridge the gap between quantum mechanics and general relativity, as gravitons are predicted by quantum mechanics but are also necessary for general relativity. Additionally, it could lead to further developments in our understanding of the fundamental forces and particles in the universe.

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