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Last modified: January 26, 2018

Quantum Gravity
---------------

Our current understanding of gravity is based on Einstein's
theory of General Relativity that describes gravity as a
curvature within space-time that is influenced by mass.
Conversely, the electromagnetic, weak and strong forces are 
described by quantum theories that depend on the emdedding of
particle fields in the flat space-time of special relativity.
Unfortunately, difficulties arise when it comes to constructing
a quantum field theory for gravity with the graviton as the
force particle.   This comes about because the theory obtained
is not renormalizable and cannot be used to make meaningful
predictions.  Let us see why this is true. We start with the
Einstein-Hilbert action given by:

S = (c4/16πG)∫d4x √(-|g|)R

Where G is Newton's constant and R is the curvature scalar.

Now G can be written quantum mechanically in terms of the 
Planck mass, mp as:

G = hc/mp2

If we set h = c = 1 we get:

G = 1/mp2

and (neglecting the factor of 16π):

L = mp2√(-|g|)R

The action becomes:

S = mp2∫d4 √(-|g|)R

  = (1/G)∫d4 √(-|g|)R

Consider small perturbations around flat Minkowski space: 

gμν = ημν + Ghμν 

We can regard ημν as the 'basic' field, hμν as the theorized 
GRAVITON field and G as the coupling constant.  The graviton 
is expected to be massless (because the gravitational force 
appears to have unlimited range like the photon) and must be 
a spin-2 boson. A spin 2 particle is also known as a tensor 
boson, compared to a spin 0 scalar boson and a spin 1 vector 
boson. The spin 2 follows from the fact that the source of 
gravitation is the stress-energy tensor which is of second 
order (compared to electromagnetism's spin-1 photon, the 
source of which is a 4 vector, a first order tensor). Also, 
it can be shown that any massless spin 2 field would give 
rise to a force indistinguishable from gravitation, because a 
massless spin 2 field would couple to the stress–energy tensor 
in the same way that gravitational interactions do. 

Taylor series review:

f(x) = f(a) + f'(a)(x - a) +  f''(a)(x - a)2/2 + ...

If we set (x - a) = ε so that x = a + ε we get:

f(a + ε) = f(a) + f'(a)ε

Where,

f'(a) = ∂f(x)/∂x|x = a

Now we can do the same for the perturbed metric, g = η + Gh.
We get:

g(η + Gh) = g(η) + Gh∂g(η)/∂g(η) + G2h2∂2g(η)/∂g(η)2)/2 + ...

          = η + Gh +  G2h2/2 +  G3h3/6 + ...

Now, schematically,

R ~ ∂Γ/∂xμ + Γ2 where Γ ~ ∂g/∂xμ ≡ ∂μg (Christoffel symbols)

Therefore,

R ~ ∂μ2g + (∂μg)2

Now (dropping constants for clarity and simplicity),

∂μg = ∂μh + Gh∂μh + G2h2∂μh + ...

∂μ2g = ∂μ2h + G2h∂μ2h

Therefore,

R ~ ∂μ2h + Gh∂μ2h + (∂μh + Gh∂μh + G2h2∂μh)2

  ~ ∂μ2h + Gh∂μ2h + (∂μh + Gh∂μh)2

  ~ ∂μ2h + Gh∂μ2h + (∂μh)2 + Gh(∂μh)2 + G2h2(∂μh)2 + Gh(∂μh)2 
    + G2h2(∂μh)2 + G3h3(∂μh)2 + G2h2(∂μh)2 + G3h3(∂μh)2 + G4h4(∂μh)2

  ~ (∂μh)2 + Gh(∂μh)2 + G2h2(∂μh)2 + G3h3(∂μh)2 + G4h4(∂μh)2 + ...        

Dimensional Comparison
----------------------

Lets now take a look at the dimensions of the Einstein-Hilbert 
action under unperturbed and perturbed conditions:

g and h are dimensionless since ds2 = gμνdxμdxν.  However, they 
are both functions of xμ so:

[∂μg] = [∂μh] = [1/L] = [M]

[d4x] = [L4] = [1/M4]

[Γ] ~ [∂g/∂xμ] ≡ [∂μg] = [1/L] = [M]

[R] ~ [∂Γ/∂xμ + Γ2] ≡ [∂μ2g + (∂μg)2] = [1/L2] = [M2]

[G] =  [1/L2] = [M2]

In the unperturbed case:

S = (1/G)∫d4x √(-|g|)R

  =  [m2][m-4][m2]

We see that S is dimensionless (as it needs to be for the Path 
Integral formulation in QFT theory).

In the perturbed case:

S = (1/G)∫d4x √(-|g|){(∂h)2 + Gh(∂h)2 + G2h2(∂h)2 + ...}

Graphically, this expansion is an expansion in numbers of 
loops in Feynman diagrams.  At each loop the whole expression 
should be dimensionless.  The first term in {} has [m2] so 
that is fine.  However, the second term has [m4].  In order 
to make it equal to [m2], we would need to add 2 more powers 
of momentum in the denominator.  Because 1/G is of the order 
of 1016 Kg, the expression becomes more and more divergent the 
higher you go in the perturtabive expansion   In order to 
cancel these divergences it would be necessary to introduce 
an infinite number of counterterms which, in terms of 
renormalization, makes no sense.  The conclusion is that,
unlike other quantum field theories s where the coupling 
constant is dimensionless, the gravitational theory is not 
renormalizable (at least at very high energies).

Effective Field Theory
----------------------

The problem of non-renormalizability of quantum gravity does 
not necessarily mean that quantum  mechanics is incompatible  
with  gravity, only that quantum gravity should be treated as 
an EFFECTIVE FIELD THEORY for energies well below the Planck 
scale of 1019 GeV.  An effective field theory appromimates 
an underlying physical theory that describes physical phenomena 
occurring at a chosen length or energy scale, while ignoring 
phenomena at shorter distances.  Intuitively, it averages over 
the behavior of the underlying theory at shorter length scales 
to derive what is hoped to be a simplified model at longer 
length scales. Effective field theories typically work best 
when there is a large separation between the length scale of 
interest and the length scale of the underlying dynamics. 

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