Natural Science Forum / Physics / Research / March 2010

The Big Bang & Variable Light Speed

I'm going to offer an interesting perspective on the FW cosmology -- a
different take that brings to light (pun intended) what's hiding in
plain sight.

Contents:
(1) Flat FW Metric = Minkowski with Variable Light Speed
(2) The Geometric Quantities
(3) Solutions
    (3.1) Exponential Case
    (3.2) Power-Law Case
(4) The Horizon (Non-?)Problem and the Light-Cone Time Reflection

(1) Flat FW Metric = Minkowski with Variable Light Speed
The FW metric is
    ds^2 = dt^2 - A^2 R^2 (dx^2 + dy^2 + dz^2)
where the spatial section has a curved metric with a conformal factor
given by
    R = 1/(1 - kr^2)
    k > 0 -> Hyperspherical,
    k = 0 -> Flat,
    k < 0 -> Hyperbolic
and A = A(t) > 0 is a function of time. As best as we can resolve, the
k = 0 case applies so that the metric can be written simply as
    ds^2 = dt^2 - (dx^2 + dy^2 + dz^2)/c(t)^2
or, simply: as the Minkowski metric with a variable speed of light, c
= c(t). Equivalently, this can be written
    ds^2 = dt^2 - alpha(t) (dx^2 + dy^2 + dz^2)
with a variable signature parameter alpha(t) = 1/c(t)^2 > 0.

The Big Bang "singularity" is a signature-changing boundary on which
alpha = 0 and across which alpha may go negative for t < 0 to a
signature comprising 4 space-like dimensions and none of time. If it
does, then the different signatures involved are:
    alpha > 0: Lorentzian ("relativistic")
    alpha = 0: Galilean ("non-relativistic")
    alpha < 0: 4-dimensional Euclidean

If alpha -> 0 as t -> 0 at least as fast as alpha(t) = O(t^2), or A(t)
= O(t), then light cones flatten out to infinity at t = 0. Otherwise,
they land on t = 0 at a FINITE radius
    H_T = integral (c(t) dt) from t = 0 to T (T = current time)
called the Cosmological Horizon.

In both cases, the t = 0 surface is a non-relativistic surface of
absolute simultaneity, and is the envelope of all past light cones.

The remaining case is where A(t) is exponential or something more
bizarre (e.g. hyperbolic cosine). Then there is no alpha = 0 or A = 0,
and no "singularity". Only the exponential occurs for the flat case k
= 0. Hyperbolic sines, hyperbolic cosines can occur for k > 0, and
sines and cosines can occur for k < 0.

(2) The Geometric Quantities
These are the details on the various geometric quantities formed of
the metric:
    g_{00} = 1,
    g_{i0} = 0 = g_{0j},
    g_{ij} = -(1/c)^2 delta_{ij}

The notation used is:
    delta_{ij} = Kroenecker delta
    i, j, k, etc. = 1, 2, 3 below;
    (x^1,x^2,x^3) = (x,y,z), x^0 = t.
    ()' denotes d()/dt.

Dual metric:
    g^{00} = 1,
    g^{i0} = 0 = g^{0j},
    g^{ij} = -c^2 delta^{ij}

Levi-Civita Connection:
    Gamma^k_{i0} = -(c'/c) delta^k_i
    Gamma^k_{0j} = -(c'/c) delta^k_j
    Gamma^0_{ij} = -alpha (c'/c) delta_{ij}
    All other components are 0

Riemann Tensor:
    R^k_{jil} = alpha (c'/c)^2 (delta^k_i delta_{lj} - delta^k_l
delta_{ij})
    R^k_{0i0} = (c''/c - 2(c'/c)^2) delta^k_i
    R^k_{00l} = -(c''/c - 2(c'/c)^2) delta^k_l
    R^0_{ji0} = alpha (c''/c - 2(c'/c)^2) delta_{ji}
    R^0_{j0l} = -alpha (c''/c - 2(c'/c)^2) delta_{jl}
    All other components are 0

Ricci Tensor:
    R^k_i = (c''/c - 4(c'/c)^2) delta^k_i
    R^k_0 = 0 = R^0_j
    R^0_0 = 3 (c''/c - 2(c'/c)^2)

Curvature Scalar:
    R = 6 (c''/c - 3(c'/c)^2) = -3 alpha''/alpha.

The curvature scalar is just the acceleration of the signature
parameter alpha, up to proportion.

Einstein Tensor:
    G^k_i = (-2(c''/c) + 5(c'/c)^2) delta^k_i = 4/3 (alpha^{3/4})''/
alpha^{3/4}
    G^k_0 = 0 = G^0_j
    G^0_0 = 3 (c'/c)^2 = 3/4 (alpha'/alpha)^2.

(3) Solutions
The stress tensor can generally be decomposed into the following form:
    T^0_0 = -mu, T^i_0 = q^i, T^0_j = q_j, T^i_j = p delta^i_j + pi^i_j
    mu = energy density, q = (q^1, q^2, q^3) = energy flux,
    (q_1, q_2, q_3) = momentum density = q,
    p = pressure, pi = shear (symmetric and trace-free)

Given the field law T^k_i = K G^k_i, with K being the (non-zero)
proportionality coefficient, it follows that the source has the form
    q = 0 (stationary fluid),
    pi = 0 (shear-free)
    mu = -3K (c'/c)^2
    p = -K (2 c''/c - 5 (c'/c)^2) = -K (2 (c'/c)' - 3 (c'/c)^2).

The order 0, 1 and 2 derivatives are also rewritten as:
    A = 1/c = Expansion Parameter
    H = -c'/c = Hubble Parameter
    q = -(c''/c)/H^2 = Acceleration Parameter
Thus
    mu = -3K H^2, p = K (2q + 5) H^2

The equations of state
    p = (g - 1) mu, g = constant (gamma is used in place of "g" in the
literature)
produces a wide class of models that are consistent with the Killing
symmetry of the underlying metric. This leads to the following:
    (c'/c)' = 3/2 g (c'/c)^2
or
    (c/c')' = -3/2 g.

(3.1) Exponential Case
The case g = 0 yields either
    c(t) = c(T) = constant, or
    c' = c/tau, c(t) = c(T) exp((t-T)/tau).
In the former case, p, mu, G^m_n are all 0. In the latter case,
    p = -mu = 3K/tau^2
    T^m_n = -(3K/tau^2) delta^m_n
    G^m_n = (3/tau^2) delta^m_n
    tau = +/- sqrt(3/Lambda), Lambda = cosmological constant

Here, there is no alpha -> 0 or c -> 0 and no horizon. Instead, if tau
< 0, then alpha -> 0 exponentially, as t -> -infinity.

(3.2) Power-Law Cases
The case g non-zero yields
    c/c' = -3/2 g t -> c(t) = c(T) (T/t)^{2/(3g)}.
The signature parameter goes as
    alpha(t) = alpha(T) (t/T)^{4/(3g)).
It also follows that
    c'/c = -2/(3gt)
    c''/c = (1 + 3g/2) (c'/c)^2
    mu = -3K (2/(3gt))^2
    p = 3K(1 - g) (2/(3gt))^2

The case of 0 pressure (the "dust model") is g = 1. That yields a time
dependence of
    c(t) = c(T) (T/t)^{2/3}
    alpha(t) = alpha(T) (t/T)^{4/3}

The case of 0 stress tensor trace (the "radiation dominant model") is
given by R = 0, with a non-zero density mu:
    0 = T^m_m = 3p - mu = (3(g - 1) - 1) mu -> g = 4/3.
Then
    c(t) = c(T) (T/t)^{1/2}
    alpha(t) = alpha(T) t/T

(4) The Horizon (Non-?)Problem and the Light-Cone Time Reflection
As per the previous discussion: the Horizon goes off to infinity only
if g < 2/3. Otherwise it lands on the t = 0 surface at a distance
    H_T = integral c(t) dt = c(T) T integral (T/t)^{2/(3g)) d(t/T)
or
    H_T = c(T) T (3g)/(3g - 2).

The case of 0 pressure, g = 1, yields:
    H_T = 3 c(T) T.
The case of 0 stress tensor trace, g = 4/3, yields:
    H_T = 2 c(T) T.

If the earliest epoch were actually radiation-dominant all the way
down to t = 0, then alpha would go linear all the way to t = 0 and
could simply continue on linearly for t < 0 into negative values.

Then the "horizon problem" becomes a red herring. This is underscored
by a closer examination of what happens to the light cones at t = 0.

In particular, look at the past light cones of t = t0 (where t0 is
inside the radiation-dominant sector which, again, we're assuming goes
all the way to t = 0), as t = 0. Solving
    dr = c(t) dt = c(t0) (t0/t)^{2/(3g)} dt
    r(t0) = 0
we get
    r(t) = 3g/(3g-2) c(t0) t0 (1 - (t/t0)^{(2-3g)/(3g)}).

For the case g = 4/3, we get
    r(t) = 2 c(t0) t0 (1 - (t/t0)^{1/2}).
or
    t t0 = (t0 - r/(2 c(t0)))^2

This is a parabola that reflects off of t = 0 at the cosmological
horizon:
    H_{t0} = 2 c(t0) t0.
and then going outwards in the FUTURE direction into space outside the
horizon. The past light cone for t > 0 is connected to:
    (a) the Galilean non-relativistic sector, t = 0, within the horizon,
    (b) the Euclidean sector, t < 0, in its entirety,
    (c) the Lorentzian sector, t > 0, OUTSIDE the outward-directed future-
pointing light cones emanating from the horizon.