The Hilbert Action

In the theory of Special Relativity we assume that space and time form a rigid frame on which observers measure intervals between events. We describe that frame fundamentally with a 4x4 tensor and the four-dimensional analogue of the Pythagorean theorem

(Eq’ns 1)

We thus have the metric tensor and the metric equation of Minkowski’s "flat" spacetime.

What happens when we take away the rigidity? It will certainly affect our derivation of the principle of least action, from which we deduce the other laws of physics, so how does that derivation play out in a non-rigid spacetime?

In the basic principle of least action we use the Lorentz
invariant dS=**p∙**d**x**-Edt=Ldt as our element of action, whose integral
must be minimized with respect to variations in the coordinates. In non-rigid
spacetime we want to keep open the option of bringing in forcefields, so we use
the Lagrangian density,

(Eq’n 2)

in which the Lagrangian density by the Lorentz-invariant element of
four-volume. In that expression g represents the determinant of the metric
tensor and dP^{σ}
represents a pressure whose value is to be determined. If we conceive points in
space as being like particles in a gas, then we can imagine a kind of pressure
pervading the Universe and being subject to an analogue of Boyle’s law where
space is deformed.

Suppose that a patch/chunk of spacetime is deformed away from the Minkowskian flat spacetime in some way that we can describe mathematically. How would that deformation change the assumed pressure?

Let’s regard the assumed pressure as the fourth (temporal)
component of a four-vector P^{σ},
whose spatial components equal zero, and look at how that component changes
across a minuscule patch of spacetime. Pick two points, O and M, on opposite
corners of a minuscule quadrangle whose sides run parallel to two of the
coordinate axes (those sides may be straight or curved). We want to know how the
four-vector changes from place to place, so we imagine parallel displacing it
around the quadrangle from O to M. Calculating the change requires the use of
the covariant derivative,

(Eq’n 3)

The upper case gamma in that equation represents the Christoffel symbol.

Imagine parallel displacing the pressure four-vector along
the line dx^{ν}
from the point O to one of the near corners of the quadrangle and then parallel
displacing it along the line dx^{μ}
to the point M. The covariant derivative that we use in the first stage of that
movement yields a mixed second-rank tensor, so the second stage uses a covariant
derivative of the form

(Eq’n 4)

Next imagine parallel displacing the pressure four-vector around the opposite
side of the quadrangle, along dx^{μ}
and then along dx^{ν}
to the point M. We can calculate a double covariant derivative for that process
and then subtract it from Equation 4, getting

(Eq’n 5)

in which

(Eq’n 6)

represents the Riemann-Christoffel curvature tensor, which has units of reciprocal area.

Equation 5 gives us a derivative of P^{σ}
with respect to the area of the quadrangle. The use of a subtraction in the
process makes the derivative analogous to a curl (a ghostly image of Stokes’
theorem looms over our result). We can envision the Riemann-Christoffel tensor
as a 4x4 matrix whose elements are 4x4 matrices whose elements are algebraic
functions of the coordinates (including constants). Because every element in
this array contains a derivative of the metric tensor, the curvature tensor
equals zero in spacetimes whose metric tensors are simply arrays of constants,
such as the flat spacetime of Minkowski. We can write from Equation 5, then,

(Eq’n 7)

The Einstein summation convention applied to the right side of that equation implies that we calculate over all possible quadrangles available at our chosen point in our four-dimensional coordinate system. We then get

(Eq’n 8)

The R-matrix is the mixed-rank Ricci tensor.

Pressure, in a fluid, is omnidirectional, but when pressing against a surface it produces a force wholly perpendicular to the surface. That fact necessitates that the shear terms in Equation 8 zero out, leaving only the terms on the matrix’s upper-left to lower-right diagonal. Equation 8 then becomes

(Eq’n 9)

in which R represents the Ricci scalar, the trace of the Ricci tensor. In the last step in that equation I have tacitly assumed that the components of the pressure tensor all have the same magnitude, in accordance with the omnidirectionality of pressure.

Substitute that result into Equation 2 and vary the result with respect to variations in the contravariant metric tensor. The result, as shown by Hendrik A. Lorentz and David Hilbert, is Einstein’s equation

(Eq’n 10)

in which T_{μν}
represents the stress-energy tensor (or the momentum-energy tensor) describing
the distribution of matter through its dynamic properties. Using that equation
to describe a thin, uniform gas enables us to evaluate the constant, which comes
out as

(Eq’n 11)

In the absence of matter we have Equation 2 as

(Eq’n 12)

which tells us that

(Eq’n 13)

That’s a force (pressure times area), so we know we have the right units.

That equation also implies the existence of a fundamental unit of area. One area that comes readily to mind is that of the imaginary spherical shell that we conceive as enclosing the Universe. With a radius of 13.7 billion lightyears, that shell has an effective area of

(Eq’n 14)

That would make the postulated pressure of raw space P=1.146x10^{-9}
newtons per square meter, about 1.1x10^{-14} atmosphere. Any further
speculations on that pressure would require another essay. For now, though, we
must be satisfied to have a reasonable derivation of Hilbert’s action.

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