Hello Urs:

This notebook sets out to be a companion for the three emails that eventually end up at the exponential metric.  In effect, it serves as a "mathematical grammar checker."

A function to test for solutions to the vacuum field equations:

test[potential_] := 
      Simplify[{D[potential[[1]], {t, 2}] - 
            D[potential[[1]], {x, 2}] - D[potential[[1]], {y, 2}] - 
            D[potential[[1]], {z, 2}], 
          D[potential[[2]], {t, 2}] - D[potential[[2]], {x, 2}] - 
            D[potential[[2]], {y, 2}] - D[potential[[2]], {z, 2}], 
          D[potential[[3]], {t, 2}] - D[potential[[3]], {x, 2}] - 
            D[potential[[3]], {y, 2}] - D[potential[[3]], {z, 2}], 
          D[potential[[4]], {t, 2}] - D[potential[[4]], {x, 2}] - 
            D[potential[[4]], {y, 2}] - 
            D[potential[[4]], {z, 2}]}]

Test the classical 1/R potential. The units of a potential are ((Sqrt[m])/(Sqrt[L])).

test[{((Sqrt[(h/c)])/(Sqrt[(x^2) + (y^2) + (z^2)])), 0, 0, 0}]

units = {γ -> 1, 
          J -> ((m^(1/2))/((L^(3/2)) t)), V -> (L^3), 
          q -> (((m^(1/2)) (L^(3/2)))/t), U -> (L/t), 
          (A^μ) -> ((m^(1/2))/(L^(1/2))), 
          (A^μ)ν -> ((m^(1/2))/(t (L^(1/2)))), (Sqrt[G]) -> 
            ((L^(3/2))/((m^(1/2)) t)), G -> ((L^3)/(m (t^2))), 
          c -> (L/t), h -> ((m (L^2))/t)};

(1/L) (Sqrt[(h/c)]) /. units

Test a potential that is a Lorentz inverse distance squared.

test[{(((Sqrt[G]) h/(c^2))/((x^2) + (y^2) + (z^2) - (t^2))), 0, 0, 
        0}]

(((Sqrt[G]) h)/((c^2) (L^2))) /. units

The units imply this potential would involve relativistic (c) quantum (h) gravity (G).

Test a potential that is a normalized linear perturbation.

a = {(((c/(Sqrt[G])))/((((1 + (((k x)/(σ^2)))))^2) + (((1 + (((k y)/(σ^2)))))^2) + (((1 + (((k z)/(σ^2)))))^2) - (((1 + (((k t)/(σ^2)))))^2))), 0, 0, 0};

test[a]

(c/(Sqrt[G])) /. units

These units imply relativistic (c) gravity (G), but not quantum mechanics (no h).  Variations on this potential are the focus of the following work.

Look at the derivatives to first order in k.

Expand[D[a, t]] /. (k^2) -> 0

For a small oscillation, the denominator will be approximately 2  (σ^2).  This substitution list will make it so (and for other cases to come).

sublist = {(k^2) -> 
            0, -(((1 + ((k t)/(σ^2))))^2) + (((1 + ((k x)/(σ^2))))^2) + (((1 + ((k y)/(σ^2))))^2) + (((1 + ((k z)/(σ^2))))^2) -> 
            2, -(((1 + ((k t)/(σ^2))))^2) + (((1 - ((k x)/(σ^2))))^2) + (((1 - ((k y)/(σ^2))))^2) + (((1 - ((k z)/(σ^2))))^2) -> 
            2, -(((1 - ((k t)/(σ^2))))^2) + (((1 + ((k x)/(σ^2))))^2) + (((1 - ((k y)/(σ^2))))^2) + (((1 - ((k z)/(σ^2))))^2) -> 
            2, -(((1 - ((k t)/(σ^2))))^2) + (((1 - ((k x)/(σ^2))))^2) + (((1 + ((k y)/(σ^2))))^2) + (((1 - ((k z)/(σ^2))))^2) -> 
            2, 
-(((1 - ((k t)/(σ^2))))^2) + (((1 - ((k x)/(σ^2))))^2) + (((1 - ((k y)/(σ^2))))^2) + (((1 + ((k z)/(σ^2))))^2) -> 2};

Expand[D[a, t]] /. sublist

Write a potential that solves the field equations, but only has derivatives along the diagonal.

diagonalSHO = {(1/((((1 + (((k x)/(σ^2)))))^2) + (((1 + (((k y)/(σ^2)))))^2) + (((1 + (((k z)/(σ^2)))))^2) - (((1 + (((k t)/(σ^2)))))^2))) + 
            (1/((((1 - (((k x)/(σ^2)))))^2) + (((1 - (((k y)/(σ^2)))))^2) + (((1 - (((k z)/(σ^2)))))^2) - (((1 + (((k t)/(σ^2)))))^2))), 
          (1/((((1 + (((k x)/(σ^2)))))^2) + (((1 + (((k y)/(σ^2)))))^2) + (((1 + (((k z)/(σ^2)))))^2) - (((1 + (((k t)/(σ^2)))))^2))) + 
            (1/((((1 + (((k x)/(σ^2)))))^2) + (((1 - (((k y)/(σ^2)))))^2) + (((1 - (((k z)/(σ^2)))))^2) - (((1 - (((k t)/(σ^2)))))^2))), 
          (1/((((1 + (((k x)/(σ^2)))))^2) + (((1 + (((k y)/(σ^2)))))^2) + (((1 + (((k z)/(σ^2)))))^2) - (((1 + (((k t)/(σ^2)))))^2))) + 
            (1/((((1 - (((k x)/(σ^2)))))^2) + (((1 + (((k y)/(σ^2)))))^2) + (((1 - (((k z)/(σ^2)))))^2) - (((1 - (((k t)/(σ^2)))))^2))), 
          (1/((((1 + (((k x)/(σ^2)))))^2) + (((1 + (((k y)/(σ^2)))))^2) + (((1 + (((k z)/(σ^2)))))^2) - (((1 + (((k t)/(σ^2)))))^2))) + 
            (1/((((1 - (((k x)/(σ^2)))))^2) + (((1 - (((k y)/(σ^2)))))^2) + (((1 + (((k z)/(σ^2)))))^2) - (((1 - (((k t)/(σ^2)))))^2)))};

test[diagonalSHO]

Take the derivative of the potential.

contraD[A_]:=(({{D[A[[1]], t], D[A[[2]], t], D[A[[3]], 
                t], D[A[[4]], t]}, {-c D[A[[1]], x], -c D[A[[2]], 
                  x], -c D[A[[3]], x], -c D[A[[4]], 
                  x]}, {-c D[A[[1]], y], -c D[A[[2]], 
                  y], -c D[A[[3]], y], -c D[A[[4]], 
                  y]}, {-c D[A[[1]], z], -c D[A[[2]], 
                  z], -c D[A[[3]], z], -c D[A[[4]], 
                  z]}}))

MatrixForm[Expand[contraD[diagonalSHO]] /. sublist]

Use this derivative of a potential in a Lorentz-like force law

contractVM[V_, M_] := {Sum[V[[i]] M[[1, i]], {i, 1, 4}], 
        Sum[V[[i]] M[[2, i]], {i, 1, 4}], 
        Sum[V[[i]] M[[3, i]], {i, 1, 4}], 
        Sum[V[[i]] M[[4, i]], {i, 1, 4}]}

covariant[
        A_] := {A[[1]], -A[[2]], -A[[3]], -A[[4]]}

v = {(Subscript[U, 0])[τ], (Subscript[U, 1])[τ], (Subscript[U, 2])[τ], 
          (Subscript[U, 3])[τ]};

Simplify[-m c contractVM[covariant[v], 
          Expand[contraD[diagonalSHO]] /. sublist]]

Replace (σ^2)with -(τ^2).  From here on, assume spherical symmetry, so (Subscript[U, R]) does the work of (Subscript[U, 1]),(Subscript[U, 2]), and (Subscript[U, 3]).

-1*{-((c k m (Subscript[U, 0])[τ])/(σ^2)), ((c k m (Subscript[U, 1])[τ])/(σ^2))} /. σ -> c τ

Apply the chain rule to the change in momentum, and assume that dm/dtau is zero.

{D[m[τ] (Subscript[U, 0])[τ, n], τ], 
        D[m[τ] (Subscript[U, R])[τ, n], τ]}/.m'[τ]->0

Solve the first order differential equations.

DSolve[{((k m (Subscript[U, 0])[τ])/(c (τ^2))) == 
          m D[(Subscript[U, 0])[τ], τ], -((k m (Subscript[U, R])[τ])/(c (τ^2))) == m D[(Subscript[U, R])[τ], τ]}, {(Subscript[U, 0])[τ], 
        (Subscript[U, R])[τ]}, τ]

Contract the relativistic velocity 4-vector solution.

{(E^(-(k/(c τ)))) C[
            2], (E^((k/(c τ)))) C[
            1]} . {(E^(-(k/(c τ)))) C[
            2], -(E^((k/(c τ)))) C[1]}

In flat spacetime, k->0

-(E^(((2 k)/(c τ)))) C[1]^2 + (E^(-((2 k)/(c τ)))) C[2]^2 == (c^2) /. 
      k -> 0

This can only be the case if the constants of integration are velocities in flat spacetime.

flatSpacetimeConstraint = { C[1] -> (dR/dtau), 
          C[2] -> c (dt/dtau)};

Apply this constraint to the contracted relativistic velocity 4-vector solution.

-(E^(((2 k)/(c τ)))) C[1]^2 + (E^(-((2 k)/(c τ)))) C[2]^2 == (c^2) /. 
      flatSpacetimeConstraint

Multiply both sides by ((dtau^2)/(c^2)).

Expand[((((c^2) (dt^2) (E^(-((2 k)/(c τ)))))/(dtau^2)) - (((dR^2) (E^(((2 k)/(c τ)))))/(dtau^2))) ((dtau^2)/(c^2))] == (c^2) ((dtau^2)/(c^2))

To apply to a weak gravitational system, substitute τ -> (R/c) and k -> (GM/(c^2)).

((dt^2) (E^(-((2 k)/(c τ)))) - (((dR^2) (E^(((2 k)/(c τ)))))/(c^2)) /. {τ -> 
              (R/c),  k -> ((G M)/(c^2))}) == (dtau^2)

This is the exponential metric.