UPDATE April 2, 2024: We forgot the factor g²¹ in the formulae.
We have to figure out how to use orthogonal coordinates.
----
We calculate separately the Ricci curvature due to the distorted metric of time, and the Ricci curvature due to the distorted radial metric. Since the Schwarzschild metric has Ricci curvature zero outside the central mass M, we expect that the sum of these separate curvatures is zero to every direction.
Let us have any configuration of masses. We only distort the metric of time according to the newtonian potential, that is, make clocks tick slower so that they explain the redshift. Generally, the metric of time makes Ricci curvature non-zero outside matter.
A difficult problem in general relativity is to find a spatial metric which resets Ricci curvature to zero. We hope that understanding the Schwarzschild metric would help us.
Schwarzschild metric
y
^
| • (x, y, z)
| r
|
●------------------> x
Axis z points out
of the screen
The mass M is at the origin. We assume weak fields. The metric signature is (- + + +).
Newtonian gravity satisfies
dg₀₀ / dr = -2 G M / c² * 1 / r²,
and the radial spatial metric
grr = 1 + 2 G M / c² * 1 / r,
dgrr / dr = -2 G M / c² * 1 / r².
We set the coefficient 2 G M / c² to 1.
We assume that r points approximately to the x direction. Let us write
ds² = grr dr² + dn₂² + dn₃²,
where dn₂ is normal to r, points approximately to the y direction and changes the "longitude" of r; dn₃ points approximately to the z direction and changes the "latitude" of r.
There
dr = (1 - 1/2 * (y² + z²) / x²) dx
+ y / x * dy
+ z / x * dz,
dn₂ = -y / x * dx
+ (1 - 1/2 * y² / x²) dy,
dn₃ = -z / x * dx
+ (1 - 1/2 * z² / x²) dz,
- y / x * dy * z / x.
We drop very small terms and write the spatial metric in terms of cartesian coordinates:
ds² = grr * dx²
+ (1 - grr) * (y² + z²) / x² * dx²
+ 2 (grr - 1) * y / x * dx dy
+ 2 (grr - 1) * z / x * dx dz
+ 2 (grr - 1) * y z / x² * dy dz
+ dy²
+ (grr - 1) y² / x² * dy²
+ dz²
+ (grr - 1) z² / x² * dz²,
where grr - 1 = 1 / r.
Spatial metric set flat: Ricci curvatures due to the metric of time
Let us first set the spatial metric flat. We calculate the Ricci curvatures which are due to the metric of time.
Γ⁰₀₀ = 0,
Γ¹₀₀ = 1/2 * -dg₀₀ / dx
= 1/2 * 1 / x²
= Γ⁰₁₀ = Γ⁰₀₁,
Γ²₀₀ = 1/2 * -dg₀₀ / dy
= 1/2 * y / x³,
Γ³₀₀ = 1/2 * z / x³,
R₀₀ = dΓ¹₀₀ / dx + dΓ²₀₀ / dy + dΓ³₀₀ / dz
= -1 / x³ + 1/2 * 1 / x³ + 1/2 * 1 / x³
= 0,
R₁₁ = -dΓ⁰₁₀ / dx
= -1 / x³,
Γ⁰₂₀ = 1/2 * -1 * dg₀₀ / dy
= 1/2 * y / x * 1 / x²
= 1/2 * y / x³,
R₂₂ = -Γ⁰₂₀ / dy
= -1/2 * 1 / x³
= R₃₃.
The metric of time set to -1: Ricci curvatures due to the spatial metric
Let us then set the metric of time -1. We want to calculate the Ricci curvatures which are due to the spatial Schwarzschild metric.
We can drop terms which are contain a product where y or z occur three or more times.
R₀₀ = dΓ¹₀₀ / dx + dΓ²₀₀ / dy + dΓ³₀₀ / dz
= 0,
Γ²₁₁ = 1/2 * 1 / g₂₁ * dg₁₁ / dx
+ dg₁₂ / dx
- 1/2 dg₁₁ / dy
= 1/2 * x / ((grr - 1) y) * -1 / x²
+ d(1 / r * y / x) / dx
- 1/2 d(grr - 1 / r * (y² + z²) / x²) / dy
= -1/2 * 1 / y
- 2 y / x³
+ 1/2 * y / x³
- 1/2 * -2 * y / x³
= -1/2 * y / x³,
Γ²₁₂ = 1/2 dg₂₂ / dx
= -3/2 * y² / x⁴.
The values for the third coordinate z are symmetric with y.
R₁₁ = dΓ²₁₁ / dy + dΓ³₁₁ / dz
- dΓ²₁₂ / dx - dΓ³₁₃ / dx
= -1 / x³.
Let us calculate R₂₂:
Γ¹₂₂ = dg₂₁ / dy - 1/2 dg₂₂ / dx
= d(1 / r * y / x) / dy
- 1/2 d(1 + 1 / r * y² / x²) / dx
= 1 / x²
- 1/2 * -3 y² / x⁴,
Γ³₂₂ = dg₂₃ / dy - 1/2 dg₂₂ / dz
= d(1 / r * y z / x²) / dy
- 1/2 d(1 + 1 / r * y² / x²) / dz
= z / x³,
Γ⁰₂₀ = -1/2 dg₀₀ / dx
= 0,
Γ¹₂₁ = 1/2 dg₁₁ / dy
= 1/2 d(grr - 1 / r * (y² + z²) / x²) / dy
= -1/2 y / x³
+ 1/2 * -2 y / x³
= -3/2 y / x³,
Γ³₂₃ = 1/2 dg₃₃ / dy
= 0,
R₂₂ = dΓ¹₂₂ / dx + dΓ³₂₂ / dz
- dΓ⁰₂₀ / dy - dΓ¹₂₁ / dy - dΓ³₂₃ / dy
= -2 / x³ + 1 / x³
+ 3/2 * 1 / x³
= 1/2 * 1 / x³
= R₃₃.
We showed that the spatial metric, indeed, resets the Ricci curvature of the temporal metric to zero.
Conclusions
We can use the calculations above to get an intuitive understanding of what "happens" in a metric, and how to make Ricci curvature zero.
We are particularly interested if we can repeat the trick for the field of a long lightweight cylinder. Close to the cylinder, the newtonian potential is something like
m c² (ε ln(r) - C),
where m is a test mass, ε > 0 is small, r is the distance from the cylinder, and C is a small positive constant. The potential corresponds to the following metric of time:
g₀₀ = -1 - 2 ε ln(r) + 2 C
Can we make Ricci curvature zero by stretching the radial metric by a factor
1 - ε ln(r) + C,
that is,
grr = 1 - 2 ε ln(r) + 2 C ?
No comments:
Post a Comment