Newtonian perturbations to Minkowski

  • Context: Graduate 
  • Thread starter Thread starter smallphi
  • Start date Start date
  • Tags Tags
    Minkowski Newtonian
Click For Summary
SUMMARY

The discussion focuses on the Newtonian gravity limit of General Relativity (GR) as represented by a perturbation to the Minkowski metric. The geodesic equations derived from this metric accurately reproduce the Newtonian equations for acceleration in a gravitational potential, described by the Poisson equation. The analysis reveals that the Einstein tensor components must be adjusted, indicating that the coefficient in the Einstein equation should be four times 8πG instead of the standard value. This discrepancy suggests that the conventional Newtonian metric may not adequately describe slowly moving dust or that a more refined treatment, possibly involving linearized gravity, is necessary.

PREREQUISITES
  • Understanding of General Relativity (GR) principles
  • Familiarity with Minkowski metric and perturbation theory
  • Knowledge of the Poisson equation in gravitational contexts
  • Experience with tensor calculus and Einstein equations
NEXT STEPS
  • Study linearized gravity and its implications in GR
  • Explore the derivation of the Einstein tensor components in perturbed metrics
  • Investigate the role of the energy momentum tensor in GR
  • Examine the conditions under which gravitational influences propagate at the speed of light
USEFUL FOR

The discussion is beneficial for theoretical physicists, researchers in gravitational physics, and students studying General Relativity who seek to deepen their understanding of the relationship between Newtonian gravity and GR.

smallphi
Messages
436
Reaction score
2
The Newtonian gravity limit of GR is represented by a perturbation to Minkowski metric:

ds^2 = - (1+ 2\Phi(x,y,z,t)) dt^2 + (1-2 \Phi(x,y,z,t))(dx^2 + dy^2 + dz^2)

The geodesic equations of this metric correctly reproduce the Newtonian equations for acceleration of a test particle in the gravitational potential Phi:

d^2 x / dt^2 = - \nabla \Phi

The Poisson equation that describes how matter generates the gravitational potential :

\nabla^2 \Phi(x,y,z,t) = 4 \pi \rho(x,y,z,t)

should be derivable from Einstein equations. In Newtonian limit of Einstein equations, we describe the gravitating matter with energy momentum tensor of dust

T_{\mu \nu} = \rho U_\mu U_\nu,

where rho is the matter density.

We have to assume the gravitating matter is moving much slower than light in the perturbed Minkowski coordinates above because only under such conditions the above Newtonian equations apply - they don't show GR or SR effects and behave as if signals propagate at infinite speed. If the matter is moving slowly, the U_0 components of the 4-velocity of gravitating matter will be ~1 and the other components suppressed by v/c, v = coordinate speed of gravitating matter. Correspondingly, the only significant component of the energy momentum tensor will be the T_00 ~ rho and the other components will be suppressed by a factor of (v/c)^2. The same applies to the components of Ricci and Einstein tensors which the Einstein equations set proportional to the energy momentum tensor.

So we expect the above Minkowski perturbed metric to produce Ricci and Einstein tensors with components supressed by (v/c)^2 compared to the 00 component. Let's see if that is really the case. I plugged the above perturbed metric in Mathematica and neglecting factors of second and higher order in phi<<1 I got:

R_{00} = 3 \partial_t \partial_t \Phi + \nabla^2 \Phi
R_{11} = R_{22} = R_{33} = - \partial_t \partial_t \Phi + \nabla^2 \Phi
R_{0i} = 2 \partial_t \partial_i \Phi

and the other components were zero or of second order in phi. The usual saying, neglect the time derivatives because gravitating matter is moving slowly, doesn't work because then R_11= R_22 = R_33 components won't be 'suppressed' but equal to the R_00 component. Requiring R_11= R_22 = R_33 = 0 imposes a wave equation condition on the gravitational potential - gravitational influence propagates at the speed of light. That is OK cause that is what we expect anyways. On the other hand, if we allow for the nonzero double time derivative, we get

R_{00} = 4 \nabla^2 \Phi

instead of

R_{00} = \nabla^2 \Phi

That means the coefficient in Einstein equation which is fixed by the Newtonian limit must be 4 times 8piG instead of 8piG.

I am also not sure how the condition R_0i = 0 should be interpreted ??

The above shows there is something improper in the Newtonian metric above - it either doesn't describe slowly moving dust (time derivatives neglected but R_11 not suppressed compared to R_00) or the coefficient in Einstein equation has to be adjusted to 4 times 8piG (time derivatives not neglected). Anyone knows what the metric must be in more proper treatment of Newtonian gravity, one that takes into consideration finite speed of gravitational influences? Is that 'linearized gravity' ?
 
Last edited:
Physics news on Phys.org
I got it. The Einstein tensor components for the above metric are

G_{00} = 2 \nabla^2 \Phi

G_{11} = G_{22} = G_{33} = 2 \partial_t \partial_t \Phi

G_{0i} = 2 \partial_t \partial_i \Phi

the other components are either zero or of second order in Phi. So dropping the time derivatives really shows the 00 component dominates and the others are zero. That was expected because the Einstein tensor is proportional to the energy momentum tensor via the Einstein equations.

The Ricci tensor 00 component is the same as the Einstein 00 after dropping the time derivative. On the other hand Ricci 11, 22, and 33 are not suppressed but equal to Ricci 00, and that is OK because Ricci doesn't have to follow the structure of the energy momentum tensor, only the Einstein tensor has to. That was the origin of the whole confusion.
 
Last edited:
My calculations with GRtensor agree. The exact expression I get for the time component of the Einstein tensor, G_{tt} is:

<br /> \frac{1}{1 -2\Phi^3} \left[ (2-8 \Phi^2)(\frac{\partial^2 \Phi}{\partial x^2}+\frac{\partial^2 \Phi}{\partial y^2}+\frac{\partial^2 \Phi}{\partial z^2} ) + (3 + 6\Phi) \left( (\frac{\partial \Phi}{\partial x})^2 + (\frac{\partial \Phi}{\partial y})^2 + (\frac{\partial \Phi}{\partial z})^2 \right) + (3 - 6 \Phi) (\frac{\partial \Phi}{\partial t})^2 \right]<br />

(I didn't look at the other components very closely).
 
Last edited:

Similar threads

  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 16 ·
Replies
16
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 8 ·
Replies
8
Views
5K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 12 ·
Replies
12
Views
3K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 16 ·
Replies
16
Views
2K
  • · Replies 37 ·
2
Replies
37
Views
5K
  • · Replies 7 ·
Replies
7
Views
3K