# Reducing General Relativity to Special Relativity in limiting case

I understand that general relativity is applicable to gravitational fields and special relativity is applicable to case when there is no gravity. But is there a derivation on how to reduce General Relativity to Special Relativity in limiting case, much like how General Relativity is reduced to Newtonian gravity in weak-gravity case?

Edit: By reducing I mean, how can we derive the Lorentz transformation from General Relativity under appropriate limits?

-
Were you looking for something beyond linear perturbations of the Minkowski metric ? – twistor59 Dec 30 '11 at 16:31
@twistor, I fail to see how your link connects the GR to SR – Graviton Dec 31 '11 at 0:20
If you neglect gravity by setting G to zero, the Minkowski metric is a solution of the Einstein equation. If you disallow gravity waves I believe it is the only solution. I don't think you need to do anything more complicated than this to reduce GR to SR. – Harry Johnston Dec 31 '11 at 0:56
@HarryJohnston, I've further clarified the question above. – Graviton Dec 31 '11 at 6:20

Given a set of coordinates in which the metric takes the standard Minkowski form

$ds^2 = dt^2 - dx^2 - dy^2 - dz^2$

you want to find another set of coordinates in which the metric also takes the same form

$ds^2 = d \bar{t}^2 - d \bar{x}^2 - d \bar{y}^2 - d \bar{z}^2$

Consider a linear boost along the x axis. We want to choose coordinates such that $\bar{x} = 0$ where $x - vt = 0$; the general solution is:

$\bar{x} = \gamma(x - vt)$

$\bar{t} = at - bx$

where $\gamma$, $a$, and $b$ are unknowns.

Then

$dt^2 - dx^2 = d\bar{t}^2 - d\bar{x}^2$

$= (a dt - b dx)^2 - \gamma^2 (dx - v dt)^2$

$= (a^2 - \gamma^2 v^2) dt^2 - (\gamma^2 - b^2) dx^2 + (\gamma^2 v - ab) dx dt$

so

$a^2 - \gamma^2 v^2 = 1$

$\gamma^2 - b^2 = 1$

$\gamma^2 v - ab = 0$

so

$a = \sqrt{1 + \gamma^2 v^2}$

$b = \sqrt{\gamma^2 - 1}$

so

$\gamma^2 v = \sqrt{\gamma^2 v^2 + 1} \sqrt{\gamma^2 - 1}$

$\implies \gamma^4 v^2 = (\gamma^2 v^2 + 1)(\gamma^2 - 1) = \gamma^4 v^2 + \gamma^2 - \gamma^2 v^2 - 1$

$\implies \gamma^2(1 - v^2) = 1$

$\implies \gamma = \sqrt{\frac{1}{1-v^2}}$

Calculating a and b (exercise left to the reader) gives $a = \gamma$ and $b = \gamma v$, so that

$\bar{t} = \gamma(t - vx)$

completing the Lorentz transformation as expected.

-

When you say "how can we derive the Lorentz transformation from General Relativity" this is really asking "how is the Minkowski metric a solution of the vacuum Einstein equation", because Special Relativity is just the geometry defined by the Minkowski metric.

If you take the Einstein equation and turn off gravity by setting G = 0 you get the vacuum Einstein equation $G_{ab}$ = 0. The Minkowski metric is a solution of this equation, but of course there are lots of others. From your question I'd guess you're hoping that the Einstein equation will simplify in the absence of gravity, and this will make it obvious how Special Relativity emerges. Sadly this isn't the case, because even in the absence of mass, or G set to zero, gravity waves are still allowed.

I don't think there is any way to simplify the Einstein equation to make the Minkowski metric the only solution. You can require that the first derivatives of the metric vanish, but this is really getting the flat space solution by requiring that space not be curved, which is a bit of a tautology. The problem is that SR the Minkowski metric is an assumption i.e. it's where you start from. In GR the Minkowski metric is just one among many solutions so there's nothing fundamental about it.

Have a look at http://en.wikipedia.org/wiki/Einstein_tensor if you want to play around with the Einstein tensor to try and extract the Minkowksi metric.

-

If you take a spacetime whose metric is a solution of Einstein's equations, then pick a point in that spacetime and introduce a locally inertial coordinate system there, the metric at that point can be brought into the Minkowski form. Effectively by working in a locally inertial frame, you are seeing what freely falling observers would see - you are removing the effects of gravity by falling along with it. To make this work fully, you also have to restrict yourself to an infinitesimally small region, otherwise, the tidal effects of gravity will become noticeable, and again the metric will deviate from the Minkowski values.

Of course the set of locally inertial coordinates is not unique, but the transformation from one set to another would have to preserve the Minkowski metric, i.e. they would be Lorentz transformations. It is in this limiting sense (inertial frame, infinitesimally small region) that GR "reduces" to SR, and the general coordinate transformations allowed by GR become restricted to the Lorentz transformations of SR.

-

A well-known technique is called post-Minkowskian approximation. It is a weak-field limit of GR and gravity shows up as a correction to Minkowski spacetime in power of Newton's constant G.

If you, in addition to powers of G, take power of (v/c)^2 expansion - that is, in addition to weak gravitational field, take slow motion, you will arrive at post-Newtonian regime.

-