How to connect Einstein's Special Relativity (SR) with General Relativity (GR)?

Question

• How Einstein's SR becomes GR?

$$ds^2=dr^2-c^2dt^2,$$

$$ds^2=g_{\mu\nu}dx^{\mu}dx^{\nu}.$$

• When the $s$ is constant $ds^2=0$, isn't it true?

• How to connect Einstein's SR with GR?

• What is the GR-differential $ds^2$?

Answer 1

The connection between general relativity and special relativity is fundamental. The principle of equivalence says that in a locally free falling system of coordinates, the effects of gravity are eliminated. This means that in a local coordinate system in free fall the metric tensor is that of special relativity. So very close to an event (i.e. in a finite neighborhood) the equations of general relativity are the same as sr. So the local structure of spacetime is minkowskian. GR is about the way you assemble sr local charts to form the riemann curved global structure of space time. That is why there exists local lorentz’s transformations, that act locally in the spacetime manifold. The GR field equations relate the local structure (sr near an event) to the global structure by means of the affine connection and spin connection. GR's metric tensor is the minkowski tensor locally. (Strong equivalence principle)

Answer 2

What is the GR-Differential $ds2$?

In SR, the interval between two events can be found by taking finite differences:

$s^2 = c^2 \Delta t^2 - \Delta x^2 - \Delta y^2 - \Delta z^2$

But, we also have:

$s^2 = c^2 \tau^2$ for $s^2 > 0$

$\tau$ is the proper time, i.e., the elapsed time along a straight (non-accelerated) worldline connecting the events.

However, if you want to find the elapsed time along a curved (accelerated) worldline between events, you must integrate the differential interval along the worldline:

$c \tau = \int_P^Q ds$

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

In GR, we not only must use the differential interval (line element), we must also consider that the line element varies from event to event:

$ds^2 = g_{00}(dx^0)^2 + g_{11}(dx^1)^2 + g_{22}(dx^2)^2 + g_{33}(dx^3)^2 + 2g_{01}dx^0 dx^1 + 2g_{02}dx^0 dx^2 + 2g_{03}dx^0 dx^3 + 2g_{12}dx^1 dx^2 + 2g_{13}dx^1 dx^3 + 2g_{23}dx^2 dx^3$

Which, using the summation convention, is much more easily written as:

$ds^2 = g_{\mu\nu}dx^{\mu}dx^{\nu}$

The $g_{\mu\nu}$ are, in general, functions of the spacetime coordinates $x^{\mu}$

When the s is constant $ds^2=0$, isn't it true

You're thinking of derivative here rather than differential. In this context, we're interested in finding the spacetime equivalent of length in space along a path. Think of arc length instead.

Answer 3

It isn't possible to explain GR as an expansion of SR, or at least not in any useful way. That's because the fundamental principles of GR are different to SR.

However what you can do is show that SR is a subset of GR i.e. that GR reduces to SR when energy density is low. This is explained in Reducing General Relativity to Special Relativity in limiting case