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  • 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 $ds^2$?

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Related: physics.stackexchange.com/q/29841/2451 –  Qmechanic Jun 29 '12 at 10:03
    
@Qmechanic Dear Newyork-City, if u don't like my changes pls roll back or use the parts u like. thanks for the edit –  user10171 Jun 29 '12 at 10:25

3 Answers 3

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)

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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.

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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

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by this is not possible you means this is impossible!? but actually einstein did it!. –  user10171 Jun 29 '12 at 8:53
    
Einstein did not formulate GR by starting with SR and enhancing it. In effect he threw away the SR work and started from scratch. Obviously he must have had in mind that his new theory must have SR as a low mass limit, but he started from completely different principles to SR. See rafimoor.com/english/GRE1.htm for a good popular science level description. –  John Rennie Jun 29 '12 at 9:18
    
@JohnRennie: How is this comment supposed to be read? GR has both local SR and asymptotic SR (for an asymptotically flat background, which Einstein considered often). He kept the local SR and kept the negative sign metric, so I can't see how you can say he started from scratch. –  Ron Maimon Jun 30 '12 at 2:44
    
@JohnRennie: Why can't you say "GR is SR plus a spin-2 fluctuating graviton field"? –  Ron Maimon Jun 30 '12 at 2:45
    
Because I think the key to understanding GR is understanding coordinate independence. This is true of SR but not emphasised in most courses. You certainly need a thorough grasp of SR before trying to learn GR, but I don't think viewing GR as an extension of SR will help you understand it. I think you need to approach GR with a new mindset. Your mileage may vary of course. –  John Rennie Jun 30 '12 at 6:26

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