# SR from GR

1. Jul 8, 2007

### quasar987

Does SR appear as the special case of GR when $T_{\mu\nu}=0$ in which case the solution of Einstein's equation is the Minkowski metric?

And what are the Ricci tensors and scalar curvature like in the case $T_{\mu\nu}=0$?

2. Jul 8, 2007

### robphy

Using the field equations (with zero cosmological constant), a zero stress tensor yields a zero ricci tensor and ricci scalar. You'll need more than requiring "vacuum" (zero stress tensor) to get SR.

You'll need a zero Weyl Tensor as well.
Strictly speaking, to get "SR", you'll need the right manifold, $$R^4$$, to start with.

3. Jul 8, 2007

### pervect

Staff Emeritus
The analogy to Maxwell's equations might be helpful here to clarify some of the reasons why a zero stress tensor doesn't guarantee a Minkowski (SR) metric.

Consider asking "Suppose you have no charges - are the E and B fields zero everywhere?". The answer is no, you could have electromagnetic radiation. Usually one specifies boundary conditions as well as a charge distribution to get a unique solution to Maxwell's equations. For Maxwell's equations, having E and B zero at infinity is a standard boundary condition, for GR the analogous boundary condition would be "asymptotic flatness".

4. Jul 8, 2007

### robphy

Are you suggesting that, for GR, vacuum and asymptotic flatness imply that the Riemann curvature is zero everywhere?

Last edited by a moderator: Jul 9, 2007
5. Jul 8, 2007

### country boy

SR is the special case of GR where the metric components $g_{\mu\nu}$ are all constants. This corresponds to a zero Riemann tensor, which is more restrictive than setting the Ricci tensor equal to zero. $T_{\mu\nu}=0$, which gives a zero Ricci tensor, corresponds to empty space. Empty space can be curved, of course.

6. Jul 9, 2007

### pervect

Staff Emeritus
No, not really.

7. Jul 9, 2007

### pmb_phy

I've used $T_{\mu\nu}$ in SR myself. In fact I learned more about it in an SR text than any GR text that I have. In my cases it was used to analyze things like the non-proportionality of energy density and inertial mass density. As far as $T_{\mu\nu}= 0$, this can lead to the Minkowski metric once you transform to an inertial frame. Otherwise the components of the metric (i.e. the set of ten gravitational potentials g_uv) may net be constant in space, i.e. there could be gravitational forces/accelerations still present.

Pete

Last edited: Jul 9, 2007