Special Relativity vs General Relativity: T_{\mu\nu} = 0

quasar987
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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?
 
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quasar987 said:
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?


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.
 
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".
 
Are you suggesting that, for GR, vacuum and asymptotic flatness imply that the Riemann curvature is zero everywhere?
 
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quasar987 said:
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?

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

No, not really.
 
quasar987 said:
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?
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
 
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