The Einstein field equations do include pressure terms and shear-stress terms.
It is the stress energy tensor on the RHS that contains these terms; that is the source term in the field equations. The LHS contains curvature only.
The Riemannian metric describes the state of space.
No it doesn't. The metric tensor ( I can only assume that is what you mean ) is defined as the inner product of basis vectors on our manifold :
$$\displaystyle{g_{\mu \nu }=\left \langle \partial _{\mu }|\partial _{\nu } \right \rangle}$$
Since the set of basis vectors in our case has four elements, the indices in the above definition run from 0...3, and we have a metric signature of (+,-,-,-), we are dealing with
space-time. And it does
not describe any "states" either, it merely defines the notions of angles as well as line/surface/volume elements on our manifold via the above inner product. This is all elementary differential geometry.
And it features pressure and shear stress.
You see, this is what happens if you rely on textual quotes without any understanding of the maths behind it. The above is
wrong.
The geometric degrees of freedom of our space-time manifold aren't the result of the metric tensor, but of the
connection used. In GR this is the Levi-Civita connection, which is an affine connection with the property that torsion vanishes everywhere on it; the only degree of freedom it permits is curvature. This connection is unique, and the direct result of the
fundamental theorem of Riemann geometry.
Now, I am telling you that what you write above is
wrong; if you are unwilling to accept that fact then I will insist on a formal mathematical derivation to show that the metric tensor on a pseudo-Riemannian (3+1) manifold "features" pressure and shear stress. In fact you will need to define those terms first in the context of a metric tensor, because so far as differential geometry is concerned these concepts are meaningless. There are no such things as "pressure" or "stress" on a connection, there is curvature and torsion
only. I will show you now that the Levi-Civita connection, which is the one used in GR, admits only curvature.
Start with the metric tensor on a general Riemann manifold
$$\displaystyle{g_{\mu \nu }=\left \langle \partial _{\mu }|\partial _{\nu } \right \rangle}$$
A connection on such a manifold is defined via the relation
$$\displaystyle{\nabla _{\partial _{\mu }}\partial _{\nu }=\sum_{k}\Gamma _{\mu \nu }^{k}\partial _{k}}$$
In order for torsion to vanish we must have
$$\displaystyle{\nabla _{\partial _{\mu }}\partial _{\nu }=\nabla _{\partial _{\nu }}\partial _{\mu }}$$
or in other words
$$\displaystyle{\Gamma _{\mu \nu }^{k}=\Gamma _{\nu \mu }^{k}}$$
Now insert back into the definition of the metric tensor
$$\displaystyle{\left \langle \nabla _{\partial _{\mu }}\partial _{\nu }|\partial _{\lambda } \right \rangle=\Gamma _{\mu \nu }^{k}g_{k\lambda }=\Gamma {_{\nu \mu }^{k}}g_{k\lambda }}$$
via permutation of indices. This means that our connection leaves the metric tensor unchanged, thereby fulfulling the vanishing torsion condition above. Since the Riemann curvature tensor is a function of this connection and its derivatives, and therefore inherits its symmetries, we get a manifold that has non-zero curvature, but everywhere vanishing torsion.
I invite you to verify this yourself for any given, specific metric.
Your assertion that a pseudo-Riemannian manifold is only "space" and has no curvature is just completely ludicrous, and even more ludicrous is that you are trying to sell your own laughable ignorance as facts.
My primary reference for all of the above is "
Differential Forms and Connections" by R.W.R. Darling, where this is all made mathematically precise and rigorous.
A scondary source then is Fecko's "
Differential Geometry and Lie Groups for Physicists", which is an excellent book and a standard text for any undergrad physics student. So don't you say I have nothing to back it up with - I can provide references for every single bit of maths and physics I presented above, if that is necessary.
I urge you once again not to comment on things which you do not understand and know nothing about, like differential geometry, making your claims look like facts. You are misleading other members or casual readers who might come here looking for real information; this behaviour of yours is what got you confined to the "Pseudo" section on other forums. What you are asserting about stress and pressure in metric tensors, and about "space" not having curvature, is plain and simply meaningless and wrong. The entire model of GR is based on mathematics which are well defined and well understood; in fact these maths were in existence long before Einstein came along, he simply made use of them. You cannot just go and reinterpret things as you please, inserting your own opinions on how things should be according to yourself. GR models only space-time, not space and not time, and that space-time has curvature as the non-vanishing curvature tensor clearly shows us. Claiming anything else about GR is wrong.
