Better Hubble Constant through Parallax?

[Subluminal]

in some methods for measuring H_o is now only a few
percent; with systematic errors, the total uncertainty is approaching
±10%. Hence, the historical factor-of-two uncertainty
in the value of the H_o is now behind us.

The above is taken from a paper by Wendy Freedman (http://www.pnas.org/content/96/20/11063.full.pdf)

In 1999, she wrote that the Space Interferometry Mission (SIM), a project primarily focused on mapping the Milky Way and finding other Earth-like worlds, would provide parallax data that could improve estimating the extragalactic distance scale and H_o. Unfortunately, the mission languished in cubicle world for years and was canceled last year.

I see why better parallax could reveal bigger triangles as years pass (as the sun follows its peculiar path relative to other galaxies) to check for error, but how does this lead to a better H_o confidence interval?

Have any other missions since provided the kind of extragalactic parallax measurements that was expected from the SIM? Has the estimate of H_o improved?

My understanding of spacetime curvature and the standard model is at a fair fraction of a Wikipedia unit (i.e. not very good), I can grasp the idea that on a cosmological scale in Cartesian coordinates, a right triangle would not quite obey the Pythagorean theorem, the hypotenuse would be longer (or shorter?) than expected due to this curvature. Dumbing down to this ballpark would be appreciated, thanks in advance. If not possible then go ahead, hit me.

 

[zhermes]

Great questions Subluminal.

I see why better parallax could reveal bigger triangles as years pass (as the sun follows its peculiar path relative to other galaxies) to check for error, but how does this lead to a better H_o confidence interval?

The Hubble `constant' relates the velocity of objects as a function of distance away. Thus to measure the constant, one must measure accurately both the velocity (usually with the doppler effect on emission/absorption lines), and the distance---by, e.g., parallax.

...I can grasp the idea that on a cosmological scale in Cartesian coordinates, a right triangle would not quite obey the Pythagorean theorem, the hypotenuse would be longer (or shorter?) than expected due to this curvature. Dumbing down to this ballpark would be appreciated...

A triangle on the surface of a sphere can have a sum of angles larger than 180 degrees (positive curvature); a triangle on a saddle can have a sum of angles smaller than 180 degrees (negative curvature); or a triangle in standard cartesian coordinates can have angles that add to exactly 180 degrees (0-curvature; minkowski spacetime / euclidean geometry). This is analogous to the overall geometry of the universe, and doesn't relate to the Hubble constant per se, but instead to the rate of change of the Hubble constant---which in tern relates to 'dark energy' and the 'cosmological constant'.