Yahoo Answers is shutting down on May 4th, 2021 (Eastern Time) and beginning April 20th, 2021 (Eastern Time) the Yahoo Answers website will be in read-only mode. There will be no changes to other Yahoo properties or services, or your Yahoo account. You can find more information about the Yahoo Answers shutdown and how to download your data on this help page.
Trending News
How can LIGO work?
LIGO measures length change by comparing how long it takes for a LASER to bounce up and down two legs. But it’s not just space that’s squeezing, it’s space-time. Any variation in length is accompanied by an exactly equivalent variation in time. If the leg gets shorter, the beam travels slower and takes the exact same amount of time to cover the distance. It should read zero even in the presence of large gravity waves. What am I missing?
2 Answers
- nebLv 78 months ago
I’ve already answered this once for you - in normal English - and you simply refused to accept it. Let’s try some math then ...
The normal metric (non-gravitational wave) has the form:
ds² = gᵤᵥdxᵘdxᵛ
A weak field gravitational waVe is modeled as a perturbation of the metric:
ds² = (hᵤᵥ+gᵤᵥ)dxᵘdxᵛ
So, hᵤᵥ represents the gravitational wave imprinted on the background metric.
The solution of hᵤᵥ in something called the transverse traceless gauge is of the form
hᵤᵥ = AᵤᵥSin(blah, blah).
Aᵤᵥ= 0 for ALL components except for the Aₓₓ and Aᵧᵧ components. That means the deviation from the background metric is ZERO for the t and z components.
- DixonLv 78 months ago
You are missing that it isn't classical space and time, it is relativistic spacetime. Gravity affects the geometry of spacetime, with the result that the space component and the time component are both different with respect to the other leg.