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?
neb
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.
Dixon
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.