matthew.evans@LIGO.ORG - posted 18:11, Monday 13 July 2015 - last comment - 20:21, Tuesday 14 July 2015(19607)
Roll mode coupling to DARM
Matt, Sheila, Eli
At some point today the bounce mode on EX got excited enough that we could see it in the PUM OSEMs as pitch motion. The RMS of the observed "pitch" was about 3 nrad, and the line in DARM was about 1e-13 m. Assuming that OSEM misalignent is providing the roll to observed pitch motion, and that this misalignment is of order 1 degree, the estimated roll motion was about 3e-7 rad.
This gives an order of magnitude estimate of the Roll to DARM coupling of 3e-7 m / rad.
Assuming a 10cm lever arm, this give a dimentionless coupling of 3e-6. Compared to the bounce to DARM coupling, which is order 1e-3, the roll coupling is tiny, which means that the roll motion is HUGE (since they both look about the same in DARM).
My 24 hours have passed, but the first sentence should read "At some point today the roll mode on EX..."
Suppose that the beam is at (X, Y)=R(cos(theta), sin(theta)) on the mirror where R=0 is the center of roll rotation and theta=0 is the horizontal line crossing the center. Though the COG is somewhat lower than the mirror center due to wedge, R should be more or less equal to the radial distance of the beam from the center of the mirror.
Mirror thickness at this position is
T(R, theta) ~ -R*sin(theta)*w + T0
where T0 is the thickness at the center and w is the wedge in radians that is about 0.08deg=1.4mrad for all ITMs and ETMs.
Roll changes the thickness by adding some small angle d_theta to theta: dT=-R*cos(theta)*w*d_theta=-X*w*d_theta.
When the rolling plane is in the middle of the front and the back surface, the light see the half of the total thickness change, so the roll-to-length coupling coefficient should be
length/roll ~ |dT/d_theta /2| = X*w/2
= (X/5mm) * 3.5E-6 [m/rad].
For Matt's estimate of 3E-7 m/rad to hold true, the horizontal centering should be 0.5mm or so, which is pretty good but not outrageously so.
What this probably means is that Matt's estimate about the roll angle was reasonable, as in it cannot be off by that much. A factor of something, not orders of magnitude.
[edit on Jul 15] However, if the roll plane is parallel to the local gravity, the above doesn't hold true.
In this case, w/2 is replaced by the angle between the local gravity and LIGO global vertical: 8urad for LHO EX, 639urad for EY, -619urad for IX and 12urad for IY (https://alog.ligo-wa.caltech.edu/aLOG/index.php?callRep=14876):
length/roll(EX)~ X*8urad = (X/5mm) * 4e-8 [m/rad],
length/roll(EY)~ X*619urad = (X/5mm) * 3E-6 [m/rad].
For EX and IY, that's two orders of magnitude smaller than what I showed yesterday, though EY and IX it didn't change.
It seems that we need a suspension model to find out the actual rotation plane.