Einstein and Gravitation — Question 10

Question 10

A differential detector rejects equal arm changes but retains a tidal pattern

An ideal gravitational-wave detector has perpendicular arms of length $L=4000\,\mathrm m$. For the chosen polarization and orientation, take $\Delta L_x=hL/2$ and $\Delta L_y=-hL/2$, with $h=10^{-21}$ at the instant considered. Define signal $D=(\Delta L_x-\Delta L_y)/L$.

1. Find both arm changes, their difference and $D$.

2. If a disturbance adds an equal change $b$ to both arms, determine its effect on $D$ and explain the role of the opposing changes.

Question 10 - Solution

A differential detector rejects equal arm changes but retains a tidal pattern

(a) Derivation and calculation

Apply the stated response convention: $$\boxed{\Delta L_x=\frac{hL}{2}=\frac{(10^{-21})(4000)}2 =2\times10^{-18}\ \mathrm m},$$ $$\boxed{\Delta L_y=-2\times10^{-18}\ \mathrm m}.$$ Therefore $$\Delta L_x-\Delta L_y=hL=4\times10^{-18}\ \mathrm m,$$ $$\boxed{D=\frac{\Delta L_x-\Delta L_y}{L}=h=10^{-21}}.$$

(b) Result and interpretation

With an identical added length change $b$, $$D'=\frac{(\Delta L_x+b)-(\Delta L_y+b)}L =\frac{\Delta L_x-\Delta L_y}L=\boxed{D}.$$ Opposite tidal changes add in the difference, while a common change cancels. Unequal disturbances would instead leave $$D'-D=\frac{b_x-b_y}L.$$

Physical check

$D$ is dimensionless; the differential response is twice either individual fractional arm change for the supplied orientation.