Here’s a derivation of the minimum detectable magnetic field fluctuation we will be able to detect as a function of our measured angular deflection. There are some limitations to this derivation:

  1. I’ll assume that we’ve perfectly aligned the torsional zero with the external magnetic field first; i.e. at t=0, the magnetic dipole moment is precisely aligned with the torsional zero.
  2. I’ll assume that the magnetic field fluctuation is completely perpendicular to the residual initial magnetic field B_0.

Within the limits of these two assumptions, this derivation is exact. So, here we go—the figure below shows the geometry of our situation.

Initially, the magnetic dipole moment \vec{\mu} is aligned with the torsional zero, and a residual magnetic field B_0 exists; then a perpendicular field component \Delta B is applied. The dipole moment experiences a magnetic torque \vec{\mu}\times B which tries to align with the net field B, but this torque is thwarted in its effort by the restoring torque from the fiber -\kappa \delta. Hence, the equilibrium position is defined by

    \[\mu B \sin\theta = \kappa \delta\]

but, since \Delta B is perpendicular to B_0, we see that

    \[\sin\phi = \frac{\Delta B}{B},\]

and by inspection, \theta = \phi - \delta, so that we have

    \[\mu B \sin(\phi - \delta) = \kappa \delta,\]

and therefore

    \[\mu B \sin(\arcsin(\frac{\Delta B}{B}) - \delta ) = \kappa \delta.\]

By standard angle addition identities for the sine, we then have

    \[\mu B (\frac{\Delta B}{B} \cos\delta - \frac{B_0}{B} \sin\delta  ) = \kappa \delta\]

simplifying and solving for \Delta B, we have

    \[\Delta B = \frac{\frac{\kappa}{\mu} \delta + B_0 \sin\delta}{\cos\delta}\]

Clearly, for a given measured shift in orientation \delta, we obtain the smallest \Delta B when the residual field B_0 is as small as possible.

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Aluminum oxide layer

On 2011-Jul-16, in experimental, Materials, physics, by paul

According to this source: http://www.finishesltd.com/anodizing.html , the naturally occurring oxide layer on aluminum is about 500 nm thick. Much more than I would have guessed.

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20110716-092802.jpg

Our ace machinist, Steve Innes, sent me this picture of the main apparatus enclosure in the center of which will hang our torsion pendulum. Steve has yet to mount the 1 meter long “chimney” onto the cube, and has to make the interior optical lever platform which will allow us to calibrate and monitor the fluctuations of our pendulum.

Here’s a question. Since the pendulum will be surrounded by aluminum (save a small opening at the bottom through which the wiring will emerge, one would expect pretty good shielding. However, aluminum oxidizes readily, and I guess then we have a layer (how thick? 1 atom? 10’s of atoms? more?) of aluminum oxide with dielectric constant of between 9-12.
(I gave a range of values, because the proper way to express the dielectric constant with a crystalline structure is with a susceptibility tensor, and aluminum oxide has diagonal elements which are not all equal.)

Now, the question: Should I be concerned about the aluminum oxide layer? Should I paint the interior of the cube with something conductive—like graphite in alcohol?

I’ll have to think about this. I don’t know the answer, but if the pendulum itself was a perfect conductor (which is likely a good approximation) and was electrically connected to the body of the aluminum enclosure, I think I know the answer, or at least how to get there.

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