I have an amplifier for a seismometer (for earthquake detection) and the attenuator dial is labeled in 6Db./Octave steps.

The seismometer has a moving coil transducer which generates a voltage proportional to velocity which is proportional to the frequency of the ground motion. The purpose of the attenuator on a seismometer amp is to filter the ground motion by its frequency content while keeping recording trace at a constant amplitude for a given amount of displacement.

Therefore with any type of generating transducer, the voltage will change by a factor of 2 when the frequency is changed by a factor of 2. For example, if the frequency of the ground motion is reduced by 1/2, but the displacement is the same, then the transducer voltage will also change by 1/2 and this represents the 6 Db./Octave ratio on the attenuator. However it seems this relationship between frequency and voltage would represent a change of 3 Db./Octave

However, since Db. correctly represents the change in power rather than voltage, I'm wondering how the 6 Db./Octave conversion is made.

 

decibel or dB is a ratio of power as well as voltage.

½ power ratio is -3dB

power increases as the square of the voltage, i.e.
power ∝ voltage squared
log(power ratio) = log (voltage ratio squared) = 2 x log (voltage ratio)
hence,
½ voltage ratio is -6dB

If you change the frequency by a factor of 2 (i.e. one octave), the voltage will change by a factor of 2 = -6dB/octave

 

For a 6dB/octave slope, the voltage changes by a factor of 2 and the power changes by a factor of 4 for each octave change in frequency.