Physiological amplifiers seem to often (always?) have both ground (G) and reference (R) electrodes along with one or more active (A) electrodes. Typical voltage at an active site is determined AFAIK as [(A-G) - (R-G)] = A-R. I haven't managed to find out why the ground electrode is needed at all and why the amplifier cannot simply do the A-R calculation directly without the need for the ground electrode? Is it because when there are multiple active electrodes that A1-R and A2-R couldn't be done without them interfering with each other?
Differential amplifiers such as instrumentation amplifiers, with or without unity gain buffers, have a certain working range for the common-mode voltage.
In general they will stop working if either input voltage gets too close to the power supply rails. For example, in the case of the INA117, both input voltages must be within the range of V+-4V to V-+4V, so if you have +/-10V supplies, each input must be within the range of +/-6V. There are also internal nodes in instrumentation amplifiers which can saturate. Here are some curves for the INA118 which show the permissible input voltages:
The reference ground makes sure that the inputs remain within the common mode range for proper operation and minimizes the common mode AC voltage. The latter will show up in the output even if the inputs remain within the common mode range, attenuated by the common mode rejection ratio (CMRR).
After looking at some schematics: (Google: "physiological amplifier circuit" and select Images tab), this is one example:
Many amplifiers do use A-R directly similar to instrumentation amplifiers. The ground connection might be needed for:
shielding around the A and R signals to prevent coupling to external noise like other equipment.
safety: since the amplifier must be mains isolated (to prevent electrocuting the patient) any charge build-up cannot escape without a ground connection. If a charge were present it could escape via the patient, not harmful but still unpleasant.
safety: if the mains isolation fails on the amplifier then the ground connection prevents the mains voltage from reaching the patient.
The reason is actually because of the buffers.
A simple differential amplifier does not care about the reference point of the input. Without any other common reference, all it cares about is the difference between the positive and negative inputs. The difference itself of course has to be within the tolerable range of the inputs.
Notice in the circuit below, the output is correctly -1V.
Why is that?
Generically, the inputs stage of all op-amps basically looks something like this...
That is the input is a balanced circuit. That balance is disturbed by any applied voltage/current difference between the two inputs. The reference point for those inputs is, in reality, the junction between those two emitters. As such the applied voltage need not be referenced to ground as long as the two inputs are referenced to each other. You should be able to see, how attaching a battery between the plus and minus pins in this circuit should be acceptable.
The buffer on the other hand needs a reference.
Obviously the circuit below, attaching a voltage with no reference, to a unity buffer makes no sense.
But how about this..
In fact what you end up with here will depend on the op-amp. Some may get you what you expect +0.5V on one -0.5V on the other, some may rail out to either or both sides. All will be very sensitive to any noise.
So what does all that mean.
It means a standard differential amplifier measures the difference voltage \$Vr\$. An instrumentation amplifier measures \$V_a - V_b\$. Mathematically that may be the same, electrically, it is not.
Touch the tip of a 10MOhm scope probe, and you'll likely see 100 or 200 volts at 50 or 60Hz; you are the antenna, the 2nd plate of the capacitor gathering charge from all the power line wiring around you. A physiological amplifier has that same input --- 100 or 200 volts--- plus the tiny signal of interest. That GND (to your ankle) greatly reduces the undesired voltage.