# Mathing the Goethe Barometer

In my home when I grew up there used to hang a weird glass fixture above the stair leading to the basement. It was always half full with water but I never really paid it much attention. At some point I asked what it was for and got some cryptic answer from my mom that it was for predicting whether it would rain. Ignoring something which was obviously magic I never did figure out how it worked until much later learning about mercury  barometers in high school at least the principle became clear.

That thing which used to hang above the stairway was a so called Goethe Barometer. Now for a long time I had maintained an image in my mind that the bulbous shape of the device was mostly for aesthetic purposes or to store excess fluid to limit the effect of evaporation and while those are probably part of the reason they’re not the most interesting one.

In thinking about this design we should think of it as a variation of a U-shaped tube with one end capped. One can imagine calibrating the device by originally having the tube be open at both ends at which point the surface level in both columns are level, and this is retained when one end of the tube is capped so long as the enclosing process doesn’t compress the volume.. If the atmospheric pressure then decreases a difference in pressure between the trapped gas pocket at one end and the atmosphere  is established and the water level at the open is raised as it is pushed up by this difference in pressure. Pascals law tells us the difference in pressure between the trapped gas and the atmosphere is related to the height difference of the two levels according to

$p' - p= \rho g h$

Thus such a device can be used as a form of barometer. One of the problems though is that it’s difficult to extract absolute pressure information from the system. Ideally we’d like to be able to derive the pressure difference relative to the pressure when the barometer was closed but unfortunately the pressure in the trapped volume is not constant as a water level change can only occur if the volume of the enclosed gas changes which necessitates a change in it’s pressure .

Thus while qualitative information about whether the pressure is higher or lower than the pressure at sealing can be gathered, the relation isn’t straight forward.

What we’ll do now is to try to arrive at a formula relating the height difference in an ideal barometer and see if we can gather some insights as to why the Goethe barometer looks the way it does in the process. We’ll use this diagram for the principal quantities during calibration and after a pressure change.

We have two kinds of relations from which to mathematically extract the quantities., geometric and physical.

Geometric constrains. Under the condition that the liquid is in-compressible constrains exist relating the height changes.

$A_1 h_1 = A_2 h_2$ (The volume of the liquid is constant)

$h = h_1 + h_2$

Physical contraints We neglect temperature variations, capillary forces, vapor pressure and other presumably secondary phenomena and assume the principal physics are Pascals law and the pressure in the trapped gas being governed by the ideal gas law.

$p_0 V_0 = p_1(V_0 + A_1 h_1)$ (Ideal gas law)

$p_1 - p = \rho g h$ (Pascals law)

Solution: Out goal will be to relate exterior pressure, $p$ to the height difference $h$. From the geometric constraints we get

$h = h_1 + h_2 = \cfrac{A_2}{A_1}h_2 + h_2 = \left (1 + \cfrac{A_2}{A_1} \right )h_1$

$p_1 = p_0 \cfrac{V_0}{V_0 + A_1 h_1} = \cfrac{p_0}{1 + \cfrac{A_1}{V_0}h_1} = \cfrac{p_0}{1 + \cfrac{A_1}{V_0 \left (1 + \cfrac{A_2}{A_1}\right )}h}$

$p = \cfrac{p_0}{1 + \cfrac{A_1}{V_0 \left (1 + \cfrac{A_2}{A_1}\right )}h} - \rho g h$

Now let us think about what characterizes the Goethe Barometer, the fact that the back containing the trapped gas is so much wider than the tubular part open to the air. Eyeballing it $A_2 / A_1 < 0.02$ so in the grand scheme of things it can be neglected which besides implying $h \approx h_2$ and $h_1 \approx 0$ also simplifies the expression

$\boxed{p = \cfrac{p_0}{1 + \cfrac{A_1}{V_0}h} - \rho g h} \quad (A_2/A_1 \approx 0)$

In effect one will have to include the volume of the trapped gas as $A_1 / V_0$ is the same order of magnitude as $h$ in the real case with normal open air pressure variations, and this equation doesn’t simplify further. Also if the surface bordering the trapped gas changes in area as it moves vertically this equation fails and you need to compute directly using the volume changes instead of going via the areas.

One could imagine using a very very large $V_0$ in which case the trapped gas volume would remain close to original pressure.

$p_2 = p_0 - \rho g h_2$

Finally, since many classrooms often contain a genuine U-tube where the area of the two openings are the same I’ll write down the equation for that case

$\boxed{p = \cfrac{p_0}{1 + \cfrac{h}{2h_0}} - \rho g h} \quad (A_2/A_1 = 1)$

Okay, I think that’s all for now though the model could use some developement principally with respect to temperature corrections, both at equilibrium and changes induced by the work involved in compressing or expanding the gas.