180 degree phase question

[water-e]

@joernc I generated the potential at lon=0 (mid-latitude of interest to me) and then ran the potential through an ordinary tide analysis program that I know well. My intuition was that the analyzed phase would be zero because the phase of the potential would be the same as the astronomical argument. Instead, what I got for all diurnal and semi-diurnal constituents was something very close to 180 degrees. Is this expected?

This is surely a convention I don't understand. Is it because the equilibrium tide is equal to potential=-g*equilibrium tide and the traditional astronomical argument is based on equilibrium tide?

Doing this reconciles the phases, but my concern in that case is about the implication that the tide should look like -potential rather than +potential. If I generate the potential for San Francisco and compare the silhouette for a year to observed data, the envelop for potential matches well. The flipped sign version does not.

[water-e]

Actually, comparing to some old JPL code for the Cartwright and Tayler (Geophys. J R. AstrSoc, 1971, 23, 45-74) development as corrected by Cartwright and Edden Geophys. J. R. Astr Soc., 1973, 33, 253-264) it seems like they match pretty well if you take the negative of what is produced here.

[joernc]

This indeed looks like a sign convention issue. I defined the gravitational potential as V = –GM/r, so the tidal force is –∇V, and the equilibrium tide is h = –V/g. Hope this helps.

[water-e]

Yes thanks!