[MITgcm-support] phihydlow without topography signal

[Katherine Quinn]

Hi Martin,
This was very helpful.  I can see now that if rho_0 isn't a good 
approximation of the mean density then there will still be quite a large 
imprint of topography on the instantaneous phihydlow.
One more question - is the density anomaly, rho', a 3D variable?  I am 
assuming yes, which means the third term in your bottom pressure 
equation will actually be an integral w.r.t depth from R_low to eta. 
That is:
bottom pressure = p_atmosphere + g*rho_0*(D+eta) + int(from R_low to 
eta){g*rho'*dz}
thanks, Katy

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Message: 2
Date: Fri, 16 Sep 2011 00:20:04 +0200
From: Martin Losch <Martin.Losch at awi.de>
To: mitgcm-support at mitgcm.org
Subject: Re: [MITgcm-support] phihydlow without topography signal
Message-ID: <08D4E79B-E5FE-4BEC-96BF-F98EC9B1D022 at awi.de>
Content-Type: text/plain; CHARSET=US-ASCII

Hi Katy,

not sure if I completely understand your question. But maybe this helps:

bottom pressure = p_atmosphere + g*rho_0*(D+eta) + g*rho'*(D+eta)
D=Water depth (=-R_low)
eta=sea surface height/"dynamic topography"
rho_0=reference density (=rhoConst)
rho'= rho-rho_0 = density anomaly

phiHydLow = [p_atmosphere + g*rho_0*eta + g*rho'*(D+eta)]/rho_0
phiHydLow = (bottom pressure)/rho_0 - g*D

 From this, you can see that  phiHydLow is the hydrostatic potential
anomaly at the depths in R_low with the constant depth-term (g*D)
removed. Add it back in, and you'll have your nearly linear fit to
R_low. There will probably always (except for special cases) be an
imprint of topography on an instantaneous phiHydLow because of
g*rho'*D/rho_0. You can remove that part, too, but that means that
you'll have to save rho' (diagnostics: RHOAnoma (?)) to remove any
terms involving D.
Does that make sense?

Martin

-- 
Katherine J. Quinn
Atmospheric and Environmental Research, Inc.   voice: 781-761-2234
131 Hartwell Avenue                              fax: 781-761-2299
Lexington, MA  02421-3126                     e-mail: kquinn at aer.com