or stupidity, depending how you look at this,
I call this lecture theory of parameterization.
There's, of course, no such thing.
What I really want to show you is
where the need for parameterization
arises not just from a conceptual and physical point
of view, which we'll do first, but also
from a mathematical point of view.
So how does this parameterization thing
come about in mathematics?
And as we do that, we'll go through a few concepts
that may be useful for the rest of the course anyway.
So let's get started.
I drew this diagram, and I had a head start,
and I felt sorry for you, because you might not
be able to draw it quickly enough.
But you don't need to, because you've seen it a million times.
This is the spatial scales of the atmosphere,
say, we are thinking about the atmosphere here.
But this is space.
This is time.
We start at a centimeter.
We go to 10,000 kilometers.
And we start at a second, and we go to 10 days and more,
more than 10 days here, right?
So let's think about, you throw a few atmospheric phenomena
at me that I should put up here.
Long spatial scales and long temporal scales.
What in the atmosphere happens up here?
Anybody?
Planetary waves, they go up here.
People who work on the tropics, things like monsoons,
MJOs, blah, blah, blah.
Really large scale phenomena that
last for several days or weeks and have scales of more
than 10,000 kilometers or something.
What goes down here?
Let's say sort of minutes and centimeters to a few meters,
tens of meters, say.
Turbulence, excellent.
So we go turbulence.
And then the whole space in between is populated.
Of course, you've seen this diagram many times
in your undergraduate education, I'm sure.
So if you're thinking about a kilometer in an hour
or something like that, this would
be a single thunderstorm, maybe.
One updraft in a thunderstorm, maybe.
Smallish thunderstorm.
Thinking 10 kilometers in a few hours,
that would be things like supercells and organized
convection, maybe.
And then that also, and here in a sort of 1,000 kilometers
and a few days, anybody?
Sorry?
Yes, an optic scale motion.
So the phenomena we would be thinking about are cyclones,
extra-tropical, and perhaps also tropical on slightly smaller
scales, obviously.
So we have all these phenomena, and they
cover a sort of range of scales.
One of the most fascinating research questions
that is always being discussed in the early days,
everybody thought it all starts here and ends down there.
And it's a one-way energy gets transferred from the larger
scales to the smaller scales.
That's not true anymore.
This is really an interaction.
There's a real interaction between those scales.
And people call this multi-scale interactions.
It's often applied in convection these days.
How does convection feed back onto the larger scales?
How do the larger scales feed back onto the convection?
And so on and so forth.
Now, why have I drawn this diagram?
Now we need a different color, which I had at some point.
Oh, here.
So I've drawn this diagram.
What has this diagram to do with parameterization?
Well, we've talked about this a little bit yesterday.
We're building a model.
And for computational reasons, mostly,
we have to limit the resolution.
And say we want to limit it at 100 kilometers, right?
So what we're effectively doing is
we're drawing a line through this diagram.
And everything that's on the right-hand side of this line,
we can consider perhaps being treated
by the resolved part of our equations, which
we looked at yesterday, right?
So everything that's on this side, the model
can actually resolve.
And everything that's on that side model can't resolve.
We could also build a model with sort of one kilometer
resolution.
So this is the one kilometer.
This is the 100 kilometer resolution.
And what we immediately see is that once we
push the resolution to here, there are new phenomena
that the model might now give us solutions for based
on its resolved scale equations.
And we might not need to worry about parameterizing it anymore.
I drew this very nicely.
There's this belief that one can deal
with thunderstorms at one kilometer resolution.
It's actually not true.
Really, only the upper edge of that scale
and more the organization of several thunderstorms
is something that one kilometer models can resolve.
And don't get me started on four kilometer models
doing real convection.
That's just complete nonsense.
And then if you run an LES, you're down here in this scale.
So the larger eddies in the turbulent spectrum
are resolved, but the smaller eddies still
need to be dealt with by a so-called parameterization.
So this is just to give you sort of a flavor of where
this idea comes from.
What we are doing by limiting our resolution
to a certain number is we include certain processes
in our set of equations, and we exclude others
by averaging over this.
So that's the concept.
Now we're going to do the math, because I
think it's important that you've seen this.
Who has seen this?
Who has ever had to derive the subgrid scale stuff from math?
Two or three, four people.
OK, that's good, because you'll be a little bit bored,
but the rest of the crowd will not be bored.
OK, and in doing this, I will introduce a few things that
are useful anyway that have nothing to do really
with the parameterization per se.
We'll just use them conveniently.
So the first thing we're going to introduce is a new variable.
Where's my blue pen?
Oh.
We're going to introduce a new variable, which some of you
will have heard about, others won't.
And I chose it for convenience.
There are others we can take.
We'll call it S, and we'll define
it to be specific heat at constant pressure times
temperature plus gravity times height.
This thing is called the dry static energy.
It's a very close relative to potential temperature,
and you will see why we are choosing it in a second.
So this is, at the moment, it's just a variable.
We have just defined it.
I could put for pure mathematicians,
I can put a third line there.
It's a definition, right?
We just chose to make this variable.
Now, why is this a useful variable?
Because it has one property that we often
use in atmospheric science and that we will be using
throughout this course, and it's under certain conditions
this variable is conserved, i.e., it doesn't change.
So let's work out under which condition this is the case.
So we want to know, when does dS dt,
when does the change in this variable, when is that 0?
So we can write this as Cp dt dt plus g dz dt, right?
And we're going to use this variable later,
just because it's convenient to look at parameterization stuff.
So that doesn't tell us much.
Luckily, for this expression, we have another law
that we can use.
We have that readily available.
Anybody who hasn't done this before know.
So we can actually use the first law of thermodynamics,
which is one form of it is Cp dt dt minus 1 over rho dp dt.
Pressure changes with time equals q.
So this is basically saying that temperature
changes are related to pressure changes and any external heating.
So q here, and we'll keep that for now,
because that will only deal with at the very end.
So that's heating, basically.
It doesn't have to be external.
In our field, things like radiation, condensation,
so on and so forth.
So just a warning, we're doing dry thermodynamics
at this point, right?
So we are not worrying about moisture.
This will happen later in the course.
It's just to keep it simple.
Well, it's a rate.
Well, heating implies that it's actually a rate of change.
Well, it's a heating.
It's the addition of heat, which will lead to changes
in temperature and or pressure in a parcel.
So that's cool, because it means now we can rewrite this.
I'll do it, because if I don't do it slowly, I'll do it wrong.
So we can rewrite this as ds dt minus g dz dt.
Then we can stick it in here, right?
So then what we get is ds dt minus g dz dt minus 1 over rho
dt dt dt is this heating rate, OK?
Now we're going to make another assumption.
We're going to make a lot of assumptions
as we go through this.
And most of the time, people forget that we've made them.
So one of the assumptions that we're
going to make to make our life simple
is we're going to make the assumption of the atmospheres
in hydrostatic equilibrium, which is a good assumption
if you're thinking about large scales.
And it's a lousy assumption if you
think about very small scales.
But since we are concerned with large scale models in this course,
mostly we can make it.
And that's actually that dp by dz is minus rho times g.
So that's the hydrostatic assumption.
In some sense, you should remember
that we're making these assumptions,
because they limit what you can do with whatever
it is we are deriving.
And things get more complicated if you can't make this assumption.
So from this, we can sort of rearrange this.
We are assuming in all we do that all the functions we are
dealing with are well-behaved.
If there's a pure mathematician in the room,
they often cringe at what we're doing.
But all the functions are well-behaved.
So there are no poorly-behaved functions in this.
If everything is well-behaved, we
can write this as dp over rho equals minus g dz.
And you can see why I want to do this,
because now what we're seeing is that ds by dt minus g dz by dt.
And then dp over rho is minus g dz.
So this becomes a plus g dz by dt equals q.
All right, very nice.
Why is this very nice?
Well, you can see that those two terms,
under the hydrostatic assumption, disappear.
And we are left with something really simple.
We are left with, and we'll use this later, an equation that's
very simply states the change in dry static energy in time
is directly related to the heating rate.
It's just the heating rate.
So if we assume what's called an adiabatic process,
and again, we are in dry land, no humidity yet.
This just makes life more complicated if we add it.
But in principle, it's exactly you
can do all this with moisture included.
And I'm sure Pierre and others will
do this when they talk about boundary layer turbulence
and clouds and things like that.
An adiabatic process, by definition, q equals 0.
And as a result, the change of s with time will be 0.
So that's what's called a conserved variable,
because s is now conserved if the process is adiabatic.
So for certain processes, variables don't change.
Now, these things, why do we invent those as meteorologists?
We invent them because they become, or in other fields
as well, because they become mathematically really nice,
and also conceptually really nice.
Because if you know a process is dry adiabatic,
suddenly the calculation goes away.
You don't need to do any calculations.
This would not be true for temperature,
because the temperature is changing
on the adiabatic processes.
So you would need to work out what the temperature change is.
But this variable, as or potential temperature,
it's close relative.
Do not change in an adiabatic process.
So it's really convenience and laziness,
perhaps to some extent.
But it also simplifies our thinking.
The difficulty with conserved variables
is that sometimes we get tempted to explain the word
by conserved variables, when really they don't always
lend themselves to explain why something is happening.
And I give you the classic example that's really simple.
If you have to calculate, that wasn't a very good idea.
If you have to calculate the velocity of a falling rock,
you have to calculate it.
How would you do it?
Take this thing I dropped, and I want
to know what the velocity is when it hits the ground is,
to first order, assuming no friction, all that.
Why would you do it?
You would use the conservation of energy to do it.
So that's a conserved variable, energy.
And you do m over 2v squared equals mgh.
And you work it out, and then you get the number.
Is the rock falling, or is this thing falling in order
to conserve energy?
Yeah.
Is what drives, what's making it fall?
Is it the fact that it needs to conserve energy that it falls?
It's the force.
So it's actually gravity.
So the explanation why the rock's falling is gravity.
You could calculate its velocity by integrating
f equals ma, which is dv dt.
And then you do a time step as it falls.
And the math is much more complicated.
And you don't need to just to do the math.
But if you want to explain why the rock's falling,
then you better use gravity as an explanation, not
the conservation of energy.
So this is a simple example.
In meteorology, we have all sorts
of conserved variables all over the place.
And we are tempted to use them to explain something,
and it's often very difficult. A good example for meteorology
is certain variables we can design
are conserved under condensation.
And then people try to explain the process of convection
with those variables.
That's kind of hard, because the very process that
makes convection what it is is condensation.
So if you're trying to describe it
with a variable that doesn't include condensation,
mathematically, this becomes very elegant.
But physically, it's not exactly very revealing,
because it's the process of condensation
that keeps the parcels buoyant as they
go up in the atmosphere, which you've removed by design
from your variables.
So we have to be always a little bit careful.
But that's just a little aside.
So what else do we need?
You all know, hopefully, that if you have a variable,
so s is a function of space and time.
And we talked about coordinate systems yesterday.
To keep our life simple, we'll just use x, y, and z today.
That's just to be simple.
Everything gets more complicated if you change the coordinates
to, say, latitude, longitude, and radius, or something
like that, which is a sensible one on a sphere.
So therefore, this function, and I'm just
hoping that everybody knows this, we can actually,
and we will use this later, we actually
need to do the partial derivatives of this function.
So this is actually ds by dx, dx dt plus ds by dy partial,
dy dt plus ds by dz, dz dt plus ds by dt, dt dt by dt.
You've all seen this.
This is going from the material derivative
to the partial derivatives and things like that.
So this is nothing else than the u velocity, the v velocity, w.
And this is obviously 1.
And so we end up with this expression
for the material derivative, which we'll use later,
that the absolute, the total derivative of s
becomes the partial derivative of s
with time plus the velocity vector,
and I write it in 3D here, times the gradient of s.
So this is an expression we will use later as well.
I might want to start collecting things
that we'll use later on this board so we can go back to it.
So that was really just a little aside on something
that I was hoping you would all know anyway.
All right, so we're in good shape.
We're getting somewhere.
You might not notice it yet, but we're getting somewhere.
So the things we've learned is we have an equation ds by dt
equals q.
That's kind of nice, simple thing.
And we have another one that says ds by dt equals ds by dt
plus velocity nabla s.
OK, we've got two bits in place to what we want to do.
Now we need to do one more trick that doesn't appear obvious
until we use it later, and that is
we need to transform our equation
into a more workable form for what we want to do later.
And by doing that, we start with this guy, ds by dt equals q.
And then we introduce a second equation
that I will not derive, but that you all know.
And that's the equation for density.
Just get it right.
I always get it wrong, so I better look it up.
Yeah, s rho times the divergence of the wind vector equals 0.
What's that one?
Continuity equation, that's right.
So it's one form, one of the many forms
of the continuity equation.
You can transform it by doing the material derivative thing
on rho, but we want it in this form.
So what we're going to do is I'm going to just try and modify
this equation using adding a 0, essentially,
because this is 0, right?
It's actually something you should also
keep in mind that most of the equations we use and write down
have already subtracted this equation often.
So we're going to add it back in,
because it will make our life easier later.
It will change the form of this equation, essentially.
So what we're going to do is we're
going to multiply this one with rho,
and then we're going to multiply this one with s,
and then we're going to add them.
So we have here, we have rho ds by dt.
The q we'll leave, that goes on the right-hand side.
So let's look at this left-hand side.
We get a plus s by d rho times d rho by dt.
We get a plus rho s divergence of the wind.
And on the right-hand side, we add q and a 0.
So that equals q.
Does that make sense?
It's pretty straightforward.
Actually, it's a rho q, because we
multiplied this equation with rho.
Good thing I checked.
OK, that's pretty straightforward.
So those two we can put into one,
so that becomes the derivative of the product of rho and s.
So we just use the product rule here.
And this one just stays rho s number v equals rho q.
So now, we take our famous material derivative bit,
this one, and we apply it to this term here.
So then what we get is d rho s by dt partial.
That becomes the partial derivative in time.
And then we have this other term that we have to put in there.
So that's plus v times the gradient of rho times s.
And that plus rho s divergence of the wind vector.
And that gives rho q.
Do you add it because of the equation?
I've added those, yeah.
So there's a big plus in here.
I've just multiplied this with rho,
multiplied this with s, and add the 2.
And I can do that nicely because there's
a 0 on the right-hand side.
So right-hand side doesn't change.
OK, in this one, use a bit of mathematics.
There's a product rule for nabla operations as well.
Simply becomes d rho s by dt plus nabla rho s v equals rho q.
So this is called the flex form of the equations.
And the reason is that what's inside the nabla operator
here is now a flex.
It's actually a flex of the variable, which in our case
is dry static energy.
So some of you will have encountered these forms.
It's just a different form of the equation.
And I use it because it becomes convenient, no other reason.
Yeah?
So it's a nabla, but are you using it in the strategy form?
No, no.
If you take this apart, you have to use the,
there's an assumption here that the density gradients are
small, by the way.
But we'll leave that as an aside.
But if you put this in here, then you're basically
using the product rule just like you would
use for a normal derivative.
You get one term, which is v times gradient of rho s,
and another term, which is rho s times divergence of v.
And you can put this together inside the nabla operator.
So it's just like the product rule of differentiation
that you use here.
That's all.
Yeah?
Do you want to see the product function?
Well, this is a product.
All this is a product, rho times s times, sorry?
They're becoming two vectors.
Yeah, this is a vector multiplied with a vector.
And this, sorry?
This one is just, you can write it like this if you want.
It's just using the product rule on this
for the nabla operator.
Pardon?
I can't hear you.
You're so far back, and you need to speak up.
I think if you need that dot, two states
will arise in a way for more.
No, this is fine.
There doesn't even need to be a dot here, because one of them
is a dot, and one of them is a gradient.
You can work this out if you do it in the actual u, v,
and w components.
You will see that this is true.
So this is just for convenience.
And we're actually going to do this.
We're going to use the u, v, and something
components in a minute.
So let's keep this equation.
Yeah, so we're treating rho s as one variable, essentially.
We're treating the density as constant,
as if we can replace it with a constant at this point in time.
It's not true.
Of course it isn't.
So in principle, we can make a horizontal term in a vertex.
If we come back to that, we're going
to make another assumption in a minute
to do this parameterization, to do the parameterization bit,
to just reveal where the parameterization terms are
mathematically.
And we're going to make an assumption.
I'm trying to keep things simple, but you are right.
So rho s has become sort of the variable here.
And there are vertical density gradients
which are very large.
But we are not, I don't think we,
we'll come back to this, OK?
I hope.
I don't know, actually, whether we will.
All right, so now finally, one more step
to do with this equation.
What are we going to do with this equation?
We're going to average it.
So one of the things that you can assume a model does,
if we're making grid boxes of sorts, you get what we are
doing, essentially, is we are describing only the average
over that grid box in our equations, right?
So essentially, making grid boxes
is like averaging the equations over certain areas, yeah?
Does that make sense?
Grid box is both in the horizontal and the vertical.
So suddenly, we are only going to start dealing
with the averages over this.
And so we're introducing this averaging procedure.
And what we do is we can write, and you
will have seen this as well in your undergraduate days,
we write every variable.
And we start with just s as an average over the grid.
And we are actually going to be lazy again
in not defining how exactly this averaging takes place.
Just think of it as your arithmetic, just
standard mean, but it doesn't have to be.
It could be a more complex averaging operator
if we wanted it to be.
But this, we just use a very simple averaging operator.
And then at every point, so this is s at every point
within the grid box.
And we break it up into the average over the grid box,
and a perturbation from that average,
which you could measure if you wanted to,
at every point within our model grid box.
We can do that with every variable in the game,
essentially, right?
There are some rules about this averaging, which we have to,
which we have to actually set.
And this defines a little bit what this averaging operator
can be.
So one of the rules we would like
is that if we average the average again,
we just get the average, right?
That's one of the rules.
This is sometimes called Reynolds averaging,
because Reynolds was the guy who sort of did this.
Well, once you have this rule, you
can immediately see if you apply this rule to this equation,
then you get s bar equals s bar bar plus s prime bar.
And this can only be true if s prime bar equals 0.
Otherwise, this doesn't hold anymore, right?
So that's another rule that we have.
And it turns out, and again, this
isn't too much to explain, any variable.
So let's call it little phi.
That's multiplied, say, with s prime.
If we have a phi bar multiplied with an s prime,
and we take the average of that, that actually becomes 0.
But if you have a phi prime multiplied by an s prime,
and average that, that's actually not 0.
And it has to do with the way the perturbations correlate
with each other within the grid box, right?
And this is going to be very important for what
we're going to do next.
So essentially, what we're going to do next
is we're going to average this equation.
We're going to put this decomposition in here.
And again, we're going to be lazy,
and we're not going to do it for rho.
So we're going to assume that for rho, which is rho bar
plus rho prime, we're going to make this assumption
that rho prime is actually much, much smaller than rho bar.
And so we're going to replace every, so everywhere
where we do this decomposition, we won't do it for density.
This is almost a Boussinesq assumption,
except that in the vertical, when the rho primes are
multiplied with g somewhere, wherever they are multiplied
with g, you keep them in.
That's the actual Boussinesq assumption, approximation.
So it's another approximation that we make here,
just again to keep our life simple.
And this will lead us to one more approximation,
and then we'll be there.
So let's do this.
So what we get is, on the left-hand side,
we got just s bar plus s prime by dt, keeping the rho in there.
And on the right-hand side, it gets messy.
So we're actually going to decompose everything
into the components now.
So this becomes, I'm going to take this term
onto the other side.
And so it's going to be a minus.
And then what we're going to do is
we're going to do the partial derivatives of this operator
here.
And so then we have u bar plus u prime times s bar plus s
prime.
That's the first term that we have to consider.
And there's a similar one, d by dy, of v bar plus v prime
times s bar.
There actually needs to be one more there.
s bar plus s prime.
You can see where that's headed.
Yeah, we lost the rho there.
So the rho stays in there.
And it's really just a rho bar everywhere now.
And then there's a w1.
It's pretty boring, dw, dz rho, w bar plus w prime.
We're just going to look at one of those.
Because it's symmetric through all the terms, s prime.
And then we finally have a plus rho bar q on the right hand
side.
Mustn't forget that one.
So we're going to look at one of them,
because otherwise it gets too boring and takes too long.
So we're going to just take this one apart
into its four components, minus d rho bar.
And then what do we get?
u bar s bar dx minus d rho bar u bar s prime dx minus d rho
bar u prime s bar dx.
I'm not going to write all of them.
Don't worry.
And then we do minus d rho bar u prime s prime dx.
OK?
There's a bar here.
All right, so this splits up into four terms.
And every one of them will do the same thing, right?
So the v component and the w component
will look exactly like this, OK, for v and w.
OK, so now we're going to just do one more thing.
And that's average of these equations again.
So we take the average of this.
So remember, what we did is we substituted the within grid.
At the moment, we haven't applied an averaging,
per se, on the equation.
We've just replaced the variable with the average
plus the perturbation.
So now we're going to take the average of our grid box,
of all this.
So if we take the average of this one,
we see there's only bars in this.
So average, what do we get?
So this term actually becomes d rho bar u bar s bar bar dx.
And this term becomes minus d rho bar u bar s prime dx bar.
This one becomes about d rho bar u prime s bar bar dx.
I should really have made my life simple and just
called rho constant, actually, u prime s prime bar dx.
And from our averaging rules, we see this one is trivial.
That just stays the mean term, d bar rho bar u s bar dx.
This one cancels out because it folds under this u bar s
prime bar equals 0, especially if rho is constant, which
I should have made it.
So that's 0.
That one's 0.
And that one, we can do nothing about.
That's not 0.
So it's just what it is, minus d rho u prime s prime bar dx.
So what we see, and we'll write down the full equation
in a minute now that we've thrown away a few terms,
because we can throw away the same terms in all
the other two components.
What you see is we end up with two terms, essentially.
One term is a term which has only averages in it,
which is kind of neat, because that
is solved by our numerical scheme applied
to our equations on the grid box scale.
And we end up with a term that has these two primes multiplied
with each other and then average.
And that's clearly related to processes
that are correlating the two deviations of the two variables,
in this case, the u component of the wind
and the dry static energy to each other.
So this is the subgrid scale term.
That's a term that is related to processes
that happen at scales much smaller,
or not much smaller, but smaller than the grid box.
OK.
And of course, by the way, if you average this,
you can see that you're left averaging this one.
Take the average.
This just gives you just this, because the s prime term
disappears.
OK.
Shall we write down the whole equations from this?
Probably we should.
I haven't written them down, so I
have to do it off the top of my head, but I think that's fine.
So on the left-hand side, we just
said d rho bar s bar by dt, partial derivative.
And on the right-hand side, we're
going to start sorting these terms.
We have the mean terms, d rho bar u bar s bar dx,
minus d rho bar u bar v bar, sorry, s bar dy,
minus d rho bar w bar s bar dz.
So that's one group of terms that we have.
These are all the large-scale terms.
The other group of terms that we have,
and I'm deliberately going to separate some out,
is the minus d rho bar u prime s prime bar dx,
and the minus d rho bar v prime s prime bar dy.
Then I'm going to single out the vertical term here for a reason
that we'll discuss in a second, w prime s prime dz.
And then there's finally our rho bar q term,
which we shouldn't forget.
So there's sort of four terms now, four groups of terms.
The first group of terms are the terms
that our equations can take care of.
The second group of terms have to do
with horizontal variations across the grid box,
because they are horizontal variations in velocity anyway
around the grid box.
So they are what you would call sub-grid scale
horizontal eddies, if you want to think of them that way,
transporting stuff around in the grid box.
Yes, so I've separated this, and I'll
come to that in a minute.
So this would normally be here.
But we're going to make another assumption,
or another assumption is made.
I haven't made it, but it is made in most large scale models.
And that is that these horizontal transport terms
on the sub-grid scale are much, much smaller
than the vertical transport on the sub-grid scale.
That is a reasonably good assumption for the atmosphere.
It tends not to be necessarily a really good assumption
for the ocean, where there is a lot of horizontal eddy motion
transporting stuff around.
On scale, it's often smaller than hundreds of kilometers.
But that's why ocean models are so keen.
For ocean models, the drive to high horizontal resolution
has often to do with these terms.
But in atmospheric models, we simply
say that these horizontal terms are much smaller
than the vertical term, and it's actually
a good assumption in general.
But it is an assumption we make in most models.
So we say these two are much smaller than this one.
So therefore, we're going to ignore them.
OK.
So if we look at that, so ignoring on a whiteboard
is easy.
So we're just going to ignore it.
This term now is the vertical term, i.e.
it's the transport of S prime by vertical variations
across the grid box in vertical velocity.
So it's actually about vertical eddies, or anything.
It doesn't have to be eddies.
That is a perturbation of vertical velocity
correlating with the perturbation
in dry static energy, which you can think of as temperature-ish
if you want, or potential temperature if you want.
This term is clearly subgrid scale.
It's actually quite large, especially in turbulent motions.
It's very large.
In convection, it tends to be very large.
This is a term that we actually need to be parameterized.
So this is actually parameterization,
one part of parameterization.
And you will hear a lot about it for the rest of the week.
It's about finding expressions for this term,
for all the variables, obviously.
So we've only looked at one variable.
Remember, our equations have the momentum.
If something related to temperature,
if something related to moisture in modern models,
and other variables we carry are the amount of liquid water
or ice in a cloud, for instance.
We carry those as prognostic variables.
And they all have terms like this in the equations,
essentially.
So this is a term that needs to be parameterized, OK?
It's certainly horizontal, because it's much in the atmosphere,
because it's much smaller than the vertical energy ignore.
Yeah, what we are saying is the transports that
are done horizontally on subgrid scales
are much smaller than the transport,
by horizontal motion, are much smaller
than the transports that are done by vertical motion,
perturbations, basically, right?
So we are basically saying that the turbulence we see
is mostly turbulence that is associated with vertical motion,
not so much with horizontal motion.
It's an error, right?
We're making an error, but we know the error is small.
You can do a scale analysis and all these things
on these terms.
You find they're relatively small in the atmosphere,
anyway, compared to this one.
There are people who actually, every now and then,
you find papers and people you talk to who are worried
about this assumption.
And we're saying, oh, can we really do this?
Especially in a cloud field, we know
we have horizontal variations between the in-cloud and out
of cloud area.
And if you have some horizontal eddy stirring things around,
but it doesn't make a big change to the mean, necessarily.
And then the other term that's to be parameterized
is this one here, right?
So the heating.
So what heating terms do we have in the atmosphere?
So in general, this becomes something
like the heating from radiation.
That's one term that heats the atmosphere, so radiative fluxes.
And so that's why you have lectures
on radiation parameterization, because radiation
does stuff to the atmosphere in the heat budget,
in particular.
Doesn't do anything in the moisture budget.
Yes, and then there is an L. So we
need to convert this into units of energy,
because it's dry static energy, times
a condensation and evaporation.
So condensation and evaporation in the atmosphere
happens on very, very small scales inside clouds,
inside convective clouds, inside also stratiform clouds.
But most of the condensation in a stratiform cloud,
and some it happens on a widespread area.
But in most of them, and even stratiform-looking clouds,
it's really small areas of the cloud
where the condensation happens, and other areas of the clouds
where evaporation happens.
So it happens certainly on scales much smaller.
If you're a real droplet enthusiast,
of course condensation happens at the microscale,
because the droplets are tiny, that condense.
But these are the things.
So basically, what I'm going to try and conclude with
is the rest of the two weeks is going
to be pretty much about these two terms.
And everything you learn from here
is how are we going to deal with these two sets of terms.
It's really for whatever you want to make it.
It's the transport term and the heating terms,
which are condensation, evaporation, and radiation,
essentially, happen on subgrid scales.
And the rest of the two weeks, people
will try and tell you how for the different processes
in the atmosphere, such as turbulence, convection,
cloud processes, and radiation, are we actually
dealing with those terms in the models.
So I just wanted to have this one lecture
to introduce the sort of concept of this, not a theory.
It's not a theory, really.
It's a concept of how these terms actually, of what it is,
we need to parameterize and why.
So the why is because we've averaged our equations, OK?
And because that reveals terms that aren't small.
If they were all small compared to this, we could ignore them.
But it turns out they are not small, all right?
And so they are so big that an army of people gets employed.
And we've gotten some of that army here
to explain to you how it's done, how
these terms are parameterized throughout the different
processes, all right?
So that makes sense.
I'm sorry about the math around rho.
I should have just used constant rho,
and then life would have been very simple.
But that happens, OK?
But then you would have criticized me
for using constant rho.
Oh, rho is not constant.
It's ridiculous.
All right?