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<publicationStmt><distributor>BASE and Oxford Text Archive</distributor>


<availability><p>The British Academic Spoken English (BASE) corpus was developed at the

Universities of Warwick and Reading, under the directorship of Hilary Nesi

(Centre for English Language Teacher Education, Warwick) and Paul Thompson

(Department of Applied Linguistics, Reading), with funding from BALEAP,

EURALEX, the British Academy and the Arts and Humanities Research Board. The

original recordings are held at the Universities of Warwick and Reading, and

at the Oxford Text Archive and may be consulted by bona fide researchers

upon written application to any of the holding bodies.

The BASE corpus is freely available to researchers who agree to the

following conditions:</p>

<p>1. The recordings and transcriptions should not be modified in any


<p>2. The recordings and transcriptions should be used for research purposes

only; they should not be reproduced in teaching materials</p>

<p>3. The recordings and transcriptions should not be reproduced in full for

a wider audience/readership, although researchers are free to quote short

passages of text (up to 200 running words from any given speech event)</p>

<p>4. The corpus developers should be informed of all presentations or

publications arising from analysis of the corpus</p><p>

Researchers should acknowledge their use of the corpus using the following

form of words:

The recordings and transcriptions used in this study come from the British

Academic Spoken English (BASE) corpus, which was developed at the

Universities of Warwick and Reading under the directorship of Hilary Nesi

(Warwick) and Paul Thompson (Reading). Corpus development was assisted by

funding from the Universities of Warwick and Reading, BALEAP, EURALEX, the

British Academy and the Arts and Humanities Research Board. </p></availability>




<recording dur="00:48:41" n="5662">


<respStmt><name>BASE team</name>



<langUsage><language id="en">English</language>



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<item n="speechevent">Lecture</item>

<item n="acaddept">Meteorology</item>

<item n="acaddiv">ps</item>

<item n="partlevel">UG2</item>

<item n="module">Radiative transfer</item>





<u who="nm0885"> going on here # this gentleman is from <pause dur="0.7"/> i think the Linguistics department </u><u who="om0886" trans="overlap"> <gap reason="inaudible" extent="1 sec"/></u><pause dur="0.2"/> <u who="nm0885" trans="pause"> right and he's he's doing study of <trunc>th</trunc> of the language that academics use and things like that <pause dur="0.5"/> # so he's he's taping a random sampling of lecturers around the university <pause dur="0.6"/> so if you're wondering why i'm replacing all technical terms with the names of <trunc>frudg</trunc> fruits and vegetables this time it's basically i'm having a go at him <pause dur="0.7"/> so <vocal desc="laugh" iterated="n"/><pause dur="0.7"/> # <pause dur="0.5"/><vocal desc="clears throat" iterated="n"/><pause dur="0.7"/> well <pause dur="2.0"/> what we are <pause dur="0.4"/> at <pause dur="0.6"/> is <pause dur="0.2"/> that <pause dur="0.2"/> formula <pause dur="0.5"/> and we spent a lot of time yesterday <pause dur="0.2"/> hacking our way up to this thing hacking that's a good word <pause dur="0.5"/> # <vocal desc="laugh" iterated="n"/><pause dur="0.7"/> and <pause dur="0.8"/> and we got it <pause dur="0.3"/> and we're really happy about it <pause dur="0.4"/> # <pause dur="0.2"/> what is it <pause dur="1.1"/><vocal desc="laugh" iterated="n"/><pause dur="6.5"/> what is D-of-L-nu </u><pause dur="1.5"/> <u who="sf0887" trans="pause"> <unclear>decrease of</unclear> the radiance </u><pause dur="1.3"/> <u who="nm0885" trans="pause"> right <pause dur="0.7"/> and some <trunc>r</trunc> if some radiation is passing through a slab of atmosphere a thin layer of atmosphere <pause dur="0.4"/> # some of it's going to get lost <pause dur="0.4"/> and the amount that gets lost is the change in radiance <pause dur="0.2"/> D-L-nu <pause dur="1.0"/> okay <pause dur="0.6"/> # <pause dur="0.4"/> what happens to the radiance that gets lost <pause dur="0.2"/> where does it go <pause dur="4.6"/> there are a couple of different

things that could happen to it </u><pause dur="2.4"/> <u who="sf0888" trans="pause"> could get scattered or absorbed </u><pause dur="0.5"/> <u who="nm0885" trans="pause"> right it can get scattered <pause dur="0.4"/> or absorbed <pause dur="2.3"/> okay <pause dur="0.3"/> and so now the amount we worked out depends on a couple of different things <pause dur="0.3"/> okay some of them are pretty obvious <pause dur="0.3"/> <trunc>y</trunc> # you lose more if you have more going in <pause dur="0.4"/> so it depends on the amount of radiance that's incident on our <pause dur="0.2"/> little slab of atmosphere <pause dur="0.7"/> it depends on <pause dur="0.9"/> something called K <pause dur="0.6"/> what <pause dur="0.3"/> does K tell us about </u><pause dur="8.6"/> <u who="sf0889" trans="pause"> <gap reason="inaudible" extent="1 sec"/> </u><pause dur="1.1"/> <u who="nm0885" trans="pause"> # <pause dur="1.0"/> scattering is one of the things that go that <trunc>g</trunc> that goes into it </u><pause dur="2.0"/> <u who="sf0890" trans="pause"> # is it the extinction coefficient </u><pause dur="0.3"/> <u who="nm0885" trans="pause"> yeah it it's it's the extinction coefficient which is the sum of the scattering coefficient plus the absorption coefficient <pause dur="1.0"/> okay <pause dur="0.2"/> and so what it's telling us about is it's telling us about the characteristics of the material <pause dur="0.3"/> is this material an effective scatterer or an effective absorber <pause dur="0.3"/> and we'll be getting into later on in the course what it is actually that makes the material a good scatterer or absorber <pause dur="0.5"/> and then it depends on <pause dur="0.4"/> rho-secant-theta-D-Z what's that <pause dur="0.6"/> that's the easy one </u><pause dur="3.8"/> <u who="sf0891" trans="pause"> pathlength </u><pause dur="0.7"/> <u who="nm0885" trans="pause">

pardon </u><pause dur="0.2"/> <u who="sf0891" trans="pause"> pathlength <gap reason="inaudible" extent="1 sec"/> </u><pause dur="0.4"/> <u who="nm0885" trans="pause"> yep this one's the pathlength why do we have a rho in there </u><pause dur="1.4"/> <u who="sf0892" trans="pause"> <gap reason="inaudible" extent="2 secs"/></u><pause dur="0.4"/> <u who="nm0885" trans="pause"> yeah it's the density of the material that's doing the interaction <pause dur="0.5"/> so <pause dur="0.2"/> if your total this thing up pathlength times the density that's the amount of material <pause dur="0.3"/> per unit area <pause dur="1.7"/><vocal desc="clears throat" iterated="n"/><pause dur="0.8"/> okay <pause dur="0.4"/> so that's for <pause dur="0.3"/> a very thin slab <pause dur="0.3"/> and that means and because it's a very thin slab we can say well things like density <pause dur="0.4"/> the amount of material <pause dur="0.3"/> extinction coefficents all all of these things don't vary much as you go across the slab <pause dur="0.4"/> i mean in principle <pause dur="0.3"/> they can vary <pause dur="0.3"/> so <pause dur="2.1"/> if we have <pause dur="0.8"/> a <pause dur="0.2"/> finite <pause dur="3.0"/> slab of material <pause dur="2.8"/> if you have a finite slab of material <pause dur="0.4"/> so <pause dur="0.2"/> instead of just <pause dur="0.3"/> a thin layer <pause dur="1.2"/> D-Z here <pause dur="0.3"/> we've actually got <pause dur="0.3"/> a thick layer <pause dur="0.4"/> which <pause dur="0.3"/> ranges between <pause dur="0.9"/> # <pause dur="0.4"/> you know <pause dur="0.5"/> a finite slab of material <pause dur="1.4"/> from Z-one <pause dur="1.0"/> to Z-two <pause dur="0.8"/> so we've got <pause dur="0.7"/> two heights involved and so if we say Z-one is the top where it's entering and Z-two is the bottom <pause dur="0.5"/> where it's going out <pause dur="1.1"/> # <pause dur="0.2"/><vocal desc="clears throat" iterated="n"/><pause dur="0.2"/> we can integrate <pause dur="0.3"/> this expression <pause dur="0.2"/> to add up <pause dur="0.7"/> the changes in the # add up the amount of depletion in

all the different layers <pause dur="0.4"/> and give us the amount coming out the bottom <pause dur="0.3"/> so if we've got an initial radiance L-nought-sub-nu <pause dur="1.3"/> in our direction here <pause dur="0.7"/> passes through the slab <pause dur="0.7"/> then <pause dur="2.2"/> what comes out <pause dur="0.5"/> is <pause dur="1.3"/> at the bottom is something less <pause dur="0.4"/> and i'll call that L with a superscript-<pause dur="0.4"/>D <pause dur="0.5"/> so that's # the depleted beam so it's lost some stuff <pause dur="0.7"/> and that's L-nu and it's <pause dur="0.2"/> of course <pause dur="0.4"/> coming out in the same direction <pause dur="0.2"/> theta-phi <pause dur="1.8"/> okay <pause dur="0.7"/> so <pause dur="0.4"/> if we have a finite slab of material from Z-one to Z-two <pause dur="0.5"/> we <pause dur="0.6"/> can <pause dur="0.8"/> integrate <pause dur="2.9"/> this <pause dur="1.1"/> expression <pause dur="2.2"/> # <pause dur="3.8"/> to give <pause dur="1.1"/> <unclear>D</unclear> <pause dur="1.2"/> direct <pause dur="2.7"/> <trunc>compo</trunc> <pause dur="2.4"/> of the <pause dur="0.2"/> emerging <pause dur="1.9"/> beam <pause dur="2.3"/> 'cause remember the emerging beam is also going to have contributions from emission <pause dur="0.3"/> and contributions from scattered stuff which we're going to have to work on a little bit later <pause dur="0.6"/> but # <pause dur="0.4"/> but the direct part of it <pause dur="1.3"/> L-D <pause dur="0.5"/> and actually yeah i <trunc>m</trunc> i made a mistake that D stands for direct <pause dur="4.2"/> we can get that by doing the integral <pause dur="0.6"/> so <pause dur="0.3"/> what we do <pause dur="0.3"/> is we <pause dur="0.5"/> take this expression <pause dur="0.6"/> and # <pause dur="0.7"/> you notice we've got <pause dur="0.3"/> L-nu on both sides so we'll take this one over to

the other side <pause dur="0.4"/> so we'll write this as D-L-nu over L-nu <pause dur="0.6"/> and that's <pause dur="0.4"/> excuse me going to be equal to <pause dur="0.4"/> minus-<pause dur="1.5"/>K-<pause dur="0.2"/>nu-E-<pause dur="1.0"/>rho-secant-<pause dur="0.2"/>theta-D-Z <pause dur="0.9"/> okay now we integrate this <pause dur="0.4"/> across <pause dur="1.7"/> the layer <pause dur="2.7"/> # <pause dur="1.0"/> so <pause dur="0.2"/> integral signs on both sides we're now going to sum up those layers <pause dur="0.7"/> so <pause dur="0.3"/> the integral on the right hand side we're integrating <pause dur="0.3"/> with respect to Z <pause dur="0.3"/> so we're integrating in height what are the limits of integration <pause dur="2.2"/> what's the bottom one </u><pause dur="0.5"/> <u who="sf0894" trans="pause"> <gap reason="inaudible" extent="1 sec"/> </u><pause dur="0.9"/> <u who="nm0885" trans="pause"> pardon </u><pause dur="0.6"/> <u who="sf0894" trans="pause"> <gap reason="inaudible" extent="1 sec"/></u><pause dur="0.2"/> <u who="nm0885" trans="pause"> Z-one <pause dur="0.2"/> that's right <pause dur="0.8"/> it's where the stuff's going in <pause dur="0.3"/> and it goes up to Z-two where the stuff's coming out <pause dur="0.4"/> now on the left hand side <pause dur="0.3"/> we're integrating over radiance <pause dur="0.4"/> D-L-nu <pause dur="0.4"/> so what's the starting radiance where we start the integration </u><pause dur="0.9"/> <u who="sm0895" trans="pause"> L-nought-sub-nu </u><pause dur="0.2"/> <u who="nm0885" trans="pause"> yeah <pause dur="0.2"/> L-nought-<pause dur="0.5"/>sub-nu <pause dur="0.4"/> and finishing up at L-D-<pause dur="0.2"/>sub-nu <pause dur="0.7"/> so <pause dur="0.9"/> that's our integral <pause dur="0.5"/> it's now a matter of doing it <pause dur="0.7"/> now <pause dur="0.9"/> left hand side <pause dur="0.6"/> is <pause dur="0.5"/> pretty simple what's the integral of D-L-over-L </u><pause dur="1.5"/>

<u who="sf0896" trans="pause"> <gap reason="inaudible" extent="1 sec"/></u> <pause dur="0.4"/> <u who="nm0885" trans="pause"> right log-L <pause dur="1.0"/> so <pause dur="0.3"/> on this side we get <pause dur="0.8"/> log <pause dur="0.3"/> of L-nu <pause dur="1.7"/> and you're <pause dur="0.2"/> ranging from L-nought to L-D-<pause dur="0.4"/>sub-nu <pause dur="0.3"/> and on this side <pause dur="1.0"/> # what do we get there <pause dur="6.6"/> this is a trick question so have a guess at it and <trunc>y</trunc> and <trunc>y</trunc> and you'll be wrong but but you won't be have to be embarrassed because it's <pause dur="0.3"/> it's a trick question </u><pause dur="4.0"/> <u who="sf0896" trans="pause"> Z </u><pause dur="1.7"/> <u who="nm0885" trans="pause"> pardon </u><u who="sf0896" trans="latching"> # <pause dur="0.3"/> Z times all that </u><pause dur="1.6"/> <u who="nm0885" trans="pause"> Z times <pause dur="0.2"/> that stuff okay <pause dur="0.2"/> and that <pause dur="0.2"/> is indeed the wrong answer <pause dur="0.4"/> and <vocal desc="laugh" iterated="n"/><pause dur="0.3"/> the <trunc>r</trunc> <trunc>r</trunc> <pause dur="0.7"/> but it's exactly what i was hoping you'd say <pause dur="0.4"/> because that would be true <pause dur="0.3"/> if all this stuff was constant <pause dur="0.4"/> then you could just pull it out of the integral sign <pause dur="0.5"/> but <pause dur="0.4"/> these properties may well the secant-theta shouldn't vary <pause dur="0.8"/> but the other stuff <pause dur="0.3"/> might potentially vary as you go through the layer and certainly you know if you imagine a beam of sunlight going down through the atmosphere <pause dur="0.4"/> # the density of just about everything <pause dur="0.2"/> is going to be increasing as as you go down to the surface <pause dur="0.5"/> and probably the only significant thing where that's not really true <pause dur="0.4"/> is

going to be something like ozone where it's really varying wildly <pause dur="0.4"/> or clouds or something <pause dur="0.3"/> so we can't do much <pause dur="0.2"/> with that side <pause dur="0.3"/> so what i'm going to do is i'm just going to leave that side as <pause dur="0.2"/> an integral <pause dur="2.0"/> from Z-one to Z-two K-nu-E-rho-secant-<pause dur="0.5"/>theta-<pause dur="1.1"/>D-Z <pause dur="1.0"/> and # <pause dur="1.6"/> i'm going <pause dur="0.6"/> to give it <pause dur="0.3"/> a name <pause dur="2.9"/> and the name is going to be <pause dur="0.6"/> optical thickness or optical depth <pause dur="0.3"/> so i'm going to write that as minus-<pause dur="0.7"/>delta lower case delta <pause dur="0.5"/> and the # and we'd better put a subscript-nu on that <pause dur="0.3"/> because this is going to be a function of frequency <pause dur="3.3"/> so <pause dur="1.2"/> where <pause dur="0.8"/> delta-nu <pause dur="0.2"/> is equal <pause dur="0.5"/> by definition to this integral Z-one to Z-two <pause dur="0.7"/> extinction coefficient <pause dur="0.2"/> times density times <pause dur="0.8"/> pathlength <pause dur="0.5"/> secant-theta-D-Z <pause dur="1.1"/> is the <pause dur="1.2"/> optical <pause dur="2.1"/> thickness <pause dur="1.6"/> or <pause dur="1.5"/> optical depth <pause dur="3.0"/> these two terms mean the same thing and <pause dur="0.4"/> people will tend to use them <trunc>in</trunc> interchangeably <pause dur="1.5"/><vocal desc="clears throat" iterated="n"/><pause dur="0.6"/> so what that is is that's a <trunc>m</trunc> <pause dur="0.4"/> is that's going to be some measure <pause dur="0.4"/> of how much <pause dur="0.2"/> stuff is getting pulled out of the beamer or what or what fraction of it <pause dur="1.0"/> and if we define the optical depth this way <pause dur="1.5"/> then <pause dur="0.3"/>

we can say that <pause dur="0.3"/> our <pause dur="1.7"/> direct beam <pause dur="0.3"/> L-D-of-nu <pause dur="0.4"/> is going to be equal to <pause dur="1.3"/> the initial beam times <pause dur="0.5"/> E-to-the-minus-<pause dur="0.3"/>delta-nu <pause dur="3.5"/> just rearranging that equation <pause dur="3.6"/> and this <pause dur="0.2"/> is a very important relationship i mean in some sense in in the theory of radiative transfer this was the first <pause dur="0.4"/> really quantitative law <pause dur="0.3"/> that was discovered <pause dur="0.7"/> and # <pause dur="1.4"/> and so it's named after its <pause dur="0.2"/> discoverer <pause dur="0.9"/> not what he was drinking at the time <pause dur="0.4"/> Beer's law <pause dur="1.4"/> # you will also find <pause dur="0.3"/> this is sometimes called <pause dur="1.3"/> Lambert's law <pause dur="2.6"/> and it is also sometimes called <pause dur="3.0"/> Bouguer's law <pause dur="2.6"/> and sometimes it's called the Beer-Lambert law and sometimes it's called the Lambert-Bouguer law and sometimes it's called the Beer-Lambert-Bouguer law <pause dur="0.3"/> or whatever <pause dur="0.3"/> # <pause dur="0.3"/> i think it <pause dur="0.2"/> probably has to do with which nation you are and which nationality this person <pause dur="0.8"/> these people are <pause dur="0.3"/> # <pause dur="1.5"/> 'cause this 'cause this relationship was probably discovered independently several times <pause dur="0.5"/> we're going to call it Beer's law <pause dur="0.5"/> in this course just for simplicity <pause dur="1.3"/> okay <pause dur="0.2"/> and so what it says is that the

amount of radiation that's coming through directly <pause dur="0.4"/> is # <pause dur="0.4"/> decreasing exponentially <pause dur="0.7"/> in <pause dur="0.5"/> the optical <pause dur="0.2"/> thickness <pause dur="0.3"/> so the fraction of energy <pause dur="5.0"/> the fraction of <pause dur="1.2"/> well radiance <pause dur="3.5"/> that <pause dur="0.8"/> emerges <pause dur="1.3"/> is <pause dur="1.8"/> given <pause dur="0.4"/> by <pause dur="0.6"/> the <pause dur="1.6"/> monochromatic <pause dur="1.6"/> transmittance <pause dur="4.2"/> and it's monochromatic because we're always doing this as a function of wave number <pause dur="0.2"/> or or sorry as a function of <pause dur="0.5"/> # frequency but we can always add up over frequency <pause dur="0.4"/> and that is defined as <pause dur="0.2"/> tau <pause dur="1.5"/> lower case Greek letter tau <pause dur="0.6"/> and that by definition is going to be the amount that's coming out the bottom L-D <pause dur="0.3"/> divided by the amount that's going in the top L-nu <pause dur="0.6"/> and you can see immediately <pause dur="0.2"/> that that's directly related to the optical depth <pause dur="0.9"/> E-to-the-minus-delta-nu <pause dur="3.4"/> so there's a couple of different ways that we can think about how much is going through we can think about a transmittance <pause dur="0.4"/> <trunc>wh</trunc> so you may have a transmittance of point-five <pause dur="0.3"/> which says half the stuff gets through <pause dur="0.6"/> # you may have an optical depth here <pause dur="0.4"/> and the <pause dur="0.2"/> thing about the optical depth is that's going

to be directly proportional <pause dur="0.4"/> to things like pathlength and density and so on so if you double the density you double the optical depth <pause dur="2.5"/> and # <pause dur="1.6"/> yeah so that that's that's the amount <pause dur="0.8"/> that gets through <pause dur="0.6"/> okay so that's transmittance but of course remember <pause dur="0.3"/> there are two other things <pause dur="0.3"/> two other ways that energy is coming out the bottom of the slab <pause dur="0.5"/>

what were they again </u><pause dur="3.3"/> <u who="sf0897" trans="pause"> scattering </u><pause dur="0.5"/> <u who="nm0885" trans="pause"> scattering <pause dur="0.9"/> and </u><pause dur="6.9"/> <u who="sf0898" trans="pause"> emission </u><pause dur="0.4"/> <u who="nm0885" trans="pause"> emission thank you <pause dur="0.5"/> it's one of these cases where <pause dur="0.3"/> everyone's whispering the answer but no one's saying it loud enough <pause dur="0.4"/> all right <pause dur="0.2"/> so <pause dur="0.4"/> so the slab can be emitting radiation <pause dur="0.6"/> and also radiation that's entering the slab from other directions can be getting bounced into this direction <pause dur="0.4"/> so let's let's write down some expressions for those ones as well <pause dur="4.3"/><kinesic desc="writes on board" iterated="y" dur="50"/> and we'll start by taking emission <pause dur="4.0"/> so <pause dur="2.4"/> energy <pause dur="2.1"/> emitted <pause dur="0.6"/> by <pause dur="1.0"/> the slab <pause dur="4.6"/> now once again <pause dur="0.4"/> this is going to depend on <pause dur="0.4"/> a couple of different factors so <pause dur="0.7"/> this <pause dur="1.3"/> depends <pause dur="0.3"/> on <pause dur="1.0"/> what's going to determine how much energy the slab is emitting <pause dur="0.7"/> at some given wavelength <pause dur="4.7"/>

temperature <pause dur="0.4"/> is going to be <pause dur="0.4"/> a big one so this depends on <pause dur="2.2"/> temperature <pause dur="0.7"/> and what's the relationship that tells us how much emission we're going to get <pause dur="0.2"/> at a given temperature </u><pause dur="4.8"/> <u who="sm0899" trans="pause"> Wein's law <gap reason="inaudible" extent="1 sec"/></u><pause dur="0.6"/> <u who="nm0885" trans="pause"> # Wein's law's going to tell us the wavelength where we get the most emission but what's going to give us the whole function at a as a function <pause dur="0.3"/> of any wavelength </u><pause dur="0.6"/> <u who="sm0900" trans="pause"> Stefan </u><pause dur="0.4"/> <u who="sf0901" trans="pause"> <gap reason="inaudible" extent="2 secs"/></u><pause dur="0.3"/> <u who="nm0885" trans="pause"> # <pause dur="2.0"/> the Stefan-Boltzmann one is isn't the one i was looking for because that's giving you the total <pause dur="0.8"/> i'm saying if you just have some arbitrary wavelength that's not at the peak what's the function that's going to tell you how much you get at that wavelength <pause dur="8.2"/> we've done it in this course we've written it out on the board <pause dur="0.8"/> i've shown plots of it on the projector </u><pause dur="0.8"/> <u who="sf0902" trans="pause"> Planck's law </u><pause dur="0.3"/> <u who="nm0885" trans="pause"> Planck's law that's right <pause dur="1.2"/> so the Planck function <pause dur="1.5"/> so <pause dur="0.6"/> the energy emitted depends on the temperature <pause dur="0.4"/> and that's given by the # Planck function <pause dur="2.7"/><kinesic desc="writes on board" iterated="y" dur="2"/> and # <pause dur="3.0"/> well <pause dur="0.8"/> what else is it going to depend on <pause dur="1.4"/> so <pause dur="0.2"/> if <pause dur="0.3"/> if it's at a given temperature <pause dur="0.7"/> what else is going to

affect the amount of emission that we get </u><pause dur="0.7"/> <u who="sf0903" trans="pause"> <gap reason="inaudible" extent="1 sec"/> </u><pause dur="0.8"/> <u who="nm0885" trans="pause"> sorry </u><pause dur="0.5"/> <u who="sf0903" trans="pause"> is it the amount of material </u><pause dur="0.4"/> <u who="nm0885" trans="pause"> the amount of material absolutely <pause dur="4.7"/><kinesic desc="writes on board" iterated="y" dur="9"/> the amount <pause dur="0.2"/> of material <pause dur="4.2"/> does it depend on anything else </u><pause dur="7.8"/> <u who="sf0904" trans="pause"> density of material </u><pause dur="1.0"/> <u who="nm0885" trans="pause"> # <pause dur="0.2"/> that's another <trunc>w</trunc> i would i'd say it's another way of saying amount <pause dur="0.3"/> okay </u><pause dur="1.6"/> <u who="sm0905" trans="pause"> the heat capacity of the material </u><pause dur="1.1"/> <u who="nm0885" trans="pause"> # <pause dur="5.4"/> probably i'm <pause dur="0.4"/> i <trunc>w</trunc> i wouldn't write it in terms of heat capacity but it does depend on the type of material <pause dur="0.4"/> different materials emit <pause dur="0.4"/> with with # <pause dur="1.0"/> <trunc>wi</trunc> with with different <pause dur="0.4"/> efficiency <pause dur="1.3"/> so it's it's also going to <pause dur="0.5"/> # <kinesic desc="writes on board" iterated="y" dur="6"/> depend on the # <pause dur="0.5"/> type <pause dur="0.8"/> of <pause dur="1.5"/> material <pause dur="0.3"/> so some materials will be better emitters <pause dur="0.2"/> than others <pause dur="0.3"/> and especially better emitters at given wavelengths and this'll depend on the molecular structure and things like that which <trunc>w</trunc> <pause dur="0.3"/> which we'll see <pause dur="0.5"/> later on <pause dur="0.8"/> # <pause dur="3.6"/> something that you <pause dur="2.3"/> well <pause dur="0.9"/> i'll come back to that in a second <pause dur="0.2"/> let's write it down <pause dur="0.2"/> okay <pause dur="0.3"/> so let's write the emitted radiance <pause dur="0.6"/> and so <kinesic desc="writes on board" iterated="y" dur="25"/> i'm going to denote that as L with a superscript-capital-E <pause dur="0.4"/> for the emitted part of the radiance subscript-nu 'cause we're at one given

frequency <pause dur="0.5"/> again <pause dur="0.7"/> and oops <pause dur="1.4"/> and let me not forget to write down the direction <pause dur="0.2"/> theta-phi <pause dur="0.7"/> so this is going to depend on <pause dur="0.7"/> these various properties <pause dur="0.4"/> so it's going to depend on the <pause dur="0.5"/> Planck function <pause dur="1.9"/> which is a function of temperature <pause dur="0.4"/> now you'll notice i've written a B here <pause dur="0.4"/> whereas before i wrote an E <pause dur="0.5"/> when i wrote down the expression before <pause dur="0.3"/> i wrote down <pause dur="0.2"/> the formula for the <pause dur="0.3"/> irradiance <pause dur="0.2"/> emitted <pause dur="0.5"/> so the total of what's coming out in all directions <pause dur="0.4"/> but in this case we're actually thinking about just one direction from the slab <pause dur="0.3"/> so we need the radiance <pause dur="0.5"/> but it's easy to go between radiance and irradiance for a black body why why is that </u><pause dur="2.7"/> <u who="sf0906" trans="pause"> isotropically </u><pause dur="0.8"/> <u who="nm0885" trans="pause"> yeah that's right a black body emits isotropically so it emits the same in any direction <pause dur="0.5"/> and that means that if we know the total <pause dur="0.4"/> then <pause dur="0.4"/> we know what the amount going in any given direction is <pause dur="0.2"/> and remember <pause dur="0.6"/> and so <pause dur="1.4"/> let's actually write this down where <pause dur="0.8"/><kinesic desc="writes on board" iterated="y" dur="32"/> B-nu-of-T <pause dur="0.2"/> is <pause dur="0.6"/> the <pause dur="1.6"/> Planck <pause dur="1.4"/> function <pause dur="3.3"/> in <pause dur="0.3"/> radiance units <pause dur="4.3"/> you use the Planck Planck function in

radiance units <pause dur="0.4"/> and that means that # <pause dur="1.4"/> if you'll recall and i don't necessarily expect you to remember this just at this moment <pause dur="0.3"/> but you might want to remember it for the exam <pause dur="0.6"/> is that # <pause dur="0.2"/> the difference between <pause dur="0.4"/> the radiance and the irradiance for an isotropic emitter is just a factor of pi <pause dur="0.7"/> so <pause dur="0.4"/> in one direction it's going to be <kinesic desc="writes on board" iterated="y" dur="5"/> E-nu-<pause dur="0.2"/>of-T <pause dur="0.3"/> divided by <pause dur="0.3"/> pi <pause dur="3.5"/> so <pause dur="1.3"/> we've got the # <pause dur="1.6"/> the temperature dependence in there because now we've got the Planck function <pause dur="1.3"/> for the radiance <pause dur="1.3"/> okay the next thing we want to do <pause dur="0.6"/> is we want to put in something about <pause dur="0.4"/> the type of material <pause dur="1.2"/> and we'll write this in terms of the <pause dur="0.8"/> mass absorption <pause dur="0.4"/> coefficient <pause dur="1.1"/> so K-A-of-nu is the mass absorption coefficient which we've already <trunc>def</trunc> <pause dur="0.5"/> which we've already <pause dur="0.3"/> mentioned <pause dur="0.4"/> why am i writing the <pause dur="0.2"/> mass absorption coefficient <pause dur="0.3"/> when i'm talking about emission <pause dur="0.3"/> here <pause dur="2.1"/> this is a question for somebody who's done the right physics course this is not something you're going to be able to figure out on the spot <pause dur="2.8"/> unless your name is

Kirchoff <pause dur="0.8"/> and you're really clever <pause dur="1.8"/> or you might be really clever but if your name isn't Kirchoff they still won't name the law after you <pause dur="0.7"/><vocal desc="laugh" iterated="n"/> </u><pause dur="6.4"/> <u who="sf0907" trans="pause"> is it because good emitters are good <trunc>abs</trunc> good absorbers </u><pause dur="0.6"/> <u who="nm0885" trans="pause"> that's <pause dur="0.2"/> exactly right <pause dur="0.7"/> and that that's Kirchoff's law <pause dur="0.3"/> and what Kirchoff's law says and this is what what you have to <trunc>wor</trunc> work your way through in a physics course <pause dur="0.5"/> is <pause dur="0.2"/> Kirchoff's law <pause dur="0.2"/> says <pause dur="0.3"/> that if you're in thermodynamic <pause dur="0.2"/> if you're in a local thermodynamic equilibrium <pause dur="0.6"/> then <pause dur="0.7"/> a property must be equally good at absorbing and emitting <pause dur="0.5"/> and the reason for that is if you're in thermodynamic equilibrium <pause dur="0.5"/> you're # you've got to be at the same temperature <pause dur="0.7"/> in in different parts that are that are close together <pause dur="0.9"/> and if you're at <pause dur="0.5"/> and if you're at the same temperature <pause dur="0.4"/> but <pause dur="0.5"/> one part of <trunc>th</trunc> <trunc>th</trunc> <trunc>b</trunc> but your substance is a good emitter <pause dur="0.7"/> # but a very bad <pause dur="0.4"/> absorber <pause dur="0.2"/> or something like that then what that means <pause dur="0.2"/> is that you'll end up with a flux of energy going from one part to the other <pause dur="0.5"/> and one part of your substance will try and

heat up relative to the other <pause dur="0.5"/> and that's a contradiction with the thermodynamic equilibrium <pause dur="0.6"/> so <pause dur="0.2"/> if you're going to have an equilibrium and have nearby stuff at the same temperature <pause dur="0.2"/> you've got to be equally good at absorbing and emitting <pause dur="0.4"/> so that there's no net flow of energy from one place to another by radiation <pause dur="0.8"/> so <pause dur="0.8"/> anyway i mean <trunc>th</trunc> that that kind of detail doesn't matter <pause dur="0.4"/> but <pause dur="1.0"/> what matters for our purposes is <pause dur="0.3"/> is that for any part of the atmosphere that we're interested in and this means not the <pause dur="0.6"/> the far outer thermosphere or whatever <pause dur="0.4"/> # <pause dur="0.4"/> we're in thermodynamic equilibrium at least locally <pause dur="0.5"/> and <pause dur="0.6"/> the <pause dur="0.3"/> absorption coefficient and the emission coefficient <pause dur="0.2"/> are going to be <pause dur="0.4"/> the same <pause dur="0.2"/> thing <pause dur="1.3"/> so # <pause dur="2.7"/> K-A-of-nu <pause dur="1.2"/> # <pause dur="2.5"/> well gives <pause dur="1.2"/> the <pause dur="1.3"/> emission <pause dur="1.1"/> properties <pause dur="4.1"/> of the material <pause dur="1.8"/> for <pause dur="1.9"/> substances <pause dur="2.1"/> in <pause dur="2.5"/> thermodynamic <pause dur="1.5"/> equilibrium <pause dur="4.1"/> and i'll just put in brackets here that this is # Kirchoff's law <pause dur="8.6"/> so <pause dur="0.2"/> it's the same for absorption <pause dur="0.6"/> so K is the same for absorption and emission <pause dur="0.9"/> okay <pause dur="0.2"/> final bit we need is the amount of

material how are we going to write the amount of material <pause dur="1.3"/> this is an easy one 'cause we've done it before </u><pause dur="6.1"/> <u who="sf0908" trans="pause"> density </u><pause dur="0.9"/> <u who="nm0885" trans="pause"> sorry </u><pause dur="0.5"/> <u who="sf0908" trans="pause"> <gap reason="inaudible" extent="1 sec"/></u><pause dur="0.4"/> <u who="nm0885" trans="pause"> density times </u><pause dur="4.3"/> <u who="sm0909" trans="pause"> volume </u><pause dur="2.3"/> <u who="nm0885" trans="pause"> volume well <pause dur="0.2"/> we're going to do it per unit area so density times pathlength <pause dur="0.8"/> and that gives you the <trunc>am</trunc> the <trunc>amou</trunc> the amount of mass per unit area <pause dur="0.4"/> and so that's going to be <pause dur="0.2"/><kinesic desc="writes on board" iterated="y" dur="2"/> rho-secant-theta-<pause dur="0.2"/>D-Z <pause dur="0.4"/> we've already worked that one out <pause dur="1.2"/> okay <pause dur="0.3"/> so that's the emission <pause dur="1.0"/> the emitted part of it <pause dur="0.3"/> and as long as we know the temperature <pause dur="0.5"/> we know the properties of the material that we're dealing with and and i've sort of glossed over how this one comes about so we're going to have to go back to this later in the course <pause dur="0.5"/> and if you know the pathlength <pause dur="0.3"/> then you can work out how much radiance is going to be emitted <pause dur="0.4"/> from a given <pause dur="0.7"/> slab of atmosphere <pause dur="0.3"/> and going off in a given direction <pause dur="0.5"/> but of course <pause dur="0.5"/> the emission's isotropic so here it's going to be the same in all the different directions anyway <pause dur="1.1"/> okay <pause dur="0.3"/> final point <pause dur="0.3"/> that we need to get in this <pause dur="0.5"/> is we need the

scattered bit <pause dur="1.1"/> so <pause dur="0.3"/> our next <pause dur="0.4"/> thing to talk about here <pause dur="0.4"/> is <pause dur="0.4"/> radiation <pause dur="0.7"/> # <pause dur="2.7"/> scattered <pause dur="1.3"/> in <pause dur="0.7"/> from <pause dur="1.0"/> other <pause dur="0.3"/> directions <pause dur="10.6"/> so if we're sitting below our slab <pause dur="1.5"/> here <pause dur="0.4"/> and we're observing <pause dur="0.5"/> a scattered radiance L-S-of-nu which is coming out in that direction <pause dur="0.6"/> # that radiance could be getting into that direction <pause dur="0.5"/> from <pause dur="0.4"/> anywhere else <pause dur="0.3"/> and it could be coming from above the slab it could be coming from below the slab <pause dur="0.4"/> or whatever <pause dur="1.0"/>

so there could be a contribution <pause dur="0.3"/> to this scattered irradiance <pause dur="0.3"/> coming from <pause dur="0.6"/> any direction so <pause dur="1.8"/> by the way is this pen fading out too much can you still read it <pause dur="1.2"/> okay give <trunc>m</trunc> give me a <trunc>sh</trunc> give me a shout if it's if it's getting bad <pause dur="0.6"/> so <pause dur="0.9"/> radiance <pause dur="2.4"/> can be <pause dur="0.8"/> scattered <pause dur="1.9"/> can be scattered <pause dur="1.9"/> into <pause dur="2.2"/> direction <pause dur="1.7"/> theta-phi <pause dur="2.1"/> from <pause dur="0.9"/> any <pause dur="1.1"/> other <pause dur="0.4"/> direction <pause dur="2.6"/> and so we'll write an arbitrary other <pause dur="0.5"/> direction <pause dur="0.5"/> as <pause dur="0.3"/> a # <pause dur="1.6"/> as a <pause dur="0.8"/> theta-primed-<pause dur="0.4"/>phi-primed <pause dur="0.8"/> so from any direction theta-prime-phi-prime <pause dur="0.5"/> the energy could potentially <pause dur="0.4"/> be scattered in into the direction that we're interested in <pause dur="15.8"/> so <pause dur="0.5"/> what's the scattered energy <pause dur="0.3"/> going to

depend on </u><pause dur="21.6"/> <u who="sf0910" trans="pause"> single scattering albedo </u><pause dur="1.1"/> <u who="nm0885" trans="pause"> right it's going to depend on the single scattering albedo <pause dur="0.6"/> can you remember what that defines what does the single scattering albedo measure </u><pause dur="4.5"/> <u who="sf0911" trans="pause"> # <gap reason="inaudible" extent="4 secs"/></u><pause dur="1.3"/> <u who="nm0885" trans="pause"> # <pause dur="0.7"/> sounds right <pause dur="0.3"/> yeah it's it's going to measure <pause dur="0.4"/> if the radiation interacts with the material <pause dur="0.3"/> what fraction of it is going to be scattered <pause dur="0.4"/> rather than absorbed so big albedo <pause dur="0.2"/> omega-equals-one <pause dur="0.8"/> everything that interacts at all is going to be scattered <pause dur="1.1"/> small albedo <pause dur="0.2"/> omega-equals-zero <pause dur="0.4"/> anything that interacts at all is going to be absorbed <pause dur="0.8"/> so <pause dur="4.5"/> so the # <pause dur="2.7"/> scattered <pause dur="1.2"/> radiation <pause dur="1.6"/> depends <pause dur="0.9"/> on <pause dur="0.9"/> a single scattering albedo <pause dur="0.2"/> omega-nu <pause dur="1.0"/> now of course it <trunc>de</trunc> if it depends on the <trunc>sing</trunc> single scattering albedo <pause dur="0.4"/> that's what happens to it <pause dur="0.2"/> if it interacts in the first place <pause dur="0.5"/> but # <pause dur="0.9"/> what's going to determine <pause dur="0.4"/> the probability that it interacts in the first place <pause dur="1.2"/> what are we going to need to worry about there </u><pause dur="1.8"/> <u who="sf0912" trans="pause"> pathlength </u><pause dur="0.5"/> <u who="nm0885" trans="pause"> yeah <pause dur="0.2"/> it's going to depend on <pause dur="0.5"/> on the pathlengths or it's going to depend on the <pause dur="0.4"/> amount of material <pause dur="4.3"/> which of course you

realize is going to be given by the density times the pathlength <pause dur="0.9"/> # what else is it going to depend on </u><pause dur="12.0"/> <u who="sf0913" trans="pause"> size of <pause dur="1.0"/> particle </u><pause dur="1.5"/> <u who="nm0885" trans="pause"> # <pause dur="1.8"/> it is <pause dur="0.4"/> actually <pause dur="0.9"/> but i wasn't going to tell you that until next week <pause dur="0.7"/> # <pause dur="0.2"/><vocal desc="laugh" iterated="n"/><pause dur="0.2"/> let's say for the moment it depends on again the properties of the material <pause dur="0.5"/> and you're right it's going to turn out that size of the particles or size of the molecules is the <trunc>p</trunc> is the property that matters most <pause dur="0.7"/> but it's going to depend on the <pause dur="0.3"/> # <pause dur="2.3"/> type <pause dur="0.3"/> of <pause dur="0.9"/> material or properties of the material <pause dur="0.4"/> and so that's going to be given by <pause dur="0.2"/> the # <pause dur="4.8"/> well we we can write it <pause dur="0.4"/> in terms of <pause dur="0.8"/> the extinction coefficient <pause dur="0.2"/> K-E-of-nu which is the probability <pause dur="0.3"/> that it's going to interact with the radiation <pause dur="1.6"/> or equivalently <pause dur="0.3"/> we can write it as the <pause dur="0.5"/> mass scattering coefficient K-S-nu <pause dur="0.7"/> and the reason for that is because if we know the single scattering albedo that's the ratio between the two of these <pause dur="0.4"/> so if <pause dur="0.3"/> all we if we have any two of these three we can work out what the third one is <pause dur="1.4"/> so it's going to depend

on the type of material it's going to depend on <pause dur="0.3"/> how much material <pause dur="0.6"/> # <pause dur="1.0"/> there's another pretty obvious one here <pause dur="2.2"/> suppose that # <pause dur="1.4"/> the sun goes very high in the sky and you now have more radiation <pause dur="0.2"/> incident <pause dur="1.4"/> what's going to happen to the scattering then </u><pause dur="1.9"/> <u who="sm0914" trans="pause"> <gap reason="inaudible" extent="1 sec"/></u><pause dur="2.4"/> <u who="nm0885" trans="pause"> yeah it's going to increase if you have more incident radiation you're going to get more scattered <pause dur="0.3"/> into any <pause dur="0.3"/> direction so it's going to depend on the <pause dur="0.7"/> incident <pause dur="0.4"/> radiance <pause dur="3.3"/> and so that's going to be <pause dur="1.4"/> L-nu <pause dur="1.4"/> in what direction <pause dur="3.2"/> is the radiation coming in <pause dur="0.2"/> that we want to worry about <pause dur="1.6"/> someone's doing this which i think means any direction <pause dur="0.6"/> yeah <pause dur="0.3"/> so any direction so if we just pick an arbitrary direction <pause dur="0.4"/> we can say <pause dur="0.5"/> in a direction <pause dur="0.2"/> theta-prime-phi-prime <pause dur="0.6"/> but <pause dur="0.5"/> this brings us on to the last factor <pause dur="0.3"/> which is really crucial here <pause dur="0.6"/> is # <pause dur="1.4"/> the contributions from different directions <pause dur="0.3"/> theta-prime-phi-prime <pause dur="0.4"/> because if radiation gets scattered off some particle <pause dur="0.2"/> a molecule or a <pause dur="0.4"/> <trunc>wa</trunc> water drop an ice crystal whatever <pause dur="0.3"/> it's not going to <pause dur="0.4"/> you're

not going to get the same amount going off in every direction the scattering is not <pause dur="0.3"/> isotropic <pause dur="0.3"/> the radiation will <pause dur="0.2"/> interact with the particle <pause dur="0.3"/> and it'll be sent off in different amounts in different directions <pause dur="0.3"/> so the other thing <pause dur="0.3"/> that we're going to have to know here <pause dur="0.3"/> is we're going to have to know <pause dur="0.2"/> that property of of of of the material we're going to have to know <pause dur="0.4"/> which direction <pause dur="0.4"/> the scattered radiation <pause dur="0.2"/> is going to go in <pause dur="0.2"/> in any <pause dur="0.5"/> given interaction so <pause dur="0.3"/> and <pause dur="0.5"/> the <pause dur="2.1"/> direction <pause dur="0.7"/> directional <pause dur="0.6"/> dependence <pause dur="4.6"/> of the scattering <pause dur="4.0"/> and so we're going to need to define <pause dur="0.4"/> an expression <pause dur="0.4"/> for this <pause dur="2.3"/> so <pause dur="0.6"/> let's <pause dur="0.6"/> define <pause dur="2.8"/> the phase function <pause dur="5.5"/> so the phase function <pause dur="0.8"/> which we'll write with a # <pause dur="0.7"/> capital-P <pause dur="0.4"/> from direction <pause dur="0.2"/> omega-prime <pause dur="0.7"/> into <pause dur="0.6"/> direction <pause dur="1.0"/> omega <pause dur="0.5"/> and i've used solid angle here so omega-prime would be theta-prime-phi-prime <pause dur="0.4"/> and omega would be theta-phi <pause dur="1.3"/> so what this is going to be <pause dur="0.4"/> is this is going to be <pause dur="1.6"/> the <pause dur="1.1"/> probability <pause dur="2.6"/> that <pause dur="2.1"/> a photon <pause dur="1.9"/> incident <pause dur="7.2"/> in direction <pause dur="1.6"/> omega-prime <pause dur="2.3"/> is scattered <pause dur="5.6"/> into <pause dur="1.1"/> the direction <pause dur="1.2"/> omega <pause dur="1.5"/>

so <pause dur="0.9"/> some radiance <pause dur="0.3"/> comes from or or is initially travelling in in direction <pause dur="1.1"/> omega-prime <pause dur="0.6"/> then <pause dur="1.1"/> in this certain pathlength and given this certain nature of material it turns out that it <pause dur="0.8"/> interacts with one of the particles there <pause dur="0.3"/> now it could in principle get sent off in any direction <pause dur="0.5"/> but the proportion that gets sent off <pause dur="0.3"/> in the direction that we're interested in <pause dur="0.4"/> omega <pause dur="0.7"/> omega <pause dur="0.4"/> is given by the phase function <pause dur="1.7"/> so the phase function says whichever angle you come from <pause dur="0.3"/> what are the odds that you're going to get scattered into the direction that we want <pause dur="2.5"/> and # again <pause dur="0.3"/> when we get on to talking about scattering in more detail <pause dur="0.3"/> we're going to have to think about what this phase function looks like so what direction does <pause dur="0.4"/> the radiance get scattered <pause dur="3.3"/> okay so # <pause dur="2.2"/> what this means then now <pause dur="0.3"/> is that if we want to get <pause dur="0.4"/> the radiance <pause dur="2.3"/> which is scattered into our direction omega <pause dur="0.4"/> what we're going to need to do <pause dur="0.3"/> is we're going to need to integrate <pause dur="0.3"/> over this phase function 'cause we're going to need to

add up the contributions <pause dur="0.3"/> from all the different possible directions <pause dur="0.7"/> that get bounced into our direction <pause dur="1.0"/> and <pause dur="0.2"/> i'm fed up with this pen so i'm going to <pause dur="1.5"/><event desc="puts pen in bin" iterated="n"/> throw it away <pause dur="4.6"/> lovely <pause dur="0.7"/> okay <pause dur="1.5"/> so # <pause dur="0.9"/> so <pause dur="2.6"/> we must integrate <pause dur="3.3"/> over <pause dur="0.9"/> all <pause dur="0.5"/> incident <pause dur="1.4"/> directions <pause dur="3.2"/> omega-prime <pause dur="1.2"/> to <pause dur="2.3"/> to get <pause dur="0.6"/> the <pause dur="1.3"/> total <pause dur="2.1"/> emerging <pause dur="2.3"/> radiance <pause dur="3.1"/> in direction <pause dur="1.5"/> omega <pause dur="4.0"/> so <pause dur="0.5"/> what does this mean well <pause dur="0.5"/> what this means is if we want to write <pause dur="0.6"/> our <pause dur="0.4"/> scattered radiance <pause dur="0.4"/> so L-superscript-capital-S-<pause dur="0.5"/>sub-nu <pause dur="0.8"/> and # <pause dur="1.1"/> forgot here let's <pause dur="1.0"/> in the direction <pause dur="0.2"/> omega <pause dur="1.1"/> we can now <pause dur="1.8"/> write this <pause dur="0.7"/> as # <pause dur="3.6"/> what are we going to have well we've got to <pause dur="0.2"/> get in <pause dur="0.4"/> all our <pause dur="0.8"/> little bits of stuff here <pause dur="0.4"/> it's going to be in terms of the # <pause dur="3.0"/> of the <pause dur="1.7"/> well let's <trunc>ca</trunc> let's write it as the mass scattering coefficient <pause dur="0.4"/> so i could also write this as <pause dur="0.2"/> omega-nu times the extinction coefficient <pause dur="0.5"/>

let's write it as as the scattering coefficient <pause dur="0.7"/> then <pause dur="0.5"/> an integral <pause dur="1.0"/> over all possible directions <pause dur="0.6"/> of # <pause dur="6.0"/> of the <pause dur="0.6"/> incident radiance <pause dur="0.2"/> in that direction <pause dur="0.9"/> so <pause dur="0.4"/> L-nu-<pause dur="1.2"/>omega-prime <pause dur="1.5"/> and # <pause dur="2.9"/> the <pause dur="1.5"/> multiply by <pause dur="0.3"/> the phase function <pause dur="1.0"/> which is the <pause dur="0.3"/> probability <pause dur="0.6"/> that # <pause dur="0.5"/> you're going to go from <pause dur="0.3"/> omega-prime <pause dur="0.4"/> into direction <pause dur="0.2"/> omega <pause dur="5.9"/> and let me <pause dur="0.3"/> clear a little bit more space here <pause dur="2.5"/> then we're going to need to write <pause dur="0.6"/> the # <pause dur="0.4"/> pathlength <pause dur="0.7"/> which is part of the probability that you're going to get scattered <pause dur="0.4"/> at all <pause dur="0.3"/> so we've got a rho-secant-theta-<pause dur="2.8"/>D-Z <pause dur="4.0"/> and # <pause dur="1.2"/> and then we're going to need to integrate over <pause dur="0.3"/> all directions <pause dur="0.2"/> omega-prime <pause dur="0.7"/> and actually i mean it's a <trunc>f</trunc> a few things like like the D-Z we can take out of that integral <pause dur="0.3"/> but the <trunc>pa</trunc> but the the <trunc>s</trunc> the secant-theta part you know the pathlength is going to depend on which direction <pause dur="0.4"/> the radiance comes in <pause dur="1.1"/> because if something's coming in at a different angle it's going to

have a different probability of getting scattered if it's coming in at a low angle <pause dur="0.4"/> it's going to get scattered <pause dur="0.7"/> much more <pause dur="1.8"/> so <pause dur="0.8"/> what we'll do is we'll integrate this <pause dur="0.3"/> over <pause dur="1.9"/> the whole sphere <pause dur="2.1"/> and that means we'll need to divide out <pause dur="0.6"/> four-pi <pause dur="0.4"/> we're integrating over the whole sphere all possible directions because radiation can <pause dur="0.3"/> can get into the slab from above and below <pause dur="0.6"/> why are we dividing out four-pi <pause dur="12.2"/> in the context of solid angle and sphere what does four-pi represent </u><pause dur="0.6"/> <u who="sf0915" trans="pause"> whole </u><pause dur="1.1"/> <u who="nm0885" trans="pause"> yeah four-pi is is the <trunc>ra</trunc> is is the solid angle <pause dur="0.2"/> of the whole sphere <pause dur="0.3"/> so we're adding up contributions from all of this <pause dur="0.3"/> and then we're dividing out the total solid angle <pause dur="0.6"/> so 'cause remember <pause dur="0.2"/> radiance is in units of per steradian <pause dur="0.4"/> so we want to keep it as per steradian <pause dur="1.7"/> okay <pause dur="0.8"/> so <pause dur="0.9"/> with that <pause dur="0.6"/> hideous mess <pause dur="0.5"/> we now have <pause dur="0.4"/> all the contributions <pause dur="0.4"/> to <pause dur="0.7"/> the # <pause dur="0.5"/> to the radiance <pause dur="0.6"/> so <pause dur="2.3"/> the radiance <pause dur="3.5"/> emerging <pause dur="2.6"/> from a slab <pause dur="1.3"/> is then going to be <pause dur="1.9"/> what is it going to be it's going to be <pause dur="1.0"/> L-nu <pause dur="1.2"/> in some direction <pause dur="1.5"/> theta-phi <pause dur="0.2"/> and it's going to be

the the sum <pause dur="0.4"/> of all these bits <pause dur="0.4"/> that we've put together <pause dur="0.4"/> so it's going to be a sum of the <pause dur="0.4"/> direct beam <pause dur="0.2"/> L-D-nu-<pause dur="0.2"/>theta-phi <pause dur="0.9"/> plus <pause dur="0.4"/> the # <pause dur="1.3"/> emitted <pause dur="0.2"/> that's right <pause dur="0.2"/> L-E-nu <pause dur="0.7"/> again what's emitted in our direction theta-phi although <pause dur="0.6"/> emission of course is going to be <pause dur="0.8"/> isotropic <pause dur="0.8"/> and it's going to be <pause dur="2.9"/> added to the scattering <pause dur="0.7"/> L-S <pause dur="0.6"/> of theta-phi <pause dur="0.7"/> so <pause dur="1.6"/> our <pause dur="0.7"/> emission at the bottom of of the slab <pause dur="0.4"/> is # <pause dur="0.4"/> is given by these three terms <pause dur="0.4"/> but remember <pause dur="0.6"/> as as we were mentioning in the last lecture <pause dur="0.3"/> these terms are not <pause dur="0.2"/> all <pause dur="0.4"/> always important <pause dur="0.3"/> at certain wavelengths <pause dur="0.4"/> the radiation is going to be <pause dur="0.3"/> dominated by # <pause dur="0.3"/> by different <pause dur="0.4"/> parts <pause dur="0.4"/> of this for example with solar radiation <pause dur="0.5"/> coming into a <trunc>fra</trunc> solar wavelengths <pause dur="0.3"/> the emission by the atmosphere is going to be negligible the atmosphere's just too cold <pause dur="0.5"/> to emit <pause dur="0.4"/> those high energy photons <pause dur="0.6"/> but on the other hand scattering is going to be <pause dur="0.2"/> pretty important <pause dur="1.1"/> and # <pause dur="0.4"/> and of course there will be some depletion of the direct beam as well <pause dur="1.7"/> okay <pause dur="5.4"/> so that's kind of cool <pause dur="0.2"/> that basically <pause dur="0.4"/>

covers the # <pause dur="0.4"/> definition <pause dur="0.2"/> section <pause dur="0.3"/> of the course we now know what we're talking about <pause dur="0.8"/> now for the rest of the course <pause dur="0.2"/> we can talk about it <pause dur="0.4"/> so what we're going to do during the rest of the course <pause dur="0.3"/> is # <pause dur="0.9"/> it's going to be divided into <pause dur="0.4"/> two <pause dur="0.4"/> big chunks <pause dur="0.3"/> okay <pause dur="0.3"/> first big chunk <pause dur="0.3"/> is we're going to do <pause dur="0.3"/> scattering <pause dur="0.9"/> and we're going to look at scattering in quite a bit of detail <pause dur="0.2"/> we're going to look at what it is in the atmosphere that does the scattering <pause dur="0.4"/> how those properties determine what the scattering does <pause dur="0.5"/> and then <pause dur="0.3"/> if you actually look up in the sky <pause dur="0.3"/> how does <pause dur="0.2"/> that knowledge of scattering actually explain the stuff <pause dur="0.3"/> that we see <pause dur="0.4"/> so we're going to do fairly obvious <pause dur="0.2"/> or <pause dur="0.8"/> reasonably obvious things like why is the sky blue <pause dur="0.4"/> why is the # <pause dur="1.0"/> why is the # well why why do rainbows form <pause dur="0.4"/> we're going to get into slightly more subtle things like why do photographers always use polarizing filters when they want to look <pause dur="0.3"/> want to photograph clouds <pause dur="0.5"/> and <pause dur="0.2"/> a few other things like that <pause dur="1.0"/> and then after that we're going

to do <pause dur="0.4"/> emission and absorption and remember emission and absorption are actually sort of two sides of the same coin so we're going to do <pause dur="0.3"/> that part together <pause dur="0.2"/> and that means the direct beam <pause dur="0.3"/> that depletion bit <pause dur="0.4"/> and the emission bit <pause dur="0.2"/> we're going to handle them together <pause dur="0.5"/> and # and there we're going to get into a little bit of the properties of the particles <pause dur="0.3"/> so we're going to get into <pause dur="0.3"/> a little bit of quantum mechanics we're not going to do it in <trunc>an</trunc> in any sort of great detail <pause dur="0.2"/> but just enough that we can see why <pause dur="0.3"/> some particles are better at certain things than some other particles <pause dur="0.5"/> and # and then we're going to get <pause dur="0.2"/> into at the end <pause dur="0.4"/> how you actually do these calculations <pause dur="0.3"/> if you happen to be interested in weather or climate or something <pause dur="0.3"/> and you want to know <pause dur="0.2"/> how to put those effects into <pause dur="0.3"/> a model or into some kind of explanation <pause dur="0.3"/> of how climate's going to change or how # <pause dur="0.6"/> or how <pause dur="0.4"/> you know is there going to be ice on the roads tomorrow morning this kind of thing we're going to get into <pause dur="0.6"/> #

how you'd actually do those calculations but we can only do that once we've actually gotten into detail of all <pause dur="0.4"/> the processes <pause dur="0.5"/> so we've just got five minutes left today <pause dur="0.3"/> so <pause dur="1.2"/> i want to # <pause dur="0.4"/> point out <pause dur="0.9"/> one slight subtlety <pause dur="0.5"/> of what we've done <pause dur="1.2"/> earlier on today <pause dur="0.2"/> now you remember <pause dur="0.4"/> optical depth <pause dur="0.4"/> is a measure <pause dur="0.3"/> of how much stuff you lose from the beam <pause dur="1.2"/> but <pause dur="0.7"/> optical depth <pause dur="0.3"/> gets used a lot you'll hear that term a lot <pause dur="0.2"/> but you want to be slightly careful with it because as we've seen <pause dur="0.3"/> optical depth is composed of two bits it's composed of absorption <pause dur="0.4"/> and scattering <pause dur="0.5"/> and those can have very different <pause dur="0.2"/> effects you know if if say you've got <pause dur="0.5"/> # a water cloud in the atmosphere with an optical depth of <pause dur="0.6"/> point-five or something like that <pause dur="0.5"/> # <pause dur="2.0"/> on a day like today which actually actually the <trunc>o</trunc> <pause dur="0.3"/> what would you guess the optical depth was going to be <pause dur="0.4"/> on a day like today <pause dur="0.9"/> that's that's that's a good question <pause dur="0.2"/> <trunc>sli</trunc> slightly subtle <pause dur="7.1"/> let me give it to you as a multiple choice <pause dur="0.7"/> ten-million <pause dur="1.0"/> one <pause dur="0.7"/> or <pause dur="0.9"/> one-over-ten-million <pause dur="1.9"/>

ten-to-the-minus-seven <pause dur="6.8"/> well <pause dur="1.5"/> think of it this way <pause dur="0.7"/> remember <pause dur="0.3"/> the transmittance the fraction of the radiation <pause dur="0.3"/> that's getting through <pause dur="0.6"/> okay <pause dur="0.3"/> suppose the optical depth <pause dur="0.2"/> were <pause dur="0.8"/> ten-to-the-minus-seven what's E-to-the-minus ten-to-the-minus-seven is that going to be a <pause dur="0.6"/> big number little number <pause dur="9.1"/> let's draw a graph <pause dur="0.3"/> so <pause dur="0.5"/> E-to-the-minus-X as a function of X <pause dur="3.7"/> what's it going to be at X-equals-zero </u><pause dur="3.5"/> <u who="sf0916" trans="pause"> one </u><pause dur="1.1"/> <u who="nm0885" trans="pause"> sorry </u><pause dur="0.4"/> <u who="sf0916" trans="pause"> one </u><pause dur="0.5"/> <u who="nm0885" trans="pause"> one <pause dur="0.7"/> okay <pause dur="0.2"/> going to be one <pause dur="0.4"/> what's it going to be as X gets big </u><pause dur="2.0"/> <u who="sm0917" trans="pause"> <gap reason="inaudible" extent="1 sec"/></u><pause dur="0.5"/> <u who="nm0885" trans="pause"> sorry </u><pause dur="2.4"/> <u who="sm0917" trans="pause"> it goes small </u><pause dur="0.6"/> <u who="nm0885" trans="pause"> yeah <pause dur="0.2"/> it's going to <pause dur="1.3"/> fade away <pause dur="2.4"/> # <pause dur="0.8"/> where is it going to get to be <pause dur="0.8"/> ten-and-one-halfish sort of thing how fast is it fading away <pause dur="6.4"/> so if there's point-one-five <pause dur="0.3"/> is that point there is that going to be closer to <pause dur="1.4"/> ten-to-the-minus-seven or ten-million or to one or <pause dur="0.5"/> which of those three </u><pause dur="0.9"/> <u who="sf0918" trans="pause"> one </u><pause dur="1.3"/> <u who="nm0885" trans="pause"> pardon </u><u who="sf0918" trans="latching"> <gap reason="inaudible" extent="1 sec"/></u><pause dur="0.5"/> <u who="nm0885" trans="pause"> yeah it's <trunc>go</trunc> it's going to be kind of oneish 'cause <pause dur="0.5"/> if if if <pause dur="0.4"/> if X is one then that's going to be E-to-the-minus-one which is going to be one-over-E <pause dur="0.3"/> which is one-over-two-point-seven whatever <pause dur="0.5"/> so <pause dur="1.5"/>

so basically <pause dur="0.4"/> if you have <pause dur="0.3"/> an optical depth which is really small sort of ten-to-the-minus-seven or something like that <pause dur="0.3"/> then basically what you're saying is the transmission is pretty much one <pause dur="0.4"/> all the light's getting through <pause dur="0.8"/> so on a day like today a lot of the light is not getting through so the optical depth is not going to be that small <pause dur="0.3"/> on the other hand if you've got an E that's something like <pause dur="0.3"/> ten-million <pause dur="0.4"/> then E is going to be very small and you're saying only a very very very small fraction of the radiation is getting through it's going to be very dark <pause dur="0.4"/> and it's not that dark <pause dur="1.3"/> # <pause dur="0.5"/> so it's going to be something of order one <pause dur="1.5"/> but <pause dur="1.4"/> that's because the radiation is scattered <pause dur="0.5"/> suppose we had an <trunc>abs</trunc> an absorbing case suppose we had a single scattering albedo in in our cloud today

of zero <pause dur="0.7"/> then what would the transmission look like <pause dur="9.7"/> i'm asking that question in a <trunc>co</trunc> <pause dur="0.2"/> kind kind of confusing way <pause dur="1.1"/> the transmission what's directly getting through <trunc>depe</trunc> depends on <pause dur="0.8"/> depends on the optical depth and that's not a problem <pause dur="0.4"/> # the problem is relating <pause dur="0.3"/> what you see when you look up at the sky <pause dur="0.4"/> to the <trunc>trans</trunc> to the transmission because what we see when we look up at the sky <pause dur="0.3"/> is actually <pause dur="0.2"/> very little <pause dur="0.2"/> direct beam <pause dur="0.2"/> and it's almost all scattered radiation <pause dur="0.5"/> so actually the optical depth today <pause dur="0.4"/> is pretty big <pause dur="1.6"/> but <pause dur="0.4"/> because of the fact that the single scattering albedo <pause dur="0.3"/> is pretty close to one <pause dur="0.4"/> most of that light <pause dur="0.3"/> that's getting taken out of the direct beam is getting bounced around and <pause dur="0.5"/> some fraction of it <pause dur="0.3"/> maybe half of it or something <pause dur="0.3"/> is getting down to us <pause dur="0.3"/> so we can still see where

we're going <pause dur="0.3"/> if we had a smoke cloud <pause dur="0.4"/> like say from a forest fire or something like that and <trunc>th</trunc> did you guys see the Australian bush fires a couple of years ago <pause dur="0.9"/> on T-V there's showing pictures of that <pause dur="0.4"/> and basically it was black <pause dur="0.6"/> under <trunc>tho</trunc> if you got close to those things and that was because <pause dur="0.7"/> you had a cloud which in terms of optical depth was probably pretty similar to what we have today <pause dur="0.3"/> but because it had a very low single scattering albedo <pause dur="0.6"/> it was very dark <pause dur="2.1"/> so <pause dur="0.3"/> you got to watch out with optical depth optical depth tells you how much is taken out of the direct beam <pause dur="0.3"/> but it doesn't tell you how much light you're actually going to see <pause dur="0.8"/> on any given day that was actually the point that i wanted to make in a <pause dur="0.3"/> slightly roundabout way <pause dur="0.3"/> okay that's it for today see you next week