Introduction to Atmospheric Dynamics

hello everybody and welcome to the first lecture in a series entitled introduction atmospheric dynamics this is based off of the book of by Holden Hakim and intends to provide some introductory lessons on learning the fundamentals of atmospheric dynamics so you'll probably notice throughout these slides that there are a number of definitions concepts and questions which have been emphasized throughout the text a definition is basically an explanation or outline of relevant jargon or terms which will be important to learn because they'll be important for later learning on this particular subject a concept is an idea that draws connection between subjects or provides an answer to a question so basically is an insight into the subject matter that should be emphasized finally questions will act as motivation for the for the course of these lectures the very first lecture here today then is going to describe forces in the atmosphere we will aim to describe the equations of motion underlying basic atmospheric dynamics and we shall see that these dynamical equations motion are a fundamental foundation for any of our future work in understanding atmospheric dynamics so the Earth's atmosphere is well understood has been understood since ancient times the radius of the earth as we know today when approximated as a sphere is approximately six thousand three hundred and seventy one point to two kilometers the atmospheric depth is a mere fraction of that and represents a thin shell around the outside of the sphere of about 100 kilometers depth of which only about 10 kilometers incorporates the troposphere the troposphere is particularly important for atmospheric dynamics because it is the component of the atmosphere in which all weather that we observe takes place it is the layer in which moisture is actually allowed to persist persist over time sufficiently long that it can be better can lead to the generation of weather phenomenon of this troposphere the highest mountain on earth is about eight point eight kilometers that's Mount Everest of course in the Himalayas and it occupies a significant depth of the troposphere but does not extend beyond the troposphere so that'll be important when we're looking at the underlying effect of topography on atmospheric dynamics so the fundamental motivating question that we're going to be looking at is how do we understand the dynamics of the atmosphere and how can we use this to explain synoptic scale features and other large-scale features which we experience in the atmosphere in in the middle attitudes of the earth so the fundamental explanation that we're going to use is going to draw on the principles of atmospheric dynamics which are in turn drawn on basic physical principles that is even a fundamental understanding from physics 101 in your undergraduate days and university should provide sufficient foundation for you to understand how exactly the atmosphere behaves that is because the fundamental principles of physics are effectively valid in any local context that we know of and so they are just as valid for two cars banging into each other as they are for fluid parcels that float float around through the atmosphere so the fundamental principle that is going to govern the derivation of the equations of motion are going to is going to be Newton's second law and Newton's second law of motion is effectively a mathematical method of stating that the change in momentum of an object is going to be equal to the sum total of all the forces acting on that object and note that this is a vector form equation that is we have a vector on the left-hand side we have a vector on the right-hand side of this equation so that implies that for each component direction that is say you have XY and z directions this fundamental relationship will hold for each component as well this is basically an in a principle which describes conservation of momentum that is if with no external forces present that is if the right-hand side of this equation is zero then following an object the momentum will be constant with time that is the time derivative of it will be equal to zero in the reviewing the basic principles of physics that we'll be relying on velocity is going to be a fundamental descriptor of the motion of an object through a coordinate space with respect to time that is the change of position of an object with respect to time and that is simply the vector derivative of the position vector X with respect to time and acceleration is the change of velocity with respect to time that is the time derivative of the velocity vector u with respect to time hence for an object of constant mass we have from Newton's second law that the time derivative of the momentum M times the velocity vector is equal to M times the time derivative of a velocity of the moving object and the reason that this holds is because the object is assumed to have constant mass and hence can be brought outside of the time derivative and since D u DT is simply equal to the acceleration we have that the time derivative of the momentum is simply equal to mass times acceleration and hence from Newton's second law we have that the acceleration is equal to 1 divided by the mass of the object times the sum total of all forces acting on that object so the question then is how do these forces induce acceleration and the fundamental answer is through Newton's law second law of motion as we saw the sum total of the forces will then be incorporated into the acceleration of the object in the atmosphere the definition of an object may not be entirely clear although whereas in undergraduate physics classes you have cars which are moving around and bumping into one another our actual physical objects you can wrap and pick up and move around in the atmosphere where you basically have a fluid the concept is not entirely clear so we're going to basically rely on this idea of an idealized parcel of fluid let's say looking like a box or something along these lines and that can be also equal to for instance a balloon which is floating around the idea that there is some sort of thin membrane which separates the flute a particular region of fluid from all of its neighboring regions and then we're going to assume that the atmosphere basically consists of a large number of these fluid parcels which are basically floating around and interacting with one another the forces on that particular object then now that we've met now that we've made the object physical and manifested it are can be then calculated on this idealized fluid parcel so regardless of what forces are acting you can then envision it analogous to how you would see it in first-year physics classes you have a parcel of fluid which is floating around and forces will be acting on it we're then going to take the limit of these parcels being infinitely small and hence that will give us our continuum approximation to the atmosphere and allow us to derive the basic equations of motion which govern this the whole fluid so the basic properties which describe our fluid then we have a density of the fluid which is basically associated with the total mass per unit volume within that domain we have a volume of the fluid that is the amount of space that it occupies and the mass is then going to be equal to the density times the volume we're going to have a temperature associated with the fluid parcel and we're going to have a pressure associated with the fluid parcel all of these quantities can then be connected to one another using fundamental principles of physics we're going to be working in a spherical coordinate system remember the earth is a sphere so it's important to do our derivations in spherical coordinates hence we will be using longitude latitude and radius in order to represent the position vector in space so the longitude basically represents the distance along the equator so moving in a longitudinal direction the latitude represents the distance from the equator to the pole so a positive latitude denotes Northern Hemisphere and a negative latitude denotes a the southern hemisphere and then radius simply denotes the distance from the center of the sphere these are have corresponding basis vectors associated with them that is vectors which point in the direction of an increase of that particular component so the eastward basis vector I then points in the direction of increasing longitude with latitude and radius held constant the northward basis vector J points in the direction of increasing latitude with longitude and radius held constant and the vertical basis vector K points in the vertical direction with longitude and radius held constant you'll notice that as you move over the surface of the sphere that these basis vectors will also move and that'll be important moving forward as we talk about apparent forces so for instance on the equator looking down at the vertical looking down the vertical basis vector the I and J components will be seen as pointing upward and to the right whereas looking down along the longitudinal basis vector one sees J and K components which are tangential to the surface of that curve or the surface of the sphere at that point we're going to use a quantity known as the material derivative here defined to be a derivative with a capital D and this will be used throughout our Stinton investigating atmospheric dynamics this simply denotes the change in a quantity following a fluid parcel so for our balloon or fluid parcel which is floating around if it undergoes heating from radiation for instance and the temperature increases then we can say that there is a change in the temperature following that fluid parcel what we're going to be interested in then is the derivative of the velocity in this context of the fluid parcel with respect to time and that is going to be via Newton's second law the quantity which is going to be affected by forces operating on the fluid parcel so here the vector you denoting the velocity can be decomposed into a 3d velocity vector with components uvw throughout the text we'll use UV and W to denote eastward velocity or zonal velocity or velocity in the positive longitudinal Direction V to denote a northward velocity radial velocity or velocity in the positive latitude direction and W to denote upward velocity or vertical velocity so now that we have our fundamental structure for how forces will affect our fluid parcel we need to basically enumerate the forces that will be acting on the fluid parcel within the atmosphere there are a number of forces which could pen should potentially be present ranging all the way from gravity for instance to electromagnetic forces but what we've discovered is that there are four fundamental forces which are going to be important for understanding large-scale atmospheric dynamics those four forces are the pressure gradient force which describe basically interactions between fluid parcels the gravitational force which describes the effect of a distant mass on our fluid parcel viscous force which is rubbing or shear action of two fluid parcels against one another and Coriolis and centrifugal force which are associated with the fact that we're in this spherical coordinate system which is rotating the total force then which determines the acceleration on the fluid parcel is then going to be the sum of all of these individual forces there are three categories of forces that can penitential EB acting on the fluid parcel we have a surface force which acts on the surface of the fluid parcel and is typically due to interactions with neighboring parcels the magnitude of the surface force is typically proportional to the surface area of the parcel or the surface area of the interaction so this includes things such as pressure force and viscous force and actually arise out of the pressure stress tensor in the Newtonian derivation of fluid annex a body force instead acts on the center of mass of the parcel of fluid the magnitude of the body force then is typically proportional to the mass of the parcel for instance gravity is a force which is proportional to both the mass of the earth and the mass of the fluid parcel when we have a coordinate system which is moving such as in our case we have a rotating earth and this varies with respect to time and/or space then there will be an apparent force which is induced due to the fact that coordinate vectors are changing following the fluid parcel so this includes Coriolis and centrifugal fourth force both of which are going to be incredibly important for large-scale middle attitudinal storm systems so going through these forces one by one then we're first going to delve into the pressure gradient force we want to come up with a mathematical description for how exactly pressure gradients will be acting on our fluid parcel so imagine that the fluid parcel is embedded in some sort of gaseous medium some sort of background flow and it experiences pressures now pressures are basically a measure of the intensity of interactions between the Manute background and the individual and the walls of the fluid parcel so this typically occurs for instance if you have a bunch of vibrating parcel particles which are at the boundary of your fluid and those interact by hitting the boundary of your fluid parcel consequently the fluid parcel will it will receive a total pressure force which is proportional to the sum total of all those individual interactions over the over the whole edge since we're not keeping track of individual particles since I'd be on too small of a scale instead we assume pressure is going to be averaged over the whole face so let's say for instance that our parcel exists in Xyz space at some point X naught Y naught Z naught we're going to be interested in determining the total pressure which is being experienced along for instance a face which is in the X direction of the are here rectangular fluid parcel and how we're going to obtain the pressure at that point is we're going to assume that the pressure is a continuum that is it is sufficiently continuous in space that we can use a Taylor series expansion in order to in order to obtain the pressure at a distant point in space hence the pressure along the right hand side experienced by the fluid parcel is going to be equal to the pressure P naught at the center of the fluid parcel plus the continuous derivative with respect to X of the pressure field times Delta x over 2 since that is the distance of half of the fluid parcel in the X direction and then plus some higher order terms which are will be neglected since Delta X will be assumed to be small we can do this at both sides of the fluid parcel both at the right hand side and the left hand side at the right hand side we have P naught plus Delta P over Delta R the dye P over DX times Delta X over 2 and at the left hand side we have P naught minus dpdx times Delta x over 2 notice the difference in sign on the two sides that simply because we're and the one side we're going in the positive Delta X direction and on the other side we're going the negative Delta X direction hence the total force experienced by the fluid parcel is equal to the pressure times the area and the area here is simply Delta Y times Delta Z and similarly on the right hand side we have minus pressure times the area here the pressure on the right hand side is inducing a force to move the fluid parcel to the left hence why we have the negative sign in front whereas the force on the left hand side here is inducing the fluid parcel to move to the right hence it is induced in the positive x-direction the total force then experience on the fluid parcel by both the left and right hand side forces is going to be equal to the sum total of these two plugging them into the equation and summing them together we obtain that the P naught terms cancel out and that the total force acting on the fluid parcel in the x-direction is simply equal to- diapey DX times Delta X Delta Y Delta Z which is the volume of the fluid parcel if we repeat this calculation in all coordinate directions we then obtain a force vector in the X direction we have dip' IX in the Y direction we have died PD Y and in the Z direction we have died Piz and all of these are multiplied by the volume of the fluid parcel and hence in vector notation we can simply say that the total force is equal to negative x nabla of the pressure times Delta X Delta Y Delta Z computing the force per unit mass that involves dividing this by the mass of the fluid parcel recall that this actually determines the acceleration experienced by the fluid parcel now if you recall that mass divided by volume is equal to the density we then observe that the force the total force divided by the mass is simply equal to negative 1 over the density times the gradient of the pressure and this then gives the total force acting on the fluid parcel in all directions from the pressure gradient the second surface force we're going to be looking at is the viscous force so the viscosity of air basically represents resisting motion from two fluid parcels rubbing up against one another so for instance if my hand represents the surface of one fluid parcel when you have another fluid parcel moving over top of that surface it experiences friction associated with that movement and that friction is then responsible for what is known as the viscous force acting on the fluid parcel here we have a fluid parcel both on top and bottom of a blue fluid parcel and both of them induce a shear stress in the X direction due to the fact that the fluid parcels are moving at different velocities with respect to one another the the shear stress or force experienced at the top of the fluid parcel is going to be approximately proportional to the difference in the velocities between the blue and red fluid parcels here UT minus u n u times the area of the interaction recall that since this is a surface force it's going to be proportional to the surface area of the interaction and a shear force or force on the bottom in the same direction associated with the difference of velocities between the bottom and the between the green fluid parcel and the blue food parcel and this one is going to be proportional to UB the bottom fluid volute parcels fluid velocity minus the fluid parcel velocity of the blue fluid parcel so observe in particular that of all the fluid parcels are traveling at the same velocity no force will be conferred because in that case we have no friction due to differential velocities between two two objects so the resistance of the fluid parcel to shearing is determined determined by a proportionality coefficient which is given by mu over Delta Y here mu is what's known as the dynamic viscosity and it is material dependent it's effectively used to describe how how willing a fluid parcel would be to undergo a shear deformation that is a deformation which would normally cause it to turn from a rectangular prism into some sort of parallelogram or parallel parallel pivot so since this quantity is inversely proportional to the thickness in particular observe that this quantity is inversely proportional to the thickness of the fluid parcel Delta Y and this is the case because the thinner object you get the more resistance that we'll have to a shear deformation because it causes a stronger deformation of the object itself if we now take the two forces the top and bottom force experienced by the fluid parcel and sum them together and add the proportionality coefficient we then obtain an expression for the total viscous force experienced by the fluid parcel which is given by mu times Delta a divided by Delta Y times UT minus two times u plus u be the force per unit mass then gives an acceleration so we define we divide the total force experience in the I direction then by the mass of the fluid parcel and substitute in the fact that mass is going to be equal to density times the volume of the fluid parcel this then gives that a this then gives that the total force experience is equal to F tow divided by Rho Delta Y Delta a and hence that the force experienced is mu divided by Rho times u t minus 2 u plus u B all divided by Delta Y squared so interestingly here we then get that the viscous force experienced by the fluid parcel is actually proportional to 1 over the square of that distance that vertical distance separating the fluid parcels and so in the limit as Delta Y goes to 0 we actually find that this difference formula reduces to a second derivative of the horizontal zonal velocity and so the total force experienced by the fluid parcel divided by the mass is equal to the is equal to MU divided by Rho times the second derivative of U so extending this derivation then on to each flow direction then yields a vector force associated with a viscosity we're going to define a kinematic viscosity denoted nu here which is simply mu divided by the density of the fluid and in particular we find that in practice nu tends to be largely constant for for for the atmosphere so the total vector force experienced by the fluid parcel divided by the mass then is equal to nu times the laplacian of u here the laplacian in Cartesian coordinates simply being the sum of the and derivatives of the particular vector quantity so this is interesting because that means that in the case of linearly sheared flow that is if we have a velocity which varies linearly in space that this quantity is actually equal to zero recall a second derivative of a linear quantity is equal to zero so basically what this means is that if we have a fluid parcel sitting here and it experiences a constant friction force in one direction and a compensating friction force in the opposite direction along the bottom then what will happen is that it experiences zero net force and that makes sense actually because if you think that you can induce if the object itself cannot be deformed and you induce a force in one direction on top and another force in the opposite direction on the bottom the overall effect will be a zero acceleration on the fluid parcel and to make sense that we actually have this quadratic derivative of the fluid velocity describing the viscous force okay so moving on then we have the gravitational force which is an example of a body force recall from Newton's law of gravity that the total gravitational force is simply given by the gravitational constant capital G times the mass of one object times the mass of the second object that is proportional to both the mass of the interacting object and the inducing object and that's divided by the square of the distance separating them here denoted by the here the distance separating them is denoted by the vector quantity R and the magnitude of that is simply the distance between them then and this is multiplied by R over the magnitude of R which is simply a directional vector saying that the force is in the direction of separation between the two objects now we know the gravitational constant this has been measured very accurately empirically and we know the mass of the earth we can assume that the fluid parcel will have almost negligible effect on the earth itself that is we're going to assume that the overall force experienced by the earth will lead to essentially zero acceleration of the earth itself but the earth will still have a very apparent effect on the fluid parcel so since the atmosphere is essentially a thin shell we make the approximation that the radius is approximately equal to the radius of the earth which we know and then we define gravity at the surface to be a little G which is equal to capital G times the mass of the earth divided by a squared now as a little exercise outside of this lecture I challenge you in order to calculate little G from the above information so given that we've calculated a local gravitational force then or a gravitational acceleration associated with the fluid parcel we can then say that the total force due to gravity associated with the fluid parcel divided by the mass is equal to negative G that is it points downwards relative to the vertical upward vil basis vector and it acts towards the surface of the earth that is it's aligned with the vertical basis vector k and has no effect on velocities in the eastward or northward direction so this then gives the total gravitational force acting on the fluid person so so far we've deferred we've derived these dynamical equations of motion for a Cartesian frame and a non rotating fluid and it incorporates three important forces pressure gradient force gravitational force and viscosity the terms are the total acceleration experienced by the fluid parcel is equal to the sum of all forces and hence we have the pressure gradient force summed with the viscous force in both the horizontal equations of motion and for the vertical velocity we also include the gravitational force all three forces are then added to so in a Cartesian frame then these will be the fundamental equations of motion describing fluid behavior we are now looking in a rotating frame and now that we've entered into a rotating frame we need to take into account apparent forces that is forces which are non inertial but simply exist because of the way that we've defined our basis vectors to follow the surface of the earth so recall the earth revolves around its axis at a certain rate Omega the Coriolis and centrifugal forces here are then known as apparent forces because they only exist in the reference frames when the reference frame is in motion and these two forces basically are used to describe two features which are associated with the motion of a fluid parcel in a rotating frame the Coriolis force is known for deflecting a fluid parcel as a consequence of the Earth's rotation whereas the centrifugal force attempts to push fluid parcels away from the axis of rotation we're going to concentrate now on the derivation of the Coriolis force and how it affects fluid parcels and then we'll briefly mention centrifugal force later so Coriolis force arises out of conservation of angular momentum that is an object with a certain angular momentum will tend to retain that angular momentum will tend to retain that rotation if it does not interact with other parcels here angular velocity is defined as the angular frequency in units of radians per second multiplied by the radius so the quantity Omega has units of 1 over seconds and R which is the radius of the fluid parcel from the axis of rotation so imagine this in a 2d frame R would then be the distance from the fluid parcel to the center of that circle or here's the center of the air here the axis of rotation and the product of those two is then going to be equal to the velocity the magnitude of the velocity vector and the velocity vector will always be tangental to the to the to the circle which represents the path of rotation so over a small time delta T we expect that the Falah arsal will subtend an angle Delta Theta and the end consequently it experiences a small adjustment in its position vector associated with that with subtending that angle so on the surface of the sphere the radius of rotation is actually equal to a perpendicular distance from the axis of rotation do not think that this is the distance from the of the fluid parcel from the center of the sphere it is the distance from the fluid parcel from the axis of rotation because at any point along the sphere it's going to be basically following a circle along line of constant latitude and can be treated then it's behavior can then be treated as a 2d plane which is perpendicular to that line to the to the axis of rotation that is along a surface of constant latitude hence the more important quantity in this case is this radial vector extending from the axis of rotation now a is the distance of the fluid parcel from the center of the earth and this is going to be multiplied by the cosine of the of the latitude Phi here so the reason why it's a cosine is because cosine represents a adjacent distance over the hypotenuse and the hypotenuse here is a and hence R the magnitude of R will be that adjacent distance at this point we have a local coordinate system analogous to the one that I discussed earlier at the red circle we're going to have a northward coordinate vector J a vertical coordinate vector K and a zonal coordinate vector I which currently points into the page so if we assume that we have a vector B at this particular location in space which is aligned with R that is a exists at constant latitude we are interested in determining the components of this vector in the local basis and we do that by saying okay it's not going to have any contribution from I because it exists perpendicular to the plane in which I is deferred perpendicular to the plane per sorry perpendicular to the vector I and hence exists in the plane of J and K which are both perpendicular to I at this point hence B will have a contribution B Y which points in the J direction and a contribution B Z which points in the vertical K direction one can quickly verify then looking at the vector B and its magnitude and using basic trigonometric principles that B Y has to be equal to negative magnitude of B times sine of latitude and that be Z has to be equal to the magnitude of B times the cosine of the latitude in particular observe that be Y squared plus B Z squared is equal to the magnitude of B squared and hence we have agreement that the that the vector defined by quantities B Y and B Z actually has the same magnitude as the vector B so we're interested then in the velocity of the food parcel at the point at the red point simp only due to the rotation of the earth that is assuming the fluid parcel is stationary with respect to a particular location on the earth it will be moving at a rate which is proportional to the to the rotation rate of the earth of that point since the velocity is purely zonal we have that UI is going to be equal to Omega times the radius R so recall from earlier this is simply the angular velocity associated with a fluid parcel at this point and that velocity points in the positive zonal direction so it's equal to UI times the eye basis vector which points each points eastwards so the velocity through the rotation of the earth is Omega times the magnitude of our angular momentum is going to be conserved following the fluid parcel that is the quantity which we denote as vector L and it is equal to the cross product of the radial vector in the linear momentum so it's equal to R cross mass times U we know what u is uhm so U is simply going to be in the I direction with magnitude UI and these are perpendicular to one another so that means that we don't have to take into account the angle between them when calculating this cross product so the magnitude of the angular momentum then is equal to the magnitude of r times mui plugging in what we know for UI we then obtain that the angular momentum of the fluid parcel is going to be equal the mass of the fluid parcel times the rotation rate of the earth times this perpendicular radius vector the magnitude of its squared so the angular momentum vector then points in the same direction as omega because it is going to be perpendicular to the plane of both I and R and those two are perpendicular to omega hence we have that the angular momentum vector is going to be aligned with the axis of rotation as expected so since we know that the magnitude of R it can be written in terms of a we then have that the for a fluid parcel at rest relative to the surface we have that the total angular momentum that is the magnitude of the angular momentum vector is going to be equal to M Omega ay squared Coast squared of latitude since angular momentum is conserved we have that following the fluid parcel we must have that the material derivative of the angular momentum is equal to zero and hence that regardless of the motion of the fluid parcel we must have that the angular momentum here is conserved so with purely zonal velocity then we now take into account we also include a zonal velocity term that is we assume that the fluid parcel itself is moving in the zone direction that is aligned with the vector I we can include that in the calculation as well in which case the angular momentum vector divided by the mass is going to be equal to Omega times the perpendicular radius vector R squared and that's the contribution to Drey D due to rotation of the earth and then plus a zonal vector Arizona velocity component U times magnitude of R so factoring out the magnitude of R squared then gives us this relationship for the angular momentum divided by the mass so since we now know what the angular momentum of the fluid parcel is we can now take into account on additional motion of the fluid parcel and then try to determine what happens in order to in order to guarantee conservation of angular moment let's imagine that the air parcel moves south or north in that case we get some small change in the radial vector R right if we move south we then observe that the radial vector extends becomes longer because we're moving farther away from that axis of rotation if we move northwards we come closer to the axis of rotation and hence R decreases but if angular momentum is conserved there is only one quantity within this angular momentum formula that can act that we actually have freedom to change and that's the zonal velocity hence we observe that any north-south displacement of the fluid parcel must lead to a corresponding change in the zonal velocity of the fluid parcel so looking back at our expression for the for the angular momentum of the fluid parcel divided by its mass we can factor out this quantity R + earth sorry we can replace the R with a small deviation R plus Delta R so here assuming that the fluid parcel moves a small distance South inducing a small placement Delta are we then observed that both that both of these expressions must be equivalent to one another that is the initial angular momentum given by the top line must be equal to the modified angular momentum given on the line below now we can solve this expression then for Delta U which is going to be the change in the zonal velocity induced by this change in Delta R and we observe that Delta U then is approximately equal to minus 2 times Omega times the displacement Delta R minus u over the magnitude of R times Delta R so this assumes small increments we're neglecting terms which may be proportional to Delta R squared in this case because we will be taking a continuum approximation later and assume and taking Delta R to be infinitely small so given this formula then for a small South where displacement we can apply trigonometry to observe that Delta R is equal to negative sine of the latitude times Delta Y and hence obtain that the total change in velocity is given by this expression here we're going to divide but through by delta T that is a small time increment and take the limit as delta T goes to 0 this will then give us the material derivative of the velocity following the fluid parcel in terms of the material derivative of the y-coordinate that is the Y here being the the displacement in the north-south direction following this formula so in particular we observe that the material derivative of Y following T that is the distance the Meridian L distance change of that with respect to time is simply the radial velocity and hence we obtain this expression which describes the acceleration of the fluid parcel which is induced due to the fact that we're in this rotating coordinate frame so hence a small Meridian a velocity will induce a acceleration in the zonal direction so this includes two terms one being the Coriolis term which is proportional to the rotation rate Omega but we also obtain a second term which you'll notice does not include the rotation rate and this exists because of curvature of the underlying frame that is the fact that the fluid parcel exists on the surface of the sphere and hence modifications and hence the zone of the definition of the zonal velocity will be modified as one moves through space hence by conservation of angular momentum we then observe that displacements of a fluid parcel in the meridian direction will lead to apparent forces as follows if the fluid parcel is displaced zonally it will also experience a acceleration in the zonal direction and this will have feedback on rudy onyl and vertical velocities if the fluid parcel is displaced vertically will also observe a increase in the distance from the axis of rotation and hence conservation of angular momentum will also lead to the indent inducement of a acceleration on the fluid parcel both of these will be left as an exercise for either discussion in class or for review from home it essentially requires repeating of a similar analogous or a similar analysis procedure in order to obtain the full derivation we combined all of the three possibilities and that will then give us a total explanation of Coriolis and curvature terms it turns out that in total we will obtain eight terms as a consequence of Coriolis force and curvature four of those terms are Coriolis related four of those terms are curvature related so as you'll see here we have a variety of interactions associated purely with the fact that we are on the surface of a sphere and that the reference frame is going to be rotating in time so these four forces must then be included in the total explanation of acceleration experienced by a fluid parcel here we're going to define a Coriolis parameter in order to simplify the notation it'll be defined as - Omega sine Phi and will be used throughout our discussion of atmospheric dynamics you'll then observe that it dramatically simplifies the X the the Coriolis terms which appear in these expressions and particularly for small for small W that is negligible vertical velocities the Coriolis force is then simply given by the material derivative of U with respect to time is equal to F times V and the material derivative of V with respect to time is equal to minus F times U so this is a very simple expression which basically arises from from the fact that we're in this rotating frame and in fact if we do a little extra analysis on this we'll find that this is simply a second order differential equation for the velocity which describes circular motion that is the Coriolis force describes deflection to the right in the northern hemisphere and deflection to the left in the southern hemisphere and with no additional forcing that is with no pressure forces with no vertical accelerations the tendency will be for fluid parcels to basically trace out a circle along the surface of the sphere centrifugal force is going to be mentioned here briefly since it is typically small and generally ignored in most analysis of the in of atmospheric dynamics it works perpendicular to the axis of rotation and hence includes terms in the meridian 'el and in the vertical direction it is analogous to the force experienced by people in a swing for instance a spinning swing whereby the spinning motion causes a force which makes you feel as though you are being thrown outwards away from the center of the axis of rotation this quantity is typically small hence the vertical component is absorbed into the gravitational term that is it's used to modify the small G that we use earlier in deriving the gravitational force and the horizontal component is absorbed into the curvature terms which we just discussed for Coriolis force hence the total dynamic equations of motion then incorporate all of the the forces that we then described we have curvature pressure gradient gravity Coriolis and viscosity curvature typically appearing on the left hand side of the evolution equations and the remaining forces appearing on the right hand side and these forces are highlighted as seen on this slide all of these forces are important for describing the fluid equations in spherical geometry for a rotating fluid and these will be the fundamental baseline which we will then use in order to describe all atmospheric dynamics going forward

Channel: Introduction to Atmospheric Dynamics

Views: 47,118

Keywords: atmospheric science, atmospheric dynamics, lectures

Id: nljud2UiWUk

Length: 47min 20sec (2840 seconds)

Published: Thu Mar 27 2014