-1=e^ipi Posted March 30, 2014 Report Share Posted March 30, 2014 (edited) I apologize for the delay in my reply. I had half of this written in 2013 (including all supporting references), but trying to decipher Waldo-speak, especially given that Waldo ‘ran-away’ from the discussion on page 29 isn’t the most appealing thing. Especially when Waldo refuses to even acknowledge something as basic as ‘the climactic conditions of Earth, during which multi-cellular life evolved and flourished, is relevant to the discussion of climate change in terms of understanding the environmental impacts of climate change’. Also, I've been really busy with real life stuff. I'll give an explanation of my understand of various claims regarding the slowing of the jet stream below. waldo, on 31 Dec 2013 - 6:07 PM, said: nice! You've now really upped your game to formally include NOAA (along with your earlier reference to NSIDC) in your alarmist labeling. Should we throw NASA into that mix, as well? How about Environment Canada? Are there any existing government organizations, or scientific based institutions, or academia, or intergovernmental organizations, etc., that you don't label as alarmist? Why do you consistently misinterpret what I say? I never said NOAA is a climate alarmist organization. I said that all climate change related organizations have been influenced by climate alarmism to some extent and individuals within these organizations may be climate alarmists. This is why you should directly evaluate the scientific models and evidence presented rather than use appeal to authority fallacies. waldo, on 31 Dec 2013 - 6:16 PM, said: I certainly have no obligation to provide you with diddly If you want to make a bunch of bogus claims without sufficient supporting evidence or reasoning then that is your choice. But it will not make a convincing argument nor be effective in making people change their positions. waldo, on 31 Dec 2013 - 6:16 PM, said: In any case, related papers have been cited... if the papers don't meet your claimed "scientific prowess", I suggest you contact the authors directly. You may also want to throw a challenge/comment in against any of the papers you feel are lacking in meeting your "expertise". You certainly have a ballsy approach for someone who couldn't even read a basic article reference and follow the direct bread-crumb trail laid out for you... repeatedly laid out for you! Blah blah, same dodge of providing sufficient evidence or explaining the scientific models behind your unclear claims. Of course the reason you keep dodging is because you do not know/understand the scientific models to explain them. Being a climate alarmist that approaches the issue of climate change dogmatically, you are only interested in out-of-context sentences from conclusions of scientific papers that support your climate alarmist conclusion and rely primarily on strawman, ad hominem and appeal-to-authority fallacies to make your arguments. waldo, on 31 Dec 2013 - 7:03 PM, said: I'd suggest, purposely ignore. Clearly, the video is a high-level presentation on the weather effects of Arctic amplification relative to the shifting jet-stream pattern. You think that presentation was 'high-level'? Then I guess some of the links & explanations I'm about to provide will be too much for your science-n00b brain. waldo, on 31 Dec 2013 - 6:36 PM, said: no - the concept of Arctic/Polar amplification speaks to the propensity for high Northern latitudes to experience enhanced warming... or cooling... in relation to the rest of the Northern Hemisphere. Yes, it's the same thing as global temperature gradient reduction. - - - - - Discussion on trying to figure out jetstream related claims by climate alarmists: Anyway, I guess that I'll continue from my last post. For the benefit of other people reading this thread, as well as trying to understand each other's claims, I'll briefly summarize what my understanding is of the relationship between the equatorial-polar temperature gradient, Rossby waves, properties of the jet streams, and resonance pressure phenomenon for the Northern mid-latitudes: Now I repeatedly asked Waldo to clarify his claims regarding the physical mechanisms behind which a decrease in the global temperature gradient (arctic amplification as he calls it) results in an increase in the frequency of resonance phenomena of planetary waves. In science, it is important to understand the physical mechanism by which physical phenomena occur. Of course Waldo could not answer this basic question since he is a science n00b that approaches the issue of climate change dogmatically. So to understand the claims I had to do my own 'research'. Below is a decent 'intro' video that briefly explains some of the relevant concepts. http://www.youtube.com/watch?v=_nzwJg4Ebzo It includes a number of relevant facts (that Waldo conveniently omits) such as the jet stream shifting poleward & stretching, but I will not get into it now. The video is self explanatory, but I will point out that it doesn't properly explain the physical mechanism by which a decrease in the global temperature gradient increases the frequency at which the jet stream forms resonance phenomena. It does explain that since warmer air has a greater volume than colder air, if one looks at the earth's isobaric (for science n00bs, isobaric means constant pressure) contour, they will notice that it is higher at the equator than at the poles; effectively the isobaric contour slopes downward towards the poles. If there is a decrease in the global temperature gradient, then this will decrease the downward slope, which will supposedly slow down the jet stream (how this is the result isn't explained in the video). The video also shows some crude correlations between zonal wind and sea ice as justification for this connection between the speed of the jet stream and arctic amplification. But other than that the video is insufficient to explain this supposed phenomena. http://arctic-news.blogspot.ca/2012/12/polar-jet-stream-appears-hugely-deformed.html Here is a somewhat useful link. It is a bit climate alarmisty and it does make some false claims such as "Accordingly, the Northern Temperate Zone used to experience only mild differences between summer and winter weather, rather than the extreme hot or cold temperatures that we've experienced recently.". This claim of course makes no sense in the context of a discussion of the resonance of planetary waves (which supposedly affect intra-seasonal temperature variation not inter-seasonal temperature variation) and the fact that a reduction in the global temperature gradient reduces (rather than increases) inter-seasonal temperature variation. They even have a ridiculous 'diagram of doom'. However the above link has a link to http://www.srh.noaa.gov/jetstream/global/jet.htm This NOAA site is less biased and is useful to effectively explain the 3 different tropospheric cells for each hemisphere and the mechanism by which a reduction in the global temperature gradient will decrease the speed of the jet stream: - Since the isobaric contour slopes downward towards the poles, air at the top of the troposphere will move poleward via gravity. - The earth has a higher centripetal velocity at the equator than at the poles. The centripetal velocity as a function of latitude is R*ω*cos(θ), where R is the radius of the earth (approx 6371 km), ω is the angular frequency of the earth (2π/(24hours)), and θ is the latitude. The effect is that as the air moves poleward, it gains eastward velocity relative to the ground below. - The acceleration at which the air moves poleward is approximately proportional to sin(arctan(M)) = M/sqrt(1 + M^2), where M is the slope of the isobaric contour, since the force that moves the air poleward is the downward force of gravity. - However, if the air is moving faster relative to the ground below it, then it will experience a friction/air-resistance force. The force of air resistance is approximately proportional to the square of the velocity. This means that if the net acceleration is zero then the poleward velocity of the air is approximately proportional to sqrt(sin(arctan(M))). - As said earlier, the centripetal velocity of the ground is R*ω*cos(θ). If we consider some air at the top of a cell/troposphere that initially has no velocity relative to the ground and moves poleward by an amount dθ in dt time, then the air's new eastward velocity relative to the ground is R*ω*sin(θ)*dθ. Since the rate at which air at the top of a cell moves poleward is simply R*dθ/dt, this means that the eastward velocity of the air relative to the ground is now proportional to ω*sin(θ)*sqrt(sin(arctan(M))). - The important thing to note here is that sqrt(sin(arctan(M))) is a positive monotonic transformation of M. This means that the eastward velocity of air at the top of the air cell relative to the ground depends positively on the isobaric slope of the air. - If we consider the air below to be an ideal gas of approximately equal temperature and constant pressure PV = nRT, then a small change in the temperature of the air is proportional to a small change in the volume of the air. And if the volume of air is some polynomial function of height, then a small change in the temperature will be proportional to a small change in the height of the air. Note that 'a small change in temperature is proportional to a small change in height' is true in more general circumstances (it's simply a linear approximation for small changes); I used the ideal gas argument for convenience. - Therefore, the poleward temperature gradient of the air is proportional to the poleward height gradient of the air (which is approximately the same thing as the isobaric slope of the air). This means that the eastward velocity of the air at the top of the air cell relative to the ground depends positively on poleward temperature gradient of the air. This implies that for small changes, a change in the temperature gradient will be proportional to a change in the eastward velocity relative to the ground below. - The areas where two tropospheric cells meet have the highest temperature gradients in the hemisphere. This is what causes polar jet streams to form between Polar & Ferrel cells and what causes subtropical jet streams to form between Ferrel & Hadely cells; jet streams are just fast eastward moving masses of air at the top of the troposphere. - If one decreases the global temperature gradient, then one will decrease the temperature gradient between two cells, which will therefore decrease the speed of the jet stream. Note that this still hasn't explained how the decrease in the speed of the jet stream results in an increase in resonance effects. It is also helpful to point out a few things: - The reason why Earth has 3 troposphere cells in each hemisphere as opposed to more or less is due to the properties of present day Earth. There is strong evidence that suggests that during the Cretaceous period on earth, there was only 1 cell per hemisphere. Jupiter on the other hand has at least 8 cells per hemisphere (you can see them as alternating light and dark bands on Jupiter). As an aside, Jupiter has had significant climate change over the past few decades (which is one of the reasons I prefer to use the term global temperature gradient rather than 'arctic amplification' as the former is more generally applicable and can be used when discussing other celestial bodies). - It is also relevant to point out that the jet streams shift poleward during the day and equatorward during the night. In addition, the jet streams shift poleward during summer and equatorward during winter. The below two links might also be useful for someone reading this and wanting to better understand the connection between a change in the global temperature gradient and resonance of planetary waves. http://skepticalscience.com/jetstream-guide.html http://blog.ucsusa.org/wacky-weather-a-warmer-arctic-and-a-slower-jet-stream-is-there-a-link-194 However, please note that neither of the above 2 links explain the physical mechanism by which a slowing of the jet stream (or other eastward moving air) will increase the rate at which we have pressure and/or jet stream resonance phenomena. They just use this as a 'given' and the closest thing to an explanation provided is something along the lines of 'slower jet stream makes it have a higher amplitude and meander more, which increases its chances of getting stuck'. Such a justification is poorly described, an oversimplification at best and misleading/incorrect at worst. However, I did manage to find a theoretical model/explanation that explains why a slowing of the jet stream 'might' increase the frequency of these pressure phenomena. http://geosci.uchicago.edu/~nnn/LAB/DEMOS/RossbyWave.html The above link is useful and you might need to enable flash player to see the video of the forced Rossby waves. They use a rotating annulus with water inside to represent the northern hemisphere, but the physics and equations used are mostly valid. The above link also helps to explain some phenomena. However, it is a bit insufficient as it doesn't explain the origin of some of its equations, and the equations are for a rotating annulus and different than for the Earth. Furthermore, many of the claims made by climate alarmists often relate wavelength, amplitude, group velocity, frequency of resonance phenomena and other things to the speed to the jet stream, so better analysis is needed. Anyway, after having consulted various sources on the subject I have a sufficiently good understanding of the connection between the zenith velocity of the jetstream relative to the Earth's surface and the wavelength, amplitude, group velocity, frequency of resonance phenomena, etc. If you are interested, Here are two decent pdfs on the subject: http://www.columbia.edu/~irs2113/3_Circulation_Vorticity_PV.pdfhttp://www.ocean.washington.edu/courses/oc512/rossby-waves-gfd109-lec5a-07.pdf But basically the physics are as follows: - Suppose you have an air mass that is moving faster than the rotation of the planet and is moving in the same direction as the rotation of the planet (example: air at the top of the troposphere that has moved pole ward via gravity). Note that this applies to both jet streams as well as air in the mid latitudes that generally move west to east (air at the equatorial regions and polar regions move east to west). And suppose that this air moves over a ridge (such as the Rocky Mountains, the Himalayas, the Ural Mountains, the Appalachian Mountains, the Alps, etc.). - As the air mass goes over the ridge, it is compressed vertically & adiabatically (since no heat is transferred to surrounding air by the compression). For an ideal gas, an adiabatic compression implies that PVγ = constant, where P is pressure, V is the volume and γ > 1 is the adiabatic index. Thus pressure must increase. However, if the pressure of the air mass is greater than the surrounding air (which is the same as the air mass before adiabatic compression), then there will be a net outward force on the air mass which will cause an adiabatic expansion in the horizontal direction until the pressure of the air mass and the surrounding air equalizes. But if the pressure is the same than the volume must also be the same since PVγ = constant. Thus the air mass will be compressed vertically but expanded horizontally and keep the same area. - Suppose this air mass is a rotating cylinder with constant density ρ, radius Q, height H, volume V, mass M and an angular frequency of ω. The moment of inertia of a rotating cylinder (about its axis of rotation) is simply 0∫Mr2dm = 0∫Vr2ρdV = 0∫H0∫Q0∫2πr2ρ r dθ dr dV = H*Q4/4*2π = 0.5MQ2. Then the angular momentum of this air mass due to the rotation of the air mass about itself is 0.5MQ2ω. However, you also have an angular momentum due to the fact that this cylindrical air mass is located on the Earth, which is rotating. If we are only concerned about the angular momentum in the radial direction then this angular momentum will consist of two components. The first is due to the fact that the air mass is spinning around the earth's axis MR2cos2(φ)sin(φ)Ω, where R is the radius of the Earth, φ is the latitude and Ω is the angular frequency of the Earth. The second is due to the fact that the earth's axis and the rotation axis of the cylinder are not necessarily perpendicular 0.5MQ2sin(φ)Ω (note that both of these components are in the radial/up direction). Note however that one of the components does not depend on Q, the radius of the air mass. - Thus, by conservation of angular momentum (Newtonian physics) in the upward/radial direction we have 0.5MQ2ω + 0.5MQ2sin(φ)Ω = constant. But M is a constant (by mass conservation) so Q2(ω + sin(φ)Ω) = constant. But volume is constant and V = πQ2H, so (ω + sin(φ)Ω)/H = constant. For this rotating air mass, we define its relative vorticity to be η = abs(curl(v)) = ∂vy/∂x - ∂vx/∂y, where x is distance in the zenith direction from the axis of rotation, y is the distance in the azimuthal direction from the axis of rotation, vy is the velocity of part of the air mass in the y direction relative to the ground and vx is the velocity of part of the air mass in the x direction relative to the ground. If we take a point in this air mass with cartesian coordinates (x,y,h), then vy = xω and vy = -yω. Thus η = 2ω. Using this we get the potential vorticity conservation equation: (η + f)/H = constant, where f = 2Ωsin(φ) is known as the coriolis parameter. - Now back to the air mass that is moving eastward, has zero initial relative vorticity and going is over a ridge. Suppose for simplicity sake that the ridge is short & wide relative to the height & radius of the air mass and that the ridge has a height of dh and a width of W. When the air mass is over the ridge, its height must decrease so by conservation of potential vorticity f/H = (η + f)/(H-dh) => f*(H-dh) =(η + f)*H => η = -fdh/H. Thus the air mass goes from zero vorticity to obtaining a negative vorticity (i.e. it starts to spin clockwise in the northern hemisphere). Suppose for simplicity that the initial eastward velocity of the air divided by the width of the ridge is large relative to this new vorticity. Then eastward velocity is approximately constant so -fdh/H = η = ∂vy/∂x - ∂vx/∂y ≈ ∂vy/∂x. Thus the acceleration is constant, so the poleward velocity of the air mass after it moves across the ridge (initial poleward velocity is zero) is W*∂vy/∂x = -f*W*dh/H, which is negative. - So basically, after some non-rotating eastward moving mass moves over a ridge, it will have zero-vorticity, the same eastward velocity and a negative poleward velocity (i.e. it will be heading to the equator). Note that this applies to all 'small' pertubations/ridges since you can just consider other shapes to be a sum of infinite ridges with constant height and varying widths. For larger perturbations/ridges the effect will still be the same but the eastward velocity after the air mass moves over the ridge will decrease to compensate for the ∂vx/∂y term. In the context of the jetstream traveling over the rockies, ignoring the ∂vx/∂y for simplicity sake is reasonable. - Anyway, after the air mass goes over the ridge, it now has zero vorticity and is moving towards the equator. However, as the air moves towards the equator, potential vorticity must be conserved but two things happen. f = 2Ωsin(φ) decreases since we are moving towards the equator, and H may increase (as gravity becomes effectively weaker due to a larger centripetal force, and the air mass may expand due to becoming warmer). If the time scale is small (which is reasonable in the context of the topic) we can ignore the warming effect. - The centripetal acceleration of the earth at the equator is v2/R = (RΩ)2/R = RΩ2 = (6371 km)*(2π/24h)2 ≈ 0.03 m/s2. The acceleration due to gravity for earth is approximately -9.81 m/s2. This means that the net radial acceleration on the earth's surface as a function of the latitude is g = -9.81 m/s2 + cos2(φ)*0.03 m/s2. For a small change in latitude dφ, the new acceleration is g = -9.81 m/s2 + cos2(φ+dφ)*0.03 m/s2 = -9.81 m/s2 + cos2(φ+dφ)*0.03 m/s2 - sin(2φ)dφ*0.03 m/s2. Note that pressure is defined as force per unit area. Thus the downward pressure of a point in the air mass must be equal to gravitational force acting on the air above it per unit area. So P α g/Q2. But we also know that for the cylindrical air mass, V = πHQ2. But since pressure and volume are constants (as explained earlier) we have H α 1/Q2 => gH = constant or g α 1/H. - For the effect of a change in the coriolis parameter, for a small change in latitude dφ, the new coriolis parameter is f = 2Ωsin(φ+dφ) = 2Ωsin(φ) + 2Ωcos(φ)*dφ. From conservation of potential vorticity we have (η + f)/H = constant. This plus g α 1/H implies that (η + f)*g = constant. Thus, for a small change in latitude, we get (η + 2Ωsin(φ))*(-9.81 m/s2 + cos2(φ)*0.03 m/s2) = (η + dη + 2Ωsin(φ) + 2Ωcos(φ)*dφ)*(-9.81 m/s2 + cos2(φ)*0.03 m/s2 - sin(2φ)dφ*0.03 m/s2). Neglecting all non-linear displacement terms gives: (η + 2Ωsin(φ))*(-9.81 m/s2 + cos2(φ)*0.03 m/s2) = (η + 2Ωsin(φ))*(-9.81 m/s2 + cos2(φ)*0.03 m/s2) + dη*(-9.81 m/s2 + cos2(φ)*0.03 m/s2) + 2Ωcos(φ)*dφ*(-9.81 m/s2 + cos2(φ)*0.03 m/s2) - sin(2φ)dφ*0.03 m/s2*(η + 2Ωsin(φ))=> 0 = dη*(-9.81 m/s2 + cos2(φ)*0.03 m/s2) + 2Ωcos(φ)*dφ*(-9.81 m/s2 + cos2(φ)*0.03 m/s2) - sin(2φ)dφ*0.03 m/s2*(η + 2Ωsin(φ))=> dη/dφ = -2Ωcos(φ) + (η + 2Ωsin(φ))*sin(2φ)*0.03 m/s2/(-9.81 m/s2 + cos2(φ)*0.03 m/s2) - Note that the first term is on the order of 2Ω, while the second term is on the order of 2Ω*0.03/9.81. Thus the second term is about 300x smaller then the first term and we can neglect it to get: dη/dφ = -2Ωcos(φ) - Similarly, we could have also introduced other effects of potential vorticity such as the fact that the isobaric surfaces are not necessarily parallel to the radial direction, the fact that the air may warm due to an increase in sunlight, the fact that the earth isn't perfectly round, etc. But by far the most dominant effect on the change in vorticity due to a change in latitude is the coriolis effect. - Integrating both sides of dη = -2Ωcos(φ)dφ gives η - η0 = -2Ωsin(φ) + 2Ωsin(φ0), where φ0 is the initial latitude and η0 = 0 is the initial vorticity. If we only consider small perturbations such that φ = φ0 + Δφ, where φ0 >> Δφ, then we get η = -2Ωcos(φ0)Δφ. - Note that η = ∂vy/∂x. Since the air mass follows it's velocity over time, we have ∂y/∂x = vy/vx => η = ∂vy/∂x = ∂(vx(∂y/∂x))/∂x. For small deviations/perturbations vx >> vy so we can treat vx as a constant. Thus vx∂2y/∂x2 = η = 2Ωcos(φ0)Δφ. If we use y = Rφ to represent the distance in the azimuthal direction and y0 = Rφ0 to be the initial 'displacement' from the equator then we get ∂2y/∂x2 = 2Ω/vx*cos(φ0)*(y-y0)/R. But this is of the form ∂2y/∂x2 = -k2*(y-y0), where k = sqrt(2Ω*cos(φ0)/(R*vx)) is the wavenumber. This differential equation has the following general solution: y = y0 + D*cos(kx) + E*sin(kx), where D,E are arbitrary constants and k = sqrt(2Ω*cos(φ0)/(R*vx)). - Taking the second derivative gives: ∂2y/∂x2 = -Dk2*cos(kx) - Ek2*sin(kx). We know that initially (after the air passes over the ridge) the relative vorticity is zero. But vx∂2y/∂x2 = η. Therefore, substituting in x = 0 for the second derivative yields 0 = η0 = -Dk2*cos(0) - Ek2*sin(0) = -Dk2*cos(0) => D = 0. - This means that the first derivative can be written as: ∂y/∂x = Ek*cos(kx). We know that initially, the velocity in the y direction is = -f*W*dh/H = -2Ωsin(φ0)*W*dh/H. Thus (-2Ωsin(φ0)*W*dh/H)/vx = (vy/vx)0 = (∂2y/∂x2)0 = Ek*cos(k*0) = Ek.=> E = (-2Ωsin(φ0)*W*dh/H)/(vx*k) = -sin(φ0)*W*dh/H*sqrt(2ΩR/(cos(φ0)*vx)). Thus the equation of the azimuthal displacement of this air mass as a function of the zenith displacement is: y = E*sin(kx), where E = -sin(φ0)*W*dh/H*sqrt(2ΩR/(cos(φ0)*vx)) and k = sqrt(2Ω*cos(φ0)/(R*vx)). - Now let us examine the effects of changing the zenith velocity vx (due to say a decrease in the temperature gradient due to global warming): - The wave number k is proportional to 1/sqrt(vx). And by definition, λ = 2π/k => the wavelength is proportional to sqrt(vx). This means that a slower velocity will lead to a shorter wavelength. - The amplitude, abs( C ), is proportional to 1/sqrt(vx). This means that a slower velocity will lead to a larger amplitude. - With respect to the frequency of resonance phenomena, you have resonance when the circle around the earth's axis at the central latitude is equal to a whole integer number of wavelengths. I.e. 2πR*cos(φ0) = 2π*sqrt(R*vx/(2Ω*cos(φ0)))*i, where i is any member of the natural numbers. Now whether there is an increase or decrease is the frequency of resonance effects is very chaotic and depends not only on the change in temperature gradient, but how this temperature gradient varies over the year, on the distribution on ridges that can cause these waves (I.e. on the earth's topology), on the interaction with other weather events, etc. But what we can do is make a statement about how the frequency of resonance effects changes in general (i.e. what happens to the density of states when we change the velocity of the air mass?). - We have a resonant state when R*cos(φ0) = sqrt(R*vx/(2Ω*cos(φ0)))*j, where j is any positive integer. Note that for a constant central latitude we have j*sqrt(vx) = constant => j = constant/sqrt(vx). To get the density of states for the velocity, we simply take the derivative of j with respect to vx, which yields (density of states) = dj/dvx = constant*(-0.5)*vx -1.5. This means that the a slower velocity will lead to a larger density of states. So it is not unreasonable to expect that a slower velocity will lead to a higher frequency of resonance phenomena. - Note that the above is the result of a standing wave (does not move in time). This explains why the jetstream usually dips equatorward after it hits a ridge (such as a mountain ridge) or dips poleward after it reaches an ocean, etc. And from the above we can determine a rough relationship between the speed of the jet stream and the frequency of resonance phenomena. However, it is also useful to determine the group velocity of a moving wave, since the jetstream is not completely stationary, to get a rough idea of how quickly the jetstream passes over an area (and to examine some of the claims by others about the speed at which the jetstream moves). - To get the speed of the jetstream, we need to look for plane-wave solutions to the conservation of potential vorticity (η + f)/H = constant. If we ignore the effect of the change in the height of the air mass (which is reasonable since, as explained earlier, the coriolis parameter dominates). This implies 0 = d(η + f)/dt = dη/dt + df/dt = ∂η/∂t + (dx/dt)*(∂η/∂x) + (dy/dt)*(∂η/∂x) + ∂f/∂t + (dx/dt)*(∂f/∂x) + (dy/dt)*(∂f/∂y). - But f = 2Ωsin(φ) = 2Ωsin(y/R) => ∂f/∂t = 0, ∂f/∂t = 0, and ∂f/∂y = 2Ωcos(y/R)/R = 2Ωcos(φ)/R. Furthermore, vx = dx/dt and vy = dy/dt. So we end up with: 0 = ∂η/∂t + vx*(∂η/∂x) + vy*(∂η/∂y) + vy*2Ωcos(φ)/R. - If we consider a wave that is flowing in the zenith direction with a mean zonal flow of u, then we can define vx = u + Δvx. Notice that for small perturbations, U >> Δvx and U >> vy. Thus we get the approximation: 0 = ∂η/∂t + u*(∂η/∂x) + vy*2Ωcos(φ)/R. - Now suppose we introduce a scalar 'stream' function where the gradient of the stream function gives the velocity of the wind at any point. I.e. we define a function Ψ(x,y,t) such that vx = ∂Ψ/∂y and vy = -∂Ψ/∂x. Then using our definition of relative vorticity, we have η = ∂vy/∂x - ∂vx/∂y = -∂Ψ2/∂x2 - ∂Ψ2/∂y2 = -∇2Ψ. Thus the differential equation becomes: 0 = ∂∇2Ψ/∂t + u*∂∇2Ψ /∂x + ∂Ψ/∂x*2Ωcos(φ)/R. If we consider small perturbations then we can treat 2Ωcos(φ)/R as a constant, so we get: 0 = ∂∇2Ψ/∂t + u*∂∇2Ψ/∂x + β*∂Ψ/∂x, where β = 2Ωcos(φ0)/R. - To look for plane-wave solutions in the zenith direction, we let Ψ = real(exp(ikx - iσt)), where i is the imaginary number, k is the wave number and σ is the angular frequency of the plane wave. Substituting this into the differential equation yields: 0 = ik2σΨ - iuk3Ψ + iβkΨ This yields the dispersion relation σ = (uk2 - β)/k. - To get the group velocity of this wave, we simply differentiate σ with respect to k to get: vg = ∂σ/∂k = u + β/k2, where u is the mean flow in the zenith direction and β = 2Ωcos(φ0)/R. - To get the wave number k, set the time derivative of the differential equation equal to zero to yield: 0 = u*∂∇2Ψ/∂x + β*∂Ψ/∂x. Substituting in stationary plane wave solutions Ψ = real(exp(ikx)) yields : 0 = -iuk3Ψ + iβkΨ => uk2 = β => k = sqrt(β/u) = sqrt(2Ωcos(φ0)/(u*R)), which is what was obtained earlier. - Thus we get: vg = u + β/sqrt(β/u)2 = 2u ≈ 2vx - Thus the group velocity is approximately proportional to the zenith velocity. I.e. if the zenith velocity is slower, then the wave pattern will move more slowly across the planet. Furthermore, not that the group velocity is in the same direction as the zenith velocity (i.e. the group velocity is eastward). - Note that the group velocity might not be that relevant for say North America where the waves are more or less stationary saves forced by the Rocky Mountains and other mountain ranges. The group velocity might have more meaning in Europe where there are less topologically significant features and the jetstream is coming from the Atlantic Ocean. However, in the case of North America, one could argue that the primary change in the shape of the jetstream is caused by variations in the zenith wind speed since we showed that the wavelength is proportional to sqrt(vx). - It is reasonable to expect that variance in the zenith wind speed should be approximately proportional to vx (i.e. just scale all wind speed values up by the same value). If this is true then the change in the wind speed over time should be approximately proportional to the mean wind speed vx. Thus (∂/∂t)λ α (∂/∂t)sqrt(vx) = 0.5*(vx)-0.5*(∂vx/∂t) α (vx)-0.5*(vx) = sqrt(vx). - Therefore, even if one claims that the group velocity has little physical meaning, we still get the result that the rate at which the jetstream and other Rossby waves change depends positively on the zenith velocity. Of course the dependence relationship is slightly different (group velocity proportional to vx compared to change in wavelength over time proportional to sqrt(vx)). - Using this information, it might be useful to examine the claim that if the wind speed of the jetstream is lowered due to a smaller temperature gradient, then this will lead to longer droughts/floods and 'prolonged extreme weather events'. The frequency at which a moving jetstream passes over a mid-latitude region is 2vg/λ. Since λ α sqrt(vx) and vgα vx, we get that the frequency is proportional to (vx)0.5. - Thus, a decrease in the wind speed of the jetstream will lead to a DECREASE in the rate at which the jet stream passes over a mid latitude region; thus longer droughts or floods could be expected. Note however, that this does not apply to resonance phenomena of standing waves. For resonance phenomena, the rate at which you move out of a resonance state will be proportional to (∂λ/∂t)/λ, but since both ∂λ/∂t and λ are proportional to sqrt(vx), we get that the rate at which the jetstream moves out of a resonant state is independent of the zonal wind velocity of the jetstream. Thus there is no reason to believe that a change in the duration of resonant phenomena will occur as a result climate change based on the physical mechanism described above. - Another thing that one should take into account when evaluating the claim that a decrease in the wind speed of the jetstream will prolong weather events is the effect of the change in the amplitude of the jetstream. As explained earlier, the amplitude is proportional to 1/sqrt(vx). This means that as wind speed decrease, amplitude will increase. However, if amplitude increases then now regions that were previously outside of the amplitude of the jet stream are now within the amplitude of the jetstream (Northern Canada, Southern Europe, etc.). Thus the jetstream will now occasionally pass over these regions, which will reduce the duration of weather events in these regions (since they will now vary more between high and low pressure). - The last thing to check is if the strengths of the high and low pressures systems change as a result of the decrease in the wind speed of the jetstream (since an increase in the strengths of the pressure systems could lead to more extreme weather events). Note that by conservation of potential vorticity we get η + 2Ωsin(φ) = constant (H was ignored because its changes are negligible as explained earlier). Since the vorticity at φ0 is 0, we get η = -2Ωcos(φ0)Δφ. Thus the vorticity at each latitude only depends on the latitude and not on the wind speed of the jetstream (due to the domination of the coriolis effect). - Since the pressure corresponds directly with the vorticity (if something rotates opposite of the rotation of the earth it becomes a region of high pressure and if something rotates in the same direction as earth it is becomes a region of low pressure), this means that the strengths of the high and low pressures at each latitude are UNCHANGED. Thus it is unreasonable to expect a change one way or another in the severity of weather events based on this change in the strength of the jetstream. - The one exception to the above is for regions that are now within the amplitude of the jetstream that were not before. These regions will now experience greater variation in their pressures over time, which one could argue could lead to more severe weather events. But this comes with the consequence of less prolonged weather events as explained earlier. - One consolation to the climate alarmists is that since the magnitude of high and low pressures remains unchanged while the wavelength gets shorter (λ α sqrt(vx)), one could argue that the magnitude of the pressure differential along the same latitude is proportional to approximately 1/sqrt(vx)). Therefore, a reduction in the global temperature gradient could lead to greater longitudinal pressure differential (and therefore maybe greater wind speeds). Of course, this is due to having slower eastward moving wind speeds in the first place... so the net effect is difficult to evaluate (we would need to look at the exact topology of the earth to determine the magnitude of pressure variation & wavelength of Rossby waves at different latitudes in order to compare the two changes in wind speeds). Summary of effects of increased CO2 concentrations on jetstream: - Increasing CO2 concentrations will lead to uneven warming between polar and equatorial regions due to the T4 nature of black body radiation power emission and the fact that polar regions will experience a greater loss in albedo due to melting ice. - This uneven heating reduces the global temperature gradient, which reduces the temperature differential between the Polar, Ferrel and Hadley cells in the troposphere. - The reduction in the temperature gradient between cells means that the height gradient at the top of these cells decreases in magnitude. Since the velocity of air in the jetstream is due to the loss of centripetal velocity as air falls from equatorial regions to polar regions along the top of the tropopause, the reduction in temperature gradient leads to a reduction in the speed at which air 'falls' towards the polar regions. - Since air resistance is proportional to the velocity squared, the speed of the jetstream is roughly proportional to sqrt(sin(arctan(M))) ≈ sqrt(M), where M is the slope of the height. Thus the speed of the jetstream is roughly proportional to sqrt(ΔT), where ΔT is the temperature differential across the two cells. - From conservation of momentum + considering the adiabatic expansion/compression of a rotating ideal gas on earth, we get the conservation of potential vorticity equation (η + f)/H = constant. - By considering an air mass moving eastward over a ridge, from conservation of vorticity we get that this air mass forms a standing wave pattern. Examining this standing wave pattern tells us that: - The wavelength is proportional to sqrt(vx), a slower velocity will lead to a shorter wavelength. - The amplitude is proportional to 1/sqrt(vx), a slower velocity will lead to a larger amplitude. - The density of states is proportional to vx -1.5, a slower velocity will lead to a larger of density of states, which means we should expect an increase in the frequency of resonance phenomena. Note that true change to resonance phenomena are extremely chaotic and to fully understand the effects one has to take into account the Earth's exact topology, variation in wind speeds over time, etc. This is especially true since for the northern polar jet stream, the perimeter around the earth at its latitude divided by its wavelength is often on the order of a single digit number. - Looking at planewave solutions to the potential vorticity equation allows us to examine a moving wave pattern. We find that: - The group velocity is proportional to vx; a slower velocity will lead to a slower group velocity. This means the wave form will move more slowly across the earth. - By comparison, the change in the wavelength of the standing wave pattern over time is proportional to sqrt(vx). This still gives the conclusion that a slower velocity will lead to a slower change in the wave pattern over time. - Looking at the frequency at which a moving jetstream passes over a region, this is proportional to (vx)0.5. Thus a slower velocity leads to decrease in the frequency at which weather events occur & an increase in the duration of weather events. - In the case of resonant phenomena due to forced standing waves, this is independent of the wind speed of the jetstream. Thus no change to the duration of resonance phenomena (corresponding to extreme droughts or floods) should be expected. - The strength of high and low pressure systems as a function of latitude will remain basically unchanged by this reduction in the speed of the jetstream, since their vorticities are primarily due to the corolis effect (i.e. changing the speed of the jetstream does not change the rotation rate of the Earth). - For areas outside of the old amplitude of the jetstream that are now within its amplitude, they should experience more weather variation, shorter duration of weather events, and an increase in the severity of weather events. - Arguably, the mean magnitude of the longitudinal pressure gradient is proportional to 1/sqrt(vx), so will increase due to a slower jetstream. This could lead to higher localized wind speeds. But since the cause is lower wind speeds in the first place, the net effect is difficult to evaluate. So my question to Waldo is 'can we at least agree upon the effects of increasing CO2 on the jetstream?'. (Note: I used various characters such as Greek letters. Tell me if you cannot see some of them.) (Edit: made several changes to make the post more readable.) (Edit 2: I made a new/better version of this post with estimates of expected change here: http://www.mapleleafweb.com/forums/topic/23461-effectsimplications-of-climate-change-on-jetstreams/) (Edit 3: Fixed error regarding the change in the frequency at which a jetstream will pass over a region.) Edited June 25, 2014 by -1=e^ipi Quote Link to comment Share on other sites More sharing options...
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