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Discussion of the Effects and Implications of an Increase in Atmospheric CO2 Concentrations on the Jetstreams and on Eastward-Moving Air in the Mid-Latitudes 1. Introduction 2. Polar-Equatorial Temperature Gradient and Speed of the Jetstream 3. Conservation of Potential Vorticity 4. Forced Standing Waves Derivation 5. Forced Standing Waves Discussion 6. Plane-Wave Solutions to Conservation of Potential Vorticity 7. Changes in the Wave-Pattern over Time 8. Summary 9. Quantification of Expected Change 10. Discussion & Conclusion 1. INTRODUCTION: Climate alarmists frequently make the claim that increasing atmospheric CO2 will cause the weather to become more 'extreme' and that weather events will become prolonged, more frequent and more severe as a result. When you ask them for a physical explanation as to why this will occur, you rarely get a response beyond 'it is general knowledge' and/or links to other climate alarmists repeating the same thing. They will continue to make this claim despite the fact that increasing atmospheric CO2 concentrations in the atmosphere will cause the polar regions to warm faster than equatorial regions due to the T4 nature of black body radiation and greater loss of albedo in polar regions from melting ice. Basic physics of heat engines tell us that reducing the temperature gradient of a heat engine will reduce its efficiency (theoretical efficiency is 1 - Tc/Th, where Tc is the temperature of the cold reservoir while Th is the temperature of the hot reservoir). Since many of the Earth's atmospheric & oceanic phenomena acts as a heat engine on this temperature gradient, wouldn't a reduction in the global temperature gradient leave less usable energy for these heat engines, thereby reducing their frequency and severity? For example, most reasonable climate models predict that the frequency and severity of tornadoes in Tornado Alley will be reduced, as a result of increasing atmospheric CO2, because there will be a reduction in the temperature gradient between warm air from the Gulf of Mexico and cold air from Canada. Furthermore, based on the physics of the radiative greenhouse effect, it is expected that the Earth's upper troposphere will warm more than the Earth's lower troposphere. Since many atmospheric phenomena such as hurricanes act as heat engines trying to reduce this temperature gradient, wouldn't this uneven heating reduce the frequency and severity of these weather events? Even IPCC has recently made revisions to their claims and now acknowledge that increasing atmospheric CO2 will either have no effect or a negative effect on the frequency and intensity of hurricanes. These facts make the climate alarmist claim, 'increasing atmospheric CO2 concentrations will result in more frequent and more severe extreme weather events', rather questionable. Recently, there has been greater interest in the effects of a reduction in the global temperature gradient on the jetstreams and on other eastward moving air in the mid latitudes. There have been many claims, some true and some false, regarding the effects of increasing atmospheric CO2 concentrations on Rossby waves and how that affects on the duration, frequency and severity of extreme weather events. Climate alarmists have jumped on this discussion and have started to use this as a 'silver bullet' to justifying their dogmatic claims that increasing atmospheric CO2 concentrations causes extreme weather events to be prolonged, more frequent and more severe. Of course, their definition of extreme weather events is poorly defined and can include floods, droughts, blizzards, storms, hurricanes, tornados, heat waves, cold snaps, etc. (i.e. everything). The purpose of this thread is to discuss the effects and implications of increases in atmospheric CO2 concentrations on the jetstreams and on other eastward moving air in the mid-latitudes and to compare these effects to other effects that result from climate change (such as the reduction in polar-equator or surface-tropopause temperature gradients, or change in rainfall patterns). I would prefer that we refrain from discussing the question of 'should humanity make changes in their behaviour to avoid climate change' and concentrate more on evaluating if the changes to the jetstreams and other Rossby waves are net good or net bad (including how they influence other weather phenomena) in order to evaluate the validity of the claims that climate alarmists frequently make regarding these phenomena and to determine if increasing atmospheric CO2 concentrations leads to longer, more frequent or more severe extreme weather events. Below I will give an explanation of the physical mechanism behind how changes in the temperature gradient affect the jetstreams and other Rossby waves, summarize the relevant results, and then discuss the implications on how it affects the duration, frequency and severity of weather phenomena. If you either do not want to read that much or lack a decent understanding of physics then please skip to the summary. 2. Polar-Equatorial Temperature Gradient and Speed of the Jetstream - Due to the T4 nature of black body radiation and greater loss of albedo in polar regions from melting ice, increasing CO2 concentrations is expected to reduce the Earth's polar-equatorial temperature gradient. - Since warmer air has a greater volume than colder air, the earth's isobaric (isobaric means constant pressure) contour 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 polar-equatorial temperature gradient, then this will decrease the downward slope in the isobaric height. - Since the isobaric contour slopes downward towards the poles, air at the top of the troposphere will move poleward via gravity. - Since the force that moves the air poleward is the downward force of gravity, the acceleration at which the air moves poleward is proportional to sin(arctan(M)) = M/sqrt(1 + M^2), where M is the poleward slope of the isobaric contour. If the slope is relatively small (M << 1, which is reasonable given the thickness of the troposphere vs the radius of the Earth) then we get that the acceleration at which the air moves poleward is approximately proportional to M. - Consider two air masses of ideal gas that are right next to each other, are in non-thermal equilibrium and are of equal mass. Being in non-thermal equilibrium means that the pressures of both air masses must be the same (otherwise one of the air masses would start to accelerate). The means that the downward pressure of the two air masses must be equal. But P = F/A α g/A, where α means 'proportional to', P is pressure, F is the downward force due to gravity on the air above, A is the cross-section area of constant height and g is the acceleration due to gravity. Thus the pressure must be proportional to the inverse of A, which means that A is constant since P is constant. - Ideal gases satisfy the equation PV = nRT, where V is the volume, n is the number of moles of gas, R is the ideal gas constant and T is the temperature. But V = hA, where h is the height of the air mass. Thus, PhA = nRT. Since the two air masses under consideration have the same mass, pressure and area, this means that the height of the air mass is proportional to the temperature. Which implies that the height differential is proportional to the temperature differential (M α ∇T). - Therefore, the acceleration at which the air moves poleward is approximately proportional to ∇T ≈ ∂T/∂y, where y = Rφ, R is the radius of the Earth and φ is the latitude. Note that 'a small change in temperature is proportional to a small change in height' is true in much more general circumstances (it's simply a linear approximation for small changes); I used the ideal gas argument for convenience. - If the air is moving faster relative to the ground below 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 (equilibrium) then the poleward velocity of the air is approximately proportional to sqrt(∇T), where ∇T is the temperature gradient. - The Earth has a higher centripetal/tangential velocity at the equator than at the poles. The centripetal velocity as a function of latitude is R*Ω*cos(φ), where R is the radius of the Earth (approx 6371 km), Ω is the angular frequency of the earth (2π/(24hours)), and φ is the latitude. The effect is that as the air moves poleward, it gains eastward velocity relative to the ground below. - 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 dφ in dt time, then the air's new eastward velocity relative to the ground is R*Ω*sin(φ)*dφ. Since the rate at which air at the top of a cell moves poleward is simply R*dφ/dt α sqrt(∇T), this means that the eastward velocity of the air relative to the ground is now proportional to Ω*sin(φ)*sqrt(∇T). - The important thing to note here is that sqrt(∇T) 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. - This means that the eastward velocity of the air at the top of the air cell relative to the ground is proportional to sqrt(∇T). In more general circumstances we will still find that the eastward velocity of the air at the top of the troposphere depends positively on ∇T. 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 (d∇T α dvx, where vx is the velocity of the air in the zenith/eastward direction relative to the ground). - The areas where two tropospheric cells meet have the highest temperature gradients in the world. 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 that occur due to high temperature gradients. - 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 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 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. This is because the location of highest temperature gradient shifts with temperature. 3. Conservation of Potential Vorticity - 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. 4. Forced Standing Waves Derivation - 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. 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. - 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. - 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. 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. - 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)). 5. Forced Standing Waves Discussion - 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)))*j, where j is any member of the natural numbers. 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. 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. - Note that to get the above result we assumed that the air was moving in the eastward direction. Westward moving air will still gain a negative vorticity when it goes over the ridge, but this will cause the air mass to accelerate poleward (rather than equatorward). As the air mass moves upwards, it will continue to gain a larger and larger negative vorticity until it starts to accelerate eastward. So westward moving wind cannot generate this standing wave pattern like eastward moving wind does. So this result is mostly relevant for the jetstreams and the Hadley cells. 6. Plane-Wave Solutions to Conservation of Potential Vorticity - To get the group velocity, 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/∂y) + ∂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. - 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, note that the group velocity is in the same direction as the zenith velocity (i.e. the group velocity is eastward). 7. Changes in the Wave-Pattern over Time - Note that the group velocity might not be that relevant for say North America where the waves are more or less stationary waves 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. 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). 8. SUMMARY - 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 troposphere, 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 of the troposphere. Thus the speed of the jetstream is roughly proportional to sqrt(ΔT), where ΔT is the temperature differential across the two tropospheric 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. vx is the velocity of the air relative to the ground in the zenith/eastward direction. - 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. This means we should expect an increase in the frequency of resonance phenomena. Note that the true changes 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. For the jet streams, this is especially true since the perimeter around the earth at the jetstream's 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 the 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 significantly change the rotation rate of the Earth). - For areas outside of the old amplitude of the jetstream that are now within the new amplitude, they should experience more weather variation, shorter duration of weather events, and arguably 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. 9. Quantification of Expected Change It is useful to try to quantify the amount of change in the properties of the jetstream that we should expect under global warming. Even more ambitious predictions on the increase in atmospheric CO2 levels under no/little mitigation policies predict approximately an increase from 400 ppm to 800 ppm in atmospheric CO2 concentrations by the end of the 21st century (a high value is picked to give the benefit of the doubt to the climate alarmists). This corresponds to an increase in global average temperatures by approximately 4K (http://www.ig.utexas.edu/people/staff/charles/uncertainties_in_model_predictio.htm) and an increase in global sea levels by approximately 0.7m (http://en.wikipedia.org/wiki/Current_sea_level_rise#Projections). Predictions on how much faster the arctic will warm compared to the rest of the word vary (between 2x and 5x). I will use the information provided here (http://earthsky.org/earth/whole-earth-is-warming-but-arctic-is-warming-fastest) to use a reasonable warming ratio of 2.0/0.6 ≈ 3.33. It is desirable to approximate the change in the temperature gradient between the Polar and Ferrel Cells in the Northern Hemisphere due to global warming in order to estimate the magnitudes of the changes to the jetstream. The Northern Polar Jetstream varies in mean latitude between approximately 50°N and 60°N (it is further north in Europe than North America). I will use the average global temperature at 40°N (which is 15°C or 288K, according to http://pielkeclimatesci.wordpress.com/2012/07/11/sea-surface-temperature-trends-as-a-function-of-latitude-bands-by-roger-a-pielke-sr-and-bob-tisdale/ , the same as the average surface temperature of Earth) to approximate the temperature of the Ferrel Cell and assume that it will warm by 4K by the end of the century. I will use the average global temperature at 70°N (which is 1.5°C or 274.5K) to approximate the temperature of the Polar Cell. Since the amount of warming we are dealing with (4K) is large compared to the temperature differential (13.5K), trying to use a linear approximation to determine the change in the temperature of the Polar Cell does not make sense (especially since for large temperature changes, a linear approximation can lead to a negative temperature differential, which is impossible). Instead, I will use an exponential form to approximate the temperature change for the Polar Cell (i.e. the temperature difference will decay as global average temperatures rise) since the exponential form approaches a temperature differential of zero with more global warming. T1(t)-T2(t) = ΔT(t) = (ΔT(2014))*exp(-τ*(T1(t)-T1(2014))), where T1(t) is the temperature of the Ferrel Cell at time t, T2(t) is the temperature of the Polar Cell at time t, ΔT(t) is the temperature differential at time t, and τ is a constant. Rearranging the equation gives T2(t) = T1(t) - (ΔT(2014))*exp(-τ*(T1(t)-T1(2014))). Taking the derivative with respect to T1(t) gives ∂T2(t)/∂T1(t) = 1 + τ*(ΔT(2014))*exp(-τ*(T1(t)-T1(2014))). Setting t = 2014 gives ∂T2(2014)/∂T1(2014) = 1 + τ*(ΔT(2014)). Using the values ∂T2(2014)/∂T1(2014) = 3.33 and ΔT(2014) = 13.5K, we find that τ = (3.33-1)/13.5K = 0.1726/K. Using this value of τ, we can estimate the temperature difference for 2100. ΔT(2100) = (ΔT(2014))*exp(-τ*(T1(2100)-T1(2014))) = (13.5K)*exp(-0.1726/K*(4K)) = 13.5K*0.50 = 6.75K. The above calculations estimate that the temperature differential between the Polar and Ferrel Cells in the Northern Hemisphere should half by 2100. Of course, I gave some benefit of the doubt to climate alarmists, in picking the values, so most likely the expected change in the temperature differential by 2100 under no/little CO2 mitigation polices is less than this. Given that the speed of the jetstream is approximately proportional to sqrt(∇T) (as shown earlier), we should expect that the speed of the Northern Polar Jetstream will be sqrt(0.50) ≈ 71% the level it is now (or a 29% decrease in speed). Using the relationships explained above, this 29% decrease in the speed of the jetstream should lead to a decrease in the wavelength of the jetstream by 16%, an increase in the amplitude of the jetstream by 19%, an increase in the frequency of resonance phenomena by 67%, a decrease in the group velocity of the jetstream by 29%, an increase in the duration of non-resonant weather events by 19%, no change to the duration of resonant phenomena, and an increase in the mean magnitude of the longitudinal pressure gradient by 19% (mean latitudinal pressure gradient is unchanged). Even though the model used isn't perfect and the numbers used are not completely accurate, the above values should still give us an idea on the order of magnitude (as well as the sign) of change that we should expect under no/little CO2 mitigation policies. So overall, we should expect significant but not catastrophic change. 9. DISCUSSION & CONCLUSION The most important thing to point out is that the effects of changes to Rossby waves are only really relevant for the mid-latitudes in the Northern Hemisphere. For the Hadley and Polar cells that cover the equatorial and polar regions, wind primarily flows from east to west, which is against the rotation direction of the Earth, so Rossby waves cannot form. For the mid-latitudes in the Southern Hemisphere, most of the Earth's surface area for these latitudes is ocean, and the land is relatively featureless topographically (except maybe the Andes), so changes to Rossby waves are not that relevant. For the northern mid-latitudes, there are many topographical features that can generate Rossby Waves including the Rocky Mountains, the Ural Mountains, the Appalachian Mountains, the Alps, Tien Shan, etc. In addition, these latitudes are mostly land and contain many of the world's geopolitically important countries including Canada, United States, Europe, Russia, China, Korea, and Japan (All G8 and 12/20 G20 countries are along these latitudes). Thus changes to Rossby waves are very relevant to humanity. However, even if changes to the jetstream increase extreme weather events in these latitudes, reductions in polar-equatorial and surface-tropopause temperature gradients (across all latitudes) might reduce extreme weather events by enough to offset these effects; comparison and discussion between these changes is needed. With regards to the claims made by climate alarmists, they are correct in their claims that increasing atmospheric CO2 concentrations will lead to a slower jetstream, a jetstream with a smaller wavelength, a jetstream with a higher amplitude, a jetstream with a slower group velocity & a jetstream that more frequently experiences resonance phenomena. Indeed the duration of non-resonant weather phenomena should increase, but for extreme droughts or floods which result from resonant phenomena of forced standing waves, this should remain unchanged. However, the magnitude of high and low pressures remains unchanged, so why should anyone expect an increase in the severity/intensity of extreme weather events? Interestingly, the longitudinal density of high and low pressures will increase which will arguably lead to an increase in the magnitude of the longitudinal pressure differential (even if the pressure differential in the latitudinal direction stays the same). This might cause greater latitudinal wind speeds to occur, but since slower moving air at the top of the troposphere is also a consequence, the net effect is unclear. Having high and low pressures closer together longitudinally may reduce their magnitudes due to increased pressure transfer. This will cause a decrease in the pressure gradient, so the latitudinal pressure differential should decrease slightly while the longitudinal pressure differential should increase slightly. A better model should include the effects of High and Low pressures on each side of the jetstreams. It is estimated that under no/little CO2 mitigation policies and using more ambitious warming predictions that by 2100 it is expected that the speed of the jetstream should decrease by 29%, the wavelength of the jetstream should decrease by 16%, the amplitude of the jetstream should increase by 19%, the frequency of resonance phenomena should increase by 67%, the group velocity of the jetstream should increase by 29%, the duration of non-resonant weather events should increase by 19%, the duration of resonant phenomena should not change, and the mean magnitude of the longitudinal pressure gradient should increase by 19%. In conclusion, the claims by climate alarmists regarding the effects on climate change on the jetstreams and on Rossby waves are ridiculous and unfounded even if there is a small amount of truth behind them. The expected changes by 2100 to the jetstream and weather patterns for the Northern Mid-Latitudes are significant but are in no way catastrophic. The climate alarmists are correct in their claim that the frequency of resonance phenomena for the jetstream will increase. But at the same time there might be other effects, such as a decrease in the frequency at which non-resonant weather phenomena occur or a reduction in the duration of weather phenomena for latitudes at the edge of the jet stream, which might offset this. For the severity/intensity of weather events, there is no good reason to believe that they will decrease or increase as a result of this phenomena. Overall, the claim that a reduction in the speed of the jetstreams and other eastward moving wind will cause prolonged, more frequent and more severe extreme weather events is not justified.
4 Fundamental Changes of Science Will Be Very Soon, My Scientific Work By Exegesisme 1 Fundamental changes of science will be very soon. 2 I am discovering the structure of atomic nuclei and a crucial progress was made recently. The structures of atomic nuclei such as Ca, Fe, Ni and so on are as clear as the structure of molecules in front of me, the positions of every proton and neutron are careful arranged in my model of atomic nuclei. 3 The modern physics has been built fundamentally on mistaking principles, their validities will be recounted on new principles. 4 Sleep meditation will become the most fundamental mothed of scientific research and other human important practises. 5 Some results now are already well prepared for journalism to make big news.
1 My Exegesis of Gravitational Constant, Aether, Dark Matter and Dark Energy, My Scientific Work By exegesisme 1, Gravitational constant G, is not a constant. It just reveals the property of the local space in which it had been measured. Its whole physical meaning is the accelerating contract rate of the local space under the action of one kilogram mass. 2, The many measured numbers are the almost same which only means the local spaces in which they are measured share almost the same property, but not every local space shares almost the same property. 3, Why can the mass even as small as one kilogram cause the contraction of the space? My answer is, there are very evenly distributed mass in the local space, the gravity between the mass and the local space with the evenly distributed mass cause the contraction of the local space. 4, The evenly distributed mass is only even at the same distance from a dominating large mass such as Earth, and is different at different distance. The evenly distributed mass is denser at a nearer distance than a longer distance, so the gravitational constant G is smaller at a nearer distance than at a longer distance, for the nearer local space with the denser evenly distributing mass is harder to contract under the action of one kilogram mass. 5, The experimental physicists are very easy to prove my exegesis, as long as the distance or height between any two measurements is large enough to make the difference of G can be measured by the method they choose. 6, The fact is, although very large masses in our universe, our universe still is in accelerating expansion, why? The reason is, the evenly distributed mass is very small particles, these particles are with holy elasticity and have a random movement, which provides pressure against gravity to make our universe expanse. The force for the universal expansion is a new fundamental force, and also is responsible for the big bang. Other four basic forces can be unified as gravity, so, our universe has only two fundamental forces. 7, This substance is the holy substance Aristotle once assumed. The holy substance with very subtle mixed magnetism is also the media for the field of macro electromagnetism, and even the media for gravity of all masses. 8, The Michelson-Morley experiment is wrong of their experimental hypothesis, and is also wrong of their explanation of the experimental result on their wrong hypothesis. The holy substance is absorbed by every mass which human can measure, and is relatively still to the mass. Earth too absorbs the holy substance around. Earth moves with the absorbed holy substance, so they did not observed a relative motion between the holy substance and Earth. The speed of light did not change which just mean there is not relative motion between the holy substance and Earth. 9, The holy substance is also the material to form the charge of electron and positive electron, and can move with electron as electron moves, which is the reason of the waving phenomena of electrons and other micro particles. 10, The holy substance is both dark energy and dark matter, is the reason of the theory of relativity, and has many other properties and functions. The size of its particles is much smaller than the Planck's magnitudes.