Put $\ell =p_{\varphi }$. The Hamiltonian has no explicit time or azimuthal dependence, so $\dot{H}=\{H,H\}=0$ and $\dot{\ell }=-H_{\varphi }=0$. Its equations of motion areWrite $C=p_R\ell \sin \varphi +({\ell }^2/R+A)\cos \varphi$. Differentiating along these equations givesEquivalently, the same cancellation evaluates $\{C,H\}$. Hence $\boxed{H,p_{\varphi },C\text{ are all constants of motion}.}$ The extra conserved component is associated with the Runge-Lenz symmetry of this inverse-distance central potential, rather than with conservation of radial momentum.

Hamilton's equations are ${\dot{q}}_i=H_{p_i}$ and ${\dot{p}}_i=-H_{q_i}$. Using the elementary brackets from the previous part givesA constant of motion is a function that stays unchanged along each physical trajectory. For a function $C(\mathbf{q},\mathbf{p})$ with no explicit time dependence, the chain rule and Hamilton's equations giveThus $\boxed{C\text{ is conserved}⟺\{C,H\}=0}$ on the accessible phase-space domain. If a function explicitly depends on time, the criterion is instead ${\partial }_{t}C+\{C,H\}=0$; the simpler statement uses the source's time-independent $C$ assumption.

For the canonical Poisson bracket, direct substitution of $F=q_i$ and $F=p_i$ givesThereforeThis expresses the bracket as a first-order derivative acting on $F$, with coefficients determined by $G$; it is the phase-space chain-rule form of the definition.

It can describe the motion per unit mass of a charged particle in the azimuthal magnetic field of a long straight current-carrying wire. Choose magnetic vector potentialThen $\mathbf{B}=\nabla \times \mathbf{A}=(mC/(qR))\hat{\mathbf{\varphi }}$ and $(T+q\mathbf{A}\cdot \mathbf{v})/m$ is the given Lagrangian. In terms of a straight-wire current $I$, $B_{\varphi }={\mu }_{0}I/(2\pi R)$, so $C=q{\mu }_{0}I/(2\pi m)$ with signs fixed by charge and current direction.

A static magnetic force does no work, consistent with the conserved kinetic energy. Its radial component is $-q\dot{z}{B}_{\varphi }$ and its axial component is $q\dot{R}{B}_{\varphi }$, reproducing the radial and axial couplings above. The reference length in the logarithm is a gauge convention, not a new physical wire radius in the ideal exterior-field model.

Write $w=a^2{\Omega }^2/C$, requiring $C\ne 0$ and $a>0$. The initial conditions fix $\ell =a^2\Omega$ and $\dot{z}=w+C\log (R/a)$. At $R=a$, the radial force is $a{\Omega }^2-Cw/a=0$. Thus the unperturbed motion is the helixTo analyse a small initial radial velocity, eliminate the cyclic velocities using the conserved quantities. The radial energy equation is ${\dot{R}}^2/2+U(R)=\mathcal{E}$, withHere $U^{\prime }(a)=0$ and $U^{\prime \prime }(a)=2{\Omega }^2+C^2/a^2>0$. The circular cylindrical radius is therefore stable. For $\eta =R-a$, the first-order equation is $\ddot{\eta }+{\omega }_r^2\eta =0$, soThe orbit acquires small radial oscillations. The other coordinates are modulated consistently with the same constants:These expressions retain the stated initial angular and axial velocities. The perturbation adds ${\delta }^2/2$ to the conserved energy. The expansion assumes $|\delta |/(a{\omega }_r)\ll 1$, which follows from the supplied smallness condition when $\Omega \ne 0$; it is not an exact finite-amplitude helix.

The Euler-Lagrange equations in cylindrical coordinates areThe last two follow because $\varphi$ and $z$ are cyclic coordinates. They give the conserved quantitiesSince the Lagrangian is autonomous, its energy function is conserved as well:The velocity-linear term cancels from this energy, but it still changes the equations of motion. A dimensionless logarithm means $\log (R/R_0)$ for some fixed reference length; replacing the printed $\log R$ by that expression changes the Lagrangian only by a total time derivative and shifts the numerical value of $k$, not the trajectory.

Hamilton's principle states that the physical path makes the action $S={\int }_{t_i}^{t_f}L(\mathbf{q},\dot{\mathbf{q}},t)dt$ stationary under differentiable variations $\delta q_i$ vanishing at the fixed endpoints. Its conventional “least action” name does not guarantee a minimum. Expanding to first order and integrating the velocity term by parts givesThe endpoint term vanishes. By the fundamental lemma of the calculus of variations, the coefficients of the independent interior variations must vanish, yieldingFor example, the zero path of a harmonic oscillator with endpoints zero can have both positive and negative second variations when the elapsed time exceeds a half-period; stationary action then need not be a least value. The Euler-Lagrange condition is the local statement that the variational principle actually supplies.

A plug-in hybrid electric vehicle can move short journeys to electricity while retaining liquid-fuel range for longer ones. It can reduce oil dependence and some local tailpipe emissions, but the carbon benefit depends on charging, trip length, electricity supply, vehicle manufacture and the fuel used when the engine runs.

For illustration, electricity consumption $0.20\mathrm{kWh}\mathrm{k}{\mathrm{m}}^{-1}$ at grid intensity $500\mathrm{g}\mathrm{C}{\mathrm{O}}_2\mathrm{kW}{\mathrm{h}}^{-1}$ gives $100\mathrm{g}\mathrm{C}{\mathrm{O}}_2\mathrm{k}{\mathrm{m}}^{-1}$ in use. A petrol vehicle using $0.07$ litres per km emits about $158\mathrm{g}\mathrm{C}{\mathrm{O}}_2\mathrm{k}{\mathrm{m}}^{-1}$ from combustion under the paper's octane-density assumptions. Ignoring other life-cycle differences, their use-phase break-even electricity intensity is therefore roughly $790\mathrm{g}\mathrm{C}{\mathrm{O}}_2\mathrm{kW}{\mathrm{h}}^{-1}$. Cleaner electricity improves the electric case; sufficiently carbon-intensive electricity can reverse it. These are conditional comparisons, not measured fleet averages.

A plug-in hybrid that is rarely charged may obtain little benefit while carrying both propulsion systems. Battery and fuel production also belong in a life-cycle assessment. Fully electric vehicles, efficient conventional vehicles, public transport, walking, cycling and reduced travel demand provide alternative or complementary routes. Plug-in hybrids can be a useful transition technology, but no physical argument makes them an essential component of every successful climate strategy.

A common minimum carbon price can make the external cost of emissions visible in investment and dispatch decisions, encourage reductions where they are cheapest, and reduce incentives to relocate emissions-intensive production to unpriced jurisdictions. A credible rising or durable price signal can be more useful for long-lived investment than a volatile short-term spot price.

For an illustrative price of $40$ currency units per tonne of carbon dioxide, generation emitting $0.9$ tonnes per MWh incurs $36$ units per MWh, while generation at $0.4$ tonnes per MWh incurs $16$. Those changes are large enough to affect some operating choices, but the actual switching point depends on fuel prices, efficiency and capacity constraints.

A price is not sufficient by itself. Measurement, enforcement, distributional effects and revenue use matter; infrastructure, innovation spillovers, information failures and landlord-tenant incentives can justify complementary policies. Nor is a uniform international minimum a logical prerequisite for any effective reduction: standards, clean-power deployment and efficiency policies can reduce emissions even before every country agrees. The strongest defensible claim is that broad, credible pricing is a valuable coordination instrument, not a uniquely necessary or automatically sufficient solution.^[ref]

The MENA solar resource could provide a useful supplement, but geographical potential does not make imports logically essential to European decarbonisation. With an illustrative annual mean insolation $200\mathrm{W}{\mathrm{m}}^{-2}$, photovoltaic conversion $15\%$ and system factor $0.8$, mean delivered-at-plant power is $24\mathrm{W}{\mathrm{m}}^{-2}$. A collector area of $10^4\mathrm{k}{\mathrm{m}}^2$ would then give $240\mathrm{GW}$ mean or about $2100\mathrm{TWh}$ annually before export losses. This is a scale calculation with stipulated assumptions, not a land-availability survey.

Long transmission corridors, converter losses, night-time supply, storage, dust, cooling or cleaning water, financing and political dependence must also be included. Solar thermal storage can shift some output but does not remove every seasonal or geopolitical risk. Projects should benefit host populations and avoid treating nominal desert area as empty infrastructure-ready land.

Europe also has domestic wind power, solar power, hydroelectricity, potential nuclear generation, demand reduction and other mitigation options. A technically credible combination of those is a counterexample to the word “essential”. MENA imports could reduce costs or improve diversity within a broader low-carbon system; whether they are preferable requires a complete system comparison rather than solar resource density alone.

Tidal power offers predictable astronomical timing, while wave power can exploit an energetic marine resource. Neither is constant: tidal streams have slack-water periods, and wave energy varies with weather and season. Diversity can help a renewable portfolio, but resource size does not by itself establish economical extractable output.

For scale, a mean marine output of $5\mathrm{GW}$ would produce $43.8\mathrm{TWh}$ per year, about $15\%$ of an assumed $300\mathrm{TWh}$ annual electricity demand. Achieving that mean at capacity factor $0.30$ would require about $17\mathrm{GW}$ installed. This is an illustrative deployment requirement, not a measured UK resource estimate or a 2012 forecast. Tidal-stream kinetic power density scales as $\rho v^3/2$, making energetic sites valuable, but channel-wide extraction also changes the flow, so isolated-rotor power densities cannot simply be multiplied by sea area.

Harsh marine conditions, access for maintenance, survivability in storms, cable connections, navigation and ecological effects constrain deployment. Barrages additionally change estuaries and sediment transport. Research and demonstration may be worthwhile for energy diversity and industrial capability, but “important to investigate” is weaker than “indispensable at any cost”. Marine investment should be compared with efficiency and other low-carbon options using delivered energy, system value and environmental impacts.

The judgement depends on the objectives and the replacement system, rather than following from a single technological fact. In the 2012 context, Germany had decided after Fukushima to accelerate its planned nuclear exit, with a timetable then extending to 2022. The contemporary official statement presented this as a reassessment of residual risk and a transition to efficiency and renewable supply.^[ref]

There is a serious climate opportunity cost if low-operational-carbon nuclear generation is replaced by unabated fossil generation. For example, replacing an illustrative $100\mathrm{TWh}\mathrm{yea}{\mathrm{r}}^{-1}$ with generation emitting $400$ or $900\mathrm{g}\mathrm{C}{\mathrm{O}}_2\mathrm{kW}{\mathrm{h}}^{-1}$ adds $40$ or $90$ million tonnes of carbon dioxide annually. Replacing that energy by wind at capacity factor $0.30$ would require about $38\mathrm{GW}$ of nameplate capacity, plus an adequate reliability plan; equal annual energy is not equal firm capacity.

Against continued operation must be weighed accident consequences, waste management, ageing, public preferences and the costs and feasibility of alternatives. Germany's specific risk assessment cannot be inferred solely from the different geography of Fukushima. A fast exit without credible low-carbon replacement can conflict with emissions goals; an exit supported by efficiency, infrastructure and low-carbon alternatives can pursue other legitimate objectives. Calling the decision intrinsically rational or irrational without those comparisons is not a technical conclusion. This answer evaluates the choice as posed in 2012, rather than importing later events as foreknowledge.

Variability reduces firm capacity and creates balancing costs, but does not impose a universal tiny upper limit on the energy fraction from wind power. As an illustrative scale calculation, $10\mathrm{GW}$ installed at capacity factor $0.30$ produces $26.3\mathrm{TWh}$ annually, about $8.8\%$ of an assumed $300\mathrm{TWh}$ annual demand. Larger installations can supply larger fractions if the rest of the system is adapted.

Geographical spread, forecasting, interconnection, flexible generation, demand response and energy storage manage fluctuations on different timescales. They are not interchangeable: a battery that balances seconds does not necessarily cover a week of low wind. Correlated weather, curtailment, network constraints and reserve requirements become more important at high penetration. Backup capacity need not burn fuel continuously at the wind plant's nameplate output; energy and capacity should be accounted for separately. The defensible conclusion is that significant penetration requires system planning and has costs, not that variability alone confines wind to a negligible contribution.^[ref]

For the liquid pump take the explicit incompressible-water estimate $v_f\simeq 10^{-3}{\mathrm{m}}^3\mathrm{k}{\mathrm{g}}^{-1}$. Its work input isso $h_2=h_1+w_p=111.998\mathrm{kJ}\mathrm{k}{\mathrm{g}}^{-1}$. With $h_3=3463$ and the preceding $h_4=1838.74$ in the same units,ThusIf pump work is neglected because no liquid density was tabulated, the corresponding approximation is $(3463-1838.74)/(3463-84)=48.1\%$. Stating the approximation explains the small difference. The temperature-extreme Carnot efficiency is about $1-293.15/873.15=66.4\%$, above both results; finite-temperature-range heat addition in the Rankine cycle does not attain that limit.

Assume an ideal isentropic turbine and use the supplied table consistently. Its inlet specific entropy is $s_3=6.282\mathrm{kJ}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-1}$. At the condenser pressure the outlet mixture has $s_4=s_f+\chi (s_g-s_f)$, where $\chi$ is the vapour mass fraction. HenceThe outlet is approximately $71.5\%$ vapour and $28.5\%$ liquid by mass. Its specific enthalpy isThe source's common unit label omits the kelvin denominator for entropy: $s$ must be interpreted in $\mathrm{kJ}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-1}$, while $h$ is in $\mathrm{kJ}\mathrm{k}{\mathrm{g}}^{-1}$. A real irreversible turbine would increase entropy and require an efficiency correction; no such correction is supplied.

Post-combustion carbon capture and storage separates carbon dioxide from an already formed flue gas, often by absorption in a regenerable solvent. The recovered gas must be purified, dehydrated, compressed, transported and injected into a suitable geological store. Caprock integrity, trapping mechanisms, monitoring and long-term liability matter as much as capture at the chimney.

Dilute carbon dioxide mixed with nitrogen requires a large gas-processing plant. Solvent regeneration consumes heat; compression consumes power; impurities can cause solvent degradation and corrosion. Retrofitting adds equipment, cooling demand and cost, and lowers net electrical output per unit fuel. Storage capacity, injection performance, leakage risk and public acceptance must be demonstrated, while upstream fuel-cycle emissions remain.

A capture fraction is not the same as the reduction per delivered kilowatt-hour. For illustrative capture $\chi =0.9$ and a plant-efficiency decrease from $0.4$ to $0.3$, residual stack emissions per delivered unit scale as $(1-\chi )(0.4/0.3)=0.133$, corresponding to an $86.7\%$ reduction rather than $90\%$. A full life-cycle assessment includes the additional fuel and upstream emissions. These figures illustrate the accounting and are not claimed as universal plant performance.

An ideal Rankine cycle has four main steps: liquid water is pumped reversibly to high pressure; heat is added at high pressure to produce hot vapour or supercritical working fluid; the fluid expands approximately isentropically through a turbine to produce work; and low-pressure heat rejection in a condenser returns it to liquid. For steady flow, the main work and heat transfers are differences of specific enthalpy.

A steam implementation of the Carnot cycle would require compression of a wet liquid-vapour mixture and reversible isothermal heat transfer, both difficult to achieve with practical machinery. Complete condensation before the Rankine pump makes compression work small because liquid specific volume is small. High-pressure heating and superheating can then occur in a boiler without demanding isothermal addition of all the heat. Superheating also reduces turbine-exit moisture, although the calculated exit may still be wet. The pump, turbine, condenser and boiler can operate continuously as separate flow devices. The Rankine cycle sacrifices some ideal thermal efficiency in return for a practicable steam-plant configuration; no real engine can exceed the Carnot efficiency for the same extreme reservoir temperatures.

Three advantages of biomass energy are:

Three disadvantages are:

Use equal combustion energy as the displacement criterion. The petrol mass and energy areThis needs ethanol mass $M_e=Q_p/(30.5\times 10^6)=1.04115\times 10^{10}\mathrm{kg}$. The ethanol fermentation stoichiometry converts $180$ mass units of glucose to $92$ of ethanol, so the ethanol yield per square kilometre is $(1500)(92/180)=766.667$ tonnes. ThereforeThe carbon mass fraction in octane is $8(12)/(8(12)+18)=96/114$. The displaced fossil carbon is thusThis is the ideal net reduction if the crop regrows and its carbon dioxide uptake balances the biogenic carbon released in fermentation and combustion, with negligible additional life-cycle emissions. Ethanol combustion is not carbon-free: it releases about $5.43$ million tonnes of carbon here, and fermentation releases further biogenic carbon. Those are balanced by crop uptake in the stated ideal accounting, rather than being ignored. Farming, fertilizer manufacture, processing heat, transport and land-use carbon losses reduce the actual benefit and may even reverse it. The equal-heat calculation also assumes equal useful-energy conversion efficiencies in the displaced application; it does not establish equal engine mileage automatically.

Take an explicitly illustrative European annual mean ground-level irradiance of $100\mathrm{W}{\mathrm{m}}^{-2}$, the stated conversion fraction $0.005$, and dry carbohydrate heating value $16\mathrm{MJ}\mathrm{k}{\mathrm{g}}^{-1}$. One square kilometre then producesHence $\boxed{M\sim 10^3\mathrm{tonnes}\mathrm{k}{\mathrm{m}}^{-2}\mathrm{yea}{\mathrm{r}}^{-1}}$, or about $10\mathrm{tonnes}\mathrm{h}{\mathrm{a}}^{-1}\mathrm{yea}{\mathrm{r}}^{-1}$, as a dry-mass order-of-magnitude estimate. The assumed irradiance is not a site-specific value supplied by the paper. The result scales linearly with irradiance and conversion efficiency and inversely with heating value; wet harvested mass would additionally depend on water content.