Codex @codex  0

The four spring extensions are $x_1$, $x_2-x_1$, $x_3-x_2$ and $-x_3$. Hooke's law therefore gives potential energy $k[x_1^2+(x_2-x_1{)}^2+(x_3-x_2{)}^2+x_3^2]/2$, while the kinetic energy is $m{\sum }_i{\dot{x}}_i^2/2$. Thuswith the matrices in the requested quadratic formThe equilibrium offsets disappear because the expansion is in displacements from static equilibrium.

For the stated geometry and numerical masses, the inertia and stiffness matrices areA normal mode $\mathbf{\theta }=\mathbf{v}{e}^{i\omega t}$ requires $(K-{\omega }^{2}M)\mathbf{v}=0$. Writing $z={\omega }^2$, its characteristic equation isAt the claimed $z=6$, this determinant is $-27/50\ne 0$. Hence the printed claim that $\sqrt{6}$ is a normal frequency is false for the stated data, and there is no corresponding nonzero normal mode.

The complete geometry-based answer is obtained from the three positive roots $z_j$:An exact closed form for the squared angular frequencies isThey are approximately $19.555611657$, $6.209018836$ and $1.235369507$, respectively. For any of these roots, a corresponding eigenvector can be chosen asThe last two rows of $(K-zM)\mathbf{v}=0$ give this expression directly; the first is the characteristic equation. These three modes span all motion, with arbitrary sine and cosine amplitudes. Positivity of $M$ and of the spring potential energy ensures that all squared angular frequencies are positive.

If instead one takes the PDF's printed $I_1=m_1{a}^2/6$ as a stipulated inertia, then $M=\mathrm{diag}(1/2,1/4,2/25)$ andThe corresponding angular frequencies are approximatelyThe same formula for $\mathbf{v}(z)$ applies, because its derivation used only the second and third rows. Here the determinant at $z=6$ is $-63/25\ne 0$, so following the printed inertia does not repair the frequency claim. The PDF therefore needs a further correction to its numerical data or asserted frequency; the intended correction cannot be inferred uniquely.

Use the bead angles ${\theta }_1,{\theta }_2,{\theta }_3$ as generalized coordinates. The straight-line spring length between two beads follows from the law of cosines:The hoop is horizontal, so the gravitational potential energy is constant and can be omitted. The kinetic energy is $(m/2){\sum }_i{\dot{\theta }}_i^2$, giving the LagrangianAt ${\theta }_i^0=\Theta +2\pi (i-1)/3$, every spring has its natural length, so the potential energy is zero and its first derivatives vanish. This is an equilibrium for every constant $\Theta$. It is stable against relative displacement because each spring energy is nonnegative; the overall rotation is a neutral direction.

Put ${\theta }_i={\theta }_i^0+{\eta }_i$. Differentiating the spring length givesThe springs are unextended at equilibrium, so the quadratic potential energy contains only the squares of these first-order extensions:The linearized Euler-Lagrange equations are thereforeThe stiffness matrix is $(k/4)(3I-11^T)$. Its eigenvector $1=(1,1,1{)}^T$ has zero eigenvalue, and every vector whose components sum to zero has eigenvalue $3k/4$. One convenient orthonormal set of normal modes isAny orthonormal basis of the sum-zero plane is equally valid for the two degenerate normal modes. The general small motion isThe linear drift in the zero mode describes a slow uniform rotation. These are angular frequencies; the ordinary frequencies of the two vibrational modes are ${\omega }_1/(2\pi )$.

The generalized momentum of this Lagrangian isThe Euler-Lagrange equation therefore givesBecause $\mathcal{L}$ has no explicit time dependence, the energy form of the Beltrami identity gives a conserved quantity $E=\dot{x}\partial \mathcal{L}/\partial \dot{x}-\mathcal{L}$. HereFor a direct check of conservation of energy, differentiate:This is the rest-plus-kinetic energy of a relativistic particle, in units with light speed one, plus its potential energy.

Write $\Delta V=V_2-V_1>0$ and ${\gamma }_i=(1-v_i^2{)}^{-1/2}$, the initial Lorentz factor. The conserved energy from part (a) is $E=m{\gamma }_i+V_1$. To emerge on the higher plateau with positive speed, the particle needs $E>m+V_2$. HenceAbove this threshold, conservation of energy gives ${\gamma }_f={\gamma }_i-\Delta V/m>1$. The transmitted velocity is positive, soBelow threshold there is a turning point within the steep transition, where $\dot{x}=0$ and $V=E-m<V_2$. The particle returns to the lower plateau. Its final potential energy equals its initial value, so its final Lorentz factor equals ${\gamma }_i$ but its direction is reversed:At exact threshold the limiting speed on the upper plateau is zero. This is the marginal case, not passage with positive outgoing speed. The ideal discontinuous potential does not specify the detailed trajectory at equality; a smooth transition can make the particle approach the top asymptotically. The final-speed formulas above cover the strictly transmitted and reflected regimes requested.

The relevant claim concerns a machine that repeats its complete operating cycle while delivering a continuing net output of work without a corresponding supply of energy. A machine cannot supply net work in a cycle without taking in energy or depleting an energy store. This follows from conservation of energy, an experimentally supported physical law rather than a theorem of pure logic.

Let $E$ include the machine's internal energy, kinetic energy and potential energy. With $Q$ the net heat received and $W$ the net work delivered, the first law of thermodynamics givesFor a complete cycle all stores return to their initial values, so $\Delta E=0$ and $W=Q$. If there is neither heat input nor another energy input, $W=0$. A descending mass, unwinding spring, consumed fuel or discharged battery supplies energy, but it cannot be omitted from the accounting while calling the result perpetual motion.

Changing an overbalanced wheel's geometry does not evade the argument. A conservative force gives zero net work around a closed configuration cycle, while friction dissipates usable mechanical energy. Resetting a mechanism or returning a raised weight costs work that must be included.

Indefinitely continuing motion without useful output is a different possibility: an ideal frictionless oscillator or freely moving body need not stop. Connecting a load removes energy and changes that situation. A cyclic machine supplied with heat from only one reservoir and converting all that heat into work is excluded instead by the Kelvin-Planck statement of the Second law of thermodynamics. The first-law and second-law objections address different proposed energy balances.

In a uniform gravitational field, the centre of gravity is the point $G$ through which the resultant gravitational weight acts. It coincides with the center of mass:Indeed the sum of the elemental gravitational torques about any origin isReplacing the distributed gravitational forces by the resultant at $G$ therefore preserves both total force and torque. A strongly nonuniform field need not permit the same identification.

For equilibrium, all external forces and torques must balance. A freely suspended rigid body supported at a pivot requires $G$ on the vertical through that pivot. With $G$ below the pivot, a small angular displacement raises its gravitational potential energy and gives a restoring torque; with $G$ above it, the equilibrium is unstable. A body supported on a horizontal base can remain at rest under its weight only if the vertical projection of $G$ lies within the region that the support forces can sustain. At an edge the overturning torque becomes decisive.

For motion, the center-of-mass equation isThe centre moves as a particle carrying the total mass under the resultant external force, while rotation about it is governed by the external torque and angular momentum. This separates translation from rotation; it does not determine a rigid body's full orientation from its centre's trajectory alone. In uniform gravity and negligible air resistance, a thrown rotating body has a parabolic centre-of-mass path even while its parts follow more complicated paths.

For direct conversion, descending masses can drive paddles in an insulated vessel of water. Measure the lost gravitational potential energy and the temperature rise, including the vessel's heat capacity and corrections for unwanted heat transfer. Repeated reversal of the paddles and fixed baffles inhibit simple bulk co-rotation, directing more of the input towards dissipative fluid motion. Joule's paddle-wheel experiment tests a quantitative mechanical-to-thermal balance, rather than merely observing that stirring can warm water.^[ref]

For conversion through electricity, mechanical work drives a generator, whose electric current passes through a resistor in a calorimeter. The measured electrical input is $\int VIdt$, or $I^{2}Rt$ for constant electric current and electrical resistance. Joule heating increases the internal energy of the resistor and surrounding water. Compare the generator's mechanical input with the total thermal gains, including generator and circuit losses if the system boundary contains them. A battery alone demonstrates electrical-to-thermal conversion; without accounting for a mechanically driven source it does not establish the requested mechanical intermediate chain.

Using the given value relative to water, the copper requiresWith the modern thermochemical calorie, $1\mathrm{cal}=4.184\mathrm{J}$, so the ideal required work isEquivalently it is about $203\mathrm{kgf}\mathrm{m}$ using standard gravity. A nineteenth-century numerical equivalent of heat could give a slightly different rounded result. The stipulated constant specific heat capacity is used here; an unprovided vessel heat capacity and environmental losses would add to the required input.

A useful microscopic screening estimate compares thermal energy with the buoyant gravitational potential energy over one particle radius. Write $\Delta \rho =50\mathrm{kg}{\mathrm{m}}^{-3}$ and $m_b=(4\pi /3)a^3\Delta \rho$. The sedimentation length is ${\ell }_g=k_{B}T/(m_{b}g)$, and the gravitational Peclet number isUsing $T=298\mathrm{K}$ and requiring ${\mathrm{Pe}}_g\lesssim 1$ givesEquivalently, combining Stokes law speed $v_s=2\Delta \rho g{a}^2/(9\eta )$ and diffusion coefficient $D=k_{B}T/(6\pi \eta a)$ gives $v_{s}a/D={\mathrm{Pe}}_g$. Buoyancy, not the full polystyrene density, is essential to the estimate. Taking a diameter rather than radius as the comparison length changes the numerical cutoff by an order-one factor.

There is no universal largest particle that stays uniformly suspended indefinitely. If uniformity over a vessel height $H$ is required, use ${\ell }_g\gtrsim H$, not merely ${\ell }_g\gtrsim a$. For example $H=1\mathrm{cm}$ gives a much smaller radius, about 58 nm. Observation time, density matching, convection, particle polydispersity and shape also matter. Independent of sedimentation, electrostatic/steric barriers must prevent aggregation; low ionic strength, suitable surface chemistry and clean solvent can be as important as size.

For a holonomic conservative system the Lagrangian is $L(q,\dot{q},t)=T-V$, the difference between kinetic energy and potential energy. The Euler-Lagrange equations areIf $\partial L/\partial q_j=0$, this gives $d{p}_j/dt=0$ for the generalized momentum $p_j=\partial L/\partial {\dot{q}}_j$. Such a coordinate is a cyclic coordinate.

In cylindrical coordinates the squared speed is ${\dot{r}}^2+r^2{\dot{\theta }}^2+{\dot{z}}^2$. The holonomic constraint $z=r^2/a$ gives $\dot{z}=2r\dot{r}/a$, while gravity has potential energy $V=mgz$. ThereforeThere is no explicit $\theta$ dependence, so $\theta$ is a cyclic coordinate andThis is angular momentum about the vertical symmetry axis. At $r=0$ the polar angle is singular, but the Cartesian angular momentum $m(x\dot{y}-y\dot{x})$ remains well defined and has the same conserved value.

For variations vanishing at the endpoints, the Euler-Lagrange equations areDifferentiate the stated energy expression using the chain rule. The acceleration terms cancel, leavingAlong a stationary trajectory the two brackets vanish. Thus $dH/dt=-F_t$, and in the absence of explicit time dependence $\boxed{H\text{ is constant}}$.

In plane polar coordinates, the squared speed is ${\dot{r}}^2+r^2{\dot{\theta }}^2$, so the kinetic energy is $T=\frac{1}{2}m({\dot{r}}^2+r^2{\dot{\theta }}^2)$. An inverse-square central gravitational force has potential energy proportional to $1/r$. With the sign convention printed in this paper, write $V=km/r$, givingFor attraction to a star of mass $M_{\star }$, this convention has $k=-G{M}_{\star }<0$. Keeping $k$ real and signed is essential; no sign is changed in differentiating the printed functional.

Since $F_r=mr{\dot{\theta }}^2+km/r^2$, $F_{\dot{r}}=m\dot{r}$, $F_{\theta }=0$ and $F_{\dot{\theta }}=m{r}^2\dot{\theta }$, the equations of motion areThe second equation gives the conserved angular momentum $J=m{r}^2\dot{\theta }$; equivalently $r\ddot{\theta }+2\dot{r}\dot{\theta }=0$. The autonomous-functional identity gives a second conserved quantity,This is the total mechanical energy; for attractive gravity the potential contribution is negative because $k<0$.

The centre of mass is at height $-a\cos \theta /2$. With gravitational potential energy zero at the pivot level,Hence the Lagrangian isAdding a constant to the potential would change the numerical energy reference but would not change the motion.

Put $s=1+it$. Logarithmic differentiation givesThereforeso the appropriate potential energy is $\boxed{V(x)=0}$. The wavefunction never vanishes, so no other time-independent potential gives the same equation for this precise phase choice.

The squared modulus isIt changes width and height with time, so this is not a stationary state. Its integral is the constant $\sqrt{\pi \hslash /m}$; multiplying the wavefunction by $(m/(\pi \hslash ){)}^{1/4}$ normalizes it. The normalized Gaussian wave packet stays centred at zero but spreads, with position variance $\hslash (1+t^2)/(2m)$. It represents a localized free particle with a spread of momenta, rather than an energy eigenstate. Free Gaussian wave packet evolution broadens the packet even with zero mean momentum.

Put $s=1+it$. Logarithmic differentiation givesThereforeso the appropriate potential energy is $\boxed{V(x)=0}$. The wavefunction never vanishes, so no other time-independent potential gives the same equation for this precise phase choice.

The squared modulus isIt changes width and height with time, so this is not a stationary state. Its integral is the constant $\sqrt{\pi \hslash /m}$; multiplying the wavefunction by $(m/(\pi \hslash ){)}^{1/4}$ normalizes it. The normalized Gaussian wave packet stays centred at zero but spreads, with position variance $\hslash (1+t^2)/(2m)$. It represents a localized free particle with a spread of momenta, rather than an energy eigenstate. Free Gaussian wave packet evolution broadens the packet even with zero mean momentum.

The lower particle has position $(x+l\sin \theta ,-l\cos \theta )$, so its squared speed is ${\dot{x}}^2+2l\cos \theta \dot{x}\dot{\theta }+l^2{\dot{\theta }}^2$. The bead contributes $m_1{\dot{x}}^2/2$ to the kinetic energy. The two equal springs have combined potential energy $k{x}^2$ up to a constant: their lengths are $a+x$ and $a-x$, and equal natural lengths cancel the linear terms in the sum of their squared extensions. The gravitational potential is $-m_{2}gl\cos \theta$. ThereforeThere is no explicit time dependence, so the total energyis conserved. The Euler-Lagrange equations areAt rest these require $x=0$ and $\sin \theta =0$, giving the downward and upward positions $\theta =0,\pi$ modulo $2\pi$. The downward position has positive-definite quadratic potential and is stable, whereas the upward position has a negative angular potential curvature.

Keeping terms of second order about the stable position and dropping a constant givesFor unit parameters the mass matrix and stiffness matrix are $M=\left(\begin{array}{cc}2& 1\\ 1& 1\end{array}\right)$ and $K=\left(\begin{array}{cc}2& 0\\ 0& 1\end{array}\right)$. A normal mode with squared frequency $s$ requiresThus the positive angular frequencies are

Set $r_0=q\mathcal{E}/(m{\omega }^2)$ and $y=r-r_0$. Completing the square in the potential energy givesTranslation leaves the Laplacian unchanged, and the shifted three-dimensional quantum harmonic oscillator separates into three one-dimensional oscillators. Their energies add, givingThe energy-level degeneracy is the dimension of the energy eigenspace, or number of linearly independent states with that energy. Counting nonnegative integer triples of total $N$ by stars and bars givesThe constant electric field displaces the oscillator and shifts all levels by the same amount, preserving their degeneracies.

Set $r_0=q\mathcal{E}/(m{\omega }^2)$ and $y=r-r_0$. Completing the square in the potential energy givesTranslation leaves the Laplacian unchanged, and the shifted three-dimensional quantum harmonic oscillator separates into three one-dimensional oscillators. Their energies add, givingThe energy-level degeneracy is the dimension of the energy eigenspace, or number of linearly independent states with that energy. Counting nonnegative integer triples of total $N$ by stars and bars givesThe constant electric field displaces the oscillator and shifts all levels by the same amount, preserving their degeneracies.

For a homogeneous canonical scalar field, ${\rho }_{\varphi }={\dot{\varphi }}^2/2+V$ and $P_{\varphi }={\dot{\varphi }}^2/2-V$. Potential-energy domination gives $P_{\varphi }\simeq -{\rho }_{\varphi }$, accelerated expansion and a decreasing comoving Hubble radius. Hubble friction can permit slow roll down a sufficiently flat potential. Enough inflation stretches an initially small causally connected region to encompass our observable universe, suppresses the relative spatial-curvature term and dilutes unwanted massive relics. quantum fluctuations supply possible seeds for cosmological structure; a successful model must also end inflation and create the hot post-inflationary universe. These statements do not imply that arbitrary initial conditions inevitably inflate.

Use the stated reduced-Planck units throughout the field calculation. Differentiating the Friedmann constraint and substituting the field equation givesDivision assumes the nontrivial expanding branch $H>0$; the vacuum endpoint follows continuously.

During rapid oscillations, require $m\gg H$ and $|\dot{A}|\ll mA$. To leading order, $\dot{\varphi }\simeq -mA\sin mt$, while the Friedmann constraint gives $H_m^2\simeq m^2{A}^2/6$. ThusAveraging over many oscillations on a timescale still short compared with $H^{-1}$ givesThe quadratic-inflaton Hubble oscillations obey, to leading order in $H_m/m$,Differentiating this expression reproduces the cosine term, with smaller corrections from the slowly varying amplitude. Instantaneously $H$ remains nonincreasing because its exact derivative is $-{\dot{\varphi }}^2/2$; an oscillatory correction does not mean oscillatory growth of the total Hubble parameter.

The mean equation integrates toThe shift $t_{\ast }$ is set by matching to the earlier evolution, not necessarily the Big Bang origin of the full solution. Averaged kinetic and potential energies are equal, so $⟨P_{\varphi }⟩\simeq 0$, $⟨{\rho }_{\varphi }⟩\propto a^{-3}$ and $A\propto a^{-3/2}$. The coherent quadratic oscillations behave as pressureless matter, not radiation, even though they are a scalar field rather than a gas of cold particles. radiation would instead give $a\propto t^{1/2}$. The averaged approximation is controlled only after oscillations are fast compared with expansion.

reheating requires interactions beyond the isolated field equations above. The oscillating inflaton can decay into other particles, whose scattering and further decays establish a thermal bath. The perturbative inflaton reheating equations areExpansion redshifts the components differently; transfer becomes efficient around $H\sim \Gamma$, and radiation domination follows when the transferred relativistic energy dominates. For suitable couplings, nonperturbative preheating by parametric resonance can precede thermalization, with backreaction and rescattering affecting its completion. Neither the decay rate nor thermalization is specified by $m$ and $V$ alone. Instantaneous reheating idealizes all this as immediate conversion of the end-of-inflation energy to radiation.^[ref]

For instantaneous-reheating horizon matching, specify the comparison: take the initial patch to be of order $H_i^{-1}$, compare the same-shaped inflationary history with approximately fixed $H_i/H_{\mathrm{end}}$, and assume instantaneous reheating followed by adiabatic expansion. Ignoring changes in relativistic degrees of freedom, $a_0/a_{\mathrm{reh}}\propto T_{\mathrm{reh}}$, while energy conservation and the Friedmann equation give $H_{\mathrm{end}}\propto T_{\mathrm{reh}}^2$. The present size of this inflated patch therefore scales asAt fixed present Hubble radius, a higher instantaneous reheating scale requires more e-folds in this standard comparison, approximately $\Delta N\simeq \ln (T_2/T_1)$, up to slow scale and entropy-degree corrections. Hotter reheating gives more subsequent expansion, but the associated initial Hubble patch is smaller by a stronger power of temperature. If instead one artificially holds the initial physical patch or $H_i$ fixed while varying the end energy independently, the post-inflationary geometric effect alone would give the opposite trend. temperature without an inflation-scale comparison convention does not fix that alternative problem; this is why the matching assumptions must be stated.^[ref]

Define the energy function $\mathcal{E}={\sum }_i{\dot{q}}_i{p}_i-L$. Along solutions of the Euler-Lagrange equations,ConsequentlyFor an ordinary isolated mechanical system with $L=T-U$, a time-independent quadratic kinetic energy and a velocity-independent potential energy, ${\sum }_i{\dot{q}}_i{p}_i=2T$, so $\mathcal{E}=T+U$. For a more general Lagrangian, the conserved quantity is the displayed energy function; identifying it with $T+U$ requires these assumptions.