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For a fixed force $\mathbf{F}$ and endpoint displacement $\Delta \mathbf{r}$, the line integral reduces to a dot product:Here $\theta$ is the angle between the force and the displacement. For straight motion, $d$ is also the distance traveled. A $3\mathrm{N}$ force acting along a $4\mathrm{m}$ displacement does $12\mathrm{J}$ of work. Turning the same force perpendicular to the displacement gives $\cos (\pi /2)=0$ and zero work; reversing it gives $\cos \pi =-1$ and $-12\mathrm{J}$. At $\theta =\pi /3$, the work is $6\mathrm{J}$. These signs describe energy transferred to or from the moving body. ^[ref]

For a point mass of constant mass $m$ observed in an inertial frame, the net work along its actual path equals its change in kinetic energy:The middle expression is a line integral; the dot product selects the component of force along the displacement. Positive net work increases speed, negative net work decreases it, and zero net work leaves the endpoint speeds equal. Equal endpoint speeds do not imply equal velocities: a perpendicular force can continually change direction while doing zero work. ^[ref]

The bridge to Newton's second law is short: take the dot product of $m\dot{\mathbf{v}}={\mathbf{F}}_{\mathrm{net}}$ with velocity. The chain rule gives $d(m|\mathbf{v}{|}^2/2)/dt={\mathbf{F}}_{\mathrm{net}}\cdot \mathbf{v}$, equating the rate of change of kinetic energy to instantaneous power. Integrating gives the theorem. No conservative force assumption entered this argument; friction also contributes to the net work.

For the harmonic oscillator with spring stiffness $k=2N/m$, release a mass $m=0.5\mathrm{kg}$ from rest at $x=0.3\mathrm{m}$. The spring exerts $F=-kx$. Its work before the first passage through $x=0$ isThus $m{v}^2/2=0.09\mathrm{J}$ gives speed $0.6m/s$ there. The velocity is negative on this passage because the position is decreasing. The energy calculation alone determines its magnitude, so the direction still requires information about the motion.

For a time-independent potential energy $V$, the conservative force contributes work $-\Delta V$. Separating all other work as $W_{\mathrm{other}}$ gives $\Delta (T+V)=W_{\mathrm{other}}$. Consequently conservation of mechanical energy requires zero net other work; the work-energy theorem itself does not require that conservation. With damping, mechanical energy decreases even though the theorem remains valid. ^[ref]

The work function is the minimum energy needed to remove an electron from a solid to the nearby vacuum level. A junction's effective barrier need not equal either electrode's isolated work function.

The drive can supply or extract work while the quantum state stays approximately fixed. For a differentiable bounded Hamiltonian operator, the Schrödinger equation givesThe state-evolution terms cancel, leaving the external drive's power. For a finite ideal step the change in mean energy is $⟨\psi |H_f-H_i|\psi ⟩$. Thus conservation of energy for a closed system with a time-independent Hamiltonian operator does not assert conservation of the driven subsystem's energy during switching.

Specify work operationally by two energy measurements: one before the switch and one after. For the prepared initial ground state, the first outcome is definite; the second samples the new Born rule probabilities. This protocol gives a distribution of measured energy differences, rather than generally the measurement distribution of the single operator $H_f-H_i$.^[ref]and henceFor ${\Delta }_i={\Delta }_f=\Delta$ and $\theta =\pi /3$, the work is $0$ or $\Delta$, with probabilities $3/4$ and $1/4$; its expectation is $\Delta /4$ and its variance is $3{\Delta }^2/16$. The initial mean energy $-\Delta /2$ becomes $-\Delta /4$. Using different initial and final scalar offsets would also shift the measured work; the displayed Hamiltonian operators fix that convention explicitly.

For a narrow beam of monoenergetic photons crossing homogeneous material of thickness $x$, the uncollided wave intensity obeys $I(x)=I_0{e}^{-\mu x}$. Here $\mu$ is the linear attenuation coefficient, counting removal from the incident beam by photon absorption and scattering of light. It depends on photon energy and material; it is not universally independent of energy.^[ref]

An original derivation makes the model explicit. If each thin layer removes fraction $\mu dx$ of the surviving beam, with the same $\mu$ throughout, thenThus equal added thicknesses multiply transmission by the same factor. The ordinary differential equation describes uncollided beam survival. To test it with a detector, keep source output, distance, selected photon energy band and detector response fixed. Use a collimator and suitable detector acceptance to exclude scattering of light into the detector. Subtract background, measure an empty-path reference, and keep the detector in its linear operating range. A changing source distance adds geometric dilution and is not an attenuation measurement at fixed geometry.

For a stipulated example, take $\mu =0.40\mathrm{m}{\mathrm{m}}^{-1}$ and $x=2.5\mathrm{mm}$. Transmission is $e^{-1}\simeq 0.368$. If the empty-path detector rate is $1010{\mathrm{s}}^{-1}$, including background $b=10{\mathrm{s}}^{-1}$, the model predicts sample rate $10+1000e^{-1}\simeq 378{\mathrm{s}}^{-1}$. Recover transmission as $(R-b)/(R_0-b)$; the raw ratio $R/R_0\simeq 0.374$ is biased. These chosen values are a calculation, not measurements of a particular material or a historical apparatus.

X-ray effective area is the area that converts an incident photon arrival rate per unit area into an accepted count rate. For a fixed direction and a sufficiently narrow energy interval, writeHere $A_{\mathrm{geom}}$ is projected geometric area, $T$ is transmittance through windows or optics, $\eta$ is quantum efficiency, ${ϵ}_s$ is the probability that a detected source event passes event selection, and $t$ is live observing time. $N_s$ is an expected value, not a guarantee of an integer count. Multiplication follows by counting incident photons over area and time, then applying the successive conditional acceptance probabilities. Live time already excludes intervals when the instrument cannot record events; multiplying by a live-time fraction again would count the same loss twice.

For an invented narrow-band example, choose $A_{\mathrm{geom}}=40\mathrm{c}{\mathrm{m}}^2$, $T=0.8$, $\eta =0.75$, and ${ϵ}_s=5/6$. Then $A_{\mathrm{eff}}=20\mathrm{c}{\mathrm{m}}^2$. A source with $F=0.30\mathrm{photons}\mathrm{c}{\mathrm{m}}^{-2}{\mathrm{s}}^{-1}$ observed for $t=100\mathrm{s}$ gives $N_s=600$. $F$ here counts photons; converting it into observed radiative flux requires their energies. If all have energy $E_0$, the corresponding radiative flux is $E_{0}F$.

A broad spectrum instead requireswhere $F_E$ is the differential photon arrival rate per area and energy. A single band-averaged X-ray effective area depends on the assumed spectrum. This is why a count rate alone cannot give an assumption-free radiative flux.

X-ray emission spectroscopy studies the frequency spectrum of emitted X-rays. The spectral line positions identify electronic energy differences; their strengths also depend on excitation, photon absorption inside the sample, geometry and detector response. An X-ray tube may excite a sample with electrons, while X-ray fluorescence uses incident photons. The distinction matters when choosing what the experiment measures: identifying a chemical element from a line position is different from determining its abundance from a line strength.

X-ray fluorescence is Characteristic X-ray emission following excitation by incident photons. In a simple initially unexcited chemical atom, photoionization can eject an inner electron if the incident photon supplies its binding energy. An outer electron can then fill the vacancy, releasing an emitted photon with the difference of shell binding energies. Nonradiative relaxation competes with this route, so one absorbed incident photon need not yield one escaping characteristic photon.

Choose a separate stipulated electron shell model with values $B_K=10.0$, $B_L=1.5$ and $B_M=0.3$, all in $\mathrm{keV}$. Ignore subshell splitting, recoil and changes of binding energy during relaxation. The possible shell differences areThese labels denote simplified shell groups, not resolved subshell electronic transitions. With $hc=1.23984\mathrm{keV}\mathrm{nm}$, their calculated wavelengths are respectively $0.145864$, $0.127819$ and $1.03320\mathrm{nm}$. An incident $9.8\mathrm{keV}$ photon exceeds both K-line energies but cannot directly create the K vacancy in this initially unexcited model. A $12.0\mathrm{keV}$ photon can, leaving approximately $2.0\mathrm{keV}$ of photoelectron kinetic energy. This comparison fixes neither the excitation probability nor the radiative branching probability.

X-ray photoelectron spectroscopy uses X-ray excitation to investigate electron removal energies, including those of core electrons. The common name electron spectroscopy for chemical analysis emphasizes its use in distinguishing chemical surroundings. The whole recorded spectrum can also contain electrons produced by processes other than direct photoionization; a peak assignment must identify the emission process as well as the chemical element.

In a modern implementation, a quartz crystal selects aluminum characteristic X-rays according to Bragg's law and focuses them onto the sample. Vacuum reduces gas collisions and contamination; a separately pumped entry chamber limits exposure to air. Lenses collect photoelectrons; an electrostatic electron energy analyzer transmits a narrow kinetic energy band. Stepping that band and counting detector pulses produces the spectrum. An electron flood source can compensate positive electric charge on insulating samples. Ion cleaning can alter vulnerable samples.^[ref]

For a scan with count $N_j$ and dwell time $t_j$ at setting $j$, compare rates $N_j/t_j$, rather than raw counts when dwell times differ. Under an independent Poisson distribution model, the count standard deviation is approximately $\sqrt{N_j}$ for a sufficiently populated bin. Longer acquisition reduces this relative counting measurement uncertainty; it does not automatically improve separation of overlapping peaks. Photoelectron peak overlap and resolution gives an example in which the same chemical separation produces either two maxima or one.

X-ray scattering techniques use the change in a scattered photon's direction and, in inelastic measurements, its energy to probe matter. Inelastic X-ray scattering transfers energy to excitations such as phonons; elastic methods probe spatial structure without that energy transfer.

For a fixed homogeneous absorber in a narrow-beam geometry, define the X-ray spectral attenuation operator on L1 space over a bounded positive energy interval. A physical input $f(E)\ge 0$ is an integrable function giving incident photon rate density. Let its linear attenuation coefficient $\mu (E)\ge 0$ be a measurable function, with $x\ge 0$. The transmitted uncollided density isHere $w(E)\ge 0$ is a fixed bounded detector response described by a measurable function; background has been subtracted. This is a positive linear operator on L1 space. Since $0\le e^{-\mu (E)x}\le 1$, it cannot increase the integrated absolute incident rate. Multiplication of exponentials proves $A_{x+y}=A_x{A}_y$ for successive layers of the same material. Ignore secondary radiation entering the detector, saturation and changes to the source. Discrete spectral lines use the finite-sum version below instead of an ordinary rate density.

For a chosen filter, stipulate ${\mu }_{\alpha }=0.20\mathrm{m}{\mathrm{m}}^{-1}$, ${\mu }_{\beta }=0.80\mathrm{m}{\mathrm{m}}^{-1}$ and input line rates $(N_{\alpha },N_{\beta })=(4,1)$ in a common arbitrary rate unit. Assume equal calibrated detector weight for both lines. At $x=2.0\mathrm{mm}$,The selected $8.5\mathrm{keV}$ line's fraction rises from $4/5=0.800$ to $0.929974$, but its absolute rate falls by about $33.0\%$. Filtering improves purity at the cost of signal. It multiplies existing components and cannot create a new emitted spectral line.

With just one integrated reading, $S=N_{\alpha }+N_{\beta }$, the two unknown line rates are underdetermined. If the source is unchanged between an empty-path reading and a filtered reading, the ideal inverse problem becomesThe determinant $e^{-1.6}-e^{-0.4}\simeq -0.468424$ is nonzero, so these two ideal readings determine the two stipulated line rates uniquely. Identical coefficients would give zero determinant and destroy that separation. This small linear algebra example explains why absorber contrast supplies information, while also showing its limit: unknown scattered components or unknown detector weights add unknowns that two readings cannot determine.

An X-ray tube accelerates electrons from a negative cathode to a positive target anode. A heated cathode supplies electrons by thermionic emission; the evacuated envelope limits collisions before the target. This differs from the gas-discharge tube in the Röntgen X-ray discovery experiment.^[ref]

In bremsstrahlung, an electron changes velocity in the target's electric fields, emitting a photon while losing kinetic energy. Different encounters allow a continuous range of transferred energies. This is distinct from the discrete electronic transitions producing Characteristic X-rays.

Changing the accelerating voltage changes the available kinetic energy per electron and the X-ray tube cutoff. Changing the number of arriving electrons changes the number of opportunities for photon emission. These are distinct controls: a larger electric current does not by itself give each electron more energy. For an original idealized budget, $V=50\mathrm{kV}$ and tube electric current $i=0.20\mathrm{mA}$ give electrical power $Vi=10\mathrm{W}$ and arrival rate $i/e\simeq 1.25\times 10^{15}{\mathrm{s}}^{-1}$, where $e$ is the elementary charge. This is input power, not emitted X-ray power; obtaining the latter requires a conversion efficiency.

For an electron starting with negligible kinetic energy and accelerated through a constant positive electric potential difference $V$, electrical work supplies $K=eV$. In the ideal single-electron bremsstrahlung model, conservation of energy bounds any one emitted photon:The endpoint corresponds to essentially all available kinetic energy going into one photon. It is a limiting bound, not a claim that every electron produces an endpoint photon. Neglect target recoil, initial kinetic energy, and losses before the emitting interaction. If the accelerating voltage varies, its largest value sets the corresponding ideal upper endpoint; a mean voltage does not substitute for that value.

For a calculated constant-$V$ example, $50\mathrm{kV}$ gives $E_{\max }=50\mathrm{keV}$ and ${\lambda }_{\min }=0.0248\mathrm{nm}$. Doubling $V$ halves ${\lambda }_{\min }$. Producing a $0.0200\mathrm{nm}$ photon requires at least $V=hc/(e\lambda )\simeq 62.0\mathrm{kV}$ in this model. A continuous frequency spectrum spans lower photon energies; the cutoff calculation predicts neither its shape nor the efficiency of photon emission.

Zeeman splitting is the separation of energy levels or mode frequencies by a background magnetic field. In atomic physics, the magnetic field couples to a magnetic dipole moment, lifting quantum degeneracy. Transitions between the shifted quantum states can turn one spectral line into several. A split level and a split spectral line are different objects: the emitted photon measures the difference between two level energies, and the electric-dipole selection rule determines which differences appear.

The Stern-Gerlach experiment also probes magnetic dipole moments, through beam deflection in a magnetic field gradient. Here the readout is a spectral line.

The pattern makes an invisible magnetic field visible through spectroscopy. It also reveals angular momentum and magnetic dipole moments inside chemical atoms. Zeeman slowers exploit a changing resonance frequency to keep moving chemical atoms absorbing laser photons; a magneto-optical trap combines position-dependent shifts with electromagnetic polarization and laser cooling. Related mode shifts occur in Zeeman splitting of an antiferromagnetic spin wave.

Zeeman splitting is the separation of energy levels or mode frequencies by a background magnetic field. For an antiferromagnetic spin wave, opposite circular spin polarizations receive opposite frequency shifts.

For potential energy homogeneous of degree $-2$, Newton's second law gives $d^2{r}^2/d{t}^2=4E/m$. Indeed, Euler's identity implies $\mathbf{r}\cdot \mathbf{F}=2U$, so the second derivative equals $2v^2+4U/m=4(K+U)/m$. Thus a trajectory with zero total energy and initially zero radial velocity remains on a sphere of constant radius. A positive electric charge released from rest on the equator of a fixed ideal electric dipole has these properties; in its meridional plane it oscillates on the semicircle on the negative-pole side. A finite two-charge electric dipole does not have exactly homogeneous potential energy and perturbs this result.

“Arrives at infinity with zero speed” is an asymptotic statement. For outward radial motion at zero specific orbital energy,Every finite distance takes finite time, but $t(r)\to \infty$ as $r\to \infty$. There is no instant of rest at an infinitely distant destination. Differentiating the displayed solution gives $\ddot{r}=-\mu /r^2$, independently checking Newton's second law. In the synthetic example, reaching $r=4\mathrm{m}$ takes exactly $(4-\sqrt{2})/3\mathrm{s}\approx 0.861929\mathrm{s}$. For nonzero specific angular momentum, the Barker equation gives the corresponding infinite-time limit along a parabolic trajectory. Positive-energy escape also takes infinite time to reach infinity; it retains nonzero asymptotic speed.

Zero-point motion is the nonzero quantum fluctuation of position and momentum in a system's ground state. For a quantum harmonic oscillator, the ground state has energy $\hslash \omega /2$ and nonzero position and momentum variances even at zero temperature. It does not represent a thermal mixture and need not carry entropy. The large zero-point motion of light helium atoms helps explain why helium can remain a liquid at low pressure even as temperature approaches absolute zero.

If two thermodynamic systems are each in thermal equilibrium with the same third system in the same equilibrium state, they are in thermal equilibrium with each other. Write $A\sim B$ for this relation. The zeroth law of thermodynamics supplies transitivity:Its symmetric relation property and self-equilibrium (reflexivity), together with transitivity, make thermal equilibrium an equivalence relation. Its equivalence classes can be labelled by a common empirical temperature; choosing an absolute numerical scale requires the further structure of thermodynamic temperature.^[ref]

The comparison assumes internally equilibrated macroscopic bodies, unchanged composition and mechanical constraints, and contact permitting only heat exchange. Use rigid, impermeable diathermal walls so that neither matter transfer nor boundary work spoils the test. The same state of $C$ matters: letting $C$ warm while measuring $A$, then cool while measuring $B$, does not supply the premise. Ordinary nonrelativistic laboratory thermodynamics is assumed; the Tolman temperature law treats gravitational qualifications. Equal temperatures establish the thermal condition, without requiring mechanical equilibrium or chemical equilibrium.

This explains why a thermometer can compare unlike materials: it supplies a reusable third system. A temperature is a property of an equilibrium state, while heat is energy transferred because of a temperature difference. Two bodies at the same temperature can have different internal energies and heat capacities. A thermometer therefore does not measure a body's stored amount of heat; the finite-body disturbance and waiting-time calculations in thermometer describe how making the comparison can change the state being measured.^[ref]

The name “zeroth” reflects logical order rather than the date of the first thermometer: this premise about thermal equilibrium precedes the already numbered first law of thermodynamics, Second law of thermodynamics and third law of thermodynamics. This is the naming explanation given in University Physics Volume 2 by Samuel J. Ling and others (§1.1, “Thermal Equilibrium”).^[ref]

The corresponding Zeroth law of black-hole mechanics states constancy of surface gravity over an equilibrium horizon; Hawking radiation identifies that quantity with physical temperature up to its universal conversion factor.