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Use $x^{\mu }=(t,x)$ and Minkowski metric ${\eta }^{\mu \nu }=\mathrm{diag}(1,-1)$, with the wave speed equal to one. Regarding $\varphi$ as a scalar field, the symmetric canonical stress-energy tensor isIt obeys ${\partial }_{\mu }{T}^{\mu \nu }=0$ on the wave equation. For a scalar field the intrinsic spin current vanishes. The total orbital Lorentz current and its integrated angular-momentum tensor areBecause there is only one spatial dimension, its only independent possibly nonzero component iswhere $P=\int pdx$. The corresponding flux is ${\mathcal{J}}^{101}=t\mathcal{E}-xp$. Symmetry and conservation of $T$ giveThus $J^{01}$ is conserved provided the boundary flux vanishes and the integrals exist. For example, localized wave packets on an infinite line have this property. One can also check directly that $d\int x\mathcal{E}dx/dt=P$, so the two terms in $J^{01}$ have cancelling derivatives.

The significance is the boost charge of a one-dimensional scalar field. There is no independent spatial rotation or ordinary spatial angular-momentum component in one dimension. Instead, the action has the mathematical Lorentz invariance of a massless wave field with invariant speed $c_s=1$, and $J^{01}$ generates its boost. Its conservation says that the center of energy moves uniformly. This is a symmetry of the ideal wave-field model, rather than an assertion of fundamental vacuum Lorentz symmetry for a material rod. Fixed boundaries can break it, and a general solution with nonvanishing boundary flux need not conserve the integrated charge.

Apply the linear transformation of the scalar fields to either time- or space-derivative product. Its coefficients become $\alpha \gamma$ multiplying the square of the first derivative, $\beta \delta$ multiplying the square of the second, and $\alpha \delta +\beta \gamma$ multiplying their cross product. Invariance for arbitrary fields therefore requiresThe complete real solution isThere are four connected components, distinguished by the diagonal/antidiagonal choice and the sign of $r$. Only the positive diagonal family is in the identity component; write $r=e^s$ to obtain ${\varphi }_1\mapsto e^s{\varphi }_1$, ${\varphi }_2\mapsto e^{-s}{\varphi }_2$. This is the reciprocal rescaling symmetry of two cross-coupled scalar fields. The antidiagonal transformations exchange the fields as well as rescale them.

A spherical three-momentum ultraviolet cutoff givesFor $m>0$ and $\Lambda \gg m$, the asymptotic expansion isIn particular, the leading ultraviolet divergence is quartic. For $m=0$, $\mathcal{E}={\Lambda }^4/(16{\pi }^2)$ exactly.

Write $\phi =⟨\psi ⟩$. Under the proposed identification, the scalar potential from a field-dependent momentum cutoff is $V_{\mathrm{ind}}(\phi )=\mathcal{E}(|\phi |)$. It is even and nonnegative, and its derivative with respect to $|\phi |$ is $|\phi {|}^2\sqrt{{\phi }^2+m^2}/(4{\pi }^2)>0$ away from zero. Consequently its only global minimum is at $\phi =0$: this contribution alone does not cause spontaneous symmetry breaking. At large $|\phi |$ it grows quartically. At small $|\phi |$ with fixed positive $m$, however,Thus the quartic sketch is an asymptotic shape, not the exact small-field behavior for a massive original field. Identifying a regulator with a new field is an additional modeling prescription; an actual effective potential also depends on the new field's dynamics and any other allowed terms.

Goldstone's theorem states that a spontaneously broken continuous internal global symmetry in field theory of a local relativistic quantum field theory produces massless Goldstone bosons. For a Lorentz invariance vacuum breaking $G$ to its stabilizer $H$, there are $\dim G-\dim H$ independent symmetry-required massless directions. Additional accidentally massless modes are possible. A gauge redundancy does not obey this physical-state counting, and nonrelativistic systems need not have one independent mode per broken generator.

For the classical proof, consider real scalar fields ${\varphi }_i$ with a positive canonical kinetic term and an invariant smooth scalar potential $V$. An infinitesimal generator $T^a$ satisfiesLet $v$ be a stationary vacuum and differentiate this identity with respect to ${\varphi }_j$. At $v$, the term proportional to ${\partial }_{i}V$ vanishes, leavingThe Hessian matrix ${\mathsf{M}}^2$ is the scalar mass matrix. An unbroken generator has $T^{a}v=0$; broken generators span the tangent space of the vacuum manifold $G/H$. Each independent nonzero tangent vector is a zero eigenvector of ${\mathsf{M}}^2$. The quadratic Euler-Lagrange equations in these directions are $\square {\pi }^a=0$, giving the massless dispersion relation ${\omega }^2={\mathbf{p}}^2$. This proves the classical result and explains why the Goldstone bosons describe motion along the degenerate minima rather than against a restoring curvature.

A scalar field has $\delta {\varphi }_a=0$ for a pure change of spacetime argument. With $\delta x^k={ϵ}^k$ and $\delta x^0=0$, its variation at a fixed argument is $\bar{\delta }{\varphi }_a=-{ϵ}^k{\partial }_k{\varphi }_a$. The supplied Noether charge becomesSince $\delta Q/\delta {\pi }_a=-{ϵ}^k{\partial }_k{\varphi }_a$, solution and antisymmetry of the Poisson bracket giveCanonical quantization replaces the Poisson bracket by a commutator divided by $i\hslash$:With this sign convention one may use $U=e^{-i\hat{Q}/\hslash }$ and ${\hat{\varphi }}_a^{\prime }=U^{†}{\hat{\varphi }}_{a}U$. Because $Q=-{ϵ}^k{P}_k$, the negative shift of the field argument follows with the required sign.

Write $d{\Pi }_p=d^{3}p/[(2\pi {)}^{3}2{\omega }_p]$, $p^{\mu }=({\omega }_p,\mathbf{p})$ and ${\omega }_p=\sqrt{{\mathbf{p}}^2+m^2}$. The Fourier expansion in the source is compatible with the oscillator canonical commutation relationsFor the free real scalar field, $\pi =\dot{\varphi }$ andSubstitution of the Fourier expansion produces spatial Dirac delta functions. Terms with two annihilation operators or two creation operators cancel: in $P_0$ their coefficient is $-{\omega }_p^2+{\mathbf{p}}^2+m^2=0$, and in $P_i$ their integral is odd under $\mathbf{p}\mapsto -\mathbf{p}$. The two mixed terms combine, with normal ordering removing the vacuum constant, to giveIn particular, $P_i$ contains the lower component $p_i=-{\mathbf{p}}_i$ in this Minkowski metric convention.

To check commutativity without manipulating a square of a Dirac delta function, define $N_f=\int d{\Pi }_{p}f(p)a_p^{†}{a}_p$. The oscillator algebra gives $[N_f,a_q]=-f(q)a_q$ and $[N_f,a_q^{†}]=f(q)a_q^{†}$. Thus $[N_f,a_q^{†}{a}_q]=0$ and $[N_f,N_g]=0$. Taking $f=p_{\mu }$, $g=p_{\nu }$ provesThese identities hold on the usual dense domain of finite-particle states, or can first be proved in a finite box and then continued to the continuum.

The real scalar field action has spacetime translations and Lorentz invariance, together giving Poincaré symmetry. With a real scalar transformation law it also has parity and time reversal. A generic polynomial potential has no continuous internal symmetry. The additional field reversal $\varphi \mapsto -\varphi$ is a symmetry precisely when every nonzero odd-power coupling vanishes. The free massless special case also admits a constant shift of $\varphi$; generic interactions remove it.

In mass units, $[{\partial }_{\mu }]=1$ and the dimensionless action requires $[\mathcal{L}]=D$. Normalizing the kinetic term fixes the classical mass dimension of the scalar field:For each interaction, $[{\lambda }_n]+n[\varphi ]=D$, soThese are classical or engineering scaling dimensions; quantum corrections can change the scaling of operators and couplings. The scale transformation prescribed later has field weight one, which equals this canonical weight only at $D=4$.

Put $V(\varphi )=m^2{\varphi }^2/2+{\sum }_{n>2}{\lambda }_n{\varphi }^n/n!$. The Euler-Lagrange equation for the scalar field is $\square \varphi +V^{\prime }(\varphi )=0$, or explicitlyThe derivative of ${\varphi }^n/n!$ is ${\varphi }^{n-1}/(n-1)!$; all interaction terms therefore enter with the displayed plus sign for the chosen Minkowski metric.

In the classically scale-invariant case from solution, use $\partial \mathcal{L}/\partial ({\partial }_{\mu }\varphi )={\partial }^{\mu }\varphi$, $\bar{\delta }\varphi =\alpha (\varphi +x^{\nu }{\partial }_{\nu }\varphi )$ and $\delta x^{\mu }=-\alpha x^{\mu }$. Dividing the supplied Noether current expression by $\alpha$ givesSince ${\theta }^{\mu \nu }={\partial }^{\mu }\varphi {\partial }^{\nu }\varphi -{\eta }^{\mu \nu }\mathcal{L}$, this isThe first term in this dilatation current comes from the intrinsic weight of the scalar field, while the second comes from moving its spacetime argument. Using the canonical energy-momentum tensor alone without the first term would not give the stated conserved current.

For a free real scalar field in four spacetime dimensions, $\mathcal{L}=(\partial \varphi {)}^2/2-m^2{\varphi }^2/2$ and $\pi =\dot{\varphi }$. Canonical quantization promotes these canonical fields to Hermitian operator-valued distributions with equal-time canonical commutation relationsExpand the solution of the Klein-Gordon equation into positive- and negative-frequency modes:Hermiticity pairs the two coefficients as adjoints. The field commutators are equivalent toFor example, inserting the expansion in $[\varphi ,\dot{\varphi }]$ yields two equal Fourier integrals, each with coefficient $i/2$, and hence $i{\delta }^{(3)}$. The vacuum obeys $a(\mathbf{k})|0⟩=0$, and repeated creation operators generate the symmetric bosonic Fock space. Normal ordering of the quadratic Hamiltonian removes its vacuum constant and gives $H=\int d{\Pi }_k{\omega }_k{a}^{†}a$; the same computation for the spatial momentum gives the stipulated four-momentum $P^{\mu }=\int d{\Pi }_k{k}^{\mu }{a}^{†}a$.

The oscillator algebra now impliesConsequentlyIndeed, ${\partial }^{\mu }{e}^{\mp ikx}=\mp i{k}^{\mu }{e}^{\mp ikx}$. For $\mu =0$, the result is $[H,\varphi ]=-i\dot{\varphi }$, which is the Heisenberg equation of motion with the correct sign. As usual, the field identities are interpreted after smearing the operator-valued distributions, or on a suitable common finite-particle domain.

A pathline records the motion of one identified fluid element: keep its starting position $\mathbf{a}$ and starting time $t_0$ fixed, and let physical time $t$ advance. In a velocity field $\mathbf{u}(\mathbf{x},t)$ its Lagrangian trajectory $\mathbf{X}(t;t_0,\mathbf{a})$ solves the initial value problemA streamline instead freezes the time argument of the velocity field; a streakline fixes the observation time and varies the release time of many fluid elements. A pathline is useful for answering where a particular parcel goes and how long it takes, rather than merely which way the fluid is moving now. ^[ref]

For a concrete analytical model, take spatial coordinates $x,y$ in metres, time $t\ge 0$ in seconds, and the spatially uniform but time-dependent velocity fieldIts numerical components in these units are $(1,t)$. A fluid element starting at the origin at $t_0=0$ has the pathlineAt $t=1\mathrm{s}$ it is at $(1,0.5)\mathrm{m}$; at $t=2\mathrm{s}$ it is at $(2,2)\mathrm{m}$. The parabola records earlier, shallower velocity directions as well as the final direction. Differentiating gives $(\dot{x},\dot{y})=(U,at)$ and acceleration $(0,a)$, so this curved pathline needs no spatial variation of the velocity field. More generally the chain rule along a pathline gives $df(\mathbf{X}(t),t)/dt={\partial }_{t}f+\mathbf{u}\cdot \nabla f$, the material derivative of a transported scalar field.

A motivating tracer demonstration appears in S. J. Kline's film notes: water passes through a contracting channel, and hydrogen bubbles mark a small element at a wire at time $t_0$. The illustrated element is shown again $4.7\mathrm{s}$ later, with its accumulated pathline superposed. Following that same marked element across successive film frames records a history; the entire bubble pattern in one frame has a different meaning. ^[ref]

For a modern version of the readout, use sparse small markers in a transparent water channel, illuminate the viewed plane, and keep an optical camera fixed with a ruler in that plane. Record timestamps and track the same isolated marker between frames; successive positions estimate its pathline and $[\mathbf{X}(t+\Delta t)-\mathbf{X}(t)]/\Delta t$ estimates its velocity. A long exposure records a pathline segment but loses the timing within it, so its total length divided by exposure time gives only mean speed. These are proposed measurement steps based on Jean Hertzberg's teaching guidance, rather than additional measurements from Kline's film. Markers leaving the viewed plane, overlapping tracks and perspective can break particle identification; repeat with a thinner viewed slice and a lower marker concentration. ^[ref]

Peter Breitenlohner was a theoretical physicist whose work with Daniel Z. Freedman established the Breitenlohner-Freedman bound for fields in Anti-de Sitter spacetime.

For a scalar field in $(d+1)$-dimensional Anti-de Sitter spacetime with curvature radius $L$, the familiar form is $m^2{L}^2\ge -d^2/4$, with stability also requiring suitable boundary conditions. The negative right-hand side explains the surprising feature: a negative mass-squared parameter alone does not establish instability in this curved background. Their result links field dynamics to the geometry and boundary conditions of spacetime.^[ref]

Phi cubed theory has interaction $-g{\varphi }^3/3!$, giving a cubic vertex $-ig$. In four spacetime dimensions the scalar field has mass dimension one and $g$ has mass dimension one.

A potential for a conservative vector field $X$ is a function with scalar values $\Phi$ whose gradient is $X$. Potentials differ by a constant on each connected component. The fundamental theorem for line integrals gives ${\int }_{\gamma }X\cdot d\mathbf{x}=\Phi (\gamma (1))-\Phi (\gamma (0))$. In mechanics a conservative force is conventionally written $-\nabla V$, so its potential energy is $V=-\Phi$. This spatial potential differs from the scalar potential $V(\varphi )$ that denotes the derivative-free energy density of a scalar field.

A function $S(x,\theta ,\bar{\theta })=\varphi (x)$ is an allowed special configuration of an unconstrained superfield. The subspace where all other components remain zero is not invariant: $\delta S=(i\theta {\sigma }^{\mu }\overline{ϵ}-iϵ{\sigma }^{\mu }\bar{\theta }){\partial }_{\mu }\varphi$ produces components depending on Grassmann variables whenever $\varphi$ is nonconstant. Thus an isolated spacetime scalar field is not closed under supersymmetry. A constant gauge-singlet value is a trivial invariant exception.

Splitting a scalar field into slow and fast components $\varphi =\phi +\eta$, the Wick contractions in $\lambda ⟨(\phi +\eta {)}^6⟩$ generate $15\lambda ⟨{\eta }^2⟩{\phi }^4$. A vanishing bare quartic coupling is therefore not generally preserved by renormalization-group flow.

Quintessence models dark energy through an evolving scalar field. For a homogeneous canonical field in natural units,A positive potential dominating the kinetic term gives negative pressure. The Friedmann acceleration equation requires $w=p_{\varphi }/{\rho }_{\varphi }<-1/3$ for scalar-dominated accelerated expansion. A nonnegative canonical kinetic term prevents $w<-1$ when ${\rho }_{\varphi }>0$.

Unlike a strict cosmological constant, the field can evolve, so $w$ need not remain $-1$. This is a model class, not an identified substance.^[ref]

A radion is a lower-dimensional scalar field describing the size or separation modulus of compact extra dimensions. In a two-brane Randall–Sundrum model, its expectation determines the inter-brane separation and the warp-generated hierarchy. A background size alone does not supply a radion potential or a stabilization mechanism.

// Target: supersymmetry.bigb

At the self-dual circle, left and right dimension-one currents generate $SU(2{)}_L\times SU(2{)}_R$. The compact-coordinate vertex $\partial Y_L\bar{\partial }{Y}_R$ is a Cartan–Cartan component of a scalar field in the $(3,3)$ representation. Its expectation value changes the circle radius and leaves $U(1{)}_L\times U(1{)}_R$ unbroken. The four extra charged vectors acquire the displayed mass, and four scalar field polarizations become their longitudinal modes. This realizes the Higgs mechanism geometrically. In a bosonic string the ever-present tachyon and additional scalar field instabilities prevent interpreting the whole family as a stable vacuum.

For the real scalar field with metric $(+,-)$ and potential $U={\kappa }^2({\varphi }^2-1{)}^2/2$, the static phi-four kink is ${\varphi }_0(x)=\tanh (\kappa x)$ and satisfies ${\varphi }_0^{\prime 2}=2U({\varphi }_0)$. Its mass is $M=\int {\varphi }_0^{\prime 2}dx=4\kappa /3$. A Lorentz boost gives $\varphi (t,x)={\varphi }_0(\gamma (x-vt))$. The stress-energy tensor gives $P=-\int {\varphi }_t{\varphi }_{x}dx=\gamma v\int {\varphi }_0^{\prime 2}dx=\gamma Mv$. Similarly, $E=\frac{1}{2}[\gamma (1+v^2)+{\gamma }^{-1}]M=\gamma M$, proving the energy–momentum relation $E^2-P^2=M^2$.