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On a connected domain, a vector field is a conservative vector field if and only if it is the gradient of a single-valued scalar potential, $\boxed{\mathbf{H}=\nabla \varphi }$. Equivalently, its line integral is path independent, or its integral around every closed curve is zero. For a continuously differentiable field on a simply connected domain, this is equivalent to $\nabla \times \mathbf{H}=0$. Without the domain hypothesis, vanishing curl alone is only a local test and is not the necessary-and-sufficient global statement.

The field $\mathbf{F}$ is not a conservative vector field: either nonzero closed line integral computed below rules out a scalar potential. Equivalently, its planar curl is ${\partial }_x{F}_y-{\partial }_y{F}_x=1-(-1)=2$.

For $\mathbf{G}$, integrating ${\Phi }_x=2x{y}^2$ gives $\Phi =x^2{y}^2+h(y)$. Matching ${\Phi }_y=2y{x}^2$ then gives $h^{\prime }=0$. ThusThis also explains why both closed line integrals of $\mathbf{G}$ vanish.

A continuously differentiable conservative vector field has zero curl. HereThis vanishes everywhere if and only if $\boxed{\beta =0}$. Since the field is smooth on the simply connected domain ${\mathbb{R}}^3$, zero curl is also sufficient; the explicit scalar potential in the next part confirms it.

With $\beta =0$, integrating ${\Phi }_x=2\alpha x+{\alpha }^{2}y+\alpha z$ givesThe other two partial derivatives require $K_y=K_z=0$. The origin condition fixes the remaining constant to zero. Hence the scalar potential isIts value at $(1,1,1)$ gives ${\alpha }^2+2\alpha =1$, so

The three components of the curl cancel pairwise:The domain is all of ${\mathbb{R}}^3$, which is simply connected, so this is consistent with a conservative vector field. To construct its scalar potential, integrate ${\Phi }_x=y+z$ to obtain $\Phi =x(y+z)+h(y,z)$. Matching ${\Phi }_y=x+z$ gives $h_y=z$, hence $h=yz+k(z)$. Finally ${\Phi }_z=x+y$ forces $k^{\prime }(z)=0$. ConsequentlyEvery scalar potential differs by a constant on this connected domain. The endpoint values differ by one, agreeing with the line integral in part (a).

For the second vector field, the curl isIt is nonzero, for example $(0,-2,2)$ at $(1,0,0)$. A differentiable scalar potential would have zero curl of its gradient, so $\mathbf{G}$ is not conservative. The fact that its curl vanishes on the line $x=y=z$ does not make it vanish throughout the domain.

For $a=0$, differentiate the proposed scalar potential directly:ThusFor $a\ne 0$, the curl is the nonzero vector $(0,0,2a)$. A gradient of a twice differentiable scalar field has zero curl, so no such scalar field exists. Equivalently the closed line integral in part (d) is nonzero, contradicting the path independence of a conservative vector field. The field is smooth on all of ${\mathbb{R}}^3$, so no excluded singular set complicates this conclusion.

In an effective field theory, higher-power terms in an inflaton scalar potential can be important when $\varphi /M_{\mathrm{Pl}}$ is large, even if $V/M_{\mathrm{Pl}}^4$ is small. An assumed simple monomial inflation potential over such a field range needs a symmetry or ultraviolet completion controlling these operators and radiative corrections. A large field value alone does not establish Planckian energy density or invalidate every possible completion.

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.

In a canonical Wess–Zumino model with cubic superpotential $g{\Phi }^3/3$, auxiliary elimination gives a complex-scalar quartic interaction with two legs of each field type. With all momenta incoming and the displayed coefficient, its Feynman rule is $-i(2!2!)|g{|}^2$. For $\phi =(A+iB)/\sqrt{2}$, the same quartic scalar potential is $|g{|}^2(A^2+B^2{)}^2/4$, with real-field rules $-6i|g{|}^2$ for four identical legs and $-2i|g{|}^2$ for two of each. Factorials depend on how the coefficient is defined, not on a new physical coupling.

A radiative correction is a quantum correction beyond the classical or tree approximation, represented in perturbative quantum field theory by Feynman diagrams of positive loop order. Such corrections can alter a scalar potential and its derivatives; symmetries and renormalization constrain which changes are allowed.

Radius stabilization fixes the size of a compactification dynamically rather than choosing it as an undetermined parameter. If the four-dimensional low-energy effective action contains a radion $R$ with positive kinetic coefficient $Z(R)$ and scalar potential $V(R)$, a stable radius $R_0$ requires $V^{\prime }(R_0)=0$ and $m_R^2=V^{\prime \prime }(R_0)/Z(R_0)>0$. This matters for the hierarchy problem: the volume or warp factor explains a hierarchy only if a consistent dynamics selects and maintains the required geometry.

Reheating transfers the inflaton energy into other particles after cosmic inflation, eventually establishing a hot thermal bath. It requires interactions in addition to a background scalar potential. The duration and averaged equation-of-state parameter of reheating affect the number of e-folds between a chosen scale's horizon exit and the end of inflation.

The scalar mass matrix is the Hessian matrix of the scalar potential at a vacuum. Its eigenvalues are the tree-level squared masses; broken symmetry directions are zero eigenvectors before a Higgs mechanism or explicit symmetry breaking removes the corresponding massless modes.

The scalar potential $V(\Phi )=\lambda ({\Phi }^T\Phi -v^2{)}^2/4$, with $\lambda >0$, is invariant under the SO(3) group. For $v>0$, its vacuum manifold is a sphere. Canonical scalar normalization gives radial mass squared $2\lambda v^2$.

A Yang-Mills theory with gauge group $SO(3)$ and a real three-component scalar field in its defining vector representation. With a symmetry-breaking scalar potential, a nonzero vector vacuum leaves its $SO(2)$ stabilizer subgroup unbroken. The two transverse internal directions produce massive gauge bosons.

Minimal ten-dimensional supersymmetric gauge theory contains a gauge field and an adjoint Majorana-Weyl spinor. They have eight propagating bosonic and eight propagating fermionic states per gauge generator and sixteen real supercharges. A toroidal zero-mode dimensional reduction to four dimensions gives one gauge field, six real adjoint scalar fields, and four adjoint Weyl spinors: a four-dimensional N=4 super Yang-Mills theory. The internal field strength $F_{mn}=-ig[X_m,X_n]$ gives the nonnegative scalar potential $V=\frac{g^2}{4}{\sum }_{m,n}\Vert [X_m,X_n]{\Vert }^2$, where the norm uses a positive invariant inner product on the Lie algebra. Its minima are commuting tuples of adjoint scalar fields.

A homogeneous canonical scalar field with positive scalar potential has energy density ${\rho }_{\varphi }={\dot{\varphi }}^2/2+V$ and pressure $P_{\varphi }={\dot{\varphi }}^2/2-V$. When its potential energy dominates its kinetic energy, the equation-of-state parameter approaches $-1$. If other components and spatial curvature are negligible, the Friedmann acceleration equation gives accelerating expansion. Small initial kinetic energy is not enough to ensure long cosmic inflation; the field must also remain on a sufficiently flat part of the scalar potential, as measured by the potential slow-roll parameters.