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Integrating ${\psi }_x=-(2xy+z)$ with respect to $x$ gives $\psi =-x^{2}y-xz+h(y,z)$. Matching ${\psi }_y=-(x^2+2y)$ requires $h=-y^2+j(z)$, and matching ${\psi }_z=-x$ requires $j$ constant. A scalar potential with the requested sign isThus the force field is a conservative vector field on all of ${\mathbb{R}}^3$.

Parametrize the straight line by $\mathbf{x}(t)=(t,t,t)$, $0\le t\le 1$. The line integral isThe change in scalar potential is $\psi (1,1,1)-\psi (0,0,0)=-3$, so the answer is $-\Delta \psi$, as required by the fundamental theorem for line integrals for $\mathbf{F}=-\nabla \psi$.

The curl is $\nabla \times {\mathbf{F}}_2=(2z,3z^2,0)$, which is not identically zero. A gradient of a smooth scalar potential has zero curl, so ${\mathbf{F}}_2$ is not conservative on ${\mathbb{R}}^3$.

Integrating the first component in $x$ gives $\varphi =x^{3}y+h(y,z)$. Matching the second requires $h_y=-2y{z}^2$, so $h=-y^2{z}^2+k(z)$. Matching the third gives $k^{\prime }=0$. Thus the field is conservative, with scalar potentialDirect differentiation verifies ${\mathbf{F}}_1=\nabla \varphi$ on all of ${\mathbb{R}}^3$.

Construct the scalar potential $F(x,y)=y{e}^{x^2}-x^2$. Its differential iswhich proves exactness directly. Multiplying the differential equation by $e^{x^2}$ shows that a solution satisfies $dF/dx=0$, so $y{e}^{x^2}-x^2=C$. The initial value at zero gives $C=-1$. ThereforeThe factor $e^{x^2}$ is also an integrating factor for the associated first-order linear equation.

Let $w=x+iy$. Then $y-ix=-iw$, and the complex trigonometric identity $\sin (-iw)=-i\sinh w$ givesThus the form is exact, with complex-valued scalar potential $F(x,y)=\cosh (x+iy)$. Differentiating that potential with respect to the two real variables checks both coefficients individually.

For $\mathbf{r}(t)=(t,t,2t^3)$, the conservative field is $(6t^7,4t^7,t^5)$ and the tangent is $(1,1,6t^2)$. ThereforeThe two different paths give the same value because their line integrals depend only on the potential difference ${\varphi }_1(B)-{\varphi }_1(A)=2$. There is no scalar potential for ${\mathbf{F}}_2$ to which the conditional request could be applied.

The coefficients follow by differentiating the same monomial:Thus this exact differential form hasExhibiting a global scalar potential proves exactness directly, without needing an additional domain argument.

Taking each partial derivative of the scalar potential givesThe factor $5y^4$ in the middle component comes from the chain rule. The potential is smooth, so the curl of a gradient identity gives $\boxed{\nabla \times \mathbf{F}=0}$.

Integrating ${\varphi }_x=2xy$ gives $\varphi =x^{2}y+h(y,z)$. Matching ${\varphi }_y=x^2+3y^2+z^2$ then gives $h=y^3+y{z}^2+k(z)$. Finally ${\varphi }_z=2yz$ forces $k$ to be constant. A scalar potential isThe initial and final points of the helix are $(b,0,0)$ and $(0,b,cL)$, where $L=5\pi /2$. The fundamental theorem for line integrals givesagreeing with the directly evaluated $I_2$.

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.

The magnetic vector potential is defined by $\mathbf{B}=\nabla \times \mathbf{A}$. The gauge transformationleaves both $\mathbf{B}$ and $\mathbf{E}=-\nabla \varphi -{\partial }_t\mathbf{A}$ unchanged. In magnetostatics choose the Coulomb gauge $\nabla \cdot \mathbf{A}=0$. Then Ampère's law and the curl-of-curl identity giveThe free-space Green function identity ${\nabla }^2|\mathbf{r}-{\mathbf{r}}^{\prime }{|}^{-1}=-4\pi {\delta }^{(3)}(\mathbf{r}-{\mathbf{r}}^{\prime })$ therefore givesThe boundary condition selects the homogeneous solution; the usual decay condition assumes suitably localized sources. An infinitely long static wire needs a reference for its logarithmically divergent potential, although its magnetic field is finite away from the wire.

For time-dependent sources use the Lorenz gauge, $\nabla \cdot \mathbf{A}+c^{-2}{\partial }_t\varphi =0$, and the outgoing retarded potentialThe scalar potential has the analogous retarded charge-density integral. The two gauges should not be conflated for a general time-dependent source.^[ref]

For the prescribed uncharged wire, $\rho =0$ and $\varphi =0$ is consistent. Translation invariance makes $\mathbf{A}=A_z(r,t)\hat{\mathbf{z}}$ with $\nabla \cdot \mathbf{A}=0$, so this particular solution satisfies both gauge conditions. With $s=z^{\prime }-z$,It is zero for $t<r/c$. For $t>r/c$, set $L=\sqrt{c^2{t}^2-r^2}$. The causal interval is $-L<s<L$, giving the switched infinite-wire retarded potentialDifferentiation gives the fields and Poynting vector, for $r>0$, $t>r/c$:Here $\mathbf{B}={\mu }_0\mathbf{H}$ and $B_{\varphi }=-{\partial }_r{A}_z$. All three fields vanish before arrival. For positive $I$, $\mathbf{E}$ opposes the imposed current and the radial energy flow after switching a line current points radially outward: $(-\hat{\mathbf{z}})\times \hat{\mathbf{\varphi }}=\hat{\mathbf{r}}$.

For the graph, let $u=ct/r$. Normalized to $E_{\ast }={\mu }_{0}Ic/(2\pi r)$, $H_{\ast }=I/(2\pi r)$ and $N_{\ast }={\mu }_0{I}^{2}c/(4{\pi }^2{r}^2)$, the curves areAt the wavefront, $|E_z|$ and $H_{\varphi }$ diverge as $(u-1{)}^{-1/2}$ and $N_r$ as $(u-1{)}^{-1}$. This abrupt line-current wavefront singularity reflects the ideal instantaneous turn-on of a line current; a finite rise time smooths the front. At late times,The vector potential grows logarithmically with time in this infinite-wire idealization even as the local magnetic field approaches its static limit. The divergent front should not be interpreted as a finite-energy pulse from a real instantaneous switching experiment.

cosmic inflation is an early interval of accelerated expansion, $\ddot{R}>0$. For expansion this means a shrinking comoving Hubble radius, $d[(RH{)}^{-1}]/dt<0$. A nearly constant positive scalar potential can supply it: in $c=1$ units,Potential domination gives ${\rho }_{\varphi }+3P_{\varphi }<0$ and approximately exponential expansion. Sustaining the interval requires sufficiently slow evolution, not just setting the initial field velocity to zero on a steep potential.

The horizon problem is the near equality of conditions in widely separated microwave-background regions that could not equilibrate in the standard decelerating hot Big Bang history. cosmic inflation permits the currently observed patch to descend from an earlier region small enough to have been causally connected, before that patch is stretched to enormous size. The claim concerns a causal history extended before the conventional hot phase; merely accelerating a collection of already disconnected regions would not explain their common temperature.

For the flatness problem, the Friedmann constraint givesA growing $RH$ drives this curvature fraction toward zero, whereas ordinary matter- or radiation-dominated deceleration makes it grow. cosmic inflation therefore suppresses observable curvature without requiring an equally extreme initial tuning within the later hot evolution.

Pre-existing monopoles and other unwanted relics are also diluted by the volume increase: a conserved particle number density falls as $e^{-3N}$ over $N$ e-folds. The transition to the hot universe must reheat efficiently, and its temperature must not regenerate an excluded relic population. The required number of e-folds is model and reheating dependent; “about sixty” is a useful scale for many scenarios rather than a universal exact requirement.

cosmic inflation also supplies a mechanism for perturbations. quantum fluctuations are stretched beyond the Hubble radius, and nearly scale-invariant comoving curvature perturbations can later seed cosmic structure. Simple single-field slow-roll models commonly predict approximately adiabatic, nearly Gaussian fluctuations, with a small spectral tilt; tensor perturbations would provide an additional test. Acoustic coherence and the observed scalar spectrum are compatible with this account, but they do not single out one inflaton potential. cosmic inflation addresses causal homogeneity, curvature and relic dilution while motivating the initial perturbations; it does not automatically solve the cosmological constant problem or determine reheating.

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.