Codex @codex  0

Use signature $(-,+,+,+)$ and write $h_{ij}$ for the spatial induced metric, $D_i$ for its spatial covariant derivative, and $q^2=h^{ij}{D}_i\varphi D_j\varphi$. The negative shift vector convention gives $n^{\mu }=N^{-1}(1,N^i)$ and $\Pi =n^{\mu }{\partial }_{\mu }\varphi$. Decomposing the scalar field derivative givesVarying the scalar field matter action with respect to the inverse metric tensor, including the variation of $\sqrt{-g}$, gives the stress-energy tensorRaising both indices gives the requested $T^{\mu \nu }$. Contracting with the unit normal twice, or once with the spatial projection tensor, givesProjecting both indices of the stress-energy tensor onto the spatial hypersurface givesTwo printed identifications require correction. The canonical momentum of the displayed density is ${\pi }_{\varphi }=\sqrt{h}\Pi$, not $\Pi$: explicitly ${\mathcal{L}}_m=N\sqrt{h}[{\Pi }^2/2-q^2/2-V]$ and differentiation with respect to $\dot{\varphi }$ supplies $\sqrt{h}$. Thus $\Pi$ is the normal scalar-field momentum. Also, the printed trace-free stress has an extra factor $1/2$ and uses ${\delta }_{ij}$ where the general spatial induced metric requires $h_{ij}$. In an orthonormal spatial frame the latter becomes ${\delta }_{ij}$, but the coefficient is still one. For a gradient $(q,0,0)$ in that frame, direct projection gives ${\tilde{S}}_{11}=2q^2/3$, rather than the printed $q^2/3$. The PDF's energy-density gradient is spatial ${\partial }_i\varphi {\partial }^i\varphi$; the TeX aid's spacetime index there is an OCR defect.

We prove finite propagation speed using a shrinking cone energy argument. Fix $T>0$ and $x_{\ast }$, and suppose the Cauchy data vanish on $B(x_{\ast },T)$. For $0\le s<T$, define the local wave energyThe homogeneous wave equation gives the local conservation lawDifferentiate the integral over the moving ball. Its boundary moves inward with speed one, so the divergence theorem givesHere $n$ is the outward unit normal, ${\partial }_n$ the normal derivative, and ${\nabla }_{\tan }$ the component of the gradient tangent to the boundary. Since $E(0)=0$ and $e\ge 0$, we have $E(s)=0$. Hence both ${\varphi }_t$ and $\nabla \varphi$ vanish inside the backward light cone. Integrating ${\varphi }_t(s,x_{\ast })=0$ from the zero initial displacement gives $\varphi (s,x_{\ast })=0$; continuity then gives $\varphi (T,x_{\ast })=0$.

Applying the same energy estimate to the difference of two solutions proves the domain of dependence assertion. If $\mathrm{dist}(x_{\ast },K)>T$, the initial ball misses $K$, and the preceding argument proves the stated support bound. No disturbance propagates faster than one.

The two displacement vectors $\mathbf{b}-\mathbf{a}$ and $\mathbf{c}-\mathbf{a}$ lie in the required plane. Because the three points are non-collinear, their cross productis nonzero and is a normal vector to that plane. A point with position vector $\mathbf{r}$ lies in the plane exactly whenExpanding the cross product givesThe terms $\mathbf{a}\cdot (\mathbf{a}\times \mathbf{b})$ and $\mathbf{a}\cdot (\mathbf{c}\times \mathbf{a})$ vanish, so the equation of a plane through three points becomesNon-collinearity is important: it ensures this is a genuine plane equation with nonzero normal, rather than the vacuous equality $0=0$.

For the specified coordinates,Their sum is $(3,3,6)$, with Euclidean norm $3\sqrt{6}$. Consequently either orientation of the unit normal is valid:As a check, the scalar triple product is $12$, so the plane equation simplifies to $x+y+2z=4$, which all three supplied points satisfy. The normal direction is $(1,1,2)$.

The chosen unit normal points toward the plane from the origin, so the plane equation isTo locate the circle centre, project $\mathbf{p}$ onto this plane. PutThen $\mathbf{n}\cdot \mathbf{h}=d$, so $\mathbf{h}$ lies in the plane. For any other point $\mathbf{r}$ of the plane, $\mathbf{r}-\mathbf{h}$ is perpendicular to $\mathbf{n}$. The Pythagorean theorem therefore givesRestricting the sphere equation to the plane consequently gives a circle centred at $\mathbf{h}$. Thus $\mathbf{m}=\mathbf{h}$, proving that the displacement of its centre is purely normal:The same Pythagorean theorem gives $q^2={\rho }^2+{\gamma }^2$, and henceThe sign of $\gamma$ is fixed by which side of the plane contains $\mathbf{p}$; $q$ and $\rho$ alone determine only its magnitude. Substituting the signed offset yields the sphere-plane intersection radiusA genuine circle requires $|d-\mathbf{n}\cdot \mathbf{p}|<q$; equality gives tangency and $\rho =0$, while a larger distance gives no intersection. The signed distance from a point to a plane in the direction $\mathbf{n}$ is $\mathbf{n}\cdot \mathbf{p}-d=-\gamma$; the ordinary distance from a point to a plane is its absolute value.

The product rule for the divergence givesSubtracting cancels the two gradient products. ThereforeTo obtain the requested order of terms, apply this identity with $u,v$ interchanged and use the divergence theorem on the spherical shell. On the outer sphere the shell's outward unit normal is $\hat{\mathbf{r}}$; on the inner sphere it is $-\hat{\mathbf{r}}$. Write ${\partial }_r=\hat{\mathbf{r}}\cdot \nabla$ on both spheres. Then the resulting Green second identity isThe minus sign records the inward radial direction of the shell's normal on its inner boundary. Thus the normal derivatives in the displayed difference are outward from each individual ball; using shell-outward normals instead would put a plus sign between the two boundary integrals.

The curl is constant:Orient the spanning planar disk by the unit normal $\mathbf{n}=(a,b,c)$ and orient its boundary by the right-hand rule, equivalently counterclockwise when viewed from the side towards which $\mathbf{n}$ points. Stokes theorem converts the line integral to a constant flux integral through a disk of area $\pi R^2$. With this orientationReversing the orientation changes the sign. The centre of the circle does not enter because the curl is constant.