Forces and Continuum — Examples

Boosted components. The boost matrix has ${\mathrm{\Lambda }}^0{}_0=\gamma$, ${\mathrm{\Lambda }}^0{}_1={\mathrm{\Lambda }}^1{}_0=-\gamma \beta$, ${\mathrm{\Lambda }}^1{}_1=\gamma$ and preserves the other two spatial directions (The passive Lorentz boost with parameter beta). Since $T^{ab}=\rho {\delta }_0^a{\delta }_0^b$, [F1] gives $T^{\mathrm{\prime }ab}=\rho {\mathrm{\Lambda }}^a{}_0{\mathrm{\Lambda }}^b{}_0$; using ${\mathrm{\Lambda }}^0{}_0=\gamma$, ${\mathrm{\Lambda }}^1{}_0=-\gamma \beta$, ${\mathrm{\Lambda }}^2{}_0={\mathrm{\Lambda }}^3{}_0=0$ this is $T^{\mathrm{\prime }00}={\gamma }^2\rho$, $T^{\mathrm{\prime }01}=T^{\mathrm{\prime }10}=-{\gamma }^2\beta \rho$, $T^{\mathrm{\prime }11}={\gamma }^2{\beta }^2\rho$ and $T^{\mathrm{\prime }22}=T^{\mathrm{\prime }33}=T^{\mathrm{\prime }02}=T^{\mathrm{\prime }03}=T^{\mathrm{\prime }12}=T^{\mathrm{\prime }13}=0$.

Factorization and conservation. For the same field $n(p)=e_0$, [F1] gives $n^{\mathrm{\prime }a}={\mathrm{\Lambda }}^a{}_b{n}^b={\mathrm{\Lambda }}^a{}_0$, hence its primed components are $n^{\mathrm{\prime }0}=\gamma$, $n^{\mathrm{\prime }1}=-\gamma \beta$, $n^{\mathrm{\prime }2}=n^{\mathrm{\prime }3}=0$. Step 1.1 therefore reads $T^{\mathrm{\prime }ab}=\rho n^{\mathrm{\prime }a}{n}^{\mathrm{\prime }b}$; the passive change preserves the fields $n$ and $T$. The primed components satisfy ${\eta }_{ab}{n}^{\mathrm{\prime }a}{n}^{\mathrm{\prime }b}=-{\gamma }^2+{\gamma }^2{\beta }^2=-1$ and $n^{\mathrm{\prime }0}>0$, consistent with the future unit normalization of $n=e_0$. All primed components are constant in the primed chart, so ${\mathrm{\partial }}_a^{\mathrm{\prime }}{T}^{\mathrm{\prime }ab}=0$, consistent with [F2] applied to the constant future unit field $n$ and constant $\rho >0$.

The primed coordinate coefficient $T^{\mathrm{\prime }00}={\gamma }^2\rho$ is the numerical factorization ${\gamma }^2\rho =\gamma \cdot (\gamma \rho )$; its source interpretation as one energy factor and one volume factor belongs to the separately adopted dust model reading and is not proved by this component computation. The example is a component computation and asserts no interaction, no dynamics and no measurement.

Rest-frame components. With $n=e_0$, $n^a={\delta }_0^a$ and $\eta =\mathrm{diag}(-1,1,1,1)$: $T^{00}=(\epsilon +P)(n^0{)}^2+P{\eta }^{00}=(\epsilon +P)-P=\epsilon$; $T^{ii}=(\epsilon +P)(n^i{)}^2+P{\eta }^{ii}=0+P=P$ for $i=1,2,3$; and $T^{0i}=(\epsilon +P)n^0{n}^i+P{\eta }^{0i}=0$.

Observer contraction in the rest frame. By [F2] applied to the observer $e_0$ the contraction is $\epsilon =T^{ab}(e_0{)}_a(e_0{)}_b=T^{00}$, consistent with step 1.1.

Boosted components. The boost matrix has ${\mathrm{\Lambda }}^0{}_0={\mathrm{\Lambda }}^1{}_1=\gamma$, ${\mathrm{\Lambda }}^0{}_1={\mathrm{\Lambda }}^1{}_0=-\gamma \beta$ and preserves the other two spatial directions. Substituting the rest-frame components of step 1.1 into [F1] and summing the nonzero terms gives $T^{\mathrm{\prime }00}={\gamma }^2\epsilon +{\gamma }^2{\beta }^{2}P={\gamma }^2(\epsilon +P)-P$, $T^{\mathrm{\prime }01}=\gamma (-\gamma \beta )\epsilon +(-\gamma \beta )\gamma P=-{\gamma }^2(\epsilon +P)\beta$, $T^{\mathrm{\prime }11}={\gamma }^2{\beta }^2\epsilon +{\gamma }^{2}P={\gamma }^2(\epsilon +P){\beta }^2+P$, $T^{\mathrm{\prime }22}=T^{\mathrm{\prime }33}=P$, $T^{\mathrm{\prime }23}=0$ and $T^{\mathrm{\prime }02}=T^{\mathrm{\prime }03}=T^{\mathrm{\prime }12}=T^{\mathrm{\prime }13}=0$.

The boosted observer contraction. Because the passive boost $B(\beta )$ is a proper orthochronous Lorentz map (The passive boost is a proper orthochronous Lorentz map with inverse B(-beta)), the primed basis is an orthonormal future-oriented frame, so the vector $e_0^{\mathrm{\prime }}$ with primed components $e_0^{\mathrm{\prime }a}={\delta }_0^a$ is future unit; by [F2] its energy contraction is ${\epsilon }^{\mathrm{\prime }}:=T^{ab}(e_0^{\mathrm{\prime }}{)}_a(e_0^{\mathrm{\prime }}{)}_b=T^{\mathrm{\prime }00}={\gamma }^2(\epsilon +P)-P$, so ${\epsilon }^{\mathrm{\prime }}-\epsilon =({\gamma }^2-1)(\epsilon +P)$: this difference vanishes exactly when $\beta =0$ (i.e. $\gamma =1$) or $\epsilon +P=0$, and ${\epsilon }^{\mathrm{\prime }}\ge \epsilon$ whenever $\epsilon +P\ge 0$; no positivity of $\epsilon +P$ is assumed. The transverse diagonal components remain $P$.

The computation uses the observer decomposition to identify the numerical contraction $T^{\mathrm{\prime }00}$ as the energy density of the boosted observer; it asserts no equation of state, no positivity of $\epsilon$ or $P$ and no measurement.

$(n,\rho )$ satisfies the dust equations. By the definitions of $\gamma$ and the Minkowski metric, $\eta (n,n)=-(n^0{)}^2+(n^1{)}^2=-{\gamma }^2(1-{\beta }^2)=-1$, and $n^0=\gamma >0$ in the chart's time orientation, so $n$ is a future unit timelike constant field; by [F3] ${\mathrm{\partial }}_a{n}^b=0$ and hence $n^a{\mathrm{\partial }}_a{n}^b=0$. The profile depends only on $x^2,x^3$, so $n^a{\mathrm{\partial }}_a\rho =\gamma ({\mathrm{\partial }}_0\rho +\beta {\mathrm{\partial }}_1\rho )=0$; and ${\mathrm{\partial }}_a{n}^a=0$, so ${\mathrm{\partial }}_a(\rho n^a)=n^a{\mathrm{\partial }}_a\rho +\rho {\mathrm{\partial }}_a{n}^a=0$. By [F1] the field $T^{ab}=\rho n^a{n}^b$ is conserved.

Nonnegative variant and support. If ${\rho }_0\ge 0$ is allowed to vanish, the same equations ${\mathrm{\partial }}_a(\rho n^a)=0$ and $n^a{\mathrm{\partial }}_a{n}^b=0$ hold, and [F1] gives conservation, with the inertial-flow statement on $\{{\rho }_0>0\}$; a compactly supported nonnegative transverse profile gives the same exact solution and the same finite-box constancy. The profile is independent of $x^1$, so a nonzero profile is not compactly supported in all three spatial variables; the flow lines are $x^1=\beta x^0+\text{const}$, $x^2,x^3$ constant.

Finite slice integrals are constant. Let $\bar{K}=[c_1,d_1]\times [c_2,d_2]\times [c_3,d_3]$ with $c_i<d_i$ and $Q^b(s)={\int }_{\bar{K}}{T}^{0b}(s,x)d^{3}x$. Since $T^{0b}=\rho n^0{n}^b=\gamma n^b{\rho }_0(x^2,x^3)$ is independent of $x^0$, each $Q^b(s)$ is constant in $s$. The same conclusion follows from [F2]: the force density vanishes by step 1.1, the transverse face terms vanish because $T^{2b}=T^{3b}=0$, and the two $x^1$-faces carry the same pointwise integrand $T^{1b}={\rho }_0\gamma \beta n^b$ and enter with opposite signs, so their contributions cancel although the individual face flux need not vanish.

The example exhibits a family of exact conserved dust solutions of the stated form with transverse profile and finite-box constancy of the slice integrals; it asserts no interaction, no dynamics of the profile beyond the displayed equations and no measurement.

Components of the coordinate divergence. The components of $T$ are $T^{00}=\rho (x^1)$, $T^{11}=P(x^1)$ and $T^{ab}=0$ for all other pairs; these are $C^1$ in $x$, so the coordinate divergence has components $D^0={\mathrm{\partial }}_0{T}^{00}+{\mathrm{\partial }}_1{T}^{10}={\mathrm{\partial }}_0\rho (x^1)=0$, $D^1={\mathrm{\partial }}_0{T}^{01}+{\mathrm{\partial }}_1{T}^{11}=P^{\mathrm{\prime }}(x^1)$, and $D^2=D^3=0$, because the omitted terms involve the zero components and $\rho ,P$ do not depend on $x^0$.

Vanishing coordinate divergence and constancy of $P$. By construction the set of values $x^1(p)$, $p\in \mathrm{\Omega }$, is exactly the interval $J$ on which $\rho ,P$ are defined. The set $J$ is open: a chart ball about any point of $x(\mathrm{\Omega })$ permits all sufficiently small changes of its $x^1$ coordinate, hence a neighbourhood of each value in $J$ remains in $J$; this is also true when $J=\mathrm{\varnothing }$. Thus every point of $J$ is interior. By step 1.1, $D=0$ on $\mathrm{\Omega }$ if and only if $P^{\mathrm{\prime }}=0$ at every interior point of $J$; by [F2] this holds if and only if $P$ is constant on $J$. For nonempty $\mathrm{\Omega }$, this is exactly local conservation by [F1]. For $\mathrm{\Omega }=J=\mathrm{\varnothing }$, $D=0$ and constancy of the empty function $P$ are both vacuous, with any real constant, for example $0$, witnessing constancy. The energy density $\rho$ is subject to no such condition: ${\rho }^{\mathrm{\prime }}$ never enters the divergence, because the only term containing $\rho$ is $T^{00}$, whose first index is contracted with ${\mathrm{\partial }}_0$ and which is independent of $x^0$.

For nonempty $\mathrm{\Omega }$, consequently, a static uniaxial stress in equilibrium has uniform stress $P$ along the stressed direction, while its energy density may vary; when $P<0$ (tension) the tension must be constant. This is a component computation for the stated field; it asserts no material model, no equation of state and no force law.

Force density. By [F2], $f^0={\mathrm{\partial }}_0{T}^{00}+{\mathrm{\partial }}_1{T}^{10}=1+1=2$, $f^1={\mathrm{\partial }}_0{T}^{01}+{\mathrm{\partial }}_1{T}^{11}=0+0=0$ and $f^2=f^3=0$; hence ${\int }_B{f}^0{d}^{4}x=2\mathrm{vol}(B)=2A(t_2-t_1)(x_2-x_1)$ by [F3], and the $b=1,2,3$ bulk integrals vanish.

The $b=0$ identity. The temporal faces contribute ${\int }_{F_0^{+}}{T}^{00}-{\int }_{F_0^{-}}{T}^{00}=A{\int }_{x_1}^{x_2}{x}^0{\mid }_{t_2}d{x}^1-A{\int }_{x_1}^{x_2}{x}^0{\mid }_{t_1}d{x}^1=A(t_2-t_1)(x_2-x_1)$; the faces with $a=1$ contribute ${\int }_{F_1^{+}}{T}^{10}-{\int }_{F_1^{-}}{T}^{10}=A{\int }_{t_1}^{t_2}{x}^1{\mid }_{x_2}d{x}^0-A{\int }_{t_1}^{t_2}{x}^1{\mid }_{x_1}d{x}^0=A(t_2-t_1)(x_2-x_1)$; the transverse faces vanish because $T^{2b}=T^{3b}=0$. The sum of face differences is $2A(t_2-t_1)(x_2-x_1)$, equal to the bulk integral of step 1.1, verifying [F1] for $b=0$.

The $b=1$ identity and the remaining components. For $b=1$, the temporal faces contribute $\pm A{\int }_{x_1}^{x_2}{x}^{1}d{x}^1$ with opposite signs, which cancel, and the faces with $a=1$ contribute $\pm A{\int }_{t_1}^{t_2}{x}^{0}d{x}^0$, which also cancel; the transverse faces vanish. Both sides of [F1] vanish, and the same is immediate for $b=2,3$ because all relevant components are zero.

Numerical specialization: for transverse area $A=6$, temporal length $t_2-t_1=2$ and axis-one length $x_2-x_1=3$, the bulk integral is $2\cdot 6\cdot 2\cdot 3=72$ and the two nonzero face differences are each $6\cdot 2\cdot 3=36$, whose sum is $72$. The example verifies the finite-box identity on a field with nonzero force density and nonzero net side flux; it asserts no matter model, no dynamics and no empirical fact.

The field solves the massless equation. By [F1], ${\mathrm{\partial }}_a\phi =f^{\mathrm{\prime }}(\theta )k_a$ and ${\mathrm{\partial }}_a{\mathrm{\partial }}_b\phi =f^{\mathrm{\prime }\mathrm{\prime }}(\theta )k_a{k}_b$; hence $\mathrm{\square }\phi ={\eta }^{ab}{\mathrm{\partial }}_a{\mathrm{\partial }}_b\phi =f^{\mathrm{\prime }\mathrm{\prime }}(\theta ){\eta }^{ab}{k}_a{k}_b=0$, because $k$ is null. With $V\equiv 0$ this is exactly the scalar-field equation $\mathrm{\square }\phi =V^{\mathrm{\prime }}(\phi )$ of [F2].

The stress-energy tensor. Substituting [F1] into the Hilbert formula of [F2] with $V\equiv 0$ gives ${\mathrm{\partial }}^a\phi {\mathrm{\partial }}^b\phi =f^{\mathrm{\prime }}(\theta {)}^2{k}^a{k}^b$ and ${\mathrm{\partial }}_d\phi {\mathrm{\partial }}^d\phi =f^{\mathrm{\prime }}(\theta {)}^2{k}_d{k}^d=0$, hence $T^{ab}=K{f}^{\mathrm{\prime }}(\theta {)}^2{k}^a{k}^b$, which is symmetric. Wherever $f^{\mathrm{\prime }}(\theta )\mathrm{\ne }0$ the coefficient $K{f}^{\mathrm{\prime }}(\theta {)}^2$ is positive and $T$ is a positive multiple of the rank-one tensor $k^{\mathrm{♯}}\otimes k^{\mathrm{♯}}$ with nonzero future null generating vector $k^{\mathrm{♯}}$; wherever $f^{\mathrm{\prime }}(\theta )=0$ all components vanish. For every future unit $n$, $T^{ab}{n}_a{n}_b=K{f}^{\mathrm{\prime }}(\theta {)}^2(k_a{n}^a{)}^2\ge 0$; in the inertial frame, $T^{00}=K{f}^{\mathrm{\prime }}(\theta {)}^2(k^0{)}^2\ge 0$ with $k^0>0$.

Conservation. Since $\phi$ solves the field equation by step 1.1, [F3] gives ${\mathrm{\partial }}_a{T}^{ab}=0$: the plane-wave tensor is conserved on the whole chart domain.

The example is an explicit conditional computation for the massless scalar model with a null plane-wave profile; it asserts no wave observation, no energy condition beyond the displayed nonnegativity and no measurement.

The field. Define the smooth symmetric field by $T^{00}=x^0$, $T^{01}=T^{10}=-x^1$, $T^{11}=x^0$, all other components zero. By [F2] its force density has components $f^0={\mathrm{\partial }}_0{x}^0+{\mathrm{\partial }}_1(-x^1)=1-1=0$, $f^1={\mathrm{\partial }}_0(-x^1)+{\mathrm{\partial }}_1{x}^0=0$ and $f^2=f^3=0$; hence ${\mathrm{\partial }}_a{T}^{ab}=0$ on all of affine Minkowski space and $T$ is locally conserved.

The slice integral is not conserved. Let $L>0$, $c_2<d_2$, $c_3<d_3$, put $A=(d_2-c_2)(d_3-c_3)>0$ and $K=[0,L]\times [c_2,d_2]\times [c_3,d_3]$, and let $Q^0(s)={\int }_K{T}^{00}(s,x)d^{3}x=A{\int }_0^L{x}^0(s)d{x}^1=ALs$. For $s_1<s_2$, [F1] applied to the sub-box with $f=0$ gives $Q^0(s_2)-Q^0(s_1)=-({\int }_{[s_1,s_2]\times K_1^{+}}{T}^{10}-{\int }_{[s_1,s_2]\times K_1^{-}}{T}^{10})-0$, and $T^{10}=-x^1$ takes the constant values $-L$ on $K_1^{+}=\{x^1=L\}$ and $0$ on $K_1^{-}=\{x^1=0\}$, so the side-flux difference equals $-A(s_2-s_1)\cdot (L-0)=-AL(s_2-s_1)$ and $Q^0(s_2)-Q^0(s_1)=AL(s_2-s_1)\mathrm{\ne }0$ whenever $s_1<s_2$; the transverse fluxes vanish because $T^{2b}=T^{3b}=0$. For example $A=6$, $L=2$ and $s_2-s_1=3$ give the value $36$.

Thus zero local divergence alone does not conserve a finite slice integral: this field has $f=0$ everywhere yet its slice integral $Q^0$ varies linearly with the slice parameter, so the side-flux (isolation) hypothesis in the finite-flux conservation statement cannot be dropped. The prescribed tensor is not claimed to arise from any matter model of this page and no choice principle is used.