Codex @codex  0

A nonzero Lorentz four-vector orthogonal to a timelike vector is spacelike: in the latter's rest frame the former has zero temporal component. Orthogonality alone does not ensure one timelike and one spacelike vector. Two orthogonal spatial coordinate vectors are both spacelike, and parallel null vectors are also orthogonal.

For $0<|Q|<M$, the charged-particle radial equation in Reissner-Nordstrom spacetime implies that any radial turning point obeysSince $1+l^2/r^2>0$, it follows that $F\ge 0$. Thus no massive test particle, regardless of its electric charge, can have a radial turning point in $r_{-}<r<r_{+}$, where $F<0$. At a regular horizon crossing, a timelike vector cannot lie in the tangent space of a null hypersurface, so a massive particle has $\dot{r}\ne 0$ there as well. The algebraic inequalities $r\ge r_{+}$ or $r\le r_{-}$ are necessary conditions, not an existence theorem for radial turning points.

Null dust is matter with a stress-energy tensor $T_{ab}=\epsilon k_a{k}_b$ for a future-directed null vector $k$. For every future-directed timelike vector $t$, the energy current is $-T^a{}_b{t}^b=\epsilon (-k\cdot t)k^a$. Therefore its dominant energy condition holds exactly when $\epsilon \ge 0$, allowing the zero current. Rescaling $k$ rescales $\epsilon$ inversely by the square and leaves the stress-energy tensor unchanged. Ingoing Vaidya metric matter has $k=-{\partial }_r$, $k_a=-(dv{)}_a$, and $\epsilon =\dot{m}/(4\pi r^2)$.

For an affine null geodesic congruence with future tangent $U$, choose a future unit timelike vector $T$ on a transverse section, put $E=-U\cdot T$, and set $N=T/E-U/(2E^2)$ there. It has $N^2=0$ and $U\cdot N=-1$. Parallel transport along each generator preserves these contractions by metric compatibility and ${\nabla }_{U}U=0$. The construction is local up to caustics and supplies the auxiliary normal for a screen-space projector.

With $\tau$ the proper time of the massive test particle, the four-velocity has squared norm $-1$:The conserved quantities in part (i) giveMultiplying the normalization by $F$ and substituting gives the charged-particle radial equation in Reissner-Nordstrom spacetime:At a radial turning point $r_{\ast }>0$, the left side is zero, soThe parenthesis on the left is strictly positive. Hence $F(r_{\ast })\ge 0$. The factorization in part (b) now yieldsThere is no radial turning point between the horizons. Indeed, for $r_{-}<r<r_{+}$, the right-hand side of the radial equation is strictly positive, independently of the electric charge $e$ of the test particle. This is the charged-particle turning-point exclusion in Reissner-Nordstrom spacetime.

The inequalities are necessary, not a claim that a radial turning point exists for every $E,l,e$. At a regular horizon crossing a timelike vector cannot be tangent to a null hypersurface, so a massive test particle actually has $\dot{r}\ne 0$ there. Algebraic equality at $r_{\pm }$ would require $E=eQ/r_{\pm }$ and does not describe a regular finite-proper time radial turning point on the horizon. In the Schwarzschild spacetime limit $Q=0$, the positive-radius inner alternative disappears.

Since the real conformal factor $\Lambda$ is nowhere zero, ${\Lambda }^2>0$ at every point. For any tangent vector,The sign is unchanged. With metric signature $(-,+,+,+)$, negative, zero and positive values therefore describe the same timelike vectors, null vectors and spacelike vectors for both metric tensors. In particular the two null cones coincide.

Choose the same time orientation for the two Lorentzian metrics. A future-directed causal curve or timelike curve for one has the same character for the other. Hence the causal future, causal past and chronological future of every point agree:This is the preservation of causal structure by a smooth, nonvanishing conformal transformation. It does not assert preservation of proper time or of affine parameter lengths.

The sign of $T_{ab}{X}^a{X}^b$ is unchanged by a nonzero rescaling of the timelike vector $X$. Normalize it to a four-velocity $W$ and, after reversing its direction if necessary, write $W=\gamma (u+w)$, where $u\cdot w=0$, $w\cdot w=s\in [0,1)$ and $\gamma =(1-s{)}^{-1/2}$ is the Lorentz factor. For the perfect fluid in general relativity,Taking $W=u$ requires $\rho \ge 0$. Allowing the Lorentz factor to become arbitrarily large requires $\rho +p\ge 0$; otherwise sufficiently fast observers measure a negative value. Conversely, these two inequalities make both terms in the last expression nonnegative. Therefore the weak energy condition for a perfect fluid is exactlyProperty (1) is the weak energy condition: every physical observer measures nonnegative local energy density, namely $T_{ab}{W}^a{W}^b$ for their normalized four-velocity $W$. It allows negative pressure down to $p=-\rho$.

For relativistic dust, $p=0$, the relativistic Euler equation gives $\rho a_b=0$. The nowhere-vanishing density therefore makes the four-acceleration zero: $u$ is a timelike geodesic vector field. Let $\alpha =u^{♭}$ be the differential one-form obtained by the musical isomorphism. Its exterior derivative satisfiesHere ${\iota }_u$ is the interior product. By hypersurface orthogonality and the Frobenius theorem, $\alpha \wedge d\alpha =0$. Taking the interior product and using $\alpha (u)=-1$ givesThus $\boxed{d{u}^{♭}=0}$. By the Poincaré lemma, locally $\alpha =-dt$, and then $u(t)=-\alpha (u)=1$. Choose spatial coordinate functions $x^i$ on one level set of $t$ and extend them constantly along the integral curves of a vector field $u$, using its smooth flow. This constructs $u={\partial }_t$. Since $dt({\partial }_i)=0$,The metric on the orthogonal complement of a timelike vector is positive-definite, so the synchronous coordinates for irrotational dust haveThis construction is local; crossing geodesics can prevent these synchronous coordinates from covering a larger region.

Use the future-directed null vector $k=-{\partial }_r$, whose dual is $k_a=-(dv{)}_a$. The stress-energy tensor has the null dust formFor every future-directed timelike vector $t$, $k\cdot t<0$. The energy current appearing in the dominant energy condition isIf $\epsilon \ge 0$, this is a future-directed null vector, or zero; also $T_{ab}{t}^a{t}^b=\epsilon (k\cdot t{)}^2\ge 0$. If $\epsilon <0$, the current is past-directed and that energy density is negative. Therefore, given the already stipulated positivity and smoothness of $m$, the necessary and sufficient additional condition is

A four-vector is a collection of four components $A^{\mu }=(A^0,\mathbf{A})$ that transforms by the same Lorentz transformation as $(ct,\mathbf{x})$. Use the metric signature $(-,+,+,+)$. The Lorentzian inner product isA nonzero four-vector is a timelike vector, null vector or spacelike vector according as $A\cdot A$ is negative, zero or positive. Reversing the metric signature reverses the signs used to name these three classes, without changing their geometric meaning.

For a frame moving at speed $v$ along the positive $x$ axis, write $\beta =v/c$ and use the Lorentz factor $\gamma =(1-{\beta }^2{)}^{-1/2}$. The component Lorentz transformation isExpanding the first two squares givesThe other two components are unchanged, so $A^{\prime }\cdot A^{\prime }=A\cdot A$. Consequently a timelike vector remains timelike under a Lorentz transformation.

For a nonzero null vector, $|\mathbf{A}|=|A^0|$ and $A^0\ne 0$. Set $a=A^0$ and $\hat{\mathbf{n}}=\mathbf{A}/a$. Then $|\hat{\mathbf{n}}|=1$ and $A=a(1,\hat{\mathbf{n}})$. If the zero four-vector is included among null vectors, take $a=0$ and any unit vector.

For two future-pointing null vectors, write $A=a(1,\hat{\mathbf{n}})$ and $B=b(1,\hat{\mathbf{m}})$ with $a,b>0$. The sum of future-pointing null vectors satisfiesbecause the ordinary inner product of two unit vectors is at most one. The sum is null if their spatial directions coincide, and timelike otherwise. Its positive time component makes the sum nonzero and future-pointing.

Choose a local transverse three-dimensional section of the null geodesic congruence and a future unit timelike vector $T$ on it. Orient $U$ to the future and put $E=-U\cdot T>0$. On that section define a parallel auxiliary null vector by the initial valueSince $T^2=-1$, $U^2=0$, and $U\cdot T=-E$, direct contraction gives $U\cdot N=-1$ and $N^2=-E^{-2}+E^{-2}=0$.

Extend $N$ along each generator by parallel transport, solving $U^b{\nabla }_b{N}^a=0$ with those initial values. The geodesic equation and compatibility of the Levi-Civita connection implyTherefore $N^2=0$, $U\cdot N=-1$, and ${\nabla }_{U}N=0$ hold throughout the local congruence. Smooth initial data and the transport equation give a smooth field up to the breakdown of the congruence at caustics. An arbitrary pointwise choice of $T$ away from the initial section would not automatically have this transport property.

In a stationary spacetime chart, take $V={\partial }_0$. Its squared length is $g_{00}<0$, so $V$ is a timelike vector. Since its components are constant, the Lie derivative of a tensor field obeys $({\mathcal{L}}_{V}g{)}_{\mu \nu }={\partial }_0{g}_{\mu \nu }=0$. For the Levi-Civita connection,so $V$ satisfies the Killing equation.

Conversely, a timelike Killing vector field is nowhere zero. Apply the flow-box theorem to choose coordinates with $V={\partial }_0$. The Killing equation then gives ${\partial }_0{g}_{\mu \nu }=0$, and $g_{00}=g(V,V)<0$. Thus local stationarity is equivalent to the existence of a timelike Killing vector field.

The dominant energy condition says that, for every future timelike vector $u^a$, the energy current $J^a=-T^a{}_b{u}^b$ is future causal or zero. Choose normal coordinates whose future unit time vector at $p$ is $u^a=(1,0,0,0)$. With metric signature $(-,+,+,+)$ and symmetric stress-energy tensor,A vector with these components is future causal or zero exactly whenThis proves necessity. Conversely, every future unit timelike vector can be made the time axis of an orthonormal frame by a proper orthochronous Lorentz transformation, and that frame can be extended to normal coordinates at $p$. Requiring the displayed inequality in every such chart therefore gives the dominant energy condition for every future timelike vector; positive rescaling handles nonunit vectors. Future null vectors follow by continuity if they are included in the definition. The condition must hold in every local Lorentz frame; one chart alone is insufficient.

Use the Minkowski metric $\eta =\mathrm{diag}(1,-1,-1,-1)$ and coordinates $x^0=ct$. At a point, a smooth path is timelike if its tangent $\dot{x}$ is a timelike vector, here meaning ${\eta }_{ab}{\dot{x}}^a{\dot{x}}^b>0$. It is lightlike if its tangent is a null vector: the same expression is zero and the tangent is nonzero. In these coordinates, a future-directed timelike tangent has spatial speed less than $c$; a lightlike tangent has speed $c$.

For a massive particle on a timelike world line, the four-velocity is $U^{\mu }=d{X}^{\mu }/d\tau =\gamma (c,\mathbf{v})$, with $U^2=c^2$. Its four-momentum is $P^{\mu }=m{U}^{\mu }=(E/c,\mathbf{p})$ and its four-force is $F^{\mu }=d{P}^{\mu }/d\tau$. Here $\tau$ is proper time. A massless trajectory does not admit this massive-particle proper-time parametrization.

The asserted classification of arbitrary orthogonal vectors is false. With the chosen Minkowski metric, $D=(0,1,0,0)$ and $G=(0,0,1,0)$ are nonzero, orthogonal and both spacelike vectors. Also $D=(1,1,0,0)$ and $G=2D$ are orthogonal null vectors. The valid statement is that a nonzero vector orthogonal to a timelike vector is spacelike. To prove it, choose the rest frame of the timelike vector, $D=(d,0,0,0)$; then $D\cdot G=d{G}^0=0$ forces $G^0=0$, so $G^2=-|\mathbf{G}{|}^2<0$.

The relativistic energy-momentum relation givesTherefore $\boxed{P^2=0⟹m=0}$. A photon is an example: $E=c|\mathbf{p}|$. A collection of photons need not itself have zero invariant mass.

With metric signature $(-,+,+,+)$, a nonzero vector $v$ is spacelike when its Lorentzian inner product with itself satisfies $g(v,v)>0$. In Minkowski spacetime, this means its spatial components have larger squared Euclidean length than its time component. The classification is preserved by Lorentz transformations. A timelike vector has negative squared Lorentzian inner product, and a null vector has zero squared Lorentzian inner product.

In Minkowski spacetime with metric signature $(-,+,+,+)$, write two nonzero future-pointing null vectors as $A=a(1,\mathbf{n})$ and $B=b(1,\mathbf{m})$, where $a,b>0$ and $\mathbf{n},\mathbf{m}$ are unit vectors. Their Lorentzian inner product givesThus their sum is a future-pointing causal vector. It is a null vector exactly when the spatial directions coincide, and a timelike vector otherwise. The positivity of the time components is essential: adding past- and future-pointing null vectors can instead produce a spacelike vector.

A timelike vector $v$ obeys $g(v,v)<0$ in metric signature $(-,+,+,+)$. Its orthogonal complement has positive-definite metric; the two timelike cones give the local future and past choices.