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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.

Assume the level set is regular and non-null. The latter hypothesis is necessary: a null hypersurface has a degenerate induced metric and hence no ordinary intrinsic Levi-Civita connection determined by that metric. Normalize the normal vector and define the non-null hypersurface projection byHere $h$ denotes the induced metric restricted to tangent vectors, while its ambient extension annihilates $n$. The intrinsic covariant derivative is $D_{X}Y=P{\nabla }_{X}Y$ for tangent vector fields. Projection of the ambient torsion-free connection and metric compatibility shows that $D$ is the Levi-Civita connection of $h$. Define the extrinsic curvature by $K(X,Y)=g({\nabla }_{X}n,Y)$, or $K_{ab}=P^c{}_a{P}^d{}_b{\nabla }_c{n}_d$. It is symmetric: differentiating the defining function twice and projecting removes the derivative of its normalization, leaving a multiple of its symmetric Hessian. This sign convention makes the scalar second fundamental form $g(n,{\nabla }_{X}Y)=-K(X,Y)$.

Let $S$ be the tangent linear map characterized by $h(SX,Y)=K(X,Y)$. Differentiating $g(n,n)=\epsilon$ and $g(n,Y)=0$ gives the Gauss formulaFix the Riemann curvature tensor conventionApplying the Gauss formula twice givesSubtracting the expression with $X,Y$ interchanged, and then subtracting the bracket derivative, provesThus the Gauss–Codazzi equations for a non-null hypersurface give the intrinsic Riemann curvature tensor asAll free indices here are tangent to the hypersurface. The normal component of the same calculation isAfter antisymmetrizing and using $D_{X}Y-D_{Y}X=[X,Y]$, this yields the curved-ambient Codazzi equationThe extrinsic curvature measures how tangent planes turn in the ambient manifold; the Gauss equation says that this turning contributes to intrinsic curvature even when the ambient Riemann curvature tensor is zero.

For the quadric in Minkowski spacetime, write its position vector as $X^a$. On the hypersurface, $g(X,X)=1$ and differentiating this identity in a tangent direction gives $g(X,Y)=0$. Hence $n=X$ is a unit normal and a spacelike vector, $\epsilon =1$, and ${\nabla }_{Y}n=Y$ in flat Cartesian coordinates. Therefore $K=h$. The ambient Riemann curvature tensor vanishes, soThis is unit-radius de Sitter spacetime. Reversing the normal changes $K$ to $-K$ but does not change this curvature. An opposite overall convention for the Riemann curvature tensor would reverse both curvature signs below.

To retain a four-dimensional Lorentzian metric while reversing the curvature, take a flat ambient metric with two negative directions and the quadricNow the position vector is a unit normal with negative squared norm, so $\epsilon =-1$ while $K=h$ remains true. The Gauss equation givesThis is unit-radius Anti-de Sitter spacetime. If a positive-definite induced metric is acceptable, an even simpler alternative is to keep the original one-time ambient Minkowski metric and change the quadric level to $g(X,X)=-1$. Each sheet then gives four-dimensional hyperbolic space with the same negative curvature, but its induced signature differs from that of de Sitter spacetime.

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.

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.

The spin of a small torque-free test gyroscope is represented by a spacelike vector orthogonal to its four-velocity, $s^{\mu }{u}_{\mu }=0$. Its temporal component vanishes in its own rest frame. Along a geodesic, $D{s}^{\mu }/d\tau =0$: parallel transport preserves the orthogonality and the invariant spin magnitude. For an accelerated observer the corresponding nonrotating transport is Fermi-Walker transport.

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.