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For a quadratic derivative source $A^{\alpha \beta }{\partial }_{\alpha }u{\partial }_{\beta }u$, the classical null condition is $A^{\alpha \beta }{\ell }_{\alpha }{\ell }_{\beta }=0$ whenever $-{\ell }_0^2+|{\ell }_{\mathrm{sp}}{|}^2=0$. It removes the leading interaction of parallel lightlike derivatives. In three dimensions this structure gives global smooth solutions for sufficiently small, localized data through the vector field method for wave equations. This PDE condition is distinct from the null condition asserting that a single vector is lightlike.

For an equatorial null geodesic of the Kerr metric, $E=-p_t$ and $L=p_{\varphi }$ are geodesic conserved quantities from Killing vectors. In units $G=c=1$, $\Delta =r^2-2Mr+a^2$, and an affine parameter gives the displayed equation. It follows by solving the stationary and axial momentum equations for $\dot{t},\dot{\varphi }$ and substituting them into the null condition. A zero of its radial polynomial is only a turning point; an exterior circular null geodesic also requires the derivative of that polynomial to vanish with $E,L$ held fixed.

The additional invariant is the squared norm of the affine tangent. Metric compatibility and the geodesic equation make it constant, and a photon has the null conditionOn the equatorial plane this givesand the conserved quantities of part (a) yieldFor a nonradial ray $\ell \ne 0$, put $u=1/r$. Sincethe orbit equation becomesHere $b=|\ell |/(c\epsilon )$ is the positive impact parameter for a future-directed ray arriving from infinity. A strictly radial ray has $\ell =0$ and must be described by the radial equation rather than by $\varphi$ as parameter.

Reflection across $\theta =\pi /2$ is an isometry; a geodesic initially tangent to its fixed equatorial plane remains in that plane by uniqueness of the geodesic equation. Thus take $\theta =\pi /2$, $\dot{\theta }=0$, and $\Sigma =r^2$. Define the algebraic abbreviationsThe conserved quantities of part (b) becomeFor $\Delta \ne 0$, inversion givesThe null condition can be written without expanding a quadratic square:Substituting the preceding expressions yields the equatorial Kerr null geodesic radial equationAn equivalent useful polynomial form isThese expressions use an affine parameter $s$; a common rescaling of $E$ and $L$ accompanies its rescaling. The inversion was performed in the exterior $r>r_{+}$, where Boyer-Lindquist coordinates are regular. The polynomial radial identity has a smooth continuation to a horizon, but the formulas for $\dot{t}$ and $\dot{\varphi }$ there must be replaced by regular-coordinate expressions. In the limit $a=0$, the result reduces to the Schwarzschild spacetime equation ${\dot{r}}^2=E^2-(1-2M/r)L^2/r^2$.

Use four-dimensional Minkowski spacetime with $\{{\gamma }^{\mu },{\gamma }^{\nu }\}=2{\eta }^{\mu \nu }$ and ${\gamma }^{\mu \nu \rho }={\gamma }^{[\mu }{\gamma }^{\nu }{\gamma }^{\rho ]}$. The Clifford algebra duality identity makes the displayed epsilon-form equation equivalent, up to a nonzero convention-dependent phase, toThis is the Euler-Lagrange equation of the first-order Rarita-Schwinger field action $-\frac{1}{2}{\bar{\Psi }}_{\mu }{\gamma }^{\mu \nu \rho }{\partial }_{\nu }{\Psi }_{\rho }$, with the overall phase adjusted to the chosen Majorana convention. Varying the Majorana spinor action accounts for both occurrences of the field and cancels the one-half normalization.

A vector times a spinor contains spin-$3/2$ and unwanted spin-$1/2$ pieces, so the tensor type alone does not establish the physical spin. The field equation has gauge invariancesince the two derivatives contract an antisymmetric pair of gamma indices. Put $\chi ={\gamma }^{\mu }{\Psi }_{\mu }$ and $d={\partial }^{\mu }{\Psi }_{\mu }$. Gamma matrix multiplication givesHence $d=/\partial \chi$. Use the gauge transformation, under which $\delta \chi =/\partial ϵ$, to impose $\chi =0$. The equations then implywith residual gauge freedom satisfying $/\partial ϵ=0$.

For a plane wave with nonzero momentum satisfying the null condition along the third axis, this residual freedom sets the temporal component to zero. Transversality then removes the longitudinal component, leaving the tensor product of two transverse vector polarizations with the two massless spinor helicities. The possible total helicities are $\pm 3/2$ and $\pm 1/2$. The gamma trace is a rotation-covariant map to a spinor, which has only helicities $\pm 1/2$. It therefore vanishes on the aligned $\pm 3/2$ combinations and removes the two mixed combinations. Equivalently, the transverse gamma trace condition has rank two on the four remaining complex positive-frequency coefficients. The massless Rarita-Schwinger polarization count is thusThe Majorana spinor reality condition relates negative-frequency coefficients to their conjugates; it does not introduce a separate antiparticle multiplet. A complex vector-spinor would also have independent antiparticle states.

A nonzero standard Rarita-Schwinger field mass term destroys the massless gauge invariance and adds the two longitudinal helicities. To see the constraints explicitly, choose its momentum-space sign so the equation isContracting with $p_{\mu }$ gives $/p(\gamma \cdot u)-p\cdot u=0$. Its gamma trace then gives $3m\gamma \cdot u=0$. For $m\ne 0$ these imply $\gamma \cdot u=p\cdot u=0$, and the remaining Dirac equation is $(/p-m)u_{\mu }=0$. In the rest frame $u_0=0$. Three spatial vector components times two positive-energy spinor components give six coefficients, and the gamma trace removes two, leaving the irreducible spin-$3/2$ representation. ThereforeIn supergravity a consistent spontaneous mass generation is the super-Higgs mechanism, which supplies those extra states from the goldstino.

The massless graviton also has two physical helicities, so the graviton and gravitino already match on shell. The reason for auxiliary fields is the off-shell component count of minimal supergravity mismatch, not a mismatch of these physical states. After removing gauge functions but before using field equations, a metric has $10-4=6$ independent bosonic components, while a Majorana vector-spinor has $16-4=12$ fermionic components. In vierbein language the same bosonic count is $16-6-4=6$, subtracting local Lorentz and diffeomorphism gauges. Old-minimal supergravity adds a complex scalar and a real vector, supplying six real nonpropagating bosonic components. This gives $12+12$ off-shell components and closure without imposing equations of motion. Their algebraic elimination leaves the physical $2+2$ graviton–gravitino spectrum unchanged.

An event horizon is a null hypersurface. For its spherical defining function $H(v,r)=r-r_H(v)$, the inverse radial metric tensor has $g^{vv}=0$, $g^{vr}=1$ and $g^{rr}=f=1-2m(v)/r$. The normal's null condition isIt follows that the event horizon follows the outgoing radial null geodesics of the Vaidya metric:This ordinary differential equation also describes other outgoing null hypersurfaces; the global event horizon is selected by its future boundary condition, not by this local equation alone.

For the round sphere, every normal has only $v,r$ components because the Reissner-Nordstrom metric has no mixed angular terms. The vector $X=-{\partial }_r$ is normal and null: $g_{rr}=0$. Write the second null vector normal as $Y=A{\partial }_v+B{\partial }_r$. The normalization gives $X\cdot Y=-A=-1$, so $A=1$. Its null condition then reads $-f+2B=0$. ThusThe specified future orientation of $X$, together with $X\cdot Y<0$, makes $Y$ future-directed as well. This choice fixes the reciprocal scaling freedom of the two null vectors on the sphere.

For radial motion, take an affine parameter $\lambda$ and use dots for $d/d\lambda$. The radial null geodesics of the Vaidya metric satisfy the two radial geodesic equationsThe radial squared norm is $-f{\dot{v}}^2+2\dot{v}\dot{r}$. If $v$ is constant it vanishes, and the geodesic equation reduces to $\ddot{r}=0$. Thus constant-$v$ radial curves are null geodesics, with $r=a\lambda +b$ an affine parametrization. Future ingoing motion has $a<0$.

For the other family let $dr/dv=f/2$. Its tangent $k={\partial }_v+(f/2){\partial }_r$ is a null vector. Using the Christoffel symbols of the Vaidya metric, direct differentiation givesHence these are also null geodesics, but $v$ is generally not affine. For completeness, put $p=dv/d\lambda$ and chooseThis makes $\ddot{v}=-(M/r^2){\dot{v}}^2$. Differentiating $\dot{r}=(f/2)\dot{v}$ then gives $\ddot{r}=(f_v/2){\dot{v}}^2=-M^{\prime }{\dot{v}}^2/r$; substitution verifies the radial $r$ equation as well. Thus the null condition and both radial geodesic equations hold, not just the null condition alone.

An outgoing radial null geodesic of the Vaidya metric prescribed as $r=r_0(v)$ determines the mass along that curve by $M(v)=r_0(v)(1-2r_0^{\prime }(v))/2$. This inverse relation comes from the radial null condition. It determines a function of advanced time wherever the prescribed curve is differentiable and has positive radius.

For a rigidly rotating prescribed vacuum path, use spatial coordinates attached to the apparatus and laboratory time $t$. At a path point $\mathbf{r}$, the local apparatus velocity is $\mathbf{v}=\mathbf{\Omega }\times \mathbf{r}$. A light displacement has instantaneous laboratory displacement $d\mathbf{r}+\mathbf{v}dt$, up to the common rotation of the spatial axes. The null condition is thereforeWrite $d\ell =|d\mathbf{r}|$. To first order in $|\mathbf{v}|/c$, the positive root isReversing the path reverses the dot product term but leaves the leading path-length term unchanged. Subtract the two return times and use the scalar triple product identity:This is a line integral version of the Sagnac area law; the vector area identity also follows from Stokes theorem. For a mirror-defined polygon the actual ray paths move slightly as the mirrors rotate; the stationary optical path at zero rotation gives the same first-order result. The prescribed-path calculation is an idealization, not an exact finite-speed formula for every moving-mirror layout.

The dot product explains the geometry: only the projection of the oriented vector area along the angular velocity contributes. A planar loop whose normal vector is perpendicular to $\mathbf{\Omega }$ has zero leading delay. For repeated traversals with the same orientation, the signed vector areas add; oppositely oriented lobes can cancel even when the total unsigned area is large. Equal perimeter alone does not imply equal response. A constant translational velocity contributes nothing to this leading closed-path line integral, since ${\oint }_{C}d\mathbf{r}=0$.