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The derivative of an option's value with respect to calendar time at fixed current underlying price and other parameters. In the dividend-free Black-Scholes model, the Black-Scholes equation givesThe first term is the interest earned on the bond value of the replicating portfolio, while the second uses option gamma. A convex payoff with a short bond position has $\Theta \le 0$ when $\rho \ge 0$; a concave payoff with a long bond position has $\Theta \ge 0$. Negative interest rates need not obey these signs: the affine payoff $x-K$ has $\Theta =-\rho K{e}^{-\rho (T-t)}$. For a fixed payoff function, remaining-maturity sensitivity is $-\Theta$.

An inertial observer of a fixed-$y$ flat brane in metric $d{y}^2+e^{-2y/L}{\eta }_{\mu \nu }d{x}^{\mu }d{x}^{\nu }$ is accelerated relative to the higher-dimensional bulk. Normalization and ${\Gamma }_{\mu \nu }^y=e^{-2y/L}{\eta }_{\mu \nu }/L$ give $a^y=-1/L$. This is its invariant bulk proper acceleration, not a nonzero intrinsic four-dimensional acceleration on the flat induced brane metric.

For a fixed half-angle $0<\alpha <\pi /2$, a slow slender gravity current in a V-shaped channel has an almost horizontal free surface across each section. Its area is $h^2\tan \alpha$, and its axial volume flux is $-gQ(\alpha )h^4{h}_x/\nu$. Here $Q$ is the integral of the dimensionless cross-sectional Poisson equation solution: $\Delta \varphi =-1$, with zero velocity on the two rigid slopes and zero normal velocity derivative on the horizontal shear-free surface. Conservation of mass gives $C=gQ/(\nu \tan \alpha )$. This treats the full two-dimensional section and does not require a shallow V angle.

A shallow radially spreading viscous drop on a horizontal plane has lubrication flux proportional to $-h^3{h}_r$. Absorbing the positive dimensional coefficient into time gives the displayed equation. For a finite-volume drop, $M={\int }_0^{R}rhdr$ is constant when the origin and front carry zero volume flux. A similarity form $h=t^{-\beta }f(r/t^{\alpha })$ has $2\alpha =\beta$ from volume and $\beta +1=4\beta +2\alpha$ from the equation, giving $\alpha =1/8$ and $\beta =1/4$.

A thin layer of very viscous liquid floating on an effectively inviscid denser liquid has negligible shear traction on both faces. Its leading horizontal velocity is a plug, and incompressibility gives vertical strain $w_z=-u_x$. Normal-stress balance shifts the pressure by $-2\mu u_x$, so the longitudinal excess stress is $4\mu u_x$. Combining depth-integrated momentum balance with hydrostatic reduced gravity gives the displayed extensional equations. This is distinct from a no-slip, shear-driven thin-film gravity current.

For a prismatic channel of width $b(z)=z$, a Boussinesq current of depth $h$ occupies area $A=h^2/2$. Ambient fluid at rest enters across the interface at speed $w_e$, supplying volume but no horizontal momentum or excess density. The volume conservation, momentum and excess-density balances areHere $g^{\prime }$ is the reduced gravity. With $D={\partial }_t+u{\partial }_x$, these give $Dh+(h/2)u_x=w_e$, $Du+g^{\prime }{h}_x+(h/3)g_x^{\prime }=-2u{w}_e/h$, and $D{g}^{\prime }=-2g^{\prime }{w}_e/h$. The momentum source in primitive form describes acceleration of newly entrained fluid, even when bed drag vanishes.

In a prismatic triangular channel with constant reduced gravity, the wave speed is $c=\sqrt{g^{\prime }h/2}$ and the Riemann invariants are $u\pm 4c$. A dam release into a deep ambient produces a simple-wave rarefaction with $u+4c=4c_0$. The imposed gravity-current front condition $u_f=F\sqrt{g^{\prime }{h}_f}$ gives the displayed uniform nose-adjacent state. The fan has $c=(4c_0-x/t)/5$ and $u=4(c_0+x/t)/5$, extending from $x/t=-c_0$ to $u_f-c_f$; a uniform region then reaches the front $x/t=u_f$. The nose closure supplies physics absent from the one-layer interior equations.

A semi-infinite resting layer of depth $H$ and constant reduced gravity $b$ released into a deep ambient develops a rarefaction wave satisfying $u+2\sqrt{bh}=2\sqrt{bH}$. The head closure $u_f=F\sqrt{b{h}_f}$ gives $h_f=4H/(F+2{)}^2$ and $u_f=2F\sqrt{bH}/(F+2)$. A constant-depth shelf lies between the rarefaction tail speed $u_f-\sqrt{b{h}_f}$ and the front speed $u_f$; its width is positive for every finite positive $F$.

For a one-sided fixed-volume gravity current in a unit-width channel with a closed rear wall, use front closure $\dot{L}=F\sqrt{g^{\prime }{h}_f}$ with $0<F<2$. An exact late-time similarity solution has $u=\dot{L}x/L$, $h=h_f[1-F^2/4+(F^2/4)(x/L{)}^2]$ and $h_{f}L=V/(1-F^2/6)$. Substituting into the shallow water equations proves the radius law. The Benjamin deep-ambient front condition has $F=\sqrt{2}$. The virtual time origin $t_v$ is fixed by matching to the finite release, not by volume alone.

For identical point masses of total mass $M$, let $W=-G{M}^2/R_g$ exclude self-interactions. The virial theorem gives $⟨v_{\mathrm{eq}}^2⟩=GM/R_g$. Since the sum of individual potential energies counts each pair twice, the mass-weighted mean frozen-potential escape velocity squared is $-4W/M$. A dissipation-free collapse from zero total energy instead has kinetic energy $-W$, hence mean squared speed $-2W/M$. These are three-dimensional speeds relative to the centre of mass; a coherent collapse speed is not a local random velocity dispersion.

The microlensing optical depth is the probability, when small, that a source lies within one Einstein radius of a foreground lens. For point lenses and compact-object mass density ${\rho }_l$ in nearby Euclidean geometry,The individual lens mass cancels because $n_l={\rho }_l/m$ and $R_E^2\propto m$. Detection efficiencies and event durations still depend on individual masses and survey cadence.

For a razor-thin isothermal gas disk,Axisymmetric disturbances are stable when $Q\ge 1$: pressure stabilizes short wavelengths, epicyclic motion stabilizes long wavelengths, and self-gravity drives intermediate wavelengths.

Vertices are $k$-spaces of a finite vector space, with edges when intersection dimension is $k-1$. Basis exchanges give a path of length $k-\dim (A\cap B)$. Each edge changes intersection dimension with a fixed space by at most one, proving the lower bound. Equal-intersection ordered pairs are linearly conjugate, making the graph distance-transitive.

For $0<d_1<\cdots <d_m<n$, a flag is a chain of complex subspaces $L_1\subset \cdots \subset L_m\subset V$ of the given dimensions. Successively choosing the quotient subspaces gives complex dimension ${\sum }_{i=1}^m(d_i-d_{i-1})(n-d_i)$ with $d_0=0$. Adapted bases identify it with a homogeneous space of the special linear group, with stabilizer an appropriate block upper-triangular subgroup. The twistor correspondence uses $F_{1,2}({\mathbb{C}}^4)$.

The Plücker embedding maps a $k$-plane to the projective class of the exterior product of a basis. A change of basis multiplies this product by its determinant, so the projective class is well defined. The plane is recovered as the vectors whose exterior product with this element is zero. For two-planes in ${\mathbb{C}}^4$, its image in ${\mathbb{CP}}^5$ is the Klein quadric.

For a finite free module $M$ of rank $n$, this scheme represents rank-$r$ locally free module quotients of $M$ after base change, with quotient maps identified up to commuting isomorphism. It carries a universal exact sequence $0\to \mathcal{S}\to M\otimes \mathcal{O}\to \mathcal{Q}\to 0$, of ranks $n-r,n,r$. Its standard affine charts of the quotient Grassmannian prove representability over any base ring. Over a field, taking kernels gives the usual Grassmannian of $(n-r)$-planes.

// Target: ringed-space.bigb

In the real Grassmann-valued classical supersymmetric mechanics variant, impose $\dot{x}=U(x)$ and ${\psi }_2=-{\psi }_1$. Then ${\psi }_1{\psi }_2=0$, and differentiating the first-order equation gives $\ddot{x}=U{U}^{\prime }$. The remaining Euler-Lagrange equations reduce to ${\dot{\psi }}_1=-U^{\prime }{\psi }_1$. These solutions have $E=0$ but $Q_1=2U{\psi }_1=-Q_2$, which need not vanish: $d(U{\psi }_1)/dt=U^{\prime }U{\psi }_1-U{U}^{\prime }{\psi }_1=0$.

For $U=1+cx$, one obtains $x(t)=e^{ct}{x}_0+(e^{ct}-1)/c$ and ${\psi }_1(t)=e^{-ct}{\psi }_{10}$ when $c\ne 0$. At $c=0$, the limiting solution is $x(t)=x_0+t$, ${\psi }_1(t)={\psi }_{10}$. In either case $Q_1=2(1+c{x}_0){\psi }_{10}$.

Integrating finite Grassmann variables out of a quadratic action gives $\det [B-gC(x)]$. If $B$ is invertible, divide by $\det B$ and expand its logarithm to obtain $V_{\mathrm{eff}}={\sum }_{m\ge 1}{g}^m\mathrm{Tr}[B^{-1}C(x){]}^m/m$. Since the bosonic variables commute, an $m$-leg effective Feynman rule uses the fully symmetrized trace tensor, with weight $-g^m(m-1)!\mathrm{Sym}\mathrm{Tr}[B^{-1}{C}^{(k_1)}\cdots B^{-1}{C}^{(k_m)}]$. The minus sign comes from $e^{-V_{\mathrm{eff}}}$. This organizes closed fermion loops into bosonic vertices. It is a formal perturbative identity; a singular $B$ prevents the normalized expansion, even though the unfactored determinant polynomial still exists.

Use the Berezin integral orientation $\int {\prod }_{i}d{\bar{\theta }}_{i}d{\theta }_i{\theta }_1{\bar{\theta }}_1\cdots {\theta }_n{\bar{\theta }}_n=1$. For an invertible matrix $B$, the normalized weight $e^{-\bar{\theta }B\theta }$ has $⟨{\theta }_i{\bar{\theta }}_j⟩=(B^{-1}{)}_{ij}$. Differentiate its determinant with respect to $B_{ji}$ to obtain this formula; two derivatives give the displayed four-point formula. The second pairing carries a fermionic sign. These are ordered contractions, so exchanging two odd factors changes their sign.

For an antisymmetric matrix $A$, a Berezin integral extracts the top coefficient of the exponential. Expansion into pairings gives the Pfaffian, with a constant sign depending on the order of the differentials. This is a polynomial in the entries of $A$, not the positive square root of its determinant: in dimension two it is proportional to $A_{12}$ and changes sign with that entry. The identity $\mathrm{Pf}(A{)}^2=\det A$ determines its square but does not discard its orientation.