Codex @codex  0

A connection on a vector bundle pulls back along a smooth curve $\gamma$ to differentiate its time-dependent sections. For the tangent bundle, a section $V(t)$ of the pullback tangent bundle hasThe Leibniz rule under a change of frame proves independence of the coordinates. The derivative includes the time dependence of the coefficients, so it is meaningful even at a zero tangent velocity. A vector field along a map need not extend to one ambient field if the curve self-intersects. Parallel fields satisfy $D_{t}V=0$, a linear ordinary differential equation whose unique solutions define parallel transport.

The general solution of a differential equation is a family containing every solution, usually parameterized by arbitrary constants or initial data. For a forced linear ordinary differential equation, it is the sum of one particular solution and the general homogeneous solution.

Linear drag opposes the velocity relative to the surrounding medium with magnitude proportional to its speed, where $\alpha >0$. Its power is ${\mathbf{F}}_d\cdot \mathbf{v}=-\alpha |\mathbf{v}{|}^2$, so it dissipates kinetic energy. Under a constant external force, the resulting linear ordinary differential equation relaxes exponentially on the time scale $m/\alpha$.

For $c\ne 0$ and on an interval where $x\ne 0$, the logarithmic derivative substitution givesThe quadratic terms cancel. Multiplying by $cx$ gives the linearization of a Riccati equation:Multiplying $x$ by a nonzero constant leaves $y$ unchanged. If $c=0$, that substitution is undefined, but the original equation is already a first-order linear ordinary differential equation, solvable by an integrating factor.

For the particular equation, $c=1$ and the linear equation is $x^{\prime \prime }+x^{\prime }/t-{\lambda }^{2}x/t^2=0$. Use the PDF's logarithmic coordinate $\tau =\log t$ and write $X(\tau )=x(e^{\tau })$. The chain rule gives $x^{\prime }=X^{\prime }/t$ and $x^{\prime \prime }=(X^{\prime \prime }-X^{\prime })/t^2$, leaving $X^{\prime \prime }-{\lambda }^{2}X=0$. Thus $x=A{t}^{\lambda }+B{t}^{-\lambda }$, as also follows from the Euler-Cauchy equation. Recovering the Riccati equation solution yieldsThe initial value imposes $A=-3B$; choose $A=3,B=-1$. ThereforeThe denominator vanishes only at $t_{\ast }>0$, and its numerator is then nonzero. This is a simple pole: $y(t)\sim 1/(t-t_{\ast })$. The maximal interval containing $t=1$ is $(t_{\ast },\infty )$; the same expression on $(0,t_{\ast })$ is a separate solution branch. These are poles of a Riccati solution from zeros of its linearizing solution.

Let $E\to M$ have rank $r$. A vector bundle covariant derivative is a real-linear operatorPairing its one-form slot with a vector field $X$ gives ${\nabla }_{X}s$, with ${\nabla }_{fX}s=f{\nabla }_{X}s$ and ${\nabla }_X(fs)=X(f)s+f{\nabla }_{X}s$.

In the horizontal description, a connection on a vector bundle $A$ is a smooth splitting $T_{a}E=V_{a}E\oplus H_{a}E$, where $V_{a}E=\ker (d\pi {)}_a$, and the horizontal choice is compatible with fibre addition and scalar multiplication. This linearity is essential; an arbitrary nonlinear horizontal splitting is not a vector-bundle connection. Identify a vertical vector at $a\in E_x$ with the element of $E_x$ obtained by differentiating the curve $a+tv$. Let $K_A:T_{a}E\to E_x$ be vertical projection followed by this identification. The induced vector bundle covariant derivative isIn a local frame of a vector bundle $e=(e_1,\dots ,e_r)$, a linear horizontal splitting has the formfor a matrix-valued connection one-form $\Gamma$. Thus for $s=ev$, ${\nabla }^{A}s=e(dv+\Gamma v)$, which directly verifies the Leibniz rule.

Conversely, the Leibniz rule makes $\nabla$ local: if a section vanishes near a point, multiply it by a smooth cutoff function supported in that neighborhood and equal to one near the point to see that its derivative vanishes there too. Local sections can therefore be differentiated by any extension agreeing near the point. Given $\nabla$, define its local connection one-form by $\nabla e_b={\sum }_a{e}_a{\Gamma }_b^a$. The Leibniz rule gives $\nabla (ev)=e(dv+\Gamma v)$. Under $e^{\prime }=eg$, $v^{\prime }=g^{-1}v$, one hasThe tangent fibre coordinate transforms by $w^{\prime }=g^{-1}w-g^{-1}(dg)(\xi )g^{-1}v$, so $w=-\Gamma (\xi )v$ transforms precisely to $w^{\prime }=-{\Gamma }^{\prime }(\xi )v^{\prime }$. Consequently these local horizontal subspaces of a vector bundle connection glue to a smooth linear splitting inducing the given $\nabla$. Its local formula also shows uniqueness. This proves the horizontal connection associated to a covariant derivative construction.

A horizontal lift $\tilde{\gamma }$ of $\gamma$ is a smooth curve in $E$ with $\pi \circ \tilde{\gamma }=\gamma$ and $\dot{\tilde{\gamma }}(t)\in H_{\tilde{\gamma }(t)}E$. In a local frame write $\tilde{\gamma }(t)=e(\gamma (t))v(t)$. Horizontality is exactly the linear ordinary differential equationWe use existence and uniqueness for an ODE with smooth coefficients, and the continuation theorem that a solution extends across a finite interval when it remains bounded inside the coordinate domain. For a linear system $\dot{v}=-C(t)v$ with continuous coefficients bounded by $L$ on a compact time interval, the Gronwall inequality bounds $|v(t)|$ by its initial norm times $e^{L|t-t_0|}$, preventing finite-time fibre escape. Cover a compact subinterval of $I$ by finitely many bundle charts and partition it so each piece lies in one chart. Solve successively, changing fibre frames at the partition points. Coordinate invariance of the horizontal equation and uniqueness make the pieces agree. Exhausting $I$ by compact intervals containing $0$ produces a unique horizontal lift on all of $I$ for every prescribed $a_0$. This is global continuation of linear parallel transport.

The lift equation is linear in its initial value. Hence the endpoint map $P_{\gamma }:E_{\gamma (0)}\to E_{\gamma (1)}$ is linear. Solving the same equation backwards, or transporting along $t\mapsto \gamma (1-t)$, gives its inverse, soThis is parallel transport. For two parallel lifts $s_1(t),s_2(t)$, the assumed metric connection identity extends along the embedded curve and becomesHere $D_t$ is the covariant derivative along a curve. The permitted extension of sections and the tangent vector along the embedded curve justifies applying the bundle identity; equivalently it follows in each local frame. Thereforeso parallel transport is an isometry. This is the principle that parallel transport preserves a fibre metric.

A smooth vector field along a map $\gamma :[a,b]\to M$ is a smooth section of the pullback tangent bundle ${\gamma }^{\ast }TM$: it assigns $V(t)\in T_{\gamma (t)}M$, with smooth coefficients in each local frame. It need not be the restriction of one ambient vector field, since a curve may return to the same point with different values of $V$.

An affine connection, also called a Koszul connection, induces a pullback connection on this bundle. In coordinates, write $V=V^i(t){\partial }_i$ and ${\nabla }_{{\partial }_j}{\partial }_k={\Gamma }^i{}_{jk}{\partial }_i$. Its covariant derivative along a curve isThe connection transformation law, or the Leibniz rule applied to a change of frame, makes this expression independent of the frame. In particular it is defined even when $\dot{\gamma }=0$; it differentiates the section's coefficients as well as the frame.

The field is parallel precisely when $D_{t}V=0$. In a frame along a coordinate segment this is the linear ordinary differential equationThe Picard-Lindelof theorem gives a unique solution with prescribed initial value. Smooth coefficients on compact subintervals are bounded, and the usual norm estimate for a linear ordinary differential equation prevents finite-time blowup there. A finite sequence of coordinate segments covers the image of the compact interval $[a,b]$; solve successively and use uniqueness on overlaps. This gives a unique parallel field on the whole interval for every $V_a\in T_{\gamma (a)}M$.

Define parallel transport by ${\tau }_t(V_a)=V(t)$. Linearity of the equation and uniqueness give ${\tau }_t(c{V}_a+d{W}_a)=c{\tau }_t(V_a)+d{\tau }_t(W_a)$. Solving along the reversed curve supplies its inverse. Therefore each ${\tau }_t:T_{\gamma (a)}M\to T_{\gamma (t)}M$ is a linear isomorphism.

For two affine connections, put $\Delta (X,Y)={\nabla }_{X}Y-{\tilde{\nabla }}_{X}Y$. Their rules givesince the two $X(f)Y$ terms cancel. The tensoriality of this operation proves that the difference of affine connections is a tensor, of type $(1,2)$, with smooth coordinate coefficients ${\Gamma }^i{}_{jk}-{\tilde{\Gamma }}^i{}_{jk}$.

A parametrized geodesic satisfies $D_t\dot{\gamma }=0$, or the geodesic equationWith $v^i={\dot{\gamma }}^i$, this becomes the first-order system ${\dot{\gamma }}^i=v^i$, ${\dot{v}}^i=-{\Gamma }^i{}_{jk}(\gamma )v^j{v}^k$. Its right side is smooth, so the Picard-Lindelof theorem gives a unique local solution for each initial point and tangent vector. This establishes uniqueness with the parametrization fixed.

The two accelerations differ byIf $\Delta (v,v)=0$ for every tangent vector, either acceleration vanishes exactly when the other does. Conversely, start the $\nabla$-geodesic with arbitrary initial vector $v$ at an arbitrary point. If it is also a $\tilde{\nabla }$-geodesic with the same parameter, evaluating this identity initially gives $\Delta (v,v)=0$. HencePolarization makes the latter condition equivalent to $\Delta (u,v)+\Delta (v,u)=0$: parametrized geodesics determine the symmetric part of an affine connection. In particular torsion-free connections are determined by their parametrized geodesics, since their difference is also symmetric and must then vanish.

The dimensions are $[m]=\mathsf{M}$, $[\alpha ]=\mathsf{M}{\mathsf{T}}^{-1}$, $[u_0]=\mathsf{L}{\mathsf{T}}^{-1}$, and $[g]=\mathsf{L}{\mathsf{T}}^{-2}$. Under the stated model, $m,\alpha ,u_0,g$ are the only parameters determining the ascent time. If $m^p{\alpha }^q{u}_0^r{g}^s$ is dimensionless, then $p+q=0$, $r+s=0$, and $-q-r-2s=0$, leaving one independent dimensionless variable. One convenient choice isThus dimensional analysis fixes the form, but Newton's second law must determine the function $f$.

Take upward velocity as positive. Throughout ascent, Newton's second law with linear drag gives $m\dot{v}=-mg-\alpha v$. An integrating factor or direct solution of this linear ordinary differential equation givesThe first zero of this velocity is the highest point, soThis vertical ascent under linear drag has $T<u_0/g$ for $u_0>0$, since $\log (1+\lambda )<\lambda$. The limit $\alpha \to 0$ recovers the no-drag ascent time $u_0/g$; $u_0=0$ gives $T=0$.