Eric Forgy Position, Velocity, and Acceleration

See the corresponding nLab page

Warning: Under Construction

Idea

Consider three paths that begin at x 0x_0 and end at x 2τx_{2\tau} enclosing the spacetime region depicted below

x 2τ v τ v τ + x τ F v 0 F + x τ + v 0 v 0 + x 0 \array{ {} & {} & {} & x_{2\tau} & {} & {} & {} \\ {} & {} & {} & \bullet & {} & {} & {} \\ {} & {} & {}^{v^-_{\tau}}\nearrow & {} & \nwarrow^{v_{\tau}^+} & {} & {} \\ x^-_{\tau} & \bullet & \stackrel{F^-}{\Uparrow} & \stackrel{v_0}{\uparr} & \stackrel{F^+}{\Uparrow} & \bullet & x^+_{\tau} \\ {} & {} & {}_{v_0^-}\nwarrow & {} & \nearrow_{v_0^+} & {} & {} \\ {} & {} & {} & \bullet & {} & {} & {} \\ {} & {} & {} & x_0 & {} & {} & {} }

Staring at the above picture, we want to understand the analogy:

category functor naturaltransformation fiber connection curvature position velocity acceleration\array{ category & functor & natural transformation \\ fiber & connection & curvature \\ position & velocity & acceleration }

For those who better understand the mathematics of the first line, this analogy is likely to help motivate the physics as we go down the ladder. For those who better understand the physics of the bottom line, it is hoped that this analogy will help to motivate the mathematics as we move up the ladder.

Formulation

First, due to the symmetry of the diagram above, we can consider the simplified region

x τ v 0 F v τ x 0 v x 2τ\array{ {} & {} & {} & x_{\tau} & {} & {} & {} \\ {} & {} & {} & \bullet & {} & {} & {} \\ {} & {} & {}^{v_0}\nearrow & {}_F\Rightarrow & \searrow^{v_{\tau}} & {} & {} \\ x_0 & \bullet & {} & \stackrel{v}{\longrightarrow} & {} & \bullet & x_{2\tau} }

This diagram represents a very simple strict 2-category we will denote by P 2(𝔻 2)P^2(\mathbb{D}^2) with

We will explicitly construct a 2-transport functor

tra:P 2(𝔻 2)VecMod,tra: P^2(\mathbb{D}^2)\to Vec-Mod,

where

xtra(x) 2x\mapsto tra(x)\in\mathcal{M}^2

Since both paths end up at the same point, the velocities must satisfy the relation

v Δt v 0 =v 0 +v Δt +,v^-_{\Delta t} - v^-_0 = v^+_0 - v^+_{\Delta t},

which implies that we can write

v Δt =v 0 +(+a)Δtv^-_{\Delta t} = v^-_0 + (+a)\Delta t

and

v Δt +=v 0 ++(a)Δtv^+_{\Delta t} = v^+_0 + (-a) \Delta t

for some constant aa.


Let VV denote the 4-velocity of a charged particle and vv denote the corresponding dual 1-form, the acceleration 1-form may be written as

a=qmi VFa=\frac{q}{m} i_V F

where FF is the curvature 2-form and the “acceleration 2-form” (corresponding to the acceleration bivector) is

va=qmv(i VF).v\wedge a=\frac{q}{m} v\wedge(i_V F).

I guess this is equal to FF. To see this, note that

i V(vF)=(i Vv)Fv(i VF).i_V(v\wedge F) = (i_V v)F - v\wedge(i_V F).

We have

i Vv=v(V)=|V| 2=1i_V v = v(V) = |V|^2 = 1

and

i V(vF)=0i_V(v\wedge F)=0

so that

F=v(i VF)=mqva.F=v\wedge(i_V F)=\frac{m}{q} v\wedge a.

I believe this is the clue that Urs mentioned, but made a bit more explicit. On the left is curvature and the right is the acceleration 2-form.

In my Galilean spacetime example, I think it is the “acceleration 2-form” vav\wedge a that specifies the 2-transport even in the absence of an electromagentic field.


Relation to Discrete Calculus

See page 68.

e ije jl=+e ijle_{ij} e_{jl} = +e_{ijl}
{}
e ike kl=e ijle_{ik} e_{kl} = -e_{ijl}
{}
H=exp(v 0 +Δt)e ij+exp(v Δt +Δt)e jl+exp(v 0 Δt)e ik+exp(v Δt Δt)e kl\mathbf{H} = \exp(v^+_0\Delta t) e_{ij} + \exp(-v^+_{\Delta t} \Delta t) e_{jl} + \exp(-v^-_0\Delta t) e_{ik} + \exp(v^-_{\Delta t} \Delta t) e_{kl}
{}
H 2=exp[(v 0 +v Δt +)Δt]e ijlexp[(v 0 v Δt )Δt]e ijl\mathbf{H}^2 = \exp[(v^+_0 - v^+_{\Delta t}) \Delta t] e_{ijl} - \exp[-(v^-_0 - v^-_{\Delta t}) \Delta t] e_{ijl}
{}
e l e kl e jl e k e ijl e j e ik e ij e i \array{ {} & {} & {} & e_l & {} & {} & {} \\ {} & {} & {} & \bullet & {} & {} & {} \\ {} & {} & {}^{e_{kl}}\nearrow & {} & \nwarrow^{e_{jl}} & {} & {} \\ e_k & \bullet & {} & \stackrel{e_{ijl}}{\Leftarrow} & {} & \bullet & e_j \\ {} & {} & {}_{e_{ik}}\nwarrow & {} & \nearrow_{e_{ij}} & {} & {} \\ {} & {} & {} & \bullet & {} & {} & {} \\ {} & {} & {} & e_i & {} & {} & {} }

dx=vdtdx = v dt
dv=adtdv = a dt
F=adxdtF = a dx\wedge dt
i + xF=+adti_{+\partial_x} F = +a dt
i xF=adti_{-\partial_x} F = -a dt
dv=i sFdv = i_{\partial_s} F

Consider two paths γ +,γ \gamma^+,\gamma^- on a directed graph GG that both begin at the same point and end at the point as depicted in the figure below

γ 1 ± exp(v t ) exp(v t +) γ t exp(a) γ t + exp(v 0 ) exp(v 0 +) γ 0 ± \array{ {} & {} & \gamma^\pm_1 & {} \\ {} & {} & \bullet & {} \\ {} & {}^{\exp(v^-_t)}\nearrow & {} & \nwarrow^{\exp(v_t^+)} \\ \gamma^-_t & \bullet & \stackrel{\exp(a)}{\Rightarrow} & \bullet & \gamma^+_t \\ {} & {}_{\exp(v_0^-)}\nwarrow & {} & \nearrow_{\exp(v_0^+)} \\ {} & {} & \bullet & {} \\ {} & {} & \gamma^\pm_0 & {} }

The edges represent velocities and the area enclosed represents the acceleration.

If the acceleration is zero, then the area collapses and any two paths that connect the same two points must correspond to the same straight line, i.e. velocity does not change.


Given an interval II and a smooth manifold MM, consider a smooth curve

γ:II×M\gamma: I\to I\times M

to be a section of the trivial bundle

π:I×MI\pi: I\times M\to I

treating II as the base space and denote γ t=γ(t)\gamma_t = \gamma(t) and M t=π 1(t)M_t = \pi^{-1}(t).


{}
γ *(ddt| t)=X tT γ t(M)\gamma_* \left(\left. \frac{d}{dt}\, \right|_t \right) = X_t \in T_{\gamma_t}(M)
{}
γ 1=Pexp( γA)γ 0\gamma_1 = Pexp(\int_\gamma A) \gamma_0

category: drafts

Revised on February 9, 2009 at 18:55:13 by Eric Forgy