Acceleration in arbitrary trajectory (differential geometry)

1. Jan 21, 2014

Zatman

1. The problem statement, all variables and given/known data
Show that for any trajectory r(t) the acceleration can be written as:

$\mathbf{a}(t)=\frac{dv}{dt}\hat{T}(t)+\frac{v^2}{\rho}\hat{N}(t)$

where v is the speed, T is a unit vector tangential to r and N is a unit vector perpendicular to T, at time t. rho is the radius of curvature.

2. The attempt at a solution

$\mathbf{a}=\frac{d\mathbf{v}}{dt}=\frac{d}{dt}(v\hat{T})$

$=\frac{dv}{dt}\hat{T}+v\frac{d\hat{T}}{dt}$

Since we can write

$\frac{1}{\rho}\hat{N}=\frac{d\hat{T}}{dt}$

This gives

$\mathbf{a}=\frac{dv}{dt}\hat{T}+\frac{v}{\rho}\hat{N}$

I can't see where to get the other 'v' from, in the second term.

2. Jan 21, 2014

LCKurtz

$v\frac{d\hat{T}}{dt}= v \frac{d\hat T}{ds}\frac{ds}{dt}= v^2\frac{d\hat T}{ds}$. But the curvature $\kappa$ is $\kappa = \left|\frac{d\hat T}{ds}\right|$ so $v^2\frac{d\hat T}{ds} = v^2\kappa \hat N(t) = \frac{v^2}{\rho}\hat N(t)$.

3. Jan 21, 2014

Zatman

Got it, thanks.

Missed that it is the derivative of $\hat{T}$ with respect to arc length that measures curvature.