1
$\begingroup$

I have so trouble following Goldsteins derivation of the time derivative in the rotating refrence frame, and its use to derive the coriolis force (sec. 4.9-10) Given an intertial frame of refrence, $S$, and a rotating one, $S'$, contected through the transformation

$$ x_i' = a_{ij}(t)x_j, \\ x_i = a_{ij}'(t)x_j' = a_{ji} (t)x_j', $$

we can get the time derivative in the rotating refrence-frame by use of the chain rule,

$$ \frac{d}{dt}x_i = a_{ij}\frac{dx_j'}{dt}+ \frac{da_{ij}}{dt} x_j'. $$

Assuming $a_{ij}(0) = \delta_{ij}$, w can rewrite the equation as

$$ \frac{d}{dt}x_i = \frac{dx_i'}{dt} + \frac{da_{ij}}{dt} x_j. $$

$\mathrm{d} a_{ij}$ is an infinitesimal rotation, and can be writen as (Goldstein in sec. 4.8)

$$ \mathrm{d}A = \begin{bmatrix} 0 & \mathrm{d}\Omega_3 & -\mathrm{d}\Omega_2\\ -\mathrm{d}\Omega_3 & 0 & \mathrm{d}\Omega_1 \\ \mathrm{d}\Omega_2 & -\mathrm{d}\Omega_1 & 0 \end{bmatrix} = \pmb{\omega} dt, $$

so that $\mathrm{d}a_{ij} = \epsilon_{ikj} \omega_k \mathrm{d}t$. This finally gives us

$$ \frac{d}{dt}x_i = \frac{dx_i'}{dt} + \epsilon_{ikj} \omega_k x_{j} = \bigg(\frac{d}{dt}a_{ji} + \epsilon_{ikj} \omega_k \bigg)x_{j}, $$

or as Goldstein writes it,

$$ \bigg(\frac{d}{dt}\bigg)_{space} = \bigg(\frac{d}{dt}\bigg)_{body} + \omega \times. $$

He then quickly uses this operator form to get

$$ \frac{d}{dt} \pmb{r} = \pmb v = \bigg(\frac{d}{dt} \pmb{r} \bigg)_{body} + \pmb\omega \times \pmb r = \pmb v' + \pmb \omega \times\pmb r', $$

(notice the switch between unprimed and primed $\pmb r$) and then

$$ \frac{d}{dt} \pmb v = \bigg(\frac{d}{dt} \pmb v \bigg)_{body} + \pmb \omega \times \pmb v = \bigg(\frac{d}{dt} (\pmb v' + \pmb \omega \times\pmb r') \bigg)_{body} + \pmb \omega \times (\pmb v' + \pmb \omega \times\pmb r') \\ = \pmb a' + 2 \pmb \omega \times \pmb v' + \pmb \omega \times(\pmb \omega \times \pmb r'). $$

I have tried to derive this using the indecies, but to no avail. If anyone is able to this, it would be greatly apriciated. I am espeacially having trouble with the term

$$ \bigg(\frac{d}{dt} (\pmb v' + \pmb \omega \times\pmb r') \bigg)_{body}. $$

I also find Goldsteins switch between the primed and unprimed system dubious, so any clarafications on this would also help.

$\endgroup$
4
  • $\begingroup$ You are trying to derive a vector expression by components? You are brave. But seriously why not work in vector form. Vector algebra is exact, and vector identities simplify complex calculations nicely. Better brush up on vector triple products. $\endgroup$ Commented Oct 15, 2019 at 15:27
  • $\begingroup$ There is an error above. The correct differential for the transformation matrix is ${\rm d}A = ( \boldsymbol{\omega}\, {\rm d} t ) \times A$ $\endgroup$ Commented Oct 15, 2019 at 18:19
  • $\begingroup$ Related : Velocity in a turning reference frame $\endgroup$
    – Frobenius
    Commented Oct 16, 2019 at 20:54
  • $\begingroup$ Thank you, @Frobenius, that seems to be the answer. I'll have to see if i can digest it. $\endgroup$ Commented Oct 17, 2019 at 19:11

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.