How Does the Coriolis Force Affect Particle Motion in a Rotating System?

[rakso]

Homework Statement

Show that the Coriolius Force can be associated with a tensor of rank 3.

Relevant Equations

$$ F_{C} = -2m \bar{\omega} \times \bar{v} $$

m = Particles mass, Omega = Systems angular frequency, v' = particles velocity.

Attempt at a Solution:

$$ F_{C} = -2m \bar{\omega} \times \bar{v}^{'} = -2 \bar{\omega} \times \bar{p} = 2 \bar{p} \times \bar{\omega} $$

Let
$$ \bar{\omega} = \frac {\bar{r} \times \bar{v}} {r^2}, \alpha = \frac {2} {r^2} $$

Then

$$ \frac {F_{C}} {\alpha} = \bar{p} \times ( \bar{r} \times \bar{v}) = \bar{e_{i}} \varepsilon_{ijk} p_{j} (\bar{r} \times \bar{v})_{k}
= \bar{e}_{i} \varepsilon_{ijk} p_{j} \varepsilon_{klm} r_{l} v_{m} = \bar{e}_{i} p_{j} r_{l} v_{m} (\delta_{il} \delta_{jm} - \delta_{im} \delta_{jl}) = \\ \bar{e}_{i} p_{j} r_{i} v_{j} - \bar{e}_{i} p_{j} r_{j} v_{i} = \bar{e}_{i} p_{j} ( r_{i} v_{j} - r_{j} v_{i}) = \bar{e}_{i} p_{j} T_{ij} $$

Where

$$ T_{ij} = r_{i} v_{j} - r_{j} v_{i} $$ is a tensor of rank 2. I don't understand how I am supposed to show it's a tensor of rank 3? Have the third rank something to do with what reference system we chose? Because I'm not quite sure the velocity of the particle is relative to the spinning system or the Newtonian one.

 

[jambaugh]

"Hey Mr Coriolis, Here's my friend Betty. She's a rank 3 tensor! You two go hang for a while and have some fun!"
...later when being interrogated by the Police..
"Mr Coriolis! Is Betty one of your associates? Remember You're under oath!"

But in all seriousness... "is associated with" is a rather vague qualification. It seems to me that there's something the inquirer is looking for but a more specific description of the question probably will give away too much toward the anser.

Looking it purely in terms of the linear algebra, the force itself is a linear function of two vectors, the angular velocity vector $\vec{\omega}$ and the test particle velocity $\vec{v}$. A mapping from two vectors to a third vector can always be expressed as a rank 3 (or more precisely rank (1,2) ) tensor.
\mathbf{f} = T[\vec{v},\vec{\omega}]
F^\alpha= T^\alpha_{\mu\lambda} v^\mu \omega^\lambda
Now since we are working in a metric space (in short since we have a dot product) we can switch between covariant and contra-variant ranks. So we can either map the force to a dual force vector:
F\bullet \vec{u} = T[ \vec{v},\vec{\omega},\vec{u}]
or map the covariant part of the rank (1,2) tensor to a contravariant equivalent,
F = T\bullet \bullet \vec{v}\vec{\omega}
(the $\bullet$ here represents the inner (dot) product which is the contraction of two vector components to scalar components via the metric or the dual equivalent. For example $\vec{v}\bullet \vec{w} = v^i g_{ij}w^j = \mathbf{g}[v,w]$)

This tensor T that I'm describing is nothing more than the combination of cross and dot products which defines the triple product of three vectors. That tensor is actually the Levi-Civita tensor for 3 dimensions. (You can better think of it as the determinant of the matrix formed by forming the component of n tensors of dimension n into an $n\times n$ matrix.) In the more general setting there tensor in question is one of the variants you might get by "raising and lowering indices", or rather by further contracting with the dot product/covariant metric and/or its dual.

I don't see (and think I can rigorously argue the impossibility of) any other relevant rank 3 tensor being involved here without bringing in some other ad hoc vectors. For example I can associate any vector with a rank 28 tensor in so far as I can use it or its dual to form a rank 28 tensor given something else with rank 27 or 29 or some 27 or 29 other somethings each with rank 1 or whatever other silly combination thereof you can imagine.

In short, by virtue of living in the full tensor algebra, any vector or tensor can "be associated with" any other vector or tensor of any rank if you're loose enough with what the heck "associated with" means. But that's a bit too loose to be practical.

In most general terms, given a homogeneous vector or tensor equation, A = T you can always express the original equations as an identity, A - T = 0 which, given it, in the most general sense, is a linear homogeneous equation can then be rewriten:
\vartheta[A,B]=0
where \vartheta is a tensor of rank rank(A)+rank(B).

(Notational note: I use brackets in functional notation, i.e. T[x] to indicate specifically homogenous linear mappings. Also a rank (n,m) tensor "can be associated with" a homogeneous linear mapping from m vectors to n vectors.)

 

[Orodruin]

A mapping from two vectors to a third vector can always be expressed as a rank 3 (or more precisely rank (1,2) ) tensor.

A linear mapping.

Fμ=Tμμλvμωλ​

A bit too many $\mu$ on the RHS. I would give you an A- on the homework (scale A-F), but it is two weeks late and the exam was a week ago (results are already reported).

 

[jambaugh]

Oops, didn't see the date... just scrolling through unanswered subs.

A linear mapping.

Right! Sorry

A bit too many $\mu$ on the RHS. I would give you an A- on the homework (scale A-F), but it is two weeks late and the exam was a week ago (results are already reported).

LOL I went to fix that, changed the first mu's to lambdas, then realized I'd already used lambda, had to change it again.