Coriolius forces tensor

In summary: D-.In summary, the conversation is discussing the rank and nature of the tensor associated with the Coriolis force, which is a linear mapping from two vectors (angular velocity and particle velocity) to a third vector (force). This tensor is a rank 3 tensor, or more precisely, a rank (1,2) tensor, and can be expressed in terms of the Levi-Civita tensor. The conversation also touches on the vague concept of being "associated with" a tensor, which can be easily manipulated in the full tensor algebra.
  • #1
rakso
18
0
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.
 
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  • #2
"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.
[tex]\mathbf{f} = T[\vec{v},\vec{\omega}][/tex]
[tex]F^\alpha= T^\alpha_{\mu\lambda} v^\mu \omega^\lambda[/tex]
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:
[tex]F\bullet \vec{u} = T[ \vec{v},\vec{\omega},\vec{u}][/tex]
or map the covariant part of the rank (1,2) tensor to a contravariant equivalent,
[tex]F = T\bullet \bullet \vec{v}\vec{\omega}[/tex]
(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, [itex]A = T[/itex] you can always express the original equations as an identity, [itex]A - T = 0 [/itex] which, given it, in the most general sense, is a linear homogeneous equation can then be rewriten:
[tex]\vartheta[A,B]=0[/tex]
where [itex]\vartheta[/itex] is a tensor of rank [itex]rank(A)+rank(B)[/itex].

(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.)
 
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  • #3
jambaugh said:
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).
 
  • #4
Oops, didn't see the date... just scrolling through unanswered subs.
Orodruin said:
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.
 

What is the Coriolius forces tensor?

The Coriolius forces tensor is a mathematical representation of the Coriolis effect, which is the apparent deflection of moving objects on a rotating surface. It describes the forces acting on an object due to its motion in a rotating reference frame.

How is the Coriolius forces tensor calculated?

The Coriolius forces tensor is calculated using the cross product of the object's velocity and the angular velocity of the rotating reference frame. It is also affected by the rotation rate and the direction of the object's motion relative to the rotation axis.

What is the significance of the Coriolius forces tensor?

The Coriolius forces tensor plays a crucial role in understanding the dynamics of objects in a rotating reference frame, such as the Earth's atmosphere and ocean currents. It helps explain the observed phenomena of curved paths and deflection of objects on a rotating surface.

Does the Coriolius forces tensor have practical applications?

Yes, the Coriolius forces tensor has several practical applications in fields such as meteorology, oceanography, and aviation. It is used to predict the movement and behavior of large-scale weather systems, ocean currents, and aircraft in flight.

How does the Coriolius forces tensor affect the Earth's climate?

The Coriolius forces tensor plays a significant role in shaping the Earth's climate by influencing large-scale atmospheric and oceanic circulation patterns. It helps distribute heat and moisture around the globe, impacting weather patterns and climate zones.

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