Question

Deriving Coriolis force using Jacobian?

Answer 1

@Kainui

Answer 2

Do you have a single idea how to do that? I am totally confused with what the lecturer wrote in his notes.

Answer 3

As far as I can tell, all fictitious force arise from an accelerating frame of reference and hence has nothing to do with physics but is purely mathematical in nature.

Answer 4

I haven't played with the Coriolis force much, but I could guess how it's done. I believe it's a pseudo force as a result of being in a nonintertial reference frame. I don't know, what's the difference between coriolis force and centrifugal force? I think if I knew what it's supposed to be I could derive it haha

Answer 5

Both arise naturally from changing reference frames and taking derivatives.

Answer 6

If we work in polar coordinates, what would be the Jacobian for changing components or changing basis? Clearly if you specified one the other is automatically fixed.

Answer 7

We were taught that \[
\underline{r}=\cos(\theta)\underline{i}+\sin(\theta)\underline{j}\\
\underline{\hat{\theta}}=-\sin(\theta)\underline{i}+\cos(\theta)\underline{j}
\]
Which is the component and which is the basis??????? In other words, tensors are important.

Answer 8

The underline denotes that it should be covariant but I don't see how the right hand side is covariant at all.

Answer 9

The derivation given by my professor is this:
\[
\underline{v}=v_x\underline{i}+v_y\underline{j}+v_z\underline{k}=v_x'\underline{i'}+v_y\underline{j'}+v_z\underline{k'}\\
\begin{align*}
&\phantom{{}={}}\frac{d\underline{v}}{dt}=\frac{dv_x'}{dt}\underline{i'}+\frac{dv_y'}{dt}\underline{j'}+\frac{dv_z'}{dt}\underline{k'}+v_x'\frac{d\underline{i'}}{dt}+v_y'\frac{d\underline{j'}}{dt}+v_z'\frac{d\underline{k'}}{dt}\\
&=\frac{dv_x'}{dt}\underline{i'}+\frac{dv_y'}{dt}\underline{j'}+\frac{dv_z'}{dt}\underline{k'}+\underline{\omega}\times \underline{v}\\
&=\left.\frac{d\underline{v}}{dt}\right|_\text{apparent}+\underline{\omega}\times \underline{v}\\
\end{align*}
\]
Do this again and you will get two terms, one is Coriolis and one is centrifugal.

Answer 10

\(\underline{v}\) is some random vector not necessarily velocity.

Answer 11

\[
\frac{d\underline{i'}}{dt}=\underline{\omega}\times\underline{i'}
\]
and similarly for the other two basis. \(\underline{\omega}\) is the angular velocity vector.

Answer 12

\(\underline{\omega}\) is assumed to be constant in my course. If \(\underline{\omega}\) is not constant then you will have an extra term called Euler force.

Answer 13

Are you dead...?

Answer 14

Yeah haha. Ok so the professor wrote this out, what's the very first line of this that gives you trouble?

Answer 15

\[
\underline{r}=\cos(\theta)\underline{i}+\sin(\theta)\underline{j}\\
\underline{\hat{\theta}}=-\sin(\theta)\underline{i}+\cos(\theta)\underline{j}
\]

Here, this is your change of basis from i and j to r and theta, you can write this as a matrix and that'll be your Jacobian between these two basis sets, literally just rewriting the same system equations as one matrix equation:

\[\begin{pmatrix} r \\ \theta \end{pmatrix} = \begin{pmatrix} \cos \theta & \sin \theta \\ - \sin \theta & \cos \theta \end{pmatrix} \begin{pmatrix} i \\ j \end{pmatrix}\]

Answer 16

How am I suppose to take the time derivative of this thing?

Answer 17

Suppose my reference frame rotates at a constant angular velocity of \(\omega\), then \[
\begin{pmatrix}
\underline{r}\\
\underline{\theta}
\end{pmatrix}=
\begin{pmatrix}
\cos(\omega t) & \sin(\omega t)\\
-sin(\omega t)& \cos(\omega t)
\end{pmatrix}
\begin{pmatrix}
\underline{i}\\
\underline{j}
\end{pmatrix}
\]Is this correct? And are those vectors or components?

Answer 18

Well it seems that your class is using a different sorta set of notation and polar coordinates than I'm accustomed to. But yeah, the things with underlines are vectors. Specifically they're your basis vectors.

Show me how your class writes Jacobians and maybe any other stuff you think might be useful and I can do it the way you are expected to do it so there's no confusion.

Answer 19

My lecturer did not use Jacobian in the derivation and I am so confused because I don't know which is the component and which is the basis.

Answer 20

The Jacobian transforms all covariant objects the same way, so it really doesn't matter what you are multiplying by at any rate, whether you're transforming the covariant basis vectors or some covariant components. To transform contravariant components (which is what has to be paired with covariant basis vectors to get an invariant, the point of tensor calculus) you have to transform by the inverse matrix. In this case, I recognized this matrix immediately since it's the rotation matrix so its inverse ends up being its transpose (This is called an orthogonal matrix, probably not too useful to you to know this right now but w/e I'll say it anyways)

Answer 21

Have you ever taken the cross product using a determinant?

Answer 22

You know how the top row is vectors and the next two rows are vector components? It's just like that, except it's going to start happening more often where you contract vectors and scalars together. This is a good thing because the framework becomes more general and you don't have to worry so much about the distinction between scalars, vectors, matrices, etc... because you can have invariant scalars and invariant vectors.

Answer 23

I routinely use determinant to calculate cross products. The problem is I got the components and the basis confused because of the notation. Now I don't even know which is contravariant and which is covariant.

Answer 24

For example, our notation for a vector in polar coordinates is \(r\underline{r}\). It is not that hard to see why I got confused.

Answer 25

What notation is commonly used in mathematics? I got drove to insanity again by Windows 10 just now... WHY MICROSOFT?????

Answer 26

What used to take me 1 minute in Windows 7 now takes me an hour to do in Windows 10 because they disabled the option in GUI and I have to google for 50 minutes to find the right registry key and change it.

Answer 27

Haha I could see why that's confusing and when I saw your version of polar coordinates I kinda frowned a bit since it didn't include the radius as part of the radial component itself. Nevertheless it's not that bad. There's like a ton of different notation to mean the same thing and I don't know why but I think everyone eventually detaches that meaning they acquired when they were a kid. Like for instance, y(x) and people solving integrals asking "that's like my dx, right?" lol

Answer 28

Everything is unitised in physics.

Answer 29

\(\underline{r}\) should be the basis vector I think?

Answer 30

Lol everything is not 'unitised' in physics. It depends on what textbook/teacher you have. Yeah that's a basis vector.

Answer 31

At least theoretical physicist unitised the Planck's constant, gravitational constant, the Boltzmann constant and a few more lol.

Answer 32

@Kainui What time is it on your side? Are you free now?