1
$\begingroup$

enter image description here$[![Picture ~from ~http://www.astro.uwo.ca/~houde/courses/PDF%20files/physics350/Noninertial_frames.pdf][1]][1]$

From the picture above, the set of coordinates $x_i$ are the ones that are not fixed and rotating, while the set of coordinates $x'_i$ are the ones that are fixed.

I have tried my best to make this question self-contained, but please refer to the link (http://www.astro.uwo.ca/~houde/courses/PDF%20files/physics350/Noninertial_frames.pdf),

My questions here are:

  1. Is $\vec r_{fixed}~=\vec r_{rotating}$ correct? To me, it seems like it should be correct since whether you are expressing a vector in fixed coordinate or rotating coordinate has nothing to do with the actual vector itself...

I asked number one because, from here on, things get complicated..

  1. If number one is true, then shouldn't $d\vec r_{fixed}= d\vec r_{rotating}$ be correct? I at first thought that it should be correct, but from equation $( d\vec r/dt)_{fixed} = ( d\vec r/dt)_{rotating}+\vec \omega\times \vec r$ (equation 8.5), it seems as though it shouldn't be correct.

    Is this because when considering dr, we have to take into account the fact that the coordinates systems themselves change? (in other words, the derivative unit vectors of the coordinate system is not zero, and therefore causes a 'change' in the vector?)

    What I find puzzling is that, the vector $d\vec r$ itself is just one thing (it shouldn't change in whatever coordinate system we are expressing at (as it is just a vector))

  2. Lastly, are $(d/dt)_{fixed}$ and $(d/dt)_{rotating}$ some sorts of a 'operator'? And if so, then does the question in number two, not make sense at all?(Since $d\vec r_{fixed}$ doesn't make sense?) Moreover, are $d\vec r_{fixed}/dt$ and $(d\vec r/dt)_{fixed}$ the same things?

Thank you for reading this question!

I hope you have a great day!

$\endgroup$
  • $\begingroup$ Unfortunately your question is not self contained since there is no image and the symbols you use are not defined. Can you try to repost the figure? $\endgroup$ – ZeroTheHero Dec 2 '18 at 13:59
  • 1
    $\begingroup$ You probably need to distinguish between the vector $\vec{r}$, its magnitude $r$, and the coordinates of the tip of the vector $r_i$. $\endgroup$ – probably_someone Dec 2 '18 at 17:35
  • $\begingroup$ @probably_someone I fixed it! Thank you for telling me :) $\endgroup$ – Danny Han Dec 3 '18 at 3:40
1
$\begingroup$

The problem with your image is that in text you use the notation $\vec{r}_{fixed}$ and $\vec{r}_{rotation}$ that don't appear in the image. The correspondence between both is as follows:

$$\vec{r}_{fixed} = \vec{r}\ ', \quad \vec{r}_{rotation} = \vec{r} \tag1$$

I will make use of the image notation because is shorter. So, from Eq. (1) you have the answer to your first question: $\vec{r}\ ' \neq \vec{r}$. This is because what you see in the rotation system is not the same as if you were out, in the fixed one. For example: imagine you are in a discus which is rotation at some angular velocity $\vec{\omega}$, and you ratate with it at that velocity. Now pick some point of the proper discus, so it moves at $\vec{\omega}$ too. Therefore, from that reference system you see some $\vec{r}$ constant in time: the point is not moving if you are in the discus.

Nevertheless, for another observer, in some fixed reference system (actually that observer is the one that mesures $\vec{\omega}$, the point is moving in circles of constant ratio.

With this, the answer to the rest of your question are natural. Let's go for the relation among $(d/dt)_{fixed}$ and $(d/dt)_{rotating}$. With your image I can write:

$$\vec{r}\ ' = \vec{R} + \vec{r}$$

Now, the velocity you can see from fixed frame is given by:

$$\frac{d\vec{r}\ '}{dt}_{fixed} = \frac{d\vec{r}}{dt}_{fixed} + \frac{d\vec{R}}{dt}_{fixed} = \frac{d\vec{r}}{dt}_{fixed} + \vec{V} $$

Now, let's focus on $\frac{d\vec{r}}{dt}_{fixed}$. To compute it, I realise that I can write $\vec{r} = x\vec{i} + y\vec{j} + z\vec{k}$, where $\{\vec{i}, \vec{j}, \vec{k}\}$ are the vector basis in the rotating system and $\{x, y, z\}$ the components of $\vec{r}$ in it. All in all,

$$\frac{d\vec{r}}{dt}_{fixed} = \dot{x}\vec{i} + \dot{y}\vec{j} + \dot{z}\vec{k} + x\dot{\vec{i}} + y\dot{\vec{j}} + z\dot{\vec{k}} \tag2$$

Where dot notation is for time derivative. So, now you can relate:

$$\dot{x}\vec{i} + \dot{y}\vec{j} + \dot{z}\vec{k} \equiv \frac{d\vec{r}}{dt}_{rotating}, \quad x\dot{\vec{i}} + y\dot{\vec{j}} + z\dot{\vec{k}} \equiv \vec{\omega}\times \vec{r} \tag3$$

The first equivalence is due to the vector (look at left hand side) is in the basis of the rotating system and the componentes are the change in time of the coordinates in THAT ROTATING system, so it's the velocity you would see from that frame. The second equivalence comes from the following expression:

$$\dot{\vec{i}} = \vec{\omega}\times \vec{i}$$

Recall that for a particle moving in circles with some angular velocity $\vec{\omega}$ and position $\vec{p}$, it's velocity is $\vec{\omega}\times \vec{p}$. But this $\vec{p}$ could be simply $\vec{i}$. Idem for the rest basis elements.

Therefore, applying Eq. (3) in Eq. (2) you get your relation between derivatives and the operator comes from rubbing out $\vec{r}$.

I hope this was what you were looking for.

$\endgroup$
  • $\begingroup$ Thank you for your answer! But there is one thing that I want to tell you: when I meant $\vec r_{fixed}$, I meant the 'vector' r with respect to the fixed axis. (In other words, in the book, $\vec r_{fixed}$ has to equal $\vec r'$ because r' is a position vector. However, what I meant to imply by saying $\vec r_{fixed}$ was that we are expressing the position vector $\vec r$ with respect to (a different coordniate system), $x'_i$ $\endgroup$ – Danny Han Dec 3 '18 at 8:40
  • $\begingroup$ But the key is that in rotating system and fixed system you don't see the same, so you have to have 2 different vector of position: $\vec{r}, \vec{r}\ '$. And the relation in order to get the equations you were trying to understand is taking in mind that fact. $\endgroup$ – Vicky Dec 3 '18 at 8:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.