Rotating frame: a question of interpretation

[Philip Wood]

I'm reading Lanczos: 'The variational Principles of Mechanics'. I need help resolving a paradox - which is probably trivial…

Lanczos (page 100 Dover edition) introduces a system, S', rotating at angular velocity \vec \Omega about an axis through a fixed point with respect to inertial system S. The radius vectors \vec R and \vec R' in the two systems are, he says, the same: \vec R = \vec R'. [I don't have trouble with this: it's fundamental to the idea of a vector that the same vector can be expressed in terms of different basis vectors, in this case, \vec i, \vec j, \vec k and \vec i', \vec j', \vec k'.]

Nevertheless, he says, the velocities and accelerations measured in both systems differ from each other because rates of change observed in the two systems are different. If a certain vector \vec B is constant in S' it rotates with the system and thus, if observed in S, undergoes in the time dt an infinitesimal change d \vec B = (\vec \Omega \times \vec B) dt. [Again, easily seen - especially for \vec R itself.]

Hence \frac{d \vec B}{dt} = (\vec \Omega \times \vec B) while at the same time, Lanczos says, \frac{d' \vec B}{dt} = 0.

Here, Lanczos has introduced the notation \frac {d'}{dt} which refers to the operation of observing the rate of change of a quantity in the moving system S'.

If you've read as far as this, well done and thank you. Now here's my problem. Regarding \vec R and \vec R' as functions of time, we can surely write:
\vec R (t) = \vec R' (t).
But because \frac{d \vec B}{dt} = (\vec \Omega \times \vec B)
\vec R (t + dt) = \vec R (t) + (\vec \Omega \times \vec R (t)) dt
while because \frac{d' \vec B}{dt} = 0
\vec R' (t + dt) = \vec R' (t) + 0.
It would therefore seem that at time (t + dt), it is no longer the case that \vec R = \vec R'. Whereas \vec R' has changed over dt, \vec R' has stayed constant.
Where is my reasoning wrong?

 

[Andrew Mason]

I am not clear on what \vec{R} \text{ and } \vec{R'} are. Are they the vectors from the inertial point to the origins S and S'? In other words, S and S' have a common origin?

AM

 

[UltrafastPED]

In other words, S and S' have a common origin?

I checked my copy of Lanczos; yes, they have a common origin. The discussion is for rotations with no translation.

 

[UltrafastPED]

Where is my reasoning wrong?

Lanczos uses the "superposition principle of infinitesimal processes", while you are not.

Also note that there is a typo on eq. (45.9) of the fourth edition; there should be a "dot" over the Ω in the final term.

 

[Philip Wood]

Thanks for replying, but with respect, UPED, I don't think either of your points solves my problem. L uses the superposition principle when moving on from a vector which is constant in S' to one which is not necessarily constant in S'. My concern is merely with a vector which is constant in S'. [Incidentally I don't much like L's appeal to a superposition principle; there are, imo, nicer treatments in Kibble or in Synge which differentiate x_i \vec e_i as a product.] Equation 45.9 is also beyond the point where my problem lies.

 

[Philip Wood]

No more ideas? Is this because my question is not understood, or is it because it's baffling you, too?

[I apologise for the typo in the last-but-one sentence of post 1. This should read: Whereas \vec R has changed over dt, \vec R' has stayed constant.]

 

[Andrew Mason]

No more ideas? Is this because my question is not understood, or is it because it's baffling you, too?

[I apologise for the typo in the last-but-one sentence of post 1. This should read: Whereas \vec R has changed over dt, \vec R' has stayed constant.]

I am still not clear on \vec R and \vec R'. I don't have a copy of Lanczos' book.


Can you post the page in question?

AM

 

[dauto]

When you say R = R' does that mean they have the same numerical values or does that mean they represent the same physical vector? You can't have it both ways. You're trying to have it both ways and that's why you're getting in trouble.

 

[Philip Wood]

AM and Dauto. Thank you for your interest.

AM Please find extracts.

Dauto: I've wondered the same thing. Yet R' = R seems to imply same value.

 

[Andrew Mason]

AM and Dauto. Thank you for your interest.

AM Please find extracts.

Dauto: I've wondered the same thing. Yet R' = R seems to imply same value.

The problem seems to be that we are not sure what \vec{R} and \vec{R}' are.

Lanczos seems to be using 2 dimensional polar co-ordinates. So \vec{R}(r,θ) differs from \vec{R}'(r,θ') only by angle: i.e. the second co-ordinate - angle θ. If that is the case, his statement:

\vec{R}= \vec{R}' cannot be true, generally.

Perhaps he meant to say |\vec{R}| = |\vec{R}'| (i.e. the length (the first co-ordinate, r) is the same for both vectors).

AM

 

[Philip Wood]

AM Thank you. Yet he has R and R' in bold and refers to them (second para in extracts) as radius vectors And he does seem to be pretty scrupulous about nomenclature. Also by starting the third para in the extract with "Nevertheless" he seems to be pointing to the peculiarity of radius vector behaving differently from velocity and acceleration vectors: there wouldn't be anything worth remarking about if radius wasn't to be considered a vector.

 

[Andrew Mason]

AM Thank you. Yet he has R and R' in bold and refers to them (second para in extracts) as radius vectors And he does seem to be pretty scrupulous about nomenclature. Also by starting the third para in the extract with "Nevertheless" he seems to be pointing to the peculiarity of radius vector behaving differently from velocity and acceleration vectors: there wouldn't be anything worth remarking about if radius wasn't to be considered a vector.

Laczos appears to be referring to \vec{R} \text{ and } \vec{R}' at a particular time e.g. \vec{R}(0) = \vec{R}'(0). If \vec{R}'(t) = \vec{R}'(0) for all t, then \vec{R}' is just a fixed vector in S'. So, while \vec{R}(0) = \vec{R}'(0), \vec{R}(t) \ne \vec{R}'(t) (unless \dot{Ω}t = n2\pi).

AM

 

[Philip Wood]

AM Yes, this interpretation is free of inconsistentency. I was loathe to accept it because to claim that \vec R = \vec R' (only) at one particular time seemed such a weak claim. Thanks to your post, I'll try to stop worrying about the issue. I have a great gift for getting stuck when trying to learn; it has not faded with advancing years.

Edit: Fickle to the last, I've now gone for a different interpretation. See post 15. Many thanks!

 

[D H]

Now here's my problem. Regarding \vec R and \vec R' as functions of time, we can surely write:
\vec R (t) = \vec R' (t).

Surely you cannot.

Imagine a merry-go-round with a hole in the middle, big enough so that you can stand on the ground in the middle of the merry-go-round and watch the horses move around you. The horses are moving. From this perspective, the position of some particular horse as a function of time is $R(t) = R\cos(\omega t)\hat x + R\sin(\omega t)\hat y$.

Alternatively, you can jump up on the merry-go-round, put a plank over that hole, and position yourself exactly as before, except now you are rotating with the merry. From this perspective, that horse is stationary: $R'(t) = R\hat x'$. The horse's velocity is zero from this rotating perspective. It's obviously not zero from the perspective of the non-rotating observer.I don't have Lanczos, but the use of "infinitesimal rotations" at the bottom of that page you posted makes it looks like he's about to do some "physics math" hand waving in his derivations. Hand waving is standard fare in this regard, even in many graduate level texts. Even Goldstein waved his hands.

 

[Philip Wood]

DH Thank you. I like your very clear scene-setting with the merry-go-round. I also like your use of unit vectors.

I'm now going to turn your argument on its head and postulate that
\textbf {R'} (t) = \textbf{R} (t)
in which case, with your notation,
R \mathbf {\hat{x'}}=R\ \text {cos} \omega t \ \mathbf {\hat{x}}+R\ \text {sin} \omega t \ \mathbf {\hat{y}}
This is the familiar business of expressing the same vector on two different basis sets. It does, though, demand that we regard vectors fixed in the rotating system, such as \mathbf {\hat{x'}}, as functions of time.

I've been labouring under the misapprehension that this is inconsistent with Lanczos's equation (in the extract in post 9):
\frac {\text{d'} \mathbf{B}}{\text{d} t} = 0
He says that \frac {\text{d'} }{\text{d} t} refers to the rate of change of a quantity in the moving system, S'. I now realize that he's not saying that \frac {\text{d} \mathbf{B'}}{\text{d} t} = 0. What he means, I think, by \frac {\text{d'}\mathbf{B}}{\text{d} t} is the rate of change of vector B' if we regard the rotating unit vectors \mathbf {\hat{x'}} etc. as constant; in other words a sort of partial differentiation.

This interpretation rings true for me. Many thanks to those who've came to my aid.

 

[D H]

Before I start delving into your last post, Philip, it's important to realize that time derivatives of vector quantities are frame dependent. It's certainly true for translation. Why wouldn't this be the case for rotation?

To illustrate that it's true for translation, suppose you and your best buddy decided to take the long weekend off with a trip to Vegas. The only problem: Who's car? You like your vintage Dodge Charger with its 426 cu. in. engine, he likes his vintage AMC Matador with its smaller but zestier 401 cu. in. engine. He comes up with the perfect solution: "Let's race!"

You're neck and neck as your cross the California-Nevada state line. From your perspective, his velocity as you cross into Nevada is nearly zero. The same is true regarding your velocity from your buddy's perspective. Unfortunately for both of you, you just zoomed past a Nevada highway patrol officer sitting on a lawn chair in the freeway median strip. That officer (along with the radar gun he was holding just before he called you in) had a slightly different perspective on your velocities than did the two of you.

DH Thank you. I like your very clear scene-setting with the merry-go-round. I also like your use of unit vectors.

I'm now going to turn your argument on its head and postulate that
\textbf {R'} (t) = \textbf{R} (t)
in which case, with your notation,
R \mathbf {\hat{x'}}=R\ \text {cos} \omega t \ \mathbf {\hat{x}}+R\ \text {sin} \omega t \ \mathbf {\hat{y}}
This is the familiar business of expressing the same vector on two different basis sets. It does, though, demand that we regard vectors fixed in the rotating system, such as \mathbf {\hat{x'}}, as functions of time.

The derivatives of those unit vectors is frame-dependent. From the perspective of the non-rotating observer, the $\hat x, \hat y, \hat z$ unit vectors are stationary (time derivatives are zero) while the $\hat x', \hat y', \hat z'$ unit vectors are rotating (time derivatives are non-zero). It's the other way around for the rotating observer.

From the above, it's obvious that $\hat x' = \cos(\omega t)\hat x + \sin(\omega t) \hat y$. Let's finish off the relationship between those unit vectors. The two observers share the same z axis: $\hat z' = \hat z$. This means that to complete a right-handed system, we must have $\hat y' = \hat z' \times \hat x' = -\sin(\omega t) \hat x + \cos(\omega t)\hat y$.

These relationships form the transformation matrix that transforms a vector as represented in the rotating frame to the representation of the same vector in the non-rotating frame:

<br /> \begin{bmatrix} x \\ y \\z \end{bmatrix} =<br /> \begin{bmatrix}<br /> \cos(\omega t) &amp; -\sin(\omega t) &amp; 0 \\<br /> \sin(\omega t) &amp; \phantom{-} \cos(\omega t) &amp; 0 \\<br /> 0 &amp; 0 &amp; 1<br /> \end{bmatrix}<br /> \;<br /> \begin{bmatrix} x&#039; \\ y&#039; \\z&#039; \end{bmatrix}<br />
Denoting that transformation matrix as $T_{R\to I}$, the above becomes $\vec R(t) = T_{R\to I}(t) \, \vec R'(t)$. Note this applies to all vectors, not just position vectors. Any vector quantity $\vec q$ can be transformed from its rotating frame representation to its non-rotating frame representation.

Taking the time derivative of above results in $\dot{\vec R}(t) = T_{R\to I}(t) \, \dot{\vec R}'(t) + \dot T_{R\to I}(t) \, \vec R'(t)$. What's the time derivative of this transformation matrix? From the above, that time derivative is
<br /> \dot T_{R\to I}(t) =<br /> \begin{bmatrix}<br /> -\omega\sin(\omega t) &amp; -\omega\cos(\omega t) &amp; 0 \\<br /> \phantom{-}\omega\cos(\omega t) &amp; -\omega\sin(\omega t) &amp; 0 \\<br /> 0 &amp; 0 &amp; 0<br /> \end{bmatrix}<br />
Pre-multiplying by the transpose of the transformation matrix yields
<br /> T_{R\to I}(t)^T \, \dot T_{R\to I}(t) =<br /> \begin{bmatrix} 0 &amp; -\omega &amp; 0 \\ \omega &amp; 0 &amp; 0 \\ 0 &amp; 0 &amp; 0 \end{bmatrix}<br />
Note that the right hand side is the skew symmetric cross product matrix corresponding to the angular velocity vector $\vec \omega = \omega \hat z'$.

My hand wave is the unproven claim (unproven here, that is) that $T_{R\to I}(t)^T \, \dot T_{R\to I}(t) = \text{Sk}(\vec \omega)$ where $\vec \omega$ is the angular velocity vector as represented in the rotating frame is always true for rotations in three dimensional space. This is a consequence that the set of all 3x3 proper real transformation matrices is the Lie group SO(3). The Lie algebra for this group is the set of 3x3 skew symmetric matrices. Proving this claim either requires a class in differential geometry and Lie groups, or five pages of a math.

Another way to write this relation between the transformation matrix, its time derivative, and the angular velocity vector is $\dot T_{R\to I}(t) = T_{R\to I}(t) \text{Sk}(\vec \omega(t))$. With this, the time derivative becomes $\dot{\vec R}(t) = T_{R\to I}(t) \, \dot{\vec R}'(t) + T_{R\to I}(t)\,\text{Sk}(\vec \omega(t)) \, \vec R'(t)$, or
\dot{\vec R}(t) = T_{R\to I}(t) \left(\dot{\vec R}&#039;(t) + \vec \omega(t) \times \vec R&#039;(t)\right)

 

[Andrew Mason]

Note that the right hand side is the skew symmetric cross product matrix corresponding to the angular velocity vector $\vec \omega = \omega \hat z'$.

My hand wave is the unproven claim (unproven here, that is) that $T_{R\to I}(t)^T \, \dot T_{R\to I}(t) = \text{Sk}(\vec \omega)$ where $\vec \omega$ is the angular velocity vector as represented in the rotating frame is always true for rotations in three dimensional space. This is a consequence that the set of all 3x3 proper real transformation matrices is the Lie group SO(3). The Lie algebra for this group is the set of 3x3 skew symmetric matrices. Proving this claim either requires a class in differential geometry and Lie groups, or five pages of a math.

Laczos must be using polar coordinates.

Using polar coordinates (where \hat{k} is the unit vector in the direction perpendicular to the plane of rotation ), the orthogonal basis vectors in S are \hat{r} \text{ and } \hat{Ω} where \hat{Ω} = \hat{k} \times \hat{r}. In S' the basis vectors are: \hat{r}&#039; \text{ and } \hat{Ω}&#039; where \hat{Ω}&#039; = \hat{k} \times \hat{r}&#039;. However, since \hat{r} = \vec{R}/|\vec{R}| = \vec{R}&#039;/|\vec{R}&#039;| = \hat{r}&#039;, \hat{Ω}&#039; = \hat{k} \times \hat{r}&#039; = \hat{k} \times \hat{r} = \hat{Ω}, so the unit basis vectors for S and S' are identical.

The only difference between \vec{R} and \vec{R}&#039; is in the values for Ω and Ω'.

\vec{R} = r\hat{r}+ Ω\hat{Ω} and

\vec{R}&#039; = r&#039;\hat{r}&#039;+ Ω&#039;\hat{Ω}&#039; = r\hat{r}+ Ω&#039;\hat{Ω}

Subtracting:

\vec{R}&#039; - \vec{R} = Ω&#039;\hat{Ω}&#039; - Ω\hat{Ω} = (Ω&#039;- Ω)\hat{Ω}

Differentiating a rotating vector R in S with respect to time results in:

\dot{\vec{R}} = \dot{r}\hat{r} + r\dot{Ω}\hat{Ω} = r\dot{Ω}\hat{Ω}

but in co-rotating frame S', \dot{Ω} = 0 so the derivative with respect to time is:

\dot{\vec{R}&#039;} = \dot{r}&#039;\hat{r} + r&#039;\dot{Ω}&#039;\hat{Ω} = 0\hat{r} + r0\hat{Ω} = 0

AM

 

[Philip Wood]

DH Thank you for this interesting piece. The case that concerned me was of a vector fixed to the rotating frame, so I think you'll agree that \left(\begin{array}{cc}x&#039;\\y&#039;\\z&#039;\ \end{array}\right) is constant, though the unit vectors \hat{\mathbf{x&#039;}} etc., are functions of time, when expressed in terms of \hat{\mathbf{x}} etc.. This, I believe, is what Lanczos means by \frac{\text{d&#039;} \mathbf{B}}{\text{d}t} = 0.

 

[D H]

Laczos must be using polar coordinates.

No, he's not. Almost assuredly. He's using two reference frames, one of which is rotating with respect to another. It doesn't really matter what kind of coordinates you use. This an issue of the time derivative of some vector quantity as observed in one reference frame versus the time derivative as observed in another reference frame.

What he's doing is developing what some call the dynamical transport theorem. You don't need coordinate systems at all to express this theorem:
\left(\frac{d\vec q}{dt}\right)_A = \left(\frac{d\vec q}{dt}\right)_B + \vec \omega \times \vec q
The subscripts on the derivatives denote the time derivative as observed by two observers, call them observer A and observer B. Observer B is rotating with respect to observer A with an angular velocity of $\vec \omega$.

 

[Andrew Mason]

No, he's not. Almost assuredly. He's using two reference frames, one of which is rotating with respect to another. It doesn't really matter what kind of coordinates you use. This an issue of the time derivative of some vector quantity as observed in one reference frame versus the time derivative as observed in another reference frame.

What he's doing is developing what some call the dynamical transport theorem. You don't need coordinate systems at all to express this theorem:
\left(\frac{d\vec q}{dt}\right)_A = \left(\frac{d\vec q}{dt}\right)_B + \vec \omega \times \vec q
The subscripts on the derivatives denote the time derivative as observed by two observers, call them observer A and observer B. Observer B is rotating with respect to observer A with an angular velocity of $\vec \omega$.

But by using polar coordinates you can use the same basis vectors for both the rotating and non-rotating reference frames. It is just much easier to do the math.

Without the pages prior to page 100 I am just guessing at what the statement \vec{R} = \vec{R}&#039; means. It just occurred to me that he may mean \vec{R} = R\hat{r} = R\hat{r}&#039; = \vec{R}&#039;. The r\hat{r} component then is always the same using polar coordinates. That may be what he is referring to.

AM

 

[UltrafastPED]

Without the pages prior to page 100 I am just guessing at what the statement \vec{R} = \vec{R}&#039; means. It just occurred to me that he may mean \vec{R} = R\hat{r} = R\hat{r}&#039; = \vec{R}&#039;. The r\hat{r} component then is always the same using polar coordinates. That may be what he is referring to.

Lanczos means that the geometric vectors are the same; that is the advantage of vector notation - you can say things that are independent of the coordinate systems!

If you find coordinates helpful, then select the coordinates that are easiest to use - polar coordinates would seem the best here.

 

[D H]

But by using polar coordinates you can use the same basis vectors for both the rotating and non-rotating reference frames. It is just much easier to do the math.

Without the pages prior to page 100 I am just guessing at what the statement \vec{R} = \vec{R}&#039; means. It just occurred to me that he may mean \vec{R} = R\hat{r} = R\hat{r}&#039; = \vec{R}&#039;. The r\hat{r} component then is always the same using polar coordinates. That may be what he is referring to.

AM

That is not what it means. $\vec R = \vec R'$ means exactly what it says, that $\vec R$ and $\vec R'$ are the same vector. It has nothing to do with polar coordinates. It has to do with two different reference frames giving two different representations of the same vector.

 

[Andrew Mason]

That is not what it means. $\vec R = \vec R'$ means exactly what it says, that $\vec R$ and $\vec R'$ are the same vector. It has nothing to do with polar coordinates. It has to do with two different reference frames giving two different representations of the same vector.

The difficulty is in defining what is mean by "the same vector" in S and S'.

Since the origins of S and S' are same, the radial displacement vector representing the displacement of a point relative to the origin is the same in both frames of reference.

If the observer in S' realizes that he is rotating and not S, then S' can do the math to subtract his rotational movement and conclude that the point is constant over time. But if he doesn't, he concludes that the point (i.e the displacement vector) keeps changing and, with it, the direction of the radial vector from the origin.

AM

 

[Philip Wood]

Thank you again for your help. The penny dropped for me back at post 15. Just a comment on...

What he's doing is developing what some call the dynamical transport theorem. You don't need coordinate systems at all to express this theorem:
\left(\frac{d\vec q}{dt}\right)_A = \left(\frac{d\vec q}{dt}\right)_B + \vec \omega \times \vec q
The subscripts on the derivatives denote the time derivative as observed by two observers, call them observer A and observer B. Observer B is rotating with respect to observer A with an angular velocity of $\vec \omega$.

I think the danger of this presentation is that it invites my original objection (in post 1) that we have the same vector in two frames but which changes at a different rate in the two frames.

The key, I think, to understanding what's going on is to realize that \left(\frac{d\vec q}{dt}\right)_A is a genuine rate of change of vector \vec q. [We need only to consider the change in the set of scalar components, because in frame A the basis vectors are fixed.] But \left(\frac{d\vec q}{dt}\right)_B is not the rate of change of \vec q. It is again the rate of change of the component set, holding the basis vectors (those fixed to frame B) constant. This makes a real difference in frame B because the basis vectors in frame B (those fixed to the frame) are actually rotating. [The real rate of change of \vec q calculated in frame B would take account of the rotation of the basis vectors, and would come out to the same value as the rate of change of \vec q in frame A!]

That's why Lanczos, in his wisdom, used a new notation, \frac{\text {d&#039;} \mathbf q}{\text {d}t} for this special sort of partial differentiation in the rotating, B, frame.

So I think tacit use IS being made of co-ordinate systems, though the notation (including that of Lanczos) hides this.

 

[D H]

I think the danger of this presentation is that it invites my original objection (in post 1) that we have the same vector in two frames but which changes at a different rate in the two frames.

Danger? What danger? That the same vector has different derivatives in different frames is entirely the point. It is key to understanding fictitious forces, key to many solutions of the three body problem (which is typically solved in the synodic frame), key to making a robotic arm behave correctly, and key to making one spacecraft safely dock with another.

 

[Philip Wood]

If the (complete) time derivative of a vector is different in frames A and B, how can it stay the same vector in A as in B?

Did you read the rest of my post 24? Here I contend that the so-called time derivative we are taking in the rotating frame treats the base vectors in that frame (and rotating with that frame) as if they were stationary. So it's only a sort of partial derivative.

 

[D H]

What do you mean by "(complete) time derivative of a vector"? You are essentially saying that doing physics in a rotating frame is invalid. It most certainly is not invalid.

Suppose two observers rotating with respect to one another watch the same event. One sees one time derivative, the other sees another. Which observer is right, which is wrong? The answer is that both are "right". What's wrong is that insisting that only one can be right.

 

[Philip Wood]

I'm certainly not saying that doing Physics in a rotating frame is invalid! Nor am I saying that one observer is right and the other wrong.

I'm merely trying to understand how to interpret the equation you write as

\left( \frac{d \vec q}{d t}\right)_A = \left( \frac{d \vec q}{d t}\right)_B +\ \vec \omega \times \vec q.
I'm arguing that \left( \frac{d \vec q}{d t}\right)_B is only a 'partial' derivative of the vector \vec q because it treats the unit vectors in the rotating (B) frame (rotating with the frame) as constants. The second term on the right is the 'missing' part of the B-frame derivative that takes account of the rotation (non-constancy) of the unit vectors. See footnote.

On the left hand side, there is only one term, \left( \frac{d \vec q}{d t}\right)_A because in the inertial frame the unit vectors are constants. So \left( \frac{d \vec q}{d t}\right)_A is a genuine, complete, derivative of \vec q.

Footnote
Mathematically, what I'm saying in the paragraph under the equation is that the 'complete' derivative of \vec q expressed in terms of the rotating unit vectors of the 'B' frame is

\frac{d}{dt}\left(q_x \hat {\vec x}+ q_y \hat {\vec y}+ q_z \hat {\vec z}\right)
in which q_x, q_y, q_z are (scalar) components of \vec q on the rotating unit vector set \hat {\vec x}, \hat {\vec y}, \hat {\vec z}.

Differentiating each term in the last equation as a product, and re-assembling:

\frac{d q_x}{dt} \hat {\vec x}+ \frac{d q_y}{dt} \hat {\vec y}+\frac{d q_z}{dt} \hat {\vec z} \ +\ q_x \frac{d \hat {\vec x}}{dt}+q_y \frac{d \hat {\vec y}}{dt}+q_z \frac{d \hat {\vec z}}{dt} \ \ \ = \ \ \ \left( \frac{d \vec q}{d t}\right)_B \ + \ \vec{\omega} \times \vec{q}
In the last step I've identified the sum of the first three terms with your \left( \frac{d \vec q}{d t}\right)_B, (what Lanczos calls \frac{d&#039; \mathbf q}{dt}), and the last three terms with \vec{\omega} \times \vec{q}. The latter can easily be justified.

 

[D H]

I'm certainly not saying that doing Physics in a rotating frame is invalid! Nor am I saying that one observer is right and the other wrong.

I'm merely trying to understand how to interpret the equation you write as

\left( \frac{d \vec q}{d t}\right)_A = \left( \frac{d \vec q}{d t}\right)_B +\ \vec \omega \times \vec q.
I'm arguing that \left( \frac{d \vec q}{d t}\right)_B is only a 'partial' derivative of the vector \vec q because it treats the unit vectors in the rotating (B) frame (rotating with the frame) as constants.

To an observer fixed with respect to that rotating frame, those unit vectors in the rotating frame most certainly are constant. By arguing that they are not you are implicitly saying that doing physics in a rotating frame is invalid. Look at it from the perspective of that rotating observer. From that person's perspective, it's the inertial unit vectors that are rotating.

 

[mattt]

Thank you again for your help. The penny dropped for me back at post 15. Just a comment on...



I think the danger of this presentation is that it invites my original objection (in post 1) that we have the same vector in two frames but which changes at a different rate in the two frames.

The key, I think, to understanding what's going on is to realize that \left(\frac{d\vec q}{dt}\right)_A is a genuine rate of change of vector \vec q. [We need only to consider the change in the set of scalar components, because in frame A the basis vectors are fixed.] But \left(\frac{d\vec q}{dt}\right)_B is not the rate of change of \vec q. It is again the rate of change of the component set, holding the basis vectors (those fixed to frame B) constant. This makes a real difference in frame B because the basis vectors in frame B (those fixed to the frame) are actually rotating. [The real rate of change of \vec q calculated in frame B would take account of the rotation of the basis vectors, and would come out to the same value as the rate of change of \vec q in frame A!]

That's why Lanczos, in his wisdom, used a new notation, \frac{\text {d&#039;} \mathbf q}{\text {d}t} for this special sort of partial differentiation in the rotating, B, frame.

So I think tacit use IS being made of co-ordinate systems, though the notation (including that of Lanczos) hides this.

D.H. is right. You will understand it all perfectly when you study Differential Geometry.

Different observers are different coordinate charts on the same Differential Manifold (Euclidean Space-time in this case).

One world-curve on the manifod is what it is, some \alpha : I \to M, but obviously the mathematical expression of the coordinates of the same curve (and its vector velocity) on two different coordinate charts, are different (in general).

The mathematical expression for the transformation from one coordinate chart to another coordinate chart is precisely what D.H. has written for this concrete case (each one observer rotating with respect to the other).

 

[Philip Wood]

To an observer fixed with respect to that rotating frame, those unit vectors in the rotating frame most certainly are constant. […] From that person's perspective, it's the inertial unit vectors that are rotating.

The relative rotation of the sets of unit vectors is undeniable, isn't it? I could agree to say that it's the inertial frame unit vectors which are rotating and the rotating frame unit vectors which are stationary. But would you really be happier with that?

In fact, there are good reasons to choose to call the inertial frame unit vectors stationary...

We can distinguish a rotating frame from a non-rotating frame. For example, on a rotating platform a string running between two balls develops a tension. In general in a rotating frame objects experience certain forces which cannot be attributed to specific objects external to themselves, and which do not, therefore come in Newton's Third Law pairs.

Granted that we can recognise a rotating frame from such observations, it seems to me far more natural, then, that we should regard axes fixed to the rotating frame as rotating, and axes in the inertial frame as stationary. It doesn't matter that the observer in the rotating frame may SEE the inertial frame as rotating and his own frame as stationary.

 

[mattt]

The relative rotation of the sets of unit vectors is undeniable. I could agree to say that it's the inertial frame vectors which are rotating and the rotating frame vectors which are stationary. But would you really be happier with that? If so, why?

That is a different thing. I tried to explain in my previous post that a geometric object (a curve, its vector velocity, etc) on a Differentiable Manifold has different coordinate expressions in different coordinate charts.

In fact, there are good reasons to choose to call the inertial frame vectors stationary...

We can distinguish a rotating frame from a non-rotating frame. For example, on a rotating platform a string running between two balls develops a tension. In general in a rotating frame objects experience certain forces which cannot be attributed to specific objects external to themselves, and which do not, therefore come in Newton's Third Law pairs.

Granted that we can recognise a rotating frame from such observations, it seems to me far more natural, then, that we should regard axes fixed to the rotating frame as rotating, and axes in the inertial frame as stationary. It doesn't matter that the observer in the rotating frame may SEE the inertial frame as rotating and his own frame as stationary.

That is simply the following: the dynamical laws (you'll see this when you study General Relativity, but it is also true in Newtonian Mechanics if you study it under Differential Geometry point of view) are geometric relations among certain geometric objects on a given differentiable manifold (space-time). You can state it in a coordinate-free way.

But, this given dynamical law has (in general) different coordinate expressions in different coordinate charts.

In Newtonian Mechanics with its Euclidean Space-time manifold, there is a subset of coordinate charts (observers) in which the dynamical law (that is a geometric relation between geometric objects, coordinate-free) has a very simple mathematical expression. We call these charts (these observers) "inertial frames".

What you call "rotating observers" are not in this subset; the same dynamical law (geometric relation) has a less simple mathematical expression in these charts.


We are not saying that "inertial frames" and "rotating frames" are the same in Newtonian Mechanics. They are not.

What we are saying is both are just valid charts on the (euclidean space-time) manifold. Any geometric object (curve, vector velocity, vector field, tensor field...) and any geometric relation among geometric objects, is what it is, but will have (in general) different coordinate expression in different coordinate charts.


What D.H. showed you is the following: if you have two different charts (each one correspond to an observer that is rotating with respect to the other observer), then the coordinate expression for the geometric object (dq/dt which is a coordinate-free concept, you will understand this when you study differential geometry) in those two charts are related in exactly that way.

 

[Philip Wood]

Is it possible, would you say, to understand the equation (dq/dt)_A = (dq/dt)_B + ω x q without using differential geometry, or the language of differential geometry?

 

[mattt]

Is it possible, would you say, to understand the equation (dq/dt)_A = (dq/dt)_B + ω x q without using differential geometry, or the language of differential geometry?

Let me start with this (later on I'll try to not use differential geometry) :

Let M be a Euclidean Space-time (a four-dimensional Differentiable Manifold with some requisites I won't go into) and let

\alpha : I \to M be some differentiable curve (that correspond to some point-particle with mass, for example).


A chart is "a way" to give space-time coordinates to every point of some region of M (again, all this has its perfect definition, I won't go into details here).

So under a chart, it will be \alpha(\tau) = (t(\tau), x(\tau), y(\tau), z(\tau))

In this particular case of Newtonian Space-time, there is a foliation that allow a reparametrization of the curve in such a way that in any chart it will be:

\alpha(t) = (t,x(t),y(t),z(t))

(i.e. the "time" is "the same" for all observers; by the way, this is not true in other Space-time manifolds, such as Lorentz Manifolds used in Special and General Relativity as Space-times).


The last three coordinates is what you call "the vector position of the particle with respect to this chart (this observer)"

\vec{r}(t) = (x(t), y(t), z(t))


x(t), y(t), z(t) don't have to be necessarily escalar components with respect to orthonormal unit vectors. It can be much more general (again, there is a perfect mathematical definition of what is a coordinate chart in a differentiable manifold, but you don't need to know these details now).


In the example of this thread, we have two different charts, each one correspond to an observer that is rotating with respect to the other observer.

Let us put some names:

\alpha(t) = (t, \vec{r}(t)) = (t, x(t), y(t), z(t) ) for one observer.

\alpha(t) = (t, \vec{s}(t)) = (t, a(t), b(t), c(t) ) for the other observer.


The curve \alpha has a vector velocity concept (a geometric object whose definition is coordinate-free) and this new object will have (in general) different coordinate representation in different coordinate charts.

For example, it happens that:

\alpha&#039;(t) = (1, x&#039;(t), y&#039;(t), z&#039;(t)) is the mathematical expression of the vector velocity of the curve for on observer (with x'(t) I mean dx(t)/dt) and

\alpha&#039;(t) = (1, a&#039;(t), b&#039;(t), c&#039;(t)) for the other observer.


Again, the last three components is what you call "vector velocity of the particle (with respect to that observer).


For the example of this thread, (x&#039;(t), y&#039;(t), z&#039;(t)) is what you call \left(\frac{d\vec{q}}{dt}\right)_A

and (a&#039;(t), b&#039;(t), c&#039;(t)) is what you call \left(\frac{d\vec{q}}{dt}\right)_B


Obviously, in general, the functions x&#039;(t) and a&#039;(t) are different, i.e. in general, x&#039;(t)\not = a&#039;(t), y&#039;(t)\not = b&#039;(t), z&#039;(t)\not = c&#039;(t)


But there is a mathematical relation between those two different coordinate expressions of THE SAME geometric object (just like in any vector space, a given vector has different coordinates in different basis, and there is a mathematical relation between these two set of coordinates; in our example is just the same, this time in the tangent vector space to each point of the manifold).


That relation is precisely what D.H. wrote for this case.


I hope it helps. :-)

 

[Philip Wood]

Hello Matt.
I'm very grateful (this is not ironic) that you've taken the trouble to try to educate me. What your post certainly does is to whet my appetite to study differential geometry. I have to say that regarding the equation (dq/dt)_A = (dq/dt)_B + ω x q, I can't see that your post shows my interpretation to be wrong. But perhaps I've missed something.

I have three textbooks which treat the equation. They are quite old and fairly elementary, but the authors are authoritative: T W Kibble, J L Synge, C Lanczos. All of them treat the equation using the mathematics of vectors; none of them mentions differential geometry. I'm not using this as an argument that DG is irrelevant, but as an indication that the insights of DG can be translated into the ordinary language of vectors.

It seems that both you and DH are trying to point out that I am making some mistake of interpretation. Is it my use of a set of rotating base vectors that you find objectionable? The best post for you to get your teeth into would probably be number 28. Maybe any insights from differential geometry could be expressed in the language of vectors; this could even be a useful exercise for you!

 

[mattt]

I'm certainly not saying that doing Physics in a rotating frame is invalid! Nor am I saying that one observer is right and the other wrong.

I'm merely trying to understand how to interpret the equation you write as

\left( \frac{d \vec q}{d t}\right)_A = \left( \frac{d \vec q}{d t}\right)_B +\ \vec \omega \times \vec q.
I'm arguing that \left( \frac{d \vec q}{d t}\right)_B is only a 'partial' derivative of the vector \vec q because it treats the unit vectors in the rotating (B) frame (rotating with the frame) as constants.

I don't know how to explain it without differential geometry concepts. But I'll try.

What you call "the (total in your wording, I guess) derivative of a vector" is in reality the coordinate expression of a certain geometric object with respect to that coordinate frame

The same object will have different coordinates in different frames.

For each frame, the derivative of that vector is just the usual derivative of its escalar components.

For example, if \vec{q}(t) = (x(t),y(t),z(t)) for a given observer, then \frac{d\vec{q}(t)}{dt} = (x&#039;(t), y&#039;(t), z&#039;(t)) for this observer.

For another different observer, the same vector may have the expression \vec{q}(t) = (a(t),b(t),c(t)), then \frac{d\vec{q}(t)}{dt} = (a&#039;(t), b&#039;(t), c&#039;(t)) for this other observer.


These two sets of coordinates can be much more general than the usual components with respect to an orthonormal vector basis. But for simplicity let us restric to the especial case where (x(t), y(t), z(t)) means

x(t)\vec{i} + y(t)\vec{j} + z(t)\vec{k}

where \vec{i} , \vec{j}, \vec{k} is an orthonormal basis.

And the same with


a(t)\vec{m} + b(t)\vec{n} + c(t)\vec{p}

where \vec{m} , \vec{n}, \vec{p} is another orthonormal basis.


So that for one observer it is

\frac{d\vec{q}(t)}{dt} = x&#039;(t)\vec{i} + y&#039;(t)\vec{j} + z&#039;(t)\vec{k}


and for the other observer it is


\frac{d\vec{q}(t)}{dt} = a&#039;(t)\vec{m} + b&#039;(t)\vec{n} + c&#039;(t)\vec{p}


Obviously I am just stating how it works. To be able to know why it is this way, you'd need to know some differential geometry concepts.

The second term on the right is the 'missing' part of the B-frame derivative that takes account of the rotation (non-constancy) of the unit vectors. See footnote.

No. In fact, from the point of view of B, it is A who is rotating with -w angular velocity. So if you make the exact same reasoning, but from B point of view, you will get to

\left( \frac{d \vec q}{d t}\right)_B = \left( \frac{d \vec q}{d t}\right)_A +\ (-\vec{\omega}) \times \vec q.

On the left hand side, there is only one term, \left( \frac{d \vec q}{d t}\right)_A because in the inertial frame the unit vectors are constants. So \left( \frac{d \vec q}{d t}\right)_A is a genuine, complete, derivative of \vec q.

Footnote
Mathematically, what I'm saying in the paragraph under the equation is that the 'complete' derivative of \vec q expressed in terms of the rotating unit vectors of the 'B' frame is

\frac{d}{dt}\left(q_x \hat {\vec x}+ q_y \hat {\vec y}+ q_z \hat {\vec z}\right)
in which q_x, q_y, q_z are (scalar) components of \vec q on the rotating unit vector set \hat {\vec x}, \hat {\vec y}, \hat {\vec z}.

Differentiating each term in the last equation as a product, and re-assembling:

\frac{d q_x}{dt} \hat {\vec x}+ \frac{d q_y}{dt} \hat {\vec y}+\frac{d q_z}{dt} \hat {\vec z} \ +\ q_x \frac{d \hat {\vec x}}{dt}+q_y \frac{d \hat {\vec y}}{dt}+q_z \frac{d \hat {\vec z}}{dt} \ \ \ = \ \ \ \left( \frac{d \vec q}{d t}\right)_B \ + \ \vec{\omega} \times \vec{q}
In the last step I've identified the sum of the first three terms with your \left( \frac{d \vec q}{d t}\right)_B, (what Lanczos calls \frac{d&#039; \mathbf q}{dt}), and the last three terms with \vec{\omega} \times \vec{q}. The latter can easily be justified.

You can do exactly the same from B point of view.

I know it may be difficult to grasp it without the proper differential geometry concepts, and in fact I have not explained why it is the way it is, I just stated how it is.

I have to go now, I'll try more later if I have time. :-)

 

[D H]

I have three textbooks which treat the equation. They are quite old and fairly elementary, but the authors are authoritative: T W Kibble, J L Synge, C Lanczos. All of them treat the equation using the mathematics of vectors; none of them mentions differential geometry. I'm not using this as an argument that DG is irrelevant, but as an indication that the insights of DG can be translated into the ordinary language of vectors.

I haven't looked at all of those texts, but I suspect they all make a huge hand-wave.

A few years ago my officemate, a former astronomy professor who came over to the dark side of working for industry, complained about the huge handwaving that inevitably arises in those textbook derivations of the kinematics transport theorem. I told him it's trivial if you know the basics of differential geometry and Lie group theory.

He went through a number of texts, from the level of Marion to Goldstein and higher. Almost all of those texts used a big hand-wave argument. He found but one that didn't use a hand-wave, but it put the derivation in a five page long appendix that was a complete math-out (Math-out: noun. A paper or presentation so encrusted with mathematical or other formal notation as to be incomprehensible.)

So what about texts on differential geometry and Lie group theory? This result is so trivial that most texts on differential geometry and Lie groups don't even bother to address it. Why bother with something so trivial and special-purpose as the group SE(3)? Mathematicians generalize, not specialize. From what I've seen, most texts on differential geometry and Lie groups go out of their way to be absolutely inscrutable to anyone but a professional mathematician. It's almost as if their job is to make higher mathematics inscrutable. The author of a mathematics text on differential geometry and Lie groups apparently has done something fundamentally wrong if the text is scrutable to a PhD physicist, PhD astronomer, or PhD engineer.

It seems that both you and DH are trying to point out that I am making some mistake of interpretation. Is it my use of a set of rotating base vectors that you find objectionable? The best post for you to get your teeth into would probably be number 28. Maybe any insights from differential geometry could be expressed in the language of vectors; this could even be a useful exercise for you!

Let's go back to the merry-go-round that I introduced in post #14. Suppose you've painted two vertical arrows pointing upward on the center point, one labeled $\hat z$ and the other, $\hat z'$. You've also painted two orthogonal horizontal arrows on the floor of the merry-go-round labeled $\hat x'$ and $\hat y'$ such that $\hat z' \times \hat x' = \hat y'$. Suppose you are currently suspended from the non-rotating ceiling of the merry-go-round and are in the finishing touches of painting two more horizontal arrows on the ceiling labeled $\hat x$ and $\hat y$ such that $\hat z \times \hat x = \hat y$.

Now suppose someone starts the merry-go-round. From your perspective, the unprimed axes are stationary while the primed axes are rotating. It's the other way around from the perspective of someone sitting on one of the horses on the merry-go-round. Who's right? The answer is that both are right. What's wrong is insisting that only one observer can be right. You have to agree to disagree here because the time derivative of a vector quantity is not an objective (frame invariant) phenomenon. This wikipedia page, http://en.wikipedia.org/wiki/Objectivity_(frame_invariance) does a fairly decent job of getting the idea across, much better than the inscrutable wikipedia pages on differential geometry and Lie group theory.

 

[mattt]

I'll try to explain it a bit more, maybe you can grasp some of it.

We have two observers, each one rotating with respect to the other at omega angular velocity (suppose the z-axis is the same for both, and it is the xy-plane of each one that is rotating with respect to the xy-plane of the other one).

Let us call (t,x,y,z) the space-time coordinates one of them assigns to each event, and (t&#039;,x&#039;,y&#039;,z&#039;) the space-time coordinates the other one assings to the same event.

If you do some math (just as I have just done here in a piece of paper), you'll get:

(t&#039;,x&#039;,y&#039;,z&#039;)\to (t,x,y,z) = (t&#039;, x&#039; cos(w t&#039;) - y&#039; sin(w t&#039;), x&#039; sin(w t&#039;) + y&#039; cos(w t&#039;), z&#039;)

That is the transformation between those two charts (i.e. the relation between the coordinates those two observers assing to the same events).

Imagine we have a particle whose space-time coordinates with respect to the second observer are:

(t&#039;,1,0,0) (that is, the particle is always at the tip of this observer's x'-axis unit vector, so in particular the particle is at rest with respect to the x'y'z'-frame of this second observer, along its x'-axis and at a distance of 1 m away from the origin).If you have understood the change of chart relation, this same particle will have the following space-time coordinates with respect to the first observer:

(t, cos(w t), sin(w t), 0)

(i.e. this same particle is moving, its trajectory is a circle of radius = 1 m, with respect to the xyz-frame of this first observer).In Differential Geometry terms, what you have is a concrete (Euclidean) 4-dimensional Riemannian Manifold M, two charts:

\phi_1 : M\to\mathbb{R}^4

where \phi_1(p) = (t(p), x(p), y(p), z(p)) for each p\in M

and

\phi_2 : M\to\mathbb{R}^4

where \phi_2(p) = (t&#039;(p), x&#039;(p), y&#039;(p), z&#039;(p)) for each p\in Mand a (differentiable) curve

\alpha : I\to MThe image of the curve (which is a subset of M) represents the set of all events where the particle exists.

For each \tau\in I

\alpha(\tau) is a point in M, that represents "a place and a time" (i.e. an event) where the particle exists.The velocity vector of the curve is a concrete differential tangent vector field along the curve, which has a perfect mathematical "coordinate-free" definition (i.e. you don't need coordinates at all to be able to define such geometric object).

It is just a concrete element of the tangent space to M at each point p\in M that belongs to the image of the curve.

It just happens that each chart allow to select a concrete vector basis in the tangent space at each point of the Manifold. So the vector velocity of the curve at each point will have some coordinates with respect to this vector basis each chart selects at each tangent space.For example:

(1, -w sin(w t), w cos(w t), 0) are the coordinates of the vector velocity of the curve, with respect to the vector basis that the chart \phi_1 selects in the tangent vector space at the point p in M of coordinates (t,x,y,z) with respect to this observer.In other words,

1 \vec{u}_{t} - w sin(w t) \vec{u}_x + w cos(w t) \vec{u}_y is the vector velocity of the curve (an element of the tangent space to M at point p of coordinates (t,x,y,z) with respect to this chart) expressed as a linear combination of the vector basis

\{\vec{u_t}, \vec{u_x}, \vec{u_y}, \vec{u_z}\} that \phi_1 allow to select in the tangent space to M at point p of coordinates (t,x,y,z).In the same way,

(1,0,0,0) are the coordinates of THE SAME vector velocity of the curve, with respect to the vector basis that the chart \phi_2 selects in the tangent vector space at the point p in M of coordinates (t',x',y',z') with respect to this second observer.

That is,

\vec{u}_{t&#039;} is the velocity vector of the curve (an element of the tangent space to M at point p of coordinates (t',x',y',z') with respect to this chart \phi_2) expressed as a linear combination of the vector basis

\{\vec{u}_{t&#039;}, \vec{u}_{x&#039;}, \vec{u}_{y&#039;}, \vec{u}_{z&#039;}\} that \phi_2 allow to select in the tangent space to M at point p of coordinates (t',x',y',z').So

1 \vec{u}_{t} - w sin(w t) \vec{u}_x + w cos(w t) \vec{u}_y

and

\vec{u}_{t&#039;}

are the SAME vector ( the same element of the tangent space to M at point p ) just expressed with respect to different vector basis (of that tangent space).
How is this all related to what you already know?The last three coordinates are the "spatial-coordinates" (the first one is the time). If you look above how I defined our example, it is obvious that what you'd call the vector velocity of the particle (the last three coordinates of the vector velocity of the curve) is just zero for the second observer (the particle does not move with respect to the second observer) but it is not zero for the first observer.And what about acceleration?If you compute the acceleration vector of the particle with respect to the first observer, you'll have (I have just done it here in a piece of paper) :

-w^2 cos(w t) \vec{u}_x - w^2 sin(w t) \vec{u}_y
That is, with respect to the first observer, the particle is describing a circle at constant speed, so for him, the acceleration vector of the particle is radial (points to the origin).If you compute the acceleration vector of the particle "naively" with respect to the second observer, you'll get zero (and you'd think it is OK, after all the particle does not move with respect to this second observer).

But if you compute the Covariant Derivative of the vector velocity of the curve in the second chart (second observer), you'll get exactly the same vector that if you compute it with the first chart. You'll get

-w^2 cos(w t) \vec{u}_x - w^2 sin(w t) \vec{u}_y = -w^2 \vec{u}_{x&#039;}Because the Covariant Derivative is again a "coordinate-free" geometric object.In this way you can formulate Newtonian Mechanics in a "coordinate-free" style, stating that the total force upon a particle is just proportional (m being the constant) to the Covariant Derivative of the vector velocity of the curve (that represent that particle)Anyway, you surely will understand this all much better after you study Differential Geometry (I will recommend "Semi-Riemannian Geometry with Applications to Relativity" (Barrett O'Neill) ).EDIT: I forgot to mention that the first observer is (by definition in this example of mine) an inertial observer (that is why with respect to his chart the Covariant Derivative reduces to the "usual" acceleration vector) and so the second observer is not an inertial observer.

 

[Philip Wood]

No. In fact, from the point of view of B, it is A who is rotating with -w angular velocity. So if you make the exact same reasoning, but from B point of view, you will get to

\left( \frac{d \vec q}{d t}\right)_B = \left( \frac{d \vec q}{d t}\right)_A +\ (-\vec{\omega}) \times \vec q.

Agreed

And on my mathematical footnote with the differentiation of (dq/dt)_B as a sum of products of scalar components and unit vectors you write...

You can do exactly the same from B point of view.

Yes. But there is relative rotation between the unit vectors and I've chosen to regard the unit vectors in A as stationary and those in B as rotating. What's wrong with that? After all we agree, I think, that we can distinguish a rotating frame from an inertial frame by experiment (see my post 31), and I'm regarding the unit vectors in the rotating frame as rotating and those in the inertial frame as stationary.

I know it may be difficult to grasp it without the proper differential geometry concepts

Well, thank you for the sympathy, but what exactly is it that I'm finding hard to grasp?

You should also extend your sympathy to all those poor blighters who studied mechanics 30 or 40 years ago (or even quite recently) using textbooks such as the ones I mentioned in post 35, which deal with the equation without using the language of differential geometry.

 

[mattt]

But there is relative rotation between the unit vectors and I've chosen to regard the unit vectors in A as stationary and those in B as rotating. What's wrong with that?

Nothing wrong with that. In the example, you can choose one of them to be an inertial frame (and so the other observer is not inertial), or you could choose both them to not be inertial (and that would be a different example).

Your "problem" was that of "total" versus "partial" derivative. You seemed to "believe" that \left(\frac{d\vec{q}(t)}{dt}\right)_A

was "right" but

\left(\frac{d\vec{q}(t)}{dt}\right)_B

was "wrong" or "incomplete".

Both are equivalent mathematical expressions for the same geometric fact.

After all we agree, I think, that we can distinguish a rotating frame from an inertial frame by experiment (see my post 31), and I'm regarding the unit vectors in the rotating frame as rotating and those in the inertial frame as stationary.

No problem with that at all. Inertial and non-Inertial frames can be distinguished by experiment and also in the geometric-theoretic part inertial and non-inertial frames are also different (their mathematical definition are different).

Well, thank you for the sympathy, but what exactly is it that I'm finding hard to grasp?

You should also extend your sympathy to all those poor blighters who studied mechanics 30 or 40 years ago (or even quite recently) using textbooks such as the ones I mentioned in post 35, which deal with the equation without using the language of differential geometry.

You seem to think that

\left(\frac{d\vec{q}(t)}{dt}\right)_A

is right, but that

\left(\frac{d\vec{q}(t)}{dt}\right)_B

is "wrong" or "incomplete".

That shows there is something you don't understand.

That is what I meant.

 

[D H]

You seem to think that

\left(\frac{d\vec{q}(t)}{dt}\right)_A

is right, but that

\left(\frac{d\vec{q}(t)}{dt}\right)_B

is "wrong" or "incomplete".

That shows there is something you don't understand.

That is what I meant.

We are on exactly the same page, mattt. That was precisely my objection to what Philip wrote.

 

[Philip Wood]

DH Many thanks for post 37. I'm quite prepared to believe that arguments of the type I've given in the mathematical footnote in my post 28, lack rigour, though I'm still not clear why they're wrong. [To repeat my defence: We can distinguish a rotating frame from an inertial frame by experiment within the frame, so why can't we consider the unit vectors anchored to the rotating frame as, err, rotating, and those anchored to the inertial frame as stationary?]

I think that you and Matt have spent a long time trying to make me see the error of my ways. Thank you for your efforts. You've certainly succeeded in making me see the urgency of a course in DG.

Matt. Many thanks for your latest post, which I shall read carefully. Thank you, too for the recommendation of Barrett O'Neill. Seems like a good place to start.

 

[Philip Wood]

Re posts 40, 41 and the 'incomplete' derivative.

In my footnote on post 28 I differentiated q in terms of the B frame representation and showed it to be a sum of two terms. The first is what I think we all agree to call (dq/dt)_B and the second is ω x q. The 'complete' derivative of q is the sum of these two terms.

This is supported by the fact that the sum of these two terms is equal to (dq/dt)_A, which is the complete derivative of q in the A frame because I'm choosing to regard the unit vectors in the A frame as stationary, so there's only one term. So the complete time derivative of q is frame-independent, which is what we'd hope for!

 

[mattt]

Matt. Many thanks for your latest post, which I shall read carefully. Thank you, too for the recommendation of Barrett O'Neill. Seems like a good place to start.

I must warn you that "Semi-Riemannian Geometry with Applications to Relativity" (Barrett O'Neill) is not a book for physicists (I know more than one physicist that did not like that book at all).

I love the book, and more so "Foundations of Differential Geometry" (Kobayashi, Nomizu, 2 volumes), but I am a mathematician, very used to the Definition-Theorem-Proof style in Mathematics.

A non-mathematician would probably hate these wonderful books, I guess :-)

 

[D H]

DH Many thanks for post 37. I'm quite prepared to believe that arguments of the type I've given in the mathematical footnote in my post 28, lack rigour, though I'm still not clear why they're wrong. [To repeat my defence: We can distinguish a rotating frame from an inertial frame by experiment within the frame, so why can't we consider the unit vectors anchored to the rotating frame as, err, rotating, and those anchored to the inertial frame as stationary?]

I gave specific examples in post #25. I'll give a couple more.

1. Try predicting the weather from the perspective of an inertial frame. You don't want to do that. You most seriously do not want to do that.
   
   
2. Consider a solid body with three distinct principal axes. It's angular momentum is given by $L=I\omega$. The rotational analog of Newton's second law is $\vec\tau_{\text{ext}} = \frac {d \vec L} {dt}$, which is valid only in an inertial frame. Now we have a problem. Freshman / sophomore physics problems nicely set up problems where the inertia tensor is constant. In real life, the inertia tensor when viewed from the perspective of an inertial frame is anything but constant. The inertia tensor for a rigid body is constant in a body fixed frame. It's much easier to look at rotational behavior from the perspective of a rotating frame than it is too look at it from the perspective of an inertial frame. Once again, you don't want to do that.

 

[D H]

A non-mathematician would probably hate these wonderful books, I guess :-)

Guessing is not needed. It's a certainty.

I've been working lately with geometric integration techniques, Lie group integrators in particular. There has been a lot of work in this area in the last fifteen years or so. My technical monitor is a PhD mathematician who did a postdoc at the Institute for Advanced Study (in other words, he's not a slouch). After grinding through one too many math journal papers on the subject, I complained to him "Why do you mathematicians seem to go out of your way to make your concepts completely inscrutable and absolutely impractical?" His reply: "That's our job! Read Mr. Hardy's apology."

 

[mattt]

Re posts 40, 41 and the 'incomplete' derivative.

In my footnote on post 28 I differentiated q in terms of the B frame representation and showed it to be a sum of two terms.

No. To "differentiate a vector \vec{q}(t) with respect to B frame" just means the following:

If \vec{q}(t) = a(t)\vec{u}_{x&#039;} + b(t)\vec{u}_{y&#039;} + c(t)\vec{u}_{z&#039;}

where \{\vec{u}_{x&#039;},\vec{u}_{y&#039;},\vec{u}_{z&#039;}\} is the vector basis in the B frame

then

\left(\frac{d\vec{q}(t)}{dt}\right)_B = \frac{da(t)}{dt} \vec{u}_{x&#039;} + \frac{db(t)}{dt} \vec{u}_{y&#039;} + \frac{dc(t)}{dt} \vec{u}_{z&#039;}

by definition.


You are not doing that. You are differentiating q(t) with respect to A frame, and that you can express as a sum of what you get when you differentiate g(t) with respect to B, plus another term.

The first is what I think we all agree to call (dq/dt)_B and the second is ω x q. The 'complete' derivative of q is the sum of these two terms.

No. That "complete" is totally yours and it is not the way these things are named.

This is supported by the fact that the sum of these two terms is equal to (dq/dt)_A, which is the complete derivative of q in the A frame because I'm choosing to regard the unit vectors in the A frame as stationary, so there's only one term. So the complete time derivative of q is frame-independent, which is what we'd hope for!

No. It is not that way.

 

[mattt]

Guessing is not needed. It's a certainty.

I've been working lately with geometric integration techniques, Lie group integrators in particular. There has been a lot of work in this area in the last fifteen years or so. My technical monitor is a PhD mathematician who did a postdoc at the Institute for Advanced Study (in other words, he's not a slouch). After grinding through one too many math journal papers on the subject, I complained to him "Why do you mathematicians seem to go out of your way to make your concepts completely inscrutable and absolutely impractical?" His reply: "That's our job! Read Mr. Hardy's apology."

hahaha. I know, but there are good reasons to the way Mathematics is done.

Some people must do the work of finding the precise mathematical definitions (suitable to represent some "fuzzy" physics concepts) and to find the rigorous mathematical proofs of the relations between them. We (mathematicians) are these (masochists) people. :-)

By the way, sometimes some mates (mathematicians) accuse me of being "a physicist", because sometimes when I am trying to explain something and I don't feel like going along the total rigorous route (I'm too lazy lately) I start explaining it in "the physics way". So physicists accuse me of being "too rigorous" sometimes, and some mathematicians (when talking about physics subjects) accuse me of being "too hand-wavy". :-)

 

[Philip Wood]

I think we may be arguing about names, here. I don't think I claimed that I was differentiating wrt frame B. I was, though, expressing q in terms of unit vectors of frame B, and then doing the differentiation.

The term, 'complete derivative' is indeed my own, and I'm quite happy to withdraw it. How about calling it the frame-independent derivative?

I await your telling me why this would be wrong.

If I ignore your responses it's because I have to be away from my desk for 36 hours.

 

[voko]

This thread reminds me of this one: https://www.physicsforums.com/showthread.php?t=750104