Rotating frame: a question of interpretation

[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'(t) = (1, x'(t), y'(t), z'(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'(t) = (1, a'(t), b'(t), c'(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'(t), y'(t), z'(t)) is what you call \left(\frac{d\vec{q}}{dt}\right)_A

and (a'(t), b'(t), c'(t)) is what you call \left(\frac{d\vec{q}}{dt}\right)_B


Obviously, in general, the functions x'(t) and a'(t) are different, i.e. in general, x'(t)\not = a'(t), y'(t)\not = b'(t), z'(t)\not = c'(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'(t), y'(t), z'(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'(t), b'(t), c'(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'(t)\vec{i} + y'(t)\vec{j} + z'(t)\vec{k}


and for the other observer it is


\frac{d\vec{q}(t)}{dt} = a'(t)\vec{m} + b'(t)\vec{n} + c'(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' \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',x',y',z') 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',x',y',z')\to (t,x,y,z) = (t', x' cos(w t') - y' sin(w t'), x' sin(w t') + y' cos(w t'), z')

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',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'(p), x'(p), y'(p), z'(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'} 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'}, \vec{u}_{x'}, \vec{u}_{y'}, \vec{u}_{z'}\} 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'}

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'}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'} + b(t)\vec{u}_{y'} + c(t)\vec{u}_{z'}

where \{\vec{u}_{x'},\vec{u}_{y'},\vec{u}_{z'}\} 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'} + \frac{db(t)}{dt} \vec{u}_{y'} + \frac{dc(t)}{dt} \vec{u}_{z'}

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

 

[Philip Wood]

Yes indeed!

I suspect that in the current thread we have a clash of language and cultures (new vs. old), compounded by my own poor terminology, rather than a direct contradiction.