User talk:Ozob/Archive 2

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Thanks for all your work on this difficult proposal. I am enjoying that a great deal of the effort into ensuring that any trial would be well-run and give us the right information is coming from editors on the oppose side of things. :) Lot 49a^talk 22:31, 6 January 2009 (UTC)

Hi,

I noticed you voted oppose in the flag revs straw pole and would like to ask if you would mind adding User:Promethean/No to your user or talk page to make your position clear to people who visit your page :) - Thanks to Neurolysis for the template   «l| Ψrometheăn ™|l»  (talk) 07:22, 8 January 2009 (UTC)

Thanks for addressing 800x600 window size of a math formula. See explanation at User_talk:Wikid77#Confusion over math formulas. -Wikid77 (talk) 12:35, 24 February 2009 (UTC)

Hi Ozob, I really don't think that "a course of study" is a good descriptor there, both on the substance and in terms of compatibility with the rest of the lead. For one thing, it's too narrow: calculus, as a mathematical discipline, consists of a collection of techniques and some proofs that go along with them. Now, there are some courses that purport to teach these techniques … (Techniques: fundamental; how/whether they are taught: secondary). Do you see what I mean? Further down, the lead talks about how calculus is widely applicable. Obviously, it's the principles and techniques of calculus that are applied, not "the course of study". I don't insist on the word "discipline", but calculus is a fairly well defined part of mathematics, and it's a lot more than the name of a course or a popular textbook title. I've made a few other changes (might as well …), but this is what triggered my involvement. Good job, otherwise! Cheers, Arcfrk (talk) 04:16, 23 April 2009 (UTC)

Having read your argument, I agree with you: "course of study" isn't as good as I thought it was. I'm going to put "discipline" back in. Thanks for the feedback! Ozob (talk) 19:19, 23 April 2009 (UTC)

Hi Ozob, I removed places that for my view have confusing grammar. however if you find some more places with poor grammar i will appreciate your help. Aleks kleyn (talk) 02:27, 26 April 2009 (UTC)

Hallo Ozob, is this list Talk:Problem_of_Apollonius#Old_sources still up to date? Is the help still needed? Have you found some of the sources? --DrJunge (talk) 17:59, 1 May 2009 (UTC)

Hi Ozob.

I like your improvement of the intro to this section, except that I disagree with this statement:

A different choice of reference frame will produce the same derivative function, but it will usually have a different formula because of the different choice of reference frame.

In general, the derivative of a vector function is different in every reference frame. Your statement seems to confuse a reference frame with set of basis vectors. If we were only talking about multiple different bases fixed in the same reference frame, the statement would be true.

The concept of relative velocity is a good counterexample, I think. If you consider that the velocity (as the derivative of displacement or position) of a pilot sitting in an airplane, it should be apparent that you don't get the same function by differentiating in different reference frames. The velocity of the pilot relative to the Earth may be very large and time-varying, but his velocity relative to the airplane is identically zero. It is not merely switching bases, but it is an entirely different function.

Do I misunderstand you? MarcusMaximus (talk) 00:50, 1 August 2009 (UTC)

No, I think you do understand me, but we have a difference in viewpoint. I think I am still right; what's confusing is that in a certain sense I think you are also right. I am considering the reference frame as a coordinate chart on a manifold, i.e., we have functions x,y,z : M → R, where M is the space under consideration (which just so happens to be Euclidean space) and the map (x,y,z) : M → R³ is a diffeomorphism. The velocity function is a function on M, so it and its derivative are absolute in some sense. I.e, assuming a choice of connection, there is only one derivative of a function, namely the one given by the connection. But the derivative can be coordinatized in many different ways, and a choice of x,y,z leads to a choice of a derivative in your sense. Your sense, as far as I understand it, is that there is no space M, but only what I would call coordinate charts and their automorphisms. From this viewpoint, one always defines things in coordinates and then checks that it behaves under coordinate chart changes; that is, that it transforms tensorially. And since there is no analog of M, there is no sense in which there is one function with many possible formulas. Instead you get a different function in every reference frame.

There is another complication, which is that you seem to always use the standard Euclidean connection in each reference frame. This explains your airplane example: "Velocity relative to Earth" is one connection, and in the reference frame in which Earth is fixed, it is the standard Euclidean connection; "Velocity relative to the plane" is a different connection, given in the reference frame of the plane by the standard Euclidean connection. Of course, when you transform one reference frame into another, you don't transform the two connections into each other!

Does that make sense? I'm not sure what to do about the article, but am I at least understanding your viewpoint? Ozob (talk) 19:02, 1 August 2009 (UTC)

No, I actually have no idea what that means. I apologize, but I'm not trained in the mathematical theory you are talking about. I'm only trying to say that for a Euclidean vector, as they are called in the article, the right hand sides of the derivatives taken in two different reference frames are not merely the same function expressed in different bases, in which case you could perform a series of substitutions and get the same formula. They are literally different functions, related by the second formula shown here defining the derivative of a vector function is one reference frame as a function of the derivative taken in another reference frame.

Perhaps you could explain how what I see as two different functions, i.e. identically zero velocity vs. a time-varying velocity, are the same function? (that is a serious question, not rhetorical) MarcusMaximus (talk) 23:48, 1 August 2009 (UTC)

OK, I'll try again. In more elementary language, what I'm saying is that your notion of derivative changes when you change reference frames. Not only is the identically zero function not the same as a non-identically zero function; they are not the same kind of derivative! Abstractly, this is described by a connection. A connection is just an abstraction of the differentiation operator. When you take velocity relative to the Earth, you're measuring a rate of change with respect to the Earth, and this is a totally different rate of change than rate of change with respect to the plane. But once you have chosen something to measure against, the rate of change is independent of the reference frame.

Let me put some formulas to things. Suppose that x(t) = t is the position of the plane relative to the Earth. If we want to find the velocity of the plane with respect to the Earth, we take the derivative and get 1 unit per unit time. Suppose now we measure the position of the plane in its own reference frame and we get y(t) = 0. To measure the velocity of the plane with respect to itself, we take the derivative and get zero. OK so far. But what if we are in the reference frame of the Earth and we want to know the velocity of the plane with respect to itself? We can't take the derivative because we are in the wrong reference frame to do that. What we would like to do is convert to the reference frame of the plane, take the derivative there, then convert back to the reference frame of the Earth. That three-step operation satisfies all the formal properties of the derivative, e.g., it has a Leibniz rule. Therefore it's a connection. But it's not the usual connection (because that's the derivative, and the derivative in the reference frame of the Earth is velocity with respect to the Earth). Instead it lets you measure things relative to the plane while you're in the reference frame of the Earth; e.g., the Earth itself is found to be moving at velocity -1 relative to the plane. And that velocity is independent of the reference frame: The Earth is always moving at velocity -1 with respect to the plane, no matter what reference frame you're in!

Does that make more sense? Connections turn up in general relativity and in the Yang-Mills equations, so you might be able to find a better physically-motivated description of them in stuff on those. Ozob (talk) 20:00, 2 August 2009 (UTC)

Yes, I think we are in agreement on the basic phenomenon,

When you take velocity relative to the Earth, you're measuring a rate of change with respect to the Earth, and this is a totally different rate of change than rate of change with respect to the plane.

which I believe is the end of the discussion. However, this part really baffles me:

But what if we are in the reference frame of the Earth and we want to know the velocity of the plane with respect to itself? We can't take the derivative because we are in the wrong reference frame to do that. What we would like to do is convert to the reference frame of the plane, take the derivative there, then convert back to the reference frame of the Earth.

With respect to the bold text, why does it matter what reference frame "we" are "in" when we take this derivative? "We" don't appear anywhere in the equations that define the position of the pilot, the airplane, or the Earth. The equation for the derivative only takes account of the reference frame in which the derivative is being taken. There appears to be no other dependency; we are only talking about the relative motion of points with respect to each other. What I mean is, the velocity of the pilot relative to his plane is merely the velocity of the pilot relative to his plane—the velocity of a pilot-fixed point moving through a field of infinitely many airplane-fixed points. It matters not how any other objects or persons in the universe are moving. Am I mistaken? MarcusMaximus (talk) 06:47, 3 August 2009 (UTC)

I think we've just about worked it out, but that part you quoted I phrased badly. Let me try yet again. This time I'll start by asking, physically, what is the derivative? It is the rate of change in a reference frame; in the reference frame of the Earth, it's the rate of change with respect to the Earth, and in the reference frame of the plane, it's the rate of change with respect to the plane. Suppose that we want to measure the velocity of a bird relative to the plane. We observe that in the reference frame of the Earth, the plane has position x(t) = t and the bird has position b(t) = 0.1t. How do we measure the velocity of the bird relative to the plane? We can't do this by taking the derivative of the position function of the bird. This is what I meant when I said that we were in the wrong reference frame to do that: "Take the derivative" means "Find the velocity with respect to the present reference frame, i.e., with respect to the Earth"--which of course is not what we want. That's where the three-step procedure (and the connection it determines) come in. In the reference frame of the plane, the bird moves like -0.9t. The derivative in this reference frame is -0.9. Finally, that -0.9 was measured in the reference frame of the plane, so to be scrupulous and pedantic we must convert back to the reference frame of the Earth, because that's where we got our measurements from. The velocity doesn't change when do this, though; this last step isn't actually necessary. (But I'm including it anyway, because if you were trying to determine a position in the reference frame of the Earth by converting to the reference frame of the plane, then this step would be important.)

Are we agreed? As I said before, I dont't know what to do about the article; what I wrote has inspired so much discussion above that there must be a better way of saying it. Ozob (talk) 03:52, 4 August 2009 (UTC)

I still don't think I agree. While the velocities of the bird and the airplane relative to Earth are interesting, they are not necessary. The velocity of the bird in the reference frame of the airplane is the derivative of the position vector connecting a point in the bird to any point fixed in the airplane (since in the airplane frame, by definition, all points fixed in the airplane have zero velocity). There is no need to worry about the Earth or any measurements taken from it--it's a problem involving only two points and one reference frame.

If, as you are suggesting, we are constrained to take measurements from a different reference frame, that is a subjective limitation, not one inherent to kinematics. MarcusMaximus (talk) 04:22, 4 August 2009 (UTC)

Such limitations can be real. What if instead of a plane and a bird we have two elementary particles or two stars? We currently have no way of taking measurements in those frames directly.

Besides your comment about necessities of measurement, I couldn't see any specific objections in what you wrote. Ozob (talk) 04:30, 4 August 2009 (UTC)

Now I am interested to know what this statement means:

There are infinitely many possible reference frames because there are many nonequivalent bases for a vector space, even after rescaling.

I think different reference frames are more than just different bases in a vector space, aren't they? MarcusMaximus (talk) 07:34, 4 August 2009 (UTC)

And by the way, let me say that I really appreciate the discussions we have had on this page. Thank you for your patience and insight. MarcusMaximus (talk) 07:48, 4 August 2009 (UTC)

The discussions have been good for me, too. It's clarified my own understanding.

Or at least, it has to the point of realizing that I'm clueless. I've decided that I don't know what a frame of reference is. What I thought at the beginning was that it was the same as choice of an orthonormal basis, what I would call a frame. That would be nice, right? Frame and frame of reference? That's what I was thinking when I made that edit. But of course you can't represent curvilinear coordinates by taking linear combinations of basis vectors. So then I decided that it was a choice of coordinates, as I said above. In that case, the statement you quoted above is obviously wrong. But a mere choice of coordinates isn't enough to distinguish an inertial frame of reference from a non-inertial one. You can't tell whether there are any fictitious forces solely on the basis of your coordinates; you can use cartesian or polar coordinates for both inertial and non-inertial reference frames.

What I think a reference frame might be now is a choice of a connection. This might sound kind of goofy, because a connection is something that lets you take derivatives. It makes some sort of sense to me because an acceleration is a second derivative, so if you get fictitious forces, then your second derivatives are bad. If I'm right, an inertial reference frame is a flat connection, and a non-inertial reference frame is a non-flat connection. This is speculation, though; someone at WP:WPPhys might actually know. Ozob (talk) 00:32, 5 August 2009 (UTC)

Ok, well, I can't comment on connections. But I can say that to confuse vector bases, coordinate systems, and reference frames is one of the most common errors in kinematics, and there is no shame in it. In my experience even most college professors don't seem to understand it clearly. Many people blend the concepts together by talking about a hybrid entity called a "coordinate frame", which I and my colleagues regard as just asking for trouble. There is actually a hierarchy among them. First you need a reference frame. Then you can fix in that reference frame a set of mutually orthogonal basis unit vectors, which are free vectors that have a direction only. Tnen you pick any point in the reference frame as the origin and establish a coordinate system by defining coordinate axes parallel to the basis vectors. Reference frames are still quite useful even without coordinate systems. MarcusMaximus (talk) 01:09, 5 August 2009 (UTC)

When you pick a reference frame, what does that mean mathematically? How does the choice of reference frame affect a calculation? Ozob (talk) 14:00, 5 August 2009 (UTC)

I'm afraid I can't give you a rigorous mathematical definition of a reference frame in terms of manifolds or connections all that stuff. But a reference frame is fundamentally a set of points whose relative positions are constant with time. Therefore, if you have a door on a hinge with a vector embedded in the door (imagine an arrow painted on the door representing the vector), there are two reference frames that are immediately relevant: a reference frame consisting of the door and all points fixed in the door, and all "imaginary" points in the universe extending in all directions that move with the door as it swings; and a reference frame fixed relative to the doorframe/house/Earth. Now imagine swinging the door open and closed--the vector you painted on the door obviously remains unchanged in the reference frame of the door, always pointing exactly parallel to the door. But relative to the Earth, that vector is swinging back and forth, pointing in a different direction relative to the house at every instant.

The calculation of the derivative is different in each reference frame, for example, because when you express the vector using a formula in terms of a basis fixed in the door and take the derivative in the reference frame of the door, the derivatives of all the door-fixed basis vectors are zero, so the second term in the product rule goes to zero and you have a derivative like this one.

However, using exactly the same formula for the vector (using basis vectors fixed in the door) and taking the derivative in the Earth-fixed frame, the basis vectors fixed in the door have nonzero derivatives as the door swings. This is shown here.

If you compare the formula at the first link to the first formula at the second link, you'll see that the only difference is the second part of the product rule for differentiation is zero in the first case.

You can also use a linear transformation (like a direction-cosine matrix) to express the vector in terms of vectors fixed in the Earth frame at any step along the way and get a different formula. However, a linear transformation cannot get you from one reference frame to the other, because they are fundamentally different functions.

This concept took me a while to wrap my mind around. MarcusMaximus (talk) 16:43, 5 August 2009 (UTC)

By the way, I just took a look at the article on frames of reference. It is terrible. It is self-contradictory. In the opening sentence it says a reference frame is a coordinate system. Then it quotes 5 or 6 guys who say the concepts of reference frame and coordinate system are distinct. No wonder nobody knows what they are talking about! MarcusMaximus (talk) 04:00, 6 August 2009 (UTC)

Well, I tried looking in the two books on mathematical physics that I happened to have handy, namely Frankel's Geometry of Physics and Dubrovin–Fomenko–Novikov's Modern Geometry, Part 1 (which isn't really on physics, but it talks about physics a bit). The first wasn't helpful at all; I didn't see any mention of moving reference frames. The second was only slightly helpful; at one point they talk about reference frames moving with respect to each other in terms of Lagrangians. I've never studied enough physics to really figure out Lagrangians, so this left me unenlightened. All I got out of it was that it may be possible to describe the rotation (or lack of rotation) of a reference frame by putting certain terms in the Lagrangian.

But now I've gotten pretty confident that a reference frame is really a connection like I was thinking before. I'm going to write this all down here, mostly for my own benefit, but maybe you'll get something out of it too.

Here's the setup I want to use: Let the manifold M be the real numbers R. Choose a coordinate t on M. Let E be the trivial vector bundle on M of rank three (for simplicity). In other words, E is just one copy of R³ for each point of M; since M is just R, E is really R¹ × R³ together with the information that the first copy of R is special and corresponds to M. A (smooth) section of E is by definition a (smooth) function s from M to E of the form t → (t, s¹(t), s²(t), s³(t)); that is, for each time t, s chooses a vector in R³, i.e., s is really a parameterized curve in R³ written in a funny way.

A connection (mathematics) can be defined in many ways—I like connection (vector bundle)—but for the present purposes it's good to think of it as a covariant derivative. A covariant derivative ∇ with coefficients in E is a function that takes two things, a vector field and a section of the vector bundle E, and determines another section of E called the derivative of the original section with respect to the vector field. It satisfies a bunch of good properties which I won't list here. The important thing for the present purposes is that a covariant derivative can always be expressed as follows: Suppose that the section is written (using the Einstein convention) as sⁱe_i, where e₁, e₂, e₃ are the standard basis sections of E (i.e., for each time t, e_i(t) is the ith standard basis vector of R³ at time t). Suppose also that d/dt represents the tangent vector field on M which points to the right with length one at all times. Then we always have the following formula:

${\displaystyle \nabla _{d/dt}(s^{i}\mathbf {e} _{i})={\frac {ds^{i}}{dt}}\mathbf {e} _{i}+s^{i}\nabla _{d/dt}(\mathbf {e} _{i}).}$

This is the Leibniz rule for connections. ${\displaystyle \nabla _{d/dt}(\mathbf {e} _{i})}$ is a section of E by definition, so it must have an expression in terms of the basis sections, too. We set:

${\displaystyle \nabla _{d/dt}(\mathbf {e} _{i})=\omega _{i}^{k}\mathbf {e} _{k}.}$

The ω^k_i are called the coefficients of the connection; it's fair to interpret them as the curvilinear partial derivative of e_i in the kth coordinate direction. Note that they have only two indices instead of the usual three because M has only one coordinate and hence there's no point in indexing over that one coordinate. In terms of these coefficients, we get, after reindexing:

${\displaystyle \nabla _{d/dt}(s^{i}\mathbf {e} _{i})={\frac {ds^{i}}{dt}}\mathbf {e} _{i}+s^{j}\omega _{j}^{i}\mathbf {e} _{i}=\left({\frac {ds^{i}}{dt}}+s^{j}\omega _{j}^{i}\right)\mathbf {e} _{i}.}$

This should be looking familiar. If the connection coefficients are zero, then we're in the case of a non-moving reference frame. If not, then we should be able to use the connection coefficients to describe how the reference frame's movement affects the derivative of s. That is, we need to replicate the formula:

${\displaystyle {\frac {{}^{\mathrm {N} }d\mathbf {a} }{dt}}=\sum _{i=1}^{3}{\frac {da_{i}}{dt}}\mathbf {e} _{i}+\sum _{i=1}^{3}a_{i}{\frac {{}^{\mathrm {N} }d\mathbf {e} _{i}}{dt}}}$

somehow. This is easy. We let ωⁱ_j equal the ith component of ${\displaystyle {}^{\mathrm {N} }d\mathbf {e} _{j}/dt}$. Tada! There we go. Ozob (talk) 22:49, 6 August 2009 (UTC)

Just out of curiosity, does this definition of angular velocity agree with your analysis?

${\displaystyle {}^{\mathrm {N} }\omega ^{\mathrm {E} }=\mathbf {e} _{1}({\frac {{}^{\mathrm {N} }d\mathbf {e} _{2}}{dt}}\cdot \mathbf {e} _{3})+\mathbf {e} _{2}({\frac {{}^{\mathrm {N} }d\mathbf {e} _{3}}{dt}}\cdot \mathbf {e} _{1})+\mathbf {e} _{3}({\frac {{}^{\mathrm {N} }d\mathbf {e} _{1}}{dt}}\cdot \mathbf {e} _{2})}$

Or in simpler notation, where the overdot replaces ${\displaystyle {}^{\mathrm {N} }d/dt}$,

${\displaystyle {}^{\mathrm {N} }\omega ^{\mathrm {E} }=\mathbf {e} _{1}{\dot {\mathbf {e} }}_{2}\cdot \mathbf {e} _{3}+\mathbf {e} _{2}{\dot {\mathbf {e} }}_{3}\cdot \mathbf {e} _{1}+\mathbf {e} _{3}{\dot {\mathbf {e} }}_{1}\cdot \mathbf {e} _{2}}$

MarcusMaximus (talk) 03:50, 7 August 2009 (UTC)

Well, let's find out if I can fit it in somehow. I don't know how one would define "angular velocity" in an intrinsic way except by trying to use the curvature of the connection, so I'm going to try that first.

I'm going to describe connections in the way that connection (vector bundle) does. That is, now ∇ is a vector bundle homomorphism E → E ⊗ Ω that satisfies the Leibniz rule; here Ω is the cotangent bundle of M. In coordinates, this means that ∇ looks like this:

${\displaystyle \nabla (\mathbf {e} _{i})=\omega _{i}^{k}\mathbf {e} _{k}\otimes dt.}$

The curvature is ∇² : E → E ⊗ Λ²Ω. But Λ²Ω is the zero bundle; in local coordinates, the unique coordinate dt on Ω determines the coordinate dt ∧ dt, which is zero.

OK, that didn't work. But one thing that's been bothering me since yesterday is that the approach I'm taking really doesn't unify space and time; it treats them as separate (time is the manifold, space is the vector bundle). So I'm going to try something else.

Let M be R⁴. Give M the standard Lorentzian metric, that is, we define the dot product of two vectors (x₀ = t, x₁, x₂, x₃) and (y₀ = u, y₁, y₂, y₃) to be tu − x₁y₁ − x₂y₂ − x₃y₃. (Note c = 1.)

The tangent bundle of M is TM = M × R⁴. A connection on the tangent bundle of M is a vector bundle homomorphism ∇ : TM → TM ⊗ Ω satisfying the Leibniz rule; in coordinates, it sends

${\displaystyle \nabla (\partial _{i})=\Gamma _{ji}^{k}\partial _{k}\otimes dx^{j}.}$

(This is really the same as the definition of a connection I gave above, but rephrased.) The Γs are the ωs from before, but now a lot more of them are zero. If we have a tangent vector field ${\displaystyle s^{i}\partial _{i}}$, then using the Leibniz rule we get:

${\displaystyle \nabla (s^{i}\partial _{i})=\partial _{i}\otimes ds^{i}+s^{i}\nabla (\partial _{i})=\partial _{i}\otimes ds^{i}+s^{i}\Gamma _{ji}^{k}\partial _{k}\otimes dx^{j}.}$

To replicate the results of my last post, we just let all the coefficients ${\displaystyle \Gamma _{ji}^{k}}$, where j is not zero, equal zero. Sort of; to handle a curve a(t), what we ought to do is pullback the tangent bundle of M along a and use the pullback connection. Unfortunately then we're in the same situation as above, where the curvature is zero for dimension reasons. I'm going to press on with the computations in the hope that this reveals something to me.

Anyway, there's a standard formula for the curvature of a connection which can be found at Riemann curvature tensor:

${\displaystyle R_{jkl}^{i}=\partial _{k}\Gamma _{lj}^{i}-\partial _{l}\Gamma _{kj}^{i}+\Gamma _{km}^{i}\Gamma _{lj}^{m}-\Gamma _{lm}^{i}\Gamma _{kj}^{m}.}$

Look at the last two terms, the ones with only Γs and no partial derivatives. Recall that ${\displaystyle \Gamma _{ji}^{k}}$ is zero when j is not zero; so unless both l and k are zero, those last two terms vanish; and if they're both zero, then upon substituting zero we see again that the last two terms also vanish. So we get:

${\displaystyle R_{jkl}^{i}=\partial _{k}\Gamma _{lj}^{i}-\partial _{l}\Gamma _{kj}^{i}.}$

By similar reasoning, we get:

${\displaystyle R_{jkl}^{i}=0,}$ when k and l are both not zero,

${\displaystyle R_{jk0}^{i}=\partial _{k}\Gamma _{0j}^{i},}$ when k is not zero,

${\displaystyle R_{j0l}^{i}=-\partial _{l}\Gamma _{0j}^{i},}$ when l is not zero,

${\displaystyle R_{j00}^{i}=\partial _{0}\Gamma _{0j}^{i}-\partial _{0}\Gamma _{0j}^{i}=0.}$

This looks like it can't replicate the angular velocity that you have above. There are no partials in the time direction, i.e., nowhere do you get a ${\displaystyle \partial _{0}}$. I guess this is consistent with the observation that it didn't work before.

For the moment, I'm out of ideas as to what the angular velocity represents in a strictly mathematical sense. Maybe this is obvious, but could you tell me where that formula comes from? It's pretty opaque to me. Ozob (talk) 00:36, 8 August 2009 (UTC)

It came from a very good dynamics textbook, Dynamics: Theory and Applications by Kane and Levinson. I work with Levinson on a daily basis and he taught me most of what I know about kinematics and dynamics. You can download the book for free here. The definition of angular velocity is presented at the top of page 16 of the text, which is page 36 of the PDF. I'm not sure if the definition of angular velocity is derived from anything, but it's a very intuitive definition--at least it makes sense physically in my head. I'll try to think of a way to explain it in text.

I was hoping this equation

${\displaystyle \nabla _{d/dt}(s^{i}\mathbf {e} _{i})={\frac {ds^{i}}{dt}}\mathbf {e} _{i}+s^{j}\omega _{j}^{i}\mathbf {e} _{i}=\left({\frac {ds^{i}}{dt}}+s^{j}\omega _{j}^{i}\right)\mathbf {e} _{i}.}$

would lead logically to this one

${\displaystyle {\frac {{}^{\mathrm {N} }d\mathbf {a} }{dt}}=\sum _{i=1}^{3}{\frac {da_{i}}{dt}}\mathbf {e} _{i}+\sum _{i=1}^{3}a_{i}{\frac {{}^{\mathrm {N} }d\mathbf {e} _{i}}{dt}}={\frac {{}^{\mathrm {E} }d\mathbf {a} }{dt}}+{}^{\mathrm {N} }\mathbf {\omega } ^{\mathrm {E} }\times \mathbf {a} }$

because somehow the indices on the ${\displaystyle s^{j}\omega _{j}^{i}\mathbf {e} _{i}}$ would work cyclically to form ${\displaystyle {}^{\mathrm {N} }\omega ^{\mathrm {E} }\times \mathbf {a} .}$

If you start with the statement you made, can you make any progress from there?

We let ωⁱ_j equal the ith component of ${\displaystyle {}^{\mathrm {N} }d\mathbf {e} _{j}/dt}$.

I don't really know what ranges those indices are supposed to cover, or I'd do it myself.

MarcusMaximus (talk) 06:41, 8 August 2009 (UTC)

The indices are supposed to go from 1 to 3; x₁, x₂, and x₃ are the x, y, and z coordinates, respectively. (The reason for using this notation rather than using x, y, and z is just that it's convenient for making computations.)

The first part of the first equation that you quoted leads to the first part of the second equation that you quoted. We have:

${\displaystyle \nabla _{d/dt}(s^{i}\mathbf {e} _{i})={\frac {{}^{\mathrm {N} }d\mathbf {a} }{dt}},}$

${\displaystyle {\frac {ds^{i}}{dt}}\mathbf {e} _{i}+s^{j}\omega _{j}^{i}\mathbf {e} _{i}=\sum _{i=1}^{3}{\frac {da_{i}}{dt}}\mathbf {e} _{i}+\sum _{i=1}^{3}a_{i}{\frac {{}^{\mathrm {N} }d\mathbf {e} _{i}}{dt}}.}$

I should reindex the left-hand side of the first equation by swapping the is and js in the second term. If I also stop using the Einstein summation convention on the left-hand side, then I get:

${\displaystyle \sum _{i=1}^{3}{\frac {ds^{i}}{dt}}\mathbf {e} _{i}+\sum _{i=1}^{3}\sum _{j=1}^{3}s^{i}\omega _{i}^{j}\mathbf {e} _{j}=\sum _{i=1}^{3}{\frac {da_{i}}{dt}}\mathbf {e} _{i}+\sum _{i=1}^{3}a_{i}{\frac {{}^{\mathrm {N} }d\mathbf {e} _{i}}{dt}},}$

which I think makes the equality a little more transparent: a_i equals sⁱ, and ω^j_i is the component of ^Nde_i/dt in the direction of the jth basis vector.

The one thing which I haven't yet worked out is where the cross product comes in. So I looked at the book you referenced. It looks very good! Unfortunately, as you say, it doesn't justify the definition of angular velocity, it simply states it. It's still helpful, though: It's convincing me that angular velocity behaves in a good way, and that the definition above isn't arbitrary but really comes from something.

I'm going to try to reason my way backwards this time, by equating the two terms that I don't understand. For each i, we can suppose that a(t) = s(t) = e_i, and then we ought to have:

${\displaystyle \omega _{i}^{j}\mathbf {e} _{j}={}^{\mathrm {N} }\omega ^{\mathrm {E} }\times \mathbf {e} _{i}.}$

Equations (9) and (10) on page 17 of the textbook tell me that the right-hand side should equal ${\displaystyle {\dot {\mathbf {e} }}_{i}.}$ But that's just the statement that we can write things in terms of ^Nde_i/dt as above, so that's not really helpful. What I think I'd like to do is to see what components ^Nω^E should have. Let's write its kth component as ^Nω^E_k. If we write out all the terms of that equation above, we get:

${\displaystyle \omega _{1}^{1}\mathbf {e} _{1}+\omega _{1}^{2}\mathbf {e} _{2}+\omega _{1}^{3}\mathbf {e} _{3}=\omega _{1}^{i}\mathbf {e} _{i}={}^{\mathrm {N} }\omega ^{\mathrm {E} }\times \mathbf {e} _{1}={}^{\mathrm {N} }\omega _{3}^{\mathrm {E} }\mathbf {e} _{2}-{}^{\mathrm {N} }\omega _{2}^{\mathrm {E} }\mathbf {e} _{3},}$

${\displaystyle \omega _{2}^{1}\mathbf {e} _{1}+\omega _{2}^{2}\mathbf {e} _{2}+\omega _{2}^{3}\mathbf {e} _{3}=\omega _{2}^{i}\mathbf {e} _{i}={}^{\mathrm {N} }\omega ^{\mathrm {E} }\times \mathbf {e} _{2}={}^{\mathrm {N} }\omega _{1}^{\mathrm {E} }\mathbf {e} _{3}-{}^{\mathrm {N} }\omega _{3}^{\mathrm {E} }\mathbf {e} _{1},}$

${\displaystyle \omega _{3}^{1}\mathbf {e} _{1}+\omega _{3}^{2}\mathbf {e} _{2}+\omega _{3}^{3}\mathbf {e} _{3}=\omega _{3}^{i}\mathbf {e} _{i}={}^{\mathrm {N} }\omega ^{\mathrm {E} }\times \mathbf {e} _{3}={}^{\mathrm {N} }\omega _{2}^{\mathrm {E} }\mathbf {e} _{1}-{}^{\mathrm {N} }\omega _{1}^{\mathrm {E} }\mathbf {e} _{2}.}$

So we deduce that:

${\displaystyle {\begin{array}{lll}\omega _{1}^{1}=0,&\omega _{1}^{2}={}^{\mathrm {N} }\omega _{3}^{\mathrm {E} },&\omega _{1}^{3}=-{}^{\mathrm {N} }\omega _{2}^{\mathrm {E} },\\\omega _{2}^{1}=-{}^{\mathrm {N} }\omega _{3}^{\mathrm {E} },&\omega _{2}^{2}=0,&\omega _{2}^{3}={}^{\mathrm {N} }\omega _{1}^{\mathrm {E} },\\\omega _{3}^{1}={}^{\mathrm {N} }\omega _{2}^{\mathrm {E} },&\omega _{3}^{2}=-{}^{\mathrm {N} }\omega _{1}^{\mathrm {E} }&\omega _{3}^{3}=0.\end{array}}}$

OK, this is looking promising! Apparently there really is some sort of good relation between the connection coefficients and the angular velocity. But I'm not really sure how to express it abstractly: All I see at the moment is "plug in the numbers in a nice-looking way". This is some sort of low-dimensional phenomenon relating three-dimensional vector spaces and three-dimensional matrices; you can't do this in two or four and higher dimensions because you don't have the right number of matrix entries. And at the moment I can't see what the relation is; I'm sure I can look it up somewhere, though. Ozob (talk) 16:09, 8 August 2009 (UTC)

Oh, wait, this is obvious. There is a linear transformation of vector spaces R³ → R³ determined by "send v to its cross product with ^Nω^E". ω^j_i is just the matrix of that linear transformation. The skew-symmetry of the matrix corresponds to the symmetry you see in the formula for a cross product (which ultimately comes from the skew-symmetry of the wedge product).

In a sense, this answers the question of the relationship between ^Nω^E and ω^j_i. But in another sense it doesn't: For a linear transformation to have the form "cross product with a vector" is very, very special. In three dimensions (which we're in), this happens exactly when the matrix is skew-symmetric. As it turns out, skew-symmetry of the matrix of connection coefficients happens exactly when the connection is "compatible with the metric", i.e., that it satisfy a certain condition relating the connection and inner products. So that explains why angular velocity is a vector and why the interesting part of the connection can be represented as a cross product: It's because the connection is compatible with the metric and we're in three dimensions.

So here is the overall picture as I see it at the moment:

• Spacetime, in classical physics, is R¹ × R³.
• A reference frame is a connection on the tangent bundle of spacetime which is:
  1. Compatible with the metric, and
  2. Zero in all the spacelike directions.
• The derivative in a certain reference frame is just an application of the connection.
• The angular velocity of a reference frame with respect to an inertial frame of reference is the connection form.
• Because the connection is zero in all the spacelike directions, the interesting part of the connection form is a 3 × 3 matrix.
• Because the connection is compatible with the metric, the 3 × 3 matrix is skew-symmetric.
• Skew-symmetric 3 × 3 matrices are determined by vectors. The vector corresponding to the connection form is the angular velocity.
• A connection determines an inertial frame of reference if and only if its torsion tensor vanishes.
• All reference frames correspond to a flat spacetime, meaning that the Riemann curvature tensor vanishes.

OK, that's a lot more complicated than I thought it would be way back when we started this discussion. But I think I've got it; do you have any comments? I hope I haven't lost you! Ozob (talk) 20:51, 8 August 2009 (UTC)

I just moved, so I've been without internet access for a while. Unfortunately a lot of what you just said is over my head, but I do know about the relationship between skew-symmetric matrices and the cross product.

The only comment I have is that you don't have to specify that one of the two reference frames is an inertial reference frame; this relationship of derivatives works for any two reference frames. This is a purely kinematical (which roughly means "geometric and mathematical") relationship, while inertia and "inertiality" only is important when you're dealing with dynamics and kinetics (the interaction of bodies and forces).

Thanks for the fantastic discussion. I'm glad to see that all of this makes sense at a fundamental level, even though my knowledge only starts somewhere in the middle of the hierarchy. MarcusMaximus (talk) 18:42, 20 August 2009 (UTC)

Hmm. Well, even if neither of the reference frames is assumed inertial, we should still be able to figure out the angular velocity somehow. I'm going to guess that all one does is compute the angular velocities of the two reference frames with respect to an inertial frame, then takes their difference.

Come to think of it, I bet there's a direct way of doing this. There was a step above where I said:

${\displaystyle \omega _{i}^{j}\mathbf {e} _{j}={}^{\mathrm {N} }\omega ^{\mathrm {E} }\times \mathbf {e} _{i}.}$

Now, while it doesn't look like it, there's an inertial frame of reference hidden here on the right hand side. The cross product is usually defined as a goofy looking differential operator with some weird stuff happening in the indices; in fact, it's not, it's very natural and comes right out of the wedge product and the Hodge dual. The Hodge dual is something to do with the ordinary differential d on Euclidean space, that is, it's related to the Euclidean derivative. I wouldn't be surprised if there's some way to define a curved Hodge dual with respect to a connection; and that using this you could define a curved cross product. If you stuck in the curved cross product on the right hand side, then you'd eliminate the implicit use of an inertial reference frame. That may not be particularly useful; after all, it should just work out to be what you'd get by finding the angular velocities of N and E with respect to an inertial frame and taking their difference. But it would be satisfying if it were true, because then all of this stuff would be kinematic like it ought to be.

Discussing these things with you is always enlightening. If you have any more questions, just ask. I'm sure I'll learn something. Ozob (talk) 22:04, 20 August 2009 (UTC)

First, I don't think I deleted any of your comments on quaternion talk. If I did I am terribly sorry, and am wondering if it might have somehow been an edit conflict, where we were both typing at the same time.

Second yes, this question that I am wondering about right now, I have been editing a bit, because the way I first typed it did not really sound as clear as I could make it after I went back and read it some more. Plus I wanted to add links to it.

Right now I am reading Jasper Jolly's 1905 Manual of Quaternions. Jolly if you don't know him well, was a royal astronomer of Ireland, a post that Hamilton had once held, and edited the 1898 version of Elements of Quaternions that Hamilton died before completing, based on the earlier 1865 version.

I will try and be more diligent in avoiding deleting comments, not sure if I actually did that, but if I did I am sorry and will not do it again.

Thanks for all your interesting comments and contributions. I always find it very interesting reading about what you have ask for. TeamQuaternion (talk) 01:34, 8 August 2009 (UTC)

Although this is relatively minor, I happened to notice the addition of, "If R is unital and is not the trivial ring {0}, the Jacobson radical is always distinct from R.", to the article. Are there any interesting non-trivial consequences of this property? In many interesting cases of the theory, it is desirable to study rings with J(R) = R; for instance "If R is nil, is the polynomial ring over R Jacobson radical?", is an equivalent form of the (open) Köthe conjecture. Apart from the fact that the quotient R/J(R) will be non-trivial for a unital ring, I do not know of any properties which can be deduced from this. Could you please tell me what you had in mind? Thanks, --PST 00:46, 20 August 2009 (UTC)

Oh, all I had in mind there was to not lose the fact that you'd removed (because it had been incorrectly stated). I noticed that the fact, properly stated, was a trivial consequence of Zorn's lemma, and while I really don't know anything about these sorts of rings, it seemed to me that J(R) equaling R or not would be pretty important. You seem to know more about these things; if it's actually not an interesting fact, then feel free to remove it. Ozob (talk) 21:44, 20 August 2009 (UTC)

Thanks, I have kept the fact but added a sentence which covers more general cases that I mentioned above. --PST 04:22, 25 August 2009 (UTC)

Dear Ozob, thanks for your advise on User talk:129.97.227.25. The edit was mine. I do not understand, why dx should be written as dx as this is to my view clearly against any common sense: variables are generally written in italics and are composed out of one (1) letter, every other use would start confusion (this notations is according to AMS style). This "d" here is not to be taken as a variable multiblied by x.

Otherwise, what is the value of ${\displaystyle \int _{0}^{\pi }\sin(t)d\rho dt}$

I wonder, whether it is ${\displaystyle \pi \sin(t)dt\,}$ or ${\displaystyle 2d\rho \,}$

Well, as it seems, you define this with spaces and so on, ok. but I disagree. However it will be like it says.

Cheers, Saippuakauppias ⇄ 01:13, 5 December 2009 (UTC)

I agree that variables are often written in italics and are only one letter long. However, dx can be considered as an infinitesimal variable, in which case it does not make sense to separate the d from the x. One can also consider d to be a function, the exterior derivative, and functions are also often written in italics. There are many ways of interpreting dx, and it often makes sense to italicize the d. I think that I have always seen the d italicized in AMS publications.

I would like to know where you learned to write an upright d. I have only ever seen it done on Wikipedia, but you seem to have learned it somewhere else. Where? Ozob (talk) 13:44, 5 December 2009 (UTC)

It is standard in the Springer style. Boris Tsirelson (talk) 21:02, 3 December 2013 (UTC)

I very much appreciate your work here. :) --Moonriddengirl ^(talk) 14:52, 20 December 2009 (UTC)

The lead section of arithmetic variety didn't seem to be copied from the source, although the section on Kazhdan's theorem was. I removed the speedy tag and stubbed the article. If you disagree, then please restore the tag. Also, I have been tagging the articles with {{copyvio}} rather than {{db-copyvio}}. Is the latter preferred in some more obvious cases? Sławomir Biały (talk) 13:09, 21 December 2009 (UTC)

Yes, I think I may have been a bit too hasty with the speedy tag on that article. I've been using {{db-copyvio}} when I felt that nothing substantial would be left. In the case of arithmetic variety, we would be left with two not obviously equivalent definitions and no context; I suppose that's something, so {{copyvio}} looks more appropriate. But the {{db-copyvio}} template I just added to noncommutative resolution is okay, for example, because the article had a single sentence, and that sentence was a copyvio. Ozob (talk) 17:38, 21 December 2009 (UTC)

Hi. A contributor to the clean-up at the CCI asks whether material can be presumptively deleted. It can, in accordance with policy. There's more about this at Wikipedia talk:WikiProject Mathematics#Copyright concerns related to your project, including a template that may prove helpful should you wish to take this approach. --Moonriddengirl ^(talk) 18:16, 21 December 2009 (UTC)

Hello Ozob, and thanks for your work patrolling new changes. I am just informing you that I declined the speedy deletion of Albert–Brauer–Hasse–Noether theorem - a page you tagged - because: Not an unambiguous copyright infringement, or there is other content to save. Please review the criteria for speedy deletion before tagging further pages. If you have any questions or problems, please let me know. JohnCD (talk) 18:50, 21 December 2009 (UTC)

Apologies - I meant to come back sooner and add to the rather curt notice that the CSDHelper script produces. My reasoning was that the copyvio passage could be removed leaving a valid stub article. I didn't realise what a swamp I was stepping into, see conversation with Moonridengirl here. Regards, JohnCD (talk) 20:39, 21 December 2009 (UTC)

That's okay. But the first sentence may also be a copyvio: See the first sentence of the source he cites for it. Given that and this user's history (and that the template lets you fill in only one url as far as I'm aware) I thought it was better to just nuke the page. Failing that I think it should be blanked and rewritten; I'm sure I could use the sources he cites to come up with appropriate content. Ozob (talk) 04:22, 22 December 2009 (UTC)