Most physics textbooks I've consulted only confused me more with regards to how to understand rotating coordinate systems so I consulted Spivak's book but I have 2 questions; in particular regarding the derivative near the bottom of the page.

$r' = B(\rho') + B'(\rho)$

Even if we write it out in uncondensed notation, the formula reads $r'(t) = B(t)(\rho'(t)) + B'(t)(\rho(t))$.

Firstly I am not sure how he gets this formula, because it seems like he used a product rule of sorts, which I am not sure how/why it even applies.

And secondly, I'm not sure how to interpret the term $B'(t)$, because from my understanding, $B$ is a function from $\mathbb{R}_{\geq0}$ into Hom$(\mathbb{R}^3)$, so how can we differentiate such an object? I know that the derivative of a function from say from $\mathbb{R}^n$ into $\mathbb{R}^m$ is a linear transformation, but what about the case where the target space is a general vector space?

So if someone could explain this step in more detail, that would be much appreciated :)

Taken from Michael Spivak's book "Physics for Mathematicans"

  • $\begingroup$ Does the $'$ sign denote derivative over time? $\endgroup$ – edm Sep 16 '18 at 9:19
  • $\begingroup$ @edm yes it does $\endgroup$ – peek-a-boo Sep 16 '18 at 9:24
  • $\begingroup$ Are you familiar with differentiation as in Spivak's Calculus on Manifolds? $\endgroup$ – edm Sep 16 '18 at 9:28
  • $\begingroup$ @edm yes that's where I learnt from $\endgroup$ – peek-a-boo Sep 16 '18 at 9:30

First we define differentiability and derivative of a map between finite-dimensional real vector spaces. You should have seen the following definition of differentiability when the vector spaces are $\Bbb R^m,\Bbb R^n$. Let $U$ be an open subset of $\Bbb R^m$. We say that $f:U\to\Bbb R^n$ is differentiable at a point $p\in U$ if there exists a necessarily unique linear transformation $L:\Bbb R^m\to\Bbb R^n$ such that

$$\lim\limits_{h\to0}\frac{\lVert f(p+h)-f(p)-L(h)\rVert_{\Bbb R^n}}{\lVert h\rVert_{\Bbb R^m}}=0,$$

where $h$ is an $\Bbb R^m$ vector. $L$ would be called the derivative of $f$ at $p$.

One thing to note about the above definition is that while we usually use the standard inner product norms on $\Bbb R^m, \Bbb R^n$ in the above definition, we can use any other norms on $\Bbb R^m, \Bbb R^n$ in place of $\lVert\cdot\rVert_{\Bbb R^m},\lVert\cdot\rVert_{\Bbb R^n}$, and the (non-)differentiability of $f$ and derivative of $f$ is the same no matter which norms you use. This is because all norms on finite-dimensional real/complex vector spaces are equivalent, and equivalent norms give us invariant notion of differentiability and derivative. See for https://math.stackexchange.com/a/2183054/356114.

Now, as for arbitrary finite-dimensional vector spaces and maps between them, you can first give each of the vector spaces a norm, and then define differentiability and derivative in the same manner as in $\Bbb R^n$ case. Again, differentiability and derivative of $f$ do not depend on the choice of norms.

As a note, the following abuse of notation will be used. For a differentiable function $f:U\to V$ whose domain $U$ is an open subset of $\Bbb R$. This is very specific. It has to be $\Bbb R$, not any non-one-dimensional vector space, and not any other one-dimensional vector space. The derivative at a point $p$ is a linear map $L:\Bbb R\to V$, which we denote $L=:f'(p)$. Consider the vector $v\in V$ defined by $f'(p)(1)=:v$. We shall abuse notation by saying that $f'(p)$ is the vector $v$ itself.

Now we proceed to product rule. The formula $r'=B(\rho')+B'(\rho)$ is indeed product rule. Product rule actually applies much more generally than what you may have seen. Whenever you have finite-dimensional real vector spaces $U,V,W$, a function $f:U\times V\to W$ that is bilinear, that is, for all $c\in\Bbb R, u_1,u_2,u\in U,v_1,v_2,v\in V$, we have $$f(cu_1+u_2,v)=cf(u_1,v)+f(u_2,v),$$and$$f(u,cv_1+v_2)=cf(u,v_1)+f(u,v_2),$$ then you can derive a product rule involving $f$, as follows.

Say you have differentiable functions $g:\Bbb R\to U, h:\Bbb R\to V$, then the function $\Bbb R\to W, t\mapsto f(g(t),h(t))$ is differentiable and the derivative is $f(g'(t),h(t))+f(g(t),h'(t))$. This is proved with chain rule applied to the composition of $t\mapsto(g(t),h(t))$ and $f$.

The map of applying a linear transformation to a vector, $(B,\rho)\mapsto B(\rho)$ is a bilinear function. So you can use product rule when differentiating $r$.

  • $\begingroup$ in your first paragraph, there are a few places with $m$ and $n$ in the wrong places... however your answer as a whole has been very helpful in answering the doubts I had!! As an aside, your answer also gave me more questions: can we define differentiability in the infinite-dimensional case analogously to the finite-dimensional case? I'm guessing there are some technicalities regarding the equivalence of norms? $\endgroup$ – peek-a-boo Sep 16 '18 at 20:56
  • 1
    $\begingroup$ @peek-a-boo You can talk about Fréchet derivative between Banach spaces: en.wikipedia.org/wiki/Fréchet_derivative. It is a generalisation of the derivative in my answer. Finite-dimensional normed spaces are Banach and linear transformations between them are always bounded, so the derivative in my answer is Fréchet derivative. $\endgroup$ – edm Sep 17 '18 at 2:46

When in "static" linear algebra we replace the standard basis $({\bf e}_i)_{1\leq i\leq 3}$ of $V={\mathbb R}^3$ by a new basis $({\bf u}_i)_{1\leq i\leq 3}$ the necessary data are stored in a matrix $B$, whereby the columns of $B$ are the coordinates of the new basis vectors in terms of the standard basis. If the new basis is again orthonormal the matrix $B$ is orthogonal, and one has $B^{-1}=B^\top$ .

The points ${\bf r}\in V$ are not moved around, but they obtain new coordinates. Let ${\bf r}$ have coordinates $(r_1,r_2,r_3)$ with respect to the standard basis and coordinates $(\rho_1,\rho_2,\rho_3)$ with respect to the new basis. It is customary to write these coordinate triples as column vectors ${\bf r}$ (one is overloading this symbol here), resp. ${\pmb \rho}$. Under this convention we have the formulas $${\bf r}=B{\pmb\rho},\qquad {\pmb\rho}=B^{-1}{\bf r}\ .\tag{1}$$

Now in the quoted passage of Spivak's book the standard basis $({\bf e}_i)_{1\leq i\leq 3}$ is viewed as a rigid frame that has been actually rotated to the new position $({\bf u}_i)_{1\leq i\leq 3}$. But the individual points of space have not been moved. Therefore the formulas $(1)$ are still valid under this view of things.

So far everything was "static". But now we set the film in motion: (i) We have a point $t\mapsto{\bf r}(t)\in V$ that describes an orbit in time, and (ii) the fixed new frame $({\bf u}_i)_{1\leq i\leq 3}$ is replaced by a rotating frame $${\bf u}_i(t)=B(t) {\bf e}_i\qquad(1\leq i\leq3)\ .$$ It follows that the formulas $(1)$ are holding for each fixed $t$: $${\bf r}(t)=B(t){\pmb\rho}(t),\qquad {\pmb\rho}(t)=B^{-1}(t){\bf r}(t)\ .\tag{2}$$ The matrix products appearing in $(2)$ are bilinear, and this implies the product rule when differentiating with respect to $t$. Therefore we obtain, e.g., $${\bf r}'(t)=B'(t){\pmb\rho}(t)+B(t){\pmb\rho}'(t)=B'(t)B^{-1}(t){\bf r}(t)+B(t){\pmb\rho}'(t)\ ,$$ whereby the exact purpose of this last formula remains unclear to me.

  • $\begingroup$ I know that on finite dimensional vector spaces linear transformations and matrices can be thought of interchangeably, but I always struggle in translating back and forth between the two pictures. Your explanation using matrices has been helpful. As for your last comment, on the next page, Spivak refers to another page in the book where he shows that the matrix $B'(t)B^{-1}(t)$ is skew symmetric; he then writes the components of this matrix in such a way that the product of this matrix with $r(t)$ can be expressed as a cross product of a vector $\omega(t)$ and $r(t)$. $\endgroup$ – peek-a-boo Sep 16 '18 at 20:47

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