How can I make the notion that a basis is fixed in time with respect to itself more precise?

Question

Let $A = (O, (e_x,e_y,e_z))$ be the standard frame in $\mathbb R^3$, such that $O = (0,0,0)$ and $(e_x,e_y,e_z)$ is the standard basis. Additionally, let $B = (O, (b_x,b_y,b_z))$ be a moving frame in $\mathbb R^3$ with respect to $A$ with the same origin as $A$ but a different basis (i.e. $(b_x,b_y,b_z)$).

Because $B$ is moving with respect to $A$, then the projections of the basis vectors $(b_x,b_y,b_z)$ onto $A$'s basis vectors $(e_x,e_y,e_z)$ are time-dependent. That is, $$ \begin{align} b_x(t) &= b_x^{x}(t) \cdot e_x + b_x^{y}(t) \cdot e_y + b_x^{z}(t) \cdot e_z \\ b_y(t) &= b_y^{x}(t) \cdot e_x + b_y^{y}(t) \cdot e_y + b_y^{z}(t) \cdot e_z \\ b_z(t) &= b_z^{x}(t) \cdot e_x + b_z^{y}(t) \cdot e_y + b_z^{z}(t) \cdot e_z \end{align} $$ where $b_i^j (t)$ represents the projection of $b_i(t)$ onto $e_j$ at time $t$. This relationship between frame $B$ and frame $A$ demonstrates the clear idea that a basis can be time-varying because its projections onto another basis are time-varying while the basis vectors themselves are not. However, I am struggling to apply a similar example when projecting a basis onto itself to show that it is always time-independent with respect to itself.

That is, intuitively, frame $B$'s basis vectors $(b_x,b_y,b_z)$ don't move over time with respect to themselves. However, I am struggling to make this idea mathematically precise. Indeed, I can express $(b_x,b_y,b_z)$ with respect to themselves as $$ \begin{align} b_x(t) &= 1 \cdot b_x(t) + 0 \cdot b_y(t) + 0 \cdot b_z(t) \\ b_y(t) &= 0 \cdot b_x(t) + 1 \cdot b_y(t) + 0 \cdot b_z(t) \\ b_z(t) &= 0 \cdot b_x(t) + 0 \cdot b_y(t) + 1 \cdot b_z(t) \\ \end{align} $$ but I am not sure how to reason from here that $b_x(t), b_y(t),$ and $b_z(t)$ are constant vectors for all time.

Answer 1

The problem that you are running into is that if you decompose an arbitrary vector $$ \mathbf v(t) = \sum_{i} v_i(t) \mathbf e_i(t) = v_i \mathbf e_i \tag{1} $$ (with an implicit Einstein sum and time arguments on the right), then the time derivative of this vector is not quite the time derivative of its components $\dot v_i$ but rather, by the product rule, $$ {\mathrm d\mathbf v\over\mathrm dt}=\dot v_i \mathbf e_i + v_i \dot{\mathbf e}_i. $$ The “correct way” to handle this, in the sense that it will probably be something you can do today to get yourself moving as fast as possible in the short term, is to use decomposition (1) to decompose these $\dot{\mathbf e}$ terms, $$ \dot{\mathbf e}_i = T_{ij} \mathbf e_j $$ and from there you just insist that all coordinates have such a T-matrix, but sometimes it is the zero matrix. Thus we have (renaming some dummy indices) $${\mathrm d\mathbf v\over\mathrm dt}=\dot v_i ~\mathbf e_i + v_k ~T_{ki} ~\mathbf e_i,\\ {\mathrm dv_i\over\mathrm dt} = {\partial v_i\over\partial t} + \sum_{k} S_{ik} v_k. $$ (Here $S = T^{\mathrm T}$ is the transpose of $T$ which is maybe easier to work with?)

This distinction between the total time derivative and the partial time derivative will immediately be clear to any physics audience you put it to, because we all had to suffer through the same thing for our fluid mechanics classes. In fact Amit in a parallel answer insists on using fixed coordinates but you have a particle that is moving through that space, and I believe that if you are in that circumstance you even get the familiar $\mathbf v\cdot\nabla$ form for $S$ if you work at it? In addition, you also see some things similar with Christoffel symbols in general relativity, so half your audience has seen something like that before, too.

But yeah, it immediately resolves your difficulties right now by just saying “oh, fixed coordinates just correspond to S=0,” taking A as just one of the things you need to have specified up-front to have a good set of coordinates. And so when someone sets up say the rotating frame of reference $$\hat p=\hat x \cos\omega t + \hat y \sin\omega t, \\\hat q = {-\hat x}\sin\omega t + \hat y\cos\omega t,$$ you just insist that one of their commitments was, $$S=\omega \begin{bmatrix}0&-1\\1&0\end{bmatrix},$$and you insist that we carry this $S$ through every expression.

Answer 2

The concept that you are reaching for is called a connection. Remember that vectors are geometric objects that are elements of a vector space. In principle, they exist independently of any basis at all. When you talk about time dependent basis sets, what you really need to do is take one step back and think about time dependent vectors first.

You need to have some method of connecting a vector at one time to a vector at another time. This is what the connection does. Once you have a connection, then you can use the connection to determine if a basis vector is time dependent or not. You very easily could choose a basis vector function that is time-varying with respect to the connection.