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I'm studying introductory vector calculus and need to confirm/clarify my concepts. The definition of the derivative of a vector (for example in $\mathbb{R}^2$) if the unit vectors are constant throughout the 2D space is in terms of its components: if we have $\mathbb{r}(t)=(x(t), y(t))$ in the standard Cartesian basis, then

$$\frac{d\mathbb{r}}{dt}=\frac{dx}{dt}\mathbb{e}_x+\frac{dy}{dt}\mathbb{e}_y$$

Now if we move to polar coordinates $\rho, \phi$, then the unit basis vectors $\mathbb{e}_{\rho},\mathbb{e}_{\phi}$ will change direction depending on the location in 2D space. To define the derivative in this case, the book that I'm studying gives the following quick method: we see that $\mathbb{r}=\rho \mathbb{e}_{\rho}$ (where $\rho$ is the distance of the vector's endpoint from the origin), which means $$\frac{d\mathbb{r}}{dt}=\frac{d\rho}{dt}\mathbb{e}_{\rho}+\rho\frac{d\mathbb{e}_{\rho}}{dt}$$ enter image description here

So far, so good: $\frac{d\rho}{dt}$ can be calculated since we can express $\rho$ in terms of $x(t)$ and $y(t)$, and differentiate that expression w.r.t. $t$. In this specific case, we can also express $\mathbb{e}_{\rho}=(\cos\phi)\mathbb{e}_x + (\sin\phi)\mathbb{e}_y$. It turns out that $$\frac{d\mathbb{e}_{\rho}}{dt}=\frac{d\phi}{dt}\mathbb{e}_{\phi}$$ because of the specific way $\mathbb{e}_{\rho}$ and $\mathbb{e}_{\phi}$ are defined in terms of $\mathbb{e}_x$ and $\mathbb{e}_y$.

Expressing the same vector $\mathbb{r}$ in a general curvilinear coordinate system $u,v$, enter image description here

To even start differentiating $\mathbb{r}$, we need to find the components of $\mathbb{r}$ in the new system. I'm assuming the way to identify $\mathbb{r}$ is to identify it as the intersection of two coordinate curves $u=c_1$ and $v=c_2$ - in this case, $u=5$ and $v=4$. Is my understanding correct? Is this the way to identify the components of a vector in a curvilinear system?

So if we have some differentiable functions $f,g$ such that $u=f(x,y)$ and $v=g(x,y)$ and $\mathbb{r}=u\mathbb{e}_u+v\mathbb{e}_v$, then $$\frac{d\mathbb{r}}{dt}=\frac{du}{dt}\mathbb{e}_u+u\frac{d\mathbb{e}_u}{dt}+\frac{dv}{dt}\mathbb{e}_v+v\frac{d\mathbb{e}_v}{dt}$$

$\frac{du}{dt}$ can be identified as $\frac{df(x(t),y(t))}{dt}$ and can be evaluated. How does one, in general, express basis vectors $\mathbb{e}_u$ and $\mathbb{e}_v$ in terms of $\mathbb{e}_x$ and $\mathbb{e}_y$? And even if we do manage to define curvilinear basis vectors in terms of $\mathbb{e}_x,\mathbb{e}_y$, it's not necessary that we'll get a nice expression for $\frac{d\mathbb{e}_u}{dt}$ and $\frac{d\mathbb{e}_v}{dt}$ in terms of $\mathbb{e}_u$ and $\mathbb{e}_v$. How do we get the curvilinear components of $\frac{d\mathbb{r}}{dt}$ in that case?

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  • $\begingroup$ A comment on your notation: Usually, the position vector is $\mathbf{r}$ while the length of the position vector is $r$. So, for example, it would be $\frac{d\mathbf{r}}{dt} = \frac{dx}{dt}\mathbf{e}_x + \frac{dy}{dt}\mathbf{e}_y$ and $\mathbf{r} = \rho\mathbf{e}_\rho$ and $r = \sqrt{x^2+y^2} = \rho$ $\endgroup$ – Hal Hollis Dec 26 '19 at 19:31
  • $\begingroup$ @HalHollis: Thanks! I'll edit the question asap $\endgroup$ – Shirish Kulhari Dec 26 '19 at 19:31
  • $\begingroup$ Math mods: Please merge. $\endgroup$ – Qmechanic Dec 29 '19 at 5:35
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This may or may not help, but we can write $$d \vec r=\frac{\partial \vec r}{\partial u}du+\frac{\partial \vec r}{\partial v}dv+\frac{\partial \vec r}{\partial w}dw$$ But by definition of the basis vectors in the $u, v, w$ system, $$d \vec r=\vec e_udu+\vec e_vdv+\vec e_wdw$$ Therefore we have $$\vec e_u=\frac{\partial \vec r}{\partial u}$$ and so on. But $$\frac{\partial \vec r}{\partial u}=\frac{\partial x}{\partial u}\frac{\partial \vec r}{\partial x}+\frac{\partial y}{\partial u}\frac{\partial \vec r}{\partial y}+\frac{\partial z}{\partial u}\frac{\partial \vec r}{\partial z}$$ So $$\vec e_u=\frac{\partial x}{\partial u}\vec e_x+\frac{\partial y}{\partial u}\vec e_y+\frac{\partial z}{\partial u}\vec e_z$$ and similarly for $\vec e_v$ and $\vec e_w$.

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  • $\begingroup$ Thanks for the answer! The math makes sense, but intuitively what does it mean for $\mathbb{e}_u$ to be equal to $\frac{\partial r}{\partial u}$. Could you explain it geometrically, if possible? $\endgroup$ – Shirish Kulhari Dec 27 '19 at 6:42
  • $\begingroup$ I think it must help! Since you can calculate those derivatives by inverting the Jacobian matrix. $\endgroup$ – Guillermo BCN Dec 27 '19 at 9:49
  • $\begingroup$ @Shirish Kulhari In my first equation, ∂𝑟⃗/∂𝑢 is clearly a vector and is the vectorial amount by which displacement changes per unit excursion in the direction such that $u$ changes without $v$ or $w$ changing, so it is the basis vector along the $u$ axis. $\endgroup$ – Philip Wood Dec 27 '19 at 9:50
  • $\begingroup$ I would add that this is in fact the definition of the basis vectors. $\endgroup$ – Guillermo BCN Dec 27 '19 at 10:05
  • $\begingroup$ @PhilipWood: So you expressed the differential vector $d\mathbf{r}$ in terms of $\mathbf{e}_u, \mathbf{e}_v, \mathbf{e}_w$. That's probably because the basis vectors themselves are only defined locally. Does that mean it's not possible to express the non-differential vector $\mathbf{r}$ (as you see in the 2nd figure of the question) in terms of basis vectors in curvilinear coordinates? i.e. Does $\mathbf{r}=u\mathbf{e}_u + v\mathbf{e}_v+w\mathbf{e}_w$ make any sense? $\endgroup$ – Shirish Kulhari Dec 27 '19 at 13:05
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To get the unit vectors in cartesian coordinates, define $\boldsymbol{r} = x e_x + y e_y$ and calculate the partial derivatives $$\frac{\partial \boldsymbol{r}}{\partial x} = e_x \quad \text{and} \quad \frac{\partial \boldsymbol{r}}{\partial y} = e_y.$$ Now suppose the coordinate transformation is given by a locally invertible functions $$\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} f^1(u,v) \\ f^2(u,v)\end{pmatrix}.$$ By plugging the transformation functions into $\boldsymbol{r}$ and taking the derivatives, one obtaines $$\frac{\partial \boldsymbol{r}}{\partial u} = \frac{\partial f^1(u,v)}{\partial u}e_x + \frac{\partial f^2(u,v)}{\partial u}e_y =\boldsymbol{r}_u \quad \text{and} \quad \frac{\partial \boldsymbol{r}}{\partial v} = \frac{\partial f^1(u,v)}{\partial v}e_x + \frac{\partial f^2(u,v)}{\partial v}e_y = \boldsymbol{r}_v.$$ Now you only need to normalize $\boldsymbol{r}_u$ and $\boldsymbol{r}_v$ to get the orthonormal basis vectors $e_u$ and $e_v$, which are expressed in terms of the canonical basis.

The curvilinear components of $\dot{\boldsymbol{r}}$ are quite easy to obtain. The trick is to use the chain rule: $$\dot{\boldsymbol{r}} = \left(\frac{\partial f^1}{\partial u} \dot{u} + \frac{\partial f^1}{\partial v}\dot{v}\right)e_x + \left(\frac{\partial f^2}{\partial u}\dot{u} + \frac{\partial f^2}{\partial v} \dot{v}\right)e_y.$$ Rearranging yields $$\dot{\boldsymbol{r}} = \left(\frac{\partial f^1}{\partial u} e_x + \frac{\partial f^2}{\partial u}e_y\right) \dot{u} + \left(\frac{\partial f^1}{\partial v} e_x + \frac{\partial f^2}{\partial v}e_y\right) \dot{v} = \boldsymbol{r}_u \dot{u} + \boldsymbol{r}_v \dot{v}.$$

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  • $\begingroup$ IMO, \dot{\mathbf{r}} looks better than using \dot{\boldsymbol r}. $\endgroup$ – Kyle Kanos Dec 27 '19 at 0:18
  • $\begingroup$ For reference, I'm on iOS 13.3 & am seeing a massive gap between the dot and the r. $\endgroup$ – Kyle Kanos Dec 27 '19 at 0:19
  • $\begingroup$ One doubt: Why do you equate $\frac{\partial \mathbb{r}}{\partial u}$ to $\mathbb{r}_u$? I can't understand the intuition behind that. $\endgroup$ – Shirish Kulhari Dec 27 '19 at 6:28
  • $\begingroup$ I just used $\mathbf{r}_u$ and $\mathbf{r}_v$ as shorthand notation for the partial derivatives. $\endgroup$ – Tobi7 Dec 27 '19 at 9:42

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