acceleration of a particle moving along a streamline using tensor calculus

Question

In a steady flow, the streamline coincides with the particle trajectory. In a book on MHD, I saw that if I pick a streamline $C$ and $s$ is a curvilinear coordinate measured along $C$, $V(s)$ is the speed $|\mathbf{u}|$, then the acceleration of a particle can be written as:

$$V\frac{\mathrm{d} V}{\mathrm{d} s} \hat{\mathbf{e}}_t - \frac{V^2}{R}\hat{\mathbf{e}}_n$$

where $R$ is the radius of curvature of the streamline, $\hat{\mathbf{e}}_t$ and $\hat{\mathbf{e}}_n$ represent unit tangent vectors tangential and normal to the streamline.

Since I do not understand the origin of the first term (second term is clear) and since I recently started learning tensor calculus more properly (using Pavel Grinfeld's book "Introduction to Tensor Analysis and the Calculus of Moving Surfaces") and i thought, i would derive this as it seemed straight forward.

I derived that in general, acceleration should have the form:

$$\mathbf{A} = \frac{\mathrm{d} \mathbf{V}}{\mathrm{d} s} = \left( \frac{\mathrm{d} V^i}{\mathrm{d} s} + V^j V^k \Gamma^i_{jk} \right)\mathbf{X}_i$$

where $\mathbf{X}_i$ represents the covariant basis. But I got stuck since I had no idea how to define the basis, I know that one basis vector has to be along the streamline and thus is parallel to the velocity vector but i have no idea how to set up the coordinates.

Since the expression should be valid for a general streamline, i believe it should be valid for a circular arc of radius $a$. Therefore, I setup my parameter dependent coordinates as: \begin{align*} x^1(s) &= a\,,\\ x^2(s) &= \varphi(s)\,. \end{align*}

and i can use polar coordinate system where I know all the necessary details (e.g. Christoffel symbols etc). I got: \begin{align*} V^1(s) &= 0\,,\\ V^2(s) &= \varphi'(s)\,,\\ A^1(s) &= \Gamma^1_{jk}V^j V^k = -a(V^2)^2 = -a\left( \frac{\mathrm{d} \varphi}{\mathrm{d} s} \right)^2 = -a\left( \frac{U^2}{a^2} \right) \\ &= -\frac{U^2}{a}\,,\\ A^2(s) &= \left( \frac{\mathrm{d} V^2}{\mathrm{d} s} + V^j V^k \Gamma^2_{jk} \right) = \left( \frac{\mathrm{d} \varphi'(s)}{\mathrm{d} s} + 2 \frac{1}{a}V^1 V^2 \right)\\ &= \frac{\mathrm{d} \omega(s)}{\mathrm{d} s} = \frac{\mathrm{d} }{\mathrm{d} s}\left( \frac{U(s)}{a}\right)\\ &= \frac{1}{a}\frac{\mathrm{d} U(s)}{\mathrm{d}s} \end{align*} Where I used that $\varphi'(s) = \omega(s) = U(s)/a$ and also denote the magnitude of velocity as $U(s)$ rather than $V(s)$ as in the picture to avoid confusion. Plugging in: \begin{align*} \mathbf{A} &= A^1 \mathbf{X_1} + A^2 \mathbf{X_2} = -\frac{U^2}{a}\mathbf{X_1} + \frac{1}{a}\frac{\mathrm{d} U(s)}{\mathrm{d}s}\mathbf{X_2}\\ &= -\frac{U^2}{a}\hat{\mathbf{X}}_1 + \frac{\mathrm{d}U(s)}{\mathrm{d}s}\hat{\mathbf{X}}_2 \end{align*} where the hatted are unit vectors. I did not get the answer even in this special case which begs the question why?

The questions are: first, how to recover the equation for the acceleration of a particle along the streamline in the special case that the path is a circular arc (or circle) --- what is the mistake I am making. Second, how to setup the problem so that I can get to the result for the general path.

Note: I think the first issue might have something to do with the notation and the curve parametrization $s(t)$ and I should be doing derivative with respect to time $t$ so the curve is parametrised in the sense as $\gamma(s(t))$ but then, i am not sure how to put together that curve is parametrized by $s$ but also by time because why cannot i directly parametrize the curve by $t$ and then call it $s$ and I am back where i started.

Answer 1

looks like you are making mistake in the calculation of the radial acceleration

A_radial = d^2r/dt^2 = -r(dθ/dt)^2

it is because the radial velocity is zero and the acceleration is due solely to the centripetal force.

For your Second Questions ,

• Computing the acceleration (a) of a particle along a general streamline in a fluid flow, one can use either a Curvilinear coordinate system or a Lagrangian approach

• The Curvilinear coordinate system is defined along the streamline, and the velocity field in the fluid can be expressed in terms of these coordinates.

• The derivatives of the velocity with respect to these coordinates then provide the acceleration of the particle.

• More , the Lagrangian approach tracks the motion of a fluid particle over time, treating the fluid as a continuum. it is particularly useful for studying unsteady fluid flows.

Answer 2

$\pmb a=a^i\pmb e_i$ where $\pmb e_i$ is a covariant basis $$ a^i=\dot v^i+\Gamma_{jk}^i v^jv^k $$ In polar coordinates $\pmb e_1=\pmb e_\rho$ and $\pmb e_2=\pmb e_\theta$ (i.e. coordinates $\rho,\theta$), so the nonzero Christoffel symbols are $$ \Gamma_{22}^1=\Gamma_{\theta\theta}^\rho=-\rho\qquad \Gamma_{12}^2=\Gamma_{\rho\theta}^\theta=\frac1\rho=\Gamma_{\theta\rho}^\theta=\Gamma_{21}^2 $$

$\pmb \rho=\rho\,\pmb e_\rho$, $\pmb v=v^i\pmb e_i$, and $v^\rho=\dot\rho,\,v^\theta=\dot\theta$. So we have $$ \begin{align*} a^\rho&=\dot v^\rho+\Gamma_{\theta\theta}^\rho v^\theta v^\theta=\ddot\rho-\rho\dot\theta^2\\ a^\theta&=\dot v^\theta+\Gamma_{\rho\theta}^\theta v^\rho v^\theta=\ddot\theta+\frac2\rho\dot\rho\dot\theta \end{align*} $$

For a uniform circular motion, $\ddot\rho=0,\,\dot\rho=0, \,\ddot\theta=0$, $$ \begin{align*} a^\rho&=-\rho\dot\theta^2=-\rho\omega^2\\ a^\theta&=0 \end{align*} $$