Proof of equilavenc between Eq. (8) and Euler-Lagrange equation (163)

Equation (8) is repeated here, i.e.,

$$\frac{d}{d t} \left( \frac{\partial \mathcal{L}}{\partial \dot{\e... ...}} \right) = \frac{\partial \mathcal{L}}{\partial \ensuremath{\boldsymbol{X}}},$$ (162)

where $\partial / \partial \dot{\ensuremath{\boldsymbol{X}}}$ and $\partial / \partial \ensuremath{\boldsymbol{X}}$ are considered as gradient operators. Note that, in Cartisian coordinates, the components of Eq. (8) are obviously equivalent to the respective Euler-Lagrange equations. We now check whether the component equations in arbitrary coordinate system obtained by evaluating the gradient of the Lagrangian $\mathcal{L}$ in Eq. (8) are equivalent to the respective Euler-Lagrange equations in that coordinates system. The Euler-Lagrange equation in any coordinates is given by

$$\frac{d}{d t} \left( \frac{\partial \mathcal{L}}{\partial \dot{q}_i} \right) = \frac{\partial \mathcal{L}}{\partial q_i},$$ (163)

which is expressed in terms of a single coordinate component, and is coordinate independent, i..e, it takes the same form for every components of any coordinate systems.

In this section, I prove that the components of Eq. (8) are equivalent to the Euler-Lagrange equation (163) in cylindrical coordinates $(R, Z, \varphi)$. First, let us consider the term $\partial \mathcal{L} / \partial \dot{\ensuremath{\boldsymbol{X}}}$, which, in Cartesian coordinators, is written as

$$\frac{\partial \mathcal{L}}{\partial \dot{\ensuremath{\boldsymbol... ...artial \mathcal{L}}{\partial \dot{z}} \widehat{\ensuremath{\boldsymbol{e}}}_z .$$ (164)

Using the chain rule, the above equation is written as

$$\frac{\partial \mathcal{L}}{\partial \dot{\ensuremath{\boldsymbol{X}}}} = \left( \frac{\partial \mathcal{L}}{\partial R} \frac{\partial R}{... ...tial \dot{Z}}{\partial \dot{x}} \right) \widehat{\ensuremath{\boldsymbol{e}}}_x$$

$$+ \left( \frac{\partial \mathcal{L}}{\partial R} \frac{\partial R}{... ...tial \dot{Z}}{\partial \dot{y}} \right) \widehat{\ensuremath{\boldsymbol{e}}}_y$$

$$+ \left( \frac{\partial \mathcal{L}}{\partial R} \frac{\partial R}{... ...tial \dot{Z}}{\partial \dot{z}} \right) \widehat{\ensuremath{\boldsymbol{e}}}_z$$

Using the transformation relation

$$\left\{ \begin{array}{l} R = \sqrt{x^2 + y^2} \varphi = \ensure... ...{ArcSin}} \left( \frac{y}{\sqrt{x^2 + y^2}} \right) Z = z \end{array} \right.$$ (165)

we obtain $\partial R / \partial \dot{x} = 0$, $\partial R / \partial \dot{y} = 0$, etc. Thus $\partial \mathcal{L} / \partial \dot{\ensuremath{\boldsymbol{X}}}$ reduces to

$$\frac{\partial \mathcal{L}}{\partial \dot{\ensuremath{\boldsymbol... ...tial \dot{Z}}{\partial \dot{z}} \right) \widehat{\ensuremath{\boldsymbol{e}}}_z$$ (166)

Using the transformation relation Eq. (165), we obtain

$$\dot{R} = \frac{x \dot{x}}{\sqrt{x^2 + y^2}} + \frac{y \dot{y}}{\sqrt{x^2 + y^2}},$$ (167)

$$\dot{\varphi} = - \frac{y}{x^2 + y^2} \dot{x} + \frac{x}{x^2 + y^2} \dot{y},$$ (168)

and

$$\dot{Z} = \dot{z}$$ (169)

Noting that $\partial \dot{Z} / \partial \dot{x} = 0$, $\partial \dot{Z} / \partial \dot{y} = 0$, $\partial \dot{R} / \partial \dot{z} = 0$, $\partial \dot{\varphi} / \partial \dot{z} = 0$, Eq. (166) is further written as

$$\frac{\partial \mathcal{L}}{\partial \dot{\ensuremath{\boldsymbol... ... \mathcal{L}}{\partial \dot{Z}} \right) \widehat{\ensuremath{\boldsymbol{e}}}_z$$ (170)

Using Eqs. (167) and (168), we obtain

$$\frac{\partial \dot{R}}{\partial \dot{x}} = \frac{x}{R} = \cos \varphi$$ (171)

$$\frac{\partial \dot{R}}{\partial \dot{y}} = \frac{y}{R} = \sin \varphi$$ (172)

$$\frac{\partial \dot{\varphi}}{\partial \dot{x}} = - \frac{y}{R^2}$$ (173)

$$\frac{\partial \dot{\varphi}}{\partial \dot{y}} = \frac{x}{R^2}$$ (174)

Using these, Eq. (170) is written as

$$\frac{\partial \mathcal{L}}{\partial \dot{\ensuremath{\boldsymbol{X}}}} = \left( \frac{\partial \mathcal{L}}{\partial \dot{R}} \cos \varphi... ... \mathcal{L}}{\partial \dot{Z}} \right) \widehat{\ensuremath{\boldsymbol{e}}}_z$$

$$= \frac{\partial \mathcal{L}}{\partial \dot{R}} \left( \cos \varphi... ... \mathcal{L}}{\partial \dot{Z}} \right) \widehat{\ensuremath{\boldsymbol{e}}}_z$$

$$= \frac{\partial \mathcal{L}}{\partial \dot{R}} \widehat{\ensuremat... ...\partial \mathcal{L}}{\partial \dot{Z}} \widehat{\ensuremath{\boldsymbol{e}}}_z$$ (175)

Using this, the left-hand side of Eq. (8) is written as

$$\frac{d}{d t} \left( \frac{\partial \mathcal{L}}{\partial \dot{\ensuremath{\boldsymbol{X}}}} \right) = \frac{d}{d t} \left( \frac{\partial \mathcal{L}}{\partial \dot{R}... ...d t} \left( \frac{1}{R} \widehat{\ensuremath{\boldsymbol{e}}}_{\varphi} \right)$$

$$= \frac{d}{d t} \left( \frac{\partial \mathcal{L}}{\partial \dot{R}... ...R - \widehat{\ensuremath{\boldsymbol{e}}}_{\varphi} \frac{\dot{R}}{R^2} \right]$$ (176)

Next, consider the right-hand side of Eq. (8), which is the space gradient of Lagrangian $\mathcal{L}$. When I at first considered this problem, I took it for granted that the space gradient in cylindrical coordinates should be

$$\frac{\partial \mathcal{L}}{\partial \ensuremath{\boldsymbol{X}}}... ...{\partial \varphi} \frac{1}{R} \widehat{\ensuremath{\boldsymbol{e}}}_{\varphi},$$ (177)

which turns out to be wrong because this formula does not take into account that $\mathcal{L}$ depends on $\dot{\ensuremath{\boldsymbol{X}}}$ which in turn depends on the spatial coordinates. The correct way to calculate the space gradient is as follows:

$$\frac{\partial \mathcal{L}}{\partial \ensuremath{\boldsymbol{X}}} = \frac{\partial \mathcal{L}}{\partial x} \widehat{\ensuremath{\bol... ...\frac{\partial \mathcal{L}}{\partial z} \widehat{\ensuremath{\boldsymbol{e}}}_z$$

$$= \left[ \frac{\partial \mathcal{L}}{\partial R} \frac{\partial R}{... ...\frac{\partial \mathcal{L}}{\partial Z} \widehat{\ensuremath{\boldsymbol{e}}}_z$$ (178)

It is important to note that $\mathcal{L}$ depends on $\dot{R}$, $\dot{\varphi}$ which in turn depend on $x$ and $y$. As a result, there exist additional terms (the last two terms in both the brackets), which would be missed if we used the formula in Eq. (177). Equation (178) is further written as

$$\frac{\partial \mathcal{L}}{\partial \ensuremath{\boldsymbol{X}}} = \left[ \frac{\partial \mathcal{L}}{\partial R} \cos \varphi + \fr... ...\frac{\partial \mathcal{L}}{\partial Z} \widehat{\ensuremath{\boldsymbol{e}}}_z$$

$$= \frac{\partial \mathcal{L}}{\partial R} \widehat{\ensuremath{\bol... ...tial \dot{\varphi}}{\partial y} \right] \widehat{\ensuremath{\boldsymbol{e}}}_y$$ (179)

Using

$$\frac{\partial \dot{R}}{\partial x} = \dot{x} \frac{y^2}{R^3} - y \dot{y} \frac{x}{R^3}$$

$$= \frac{\dot{x}}{R} \sin^2 \varphi - \frac{\dot{y}}{R} \sin \varphi \cos \varphi$$

$$= - \dot{\varphi} \sin \varphi$$ (180)

$$\frac{\partial \dot{R}}{\partial y} = \dot{y} \frac{x^2}{R^3} - x \dot{x} \frac{y}{R^3}$$

$$= \frac{\dot{y}}{R} \cos^2 \varphi - \frac{\dot{x}}{R} \sin \varphi \cos \varphi$$

$$= \dot{\varphi} \cos \varphi,$$ (181)

$$\frac{\partial \dot{\varphi}}{\partial x} = - y \dot{x} \left( - \frac{2 x}{R^4} \right) + \dot{y} \left( \frac{1}{R^2} - \frac{2 x^2}{R^4} \right)$$

$$= \frac{2 \dot{x}}{R^2} \cos \varphi \sin \varphi + \frac{\dot{y}}{R^2} \left( 1 - 2 \cos^2 \varphi \right)$$

$$= \frac{2 \dot{x}}{R^2} \cos \varphi \sin \varphi + \frac{\dot{y}}{R^2} \left( \sin^2 \varphi - \cos^2 \varphi \right)$$

$$= - \frac{\cos \varphi}{R} \dot{\varphi} + \frac{\sin \varphi}{R^2} \dot{R},$$ (182)

$$\frac{\partial \dot{\varphi}}{\partial y} = - \dot{x} \left( - \frac{2 y^2}{R^4} + \frac{1}{R^2} \right) - x \dot{y} \frac{2 y}{R^4}$$

$$= - \frac{2 \dot{y}}{R^2} \cos \varphi \sin \varphi + \frac{\dot{x}}{R^2} \left( 2 \sin^2 \varphi - 1 \right)$$

$$= - \frac{2 \dot{y}}{R^2} \cos \varphi \sin \varphi + \frac{\dot{x}}{R^2} \left( \sin^2 \varphi - \cos^2 \varphi \right)$$

$$= - \frac{\dot{y}}{R^2} \cos \varphi \sin \varphi - \frac{\dot{x}}{... ...ac{\dot{y}}{R^2} \cos \varphi \sin \varphi + \frac{\dot{x}}{R^2} \sin^2 \varphi$$

$$= - \frac{\cos \varphi}{R^2} \dot{R} - \frac{\sin \varphi}{R} \dot{\varphi}$$ (183)

Using these results, the last two terms in Eq. (179) is written as

$$\left[ - \frac{\partial \mathcal{L}}{\partial \dot{R}} \dot{\varp... ...arphi}{R} \dot{\varphi} \right) \right] \widehat{\ensuremath{\boldsymbol{e}}}_y$$

$$= \frac{\partial \mathcal{L}}{\partial \dot{R}} \dot{\varphi} \wi... ...- \frac{\dot{R}}{R^2} \widehat{\ensuremath{\boldsymbol{e}}}_{\varphi} \right) .$$ (184)

Using this, Eq. (179) is written as

$$\frac{\partial \mathcal{L}}{\partial \ensuremath{\boldsymbol{X}}}... ...- \frac{\dot{R}}{R^2} \widehat{\ensuremath{\boldsymbol{e}}}_{\varphi} \right] .$$ (185)

Note that, compared with Eq. (177), the above expression contains additional terms. The additional terms are the source of confusion when I at first tried to prove the equivalence between Eqs. (8) and (163). (I was confused for many days before I fininaly found the solution given here.) Using Eqs. (185) and (176) in Eq. (162), we recover the Euler-Lagrange equation in cylindrical coordinates, i.e.,

$$\frac{d}{d t} \left( \frac{\partial \mathcal{L}}{\partial \dot{R}} \right) = \frac{\partial \mathcal{L}}{\partial R},$$ (186)

$$\frac{d}{d t} \left( \frac{\partial \mathcal{L}}{\partial \dot{\varphi}} \right) = \frac{\partial \mathcal{L}}{\partial \varphi},$$ (187)

$$\frac{d}{d t} \left( \frac{\partial \mathcal{L}}{\partial \dot{Z}} \right) = \frac{\partial \mathcal{L}}{\partial Z} .$$ (188)