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Suppose I have an inertial frame with coordinate $\{q\}$. Now I define another reference frame with coordinate $\{q'(q,\dot q,t)\}$. I obtain the equation of motion in $\{q'\}$ in two different ways:

  1. First obtain the equation of motion in $\{q\}$ by the Euler Lagrange equation $$\frac{d}{dt}\left(\frac{\partial L}{\partial \dot q}\right)-\frac{\partial L}{\partial q}=0$$ and then rewrite the equation in terms of $\{q'\}$.

  2. First transform $L(q,t)$ to $L'(q',t)=L(q(q',t),t)$ and then obtain the equation of motion $$\frac{d}{dt}\left(\frac{\partial L'}{\partial \dot q'}\right)-\frac{\partial L'}{\partial q'}=0.$$

Are the two answers just the same?

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I) The Euler-Lagrange (EL) equations behave covariantly under reparametrizations$^1$ of the form

$$ \tag{1} q^{\prime i}=f^i(q,t),$$

i.e. it is equivalent to reparametrize before or after forming the EL equations.

II) The above property even holds for a Lagrangian $L(q,\dot{q},\ddot{q},\ldots, \frac{d^Nq}{dt^N};t)$ that depends on higher-order time-derivatives, although a higher-order version of Euler-Lagrange equations with higher-order derivatives is needed in such case.

III) However, for a velocity-dependent reparametrization $q^{\prime }=f(q,\dot q,t)$, which OP mentions in his second line (v2), the substitution before or after in general leads to EL eqs. of different orders. We expect that the higher-order EL eqs. to always factorize via the corresponding lower-order EL eqs., so that solutions to the lower-order EL eqs. are also solutions to the higher-order EL eqs. but not vice-versa.

Similarly for acceleration-dependent reparametrizations, etc.

IV) Example: Consider the velocity-dependent reparametrization

$$\tag{2} q^{\prime}~=~q+A \dot{q}, \qquad A>0,$$

of the Lagrangian$^2$

$$\tag{3} L^{\prime}~=~ \frac{1}{2} q^{\prime 2}~=~\frac{1}{2}(q+A \dot{q})^2~\sim~ \frac{1}{2}q^2 +\frac{A^2}{2} \dot{q}^2. $$

(We call $q^{\prime}$ and $q$ the old and new variables, respectively.) Before, the EL equation is of first order in the new variables$^3$

$$\tag{4} 0\approx q^{\prime}~=~q+A \dot{q},$$

with only exponentially decaying solutions. After the reparametrization, the EL equation is of second order

$$\tag{5} 0\approx q- A^2 \ddot{q}~=~(1-A\frac{d}{dt})(q+A \dot{q}),$$

so that it has more solutions. Note however that eq. (5) factorize via (=can be obtained from) eq. (4) by applying a differential operator $1-A\frac{d}{dt}$.

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$^1$ There are various standard regularity conditions on a reparametrization (1) such as e.g. invertibility and sufficiently differentiability. The higher jets (velocity, acceleration, jerk, etc) are implicitly assumed to transform in the natural way.

$^2$ The $\sim$ sign means here equal modulo total derivative terms.

$^3$ The $\approx$ sign means here equal modulo the EL equations.

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  • $\begingroup$ @Qmechanics: What does it mean by higher-order version E-L equations? $\endgroup$ – velut luna Jun 11 '14 at 4:27
  • $\begingroup$ @Qmechanics: So does this contradict the fact that physics should be independent of parametrization? $\endgroup$ – velut luna Jun 11 '14 at 7:07
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    $\begingroup$ @user139981: it's independent of parametrization, but not independent of geometry: the derivation of the EL equations assumes coordinates that respect the bundle structure, and if you do a transformation that doesn't, you'll get nonsense $\endgroup$ – Christoph Jun 11 '14 at 7:28
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When you change Lagrangian coordinates (but it does not necessarily mean changing reference frame!), as you are dealing with a jet bundle over the real temporal line $\mathbb R$ (equipped with a preferred coordinate $t$ defined up to an additive constant), you have $$t' = t+c\quad, q'^k = q'^k(t,q)\:,\quad \dot{q}'^k = \sum_j\frac{\partial q'^k}{\partial q^j} \dot{q}^j + \frac{\partial q'^k}{\partial t}\:.\tag{1}$$ In particular, as this transformation of coordinates is required to be smooth, invertible, with smooth inverse, it also arises $$\det \left[ \frac{\partial q'^k}{\partial q^j} \right] \neq 0\:,\quad \det \left[ \frac{\partial q^j}{\partial q'^k} \right] \neq 0 \tag{2'}\:.$$ If, as you did, you assume that the Lagrangian function is a scalar, i.e., $${\cal L}'(t',q', \dot{q}') = {\cal L}(t,q, \dot{q})\quad \mbox{where (1) hold,}\tag{2}$$ you can verify the following identity valid on any point of a generic curve (section) $t \mapsto \gamma(t):= (t, q(t), \dot{q}(t))$ (also described with the other coordinate system) $$\left.\left(\frac{d}{dt}\frac{\partial {\cal L}'}{\partial \dot{q}'^k} -\frac{\partial {\cal L}'}{\partial q'^k}\right)\right|_{\gamma(t)} = \sum_j\left.\frac{\partial q^j}{\partial q'^k}\right|_{\gamma(t)}\left.\left(\frac{d}{dt}\frac{\partial {\cal L}}{\partial \dot{q}^j} -\frac{\partial {\cal L}}{\partial q^j} \right)\right|_{\gamma(t)} \:,$$ where (2)' holds true.

As a consequence, the curve $t \mapsto \gamma(t):= (t, q(t), \dot{q}(t))$ satisfies Euler-Lagrange equations with respect to ${\cal L}'$ and the coordinates $(t',q', \dot{q}')$ if and only if it verifies Euler-Lagrange equations with respect to ${\cal L}$ and the coordinates $(t,q, \dot{q})$.

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