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:
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 $\{q'\}$.
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?
Answer
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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