Benefits of Lagrangian mechanics with generalised coordinates

[lriuui0x0]

I have sometimes seen the claim that one advantage of Lagrangian mechanics is that it works in any frame of reference, instead of like Newtonian mechanics which will hold only in the inertial frame of reference. However isn't this gain only at the sacrifice that the Lagrangian will need to take a more complicated form? If this can be considered as frame independent, we can also claim that Newton's second law $F = ma$ works in any frame of reference, if we allow the force to take a more complicated form, i.e. with the addition of the fictitious forces.

Just want to double check that if this reasoning is correct and is there any more fundamental advantages of Lagrangian mechanics in this regard, or is this just that the Lagrangian being a scalar making the coordinate transformation easier than force vectors?

 

[ergospherical]

That is one aspect, yes. That a particular trajectory is an extremal of the action functional does not depend on the coordinates, and indeed you should try to show that if the Euler-Lagrange equation is satisfied in coordinates $q^i$ then it is also satisfied in coordinates $\tilde{q}^i = \tilde{q}^i(q^1, \dots, q^n, t)$.

There are other reasons! Importantly, the Lagrangian method allows you to solve problems without worrying about the unknown holonomic constraint forces (such as the contact force that a hoop exerts on a bead, etc.).

 

[lriuui0x0]

That is one aspect, yes. That a particular trajectory is an extremal of the action functional does not depend on the coordinates, and indeed you should try to show that if the Euler-Lagrange equation is satisfied in coordinates qi then it is also satisfied in coordinates q~i=q~i(q1,…,qn,t).

About this particular benefit, can't we say that Newton's second law is on the same footing here? Because we can also transform the coordinates to get Newton's equation of motion into a non-inertial frame? Similar to transforming coordinates for the Lagrangian?

 

[ergospherical]

Sort of, but not really. In it's original form, Newton $\mathrm{II}$ holds only in an inertial frame, $m\boldsymbol{a}_{\mathrm{in}} = \boldsymbol{F}$. To transform this equation to a non-inertial frame requires some facts about the derivatives of vectors with respect to different frames.

Viz, given a frame $S = Oxyz$, the derivative of a vector $\boldsymbol{u}(t) = u^x(t) \boldsymbol{e}_x + u^y(t) \boldsymbol{e}_y + u^z(t) \boldsymbol{e}_z$ with respect to time is defined as\begin{align*}\dfrac{d\boldsymbol{u}}{dt} \bigg{|}_S = \dfrac{du^x}{dt}(t) \boldsymbol{e}_x + \dfrac{du^y}{dt}(t) \boldsymbol{e}_y + \dfrac{du^z}{dt}(t) \boldsymbol{e}_z\end{align*}Given a different frame $S' = O'x'y'z'$, rotating at angular velocity $\boldsymbol{\omega}$ with respect to $S$, one can prove that\begin{align*}\dfrac{d\boldsymbol{u}}{dt} \bigg{|}_{S} = \dfrac{d\boldsymbol{u}}{dt} \bigg{|}_{S'} + \boldsymbol{\omega} \times \boldsymbol{u}\end{align*}Let $S$ be the inertial frame and $S'$ be the non-inertial frame. For simplicity, assume that the origins coincide at all times. Then \begin{align*}\boldsymbol{a}_{\mathrm{in}} &= \dfrac{d^2 \boldsymbol{r}}{dt^2} \bigg{|}_S \\&= \dfrac{d}{dt} \bigg{|}_S \left(\dfrac{d\boldsymbol{r}}{dt} \bigg{|}_{S'} + \boldsymbol{\omega} \times \boldsymbol{r} \right) \\&= \dfrac{d^2 \boldsymbol{r}}{dt^2} \bigg{|}_{S'} + \boldsymbol{\omega} \times \dfrac{d\boldsymbol{r}}{dt} \bigg{|}_{S'} + \dfrac{d}{dt} \bigg{|}_S \left( \boldsymbol{\omega} \times \boldsymbol{r} \right)\end{align*}The last term is\begin{align*}\dfrac{d}{dt} \bigg{|}_S \left( \boldsymbol{\omega} \times \boldsymbol{r} \right) &= \dfrac{d\boldsymbol{\omega}}{dt} \bigg{|}_S \times \boldsymbol{r} + \boldsymbol{\omega} \times \dfrac{d\boldsymbol{r}}{dt} \bigg{|}_S \\&= \boldsymbol{\alpha} \times \boldsymbol{r} + \boldsymbol{\omega} \times \left( \dfrac{d\boldsymbol{r}}{dt} \bigg{|}_{S'} + \boldsymbol{\omega} \times \boldsymbol{r} \right) \\&= \boldsymbol{\alpha} \times \boldsymbol{r} + \boldsymbol{\omega} \times \dfrac{d\boldsymbol{r}}{dt} \bigg{|}_{S'} + \boldsymbol{\omega} \times (\boldsymbol{\omega} \times \boldsymbol{r})\end{align*}To simplify the notation, one calls $\boldsymbol{a}' \equiv \dfrac{d^2 \boldsymbol{r}}{dt^2} \bigg{|}_{S'}$ the acceleration relative to $S'$ and $\boldsymbol{v'} \equiv \dfrac{d\boldsymbol{r}}{dt} \bigg{|}_{S'}$ the velocity relative to $S'$. You have\begin{align*}\boldsymbol{a}_{\mathrm{in}}= \boldsymbol{a}' + 2\boldsymbol{\omega} \times \boldsymbol{v}' +\boldsymbol{\omega} \times (\boldsymbol{\omega} \times \boldsymbol{r}) + \boldsymbol{\alpha} \times \boldsymbol{r}\end{align*}Newton $\mathrm{II}$ is transformed to\begin{align*}m \boldsymbol{a}' = \boldsymbol{F} -2m\boldsymbol{\omega} \times \boldsymbol{v}' -m \boldsymbol{\omega} \times (\boldsymbol{\omega} \times \boldsymbol{r}) -m \boldsymbol{\alpha} \times \boldsymbol{r}\end{align*}Heuristically, the three correction terms on the right hand side can be viewed as forces.

On the other hand, try starting with a general Lagrangian $L = \dfrac{1}{2}m \dot{\boldsymbol{r}}^2 - V(\boldsymbol{r})$ in an inertial frame. Then transform this Lagrangian to a non-inertial frame by means of a coordinate transformation, using the fact that the Lagrangian approach is valid in any coordinates. You'll notice that the correction terms arise, as expected.

 

[Dale]

About this particular benefit, can't we say that Newton's second law is on the same footing here? Because we can also transform the coordinates to get Newton's equation of motion into a non-inertial frame? Similar to transforming coordinates for the Lagrangian?

In addition to what @ergospherical mentioned, one other consideration is that the Lagrangian is a scalar while Newton’s 2nd is a vector equation.

 

[lriuui0x0]

In addition to what @ergospherical mentioned, one other consideration is that the Lagrangian is a scalar while Newton’s 2nd is a vector equation.

So this means the calculation with scalar is easier?

 

[Dale]

So this means the calculation with scalar is easier?

Yes, by about a factor of 3.