The problem of the mass in inclined planed solved using the Lagrangian mechanics is well known, by example on this page.

However, according to my information, method is valid for any basis and expressions of the involved concepts. Thus, if I define a 2D euclidean basis, being $x$ the horizontal distance and $y$ the heights, I can say that:

  • potential energy of the mass, $V(x,y) = mgy$ (fixing $V(0,0)=0$)
  • kinetic energy, $T(x,y) = T_0 + mg(y_0-y)$, being $y_0,T_0$ the values at $t=0$.
  • Lagrangian, $L(x,y) = T(x,y)-V(x,y) = T_0 + mg(y_0-2y)$
  • Lagrange equations: $$ \frac{\partial L}{\partial x} = 0 = \frac{d}{dt}\frac{\partial L}{\partial \dot{x}} = 0 $$ $$ \frac{\partial L}{\partial y} = -2mgy = \frac{d}{dt}\frac{\partial L}{\partial \dot{y}} = 0 $$

being both a probable error, because first one say nothing about $x(t)$ and second one implies $y(t)=0$.

Could someone say me where is the error in this development?

Remark: I'm not looking about how to solve the problem of the inclined plane using Lagrange mechanics, something easy to find in the bibliography. Please, answers must be as near as possible to this approach to the problem (this basis, these expressions for $V$ and $T$) or show why this approach is a no way, against the Lagrange postulates.


About the expression for kinetic energy:

a) using second Law (being $\theta$ plane angle and "s" position along plane):

total force, $F=-mg\sin(\theta)\hat{s}$ where $\hat{s}=cos(\theta)\hat{x}+sin(\theta)\hat{y}$

speed, $v=(-g\sin(\theta)t+v_0)\hat{s}$

position $\mathbf{r}=(-\frac{1}{2}g\sin(\theta)t+v_0)t\hat{s}+\mathbf{r}_0$

y component of position, $y=y_0+(-\frac{1}{2}g\sin(\theta)t+v_0)t\sin(\theta)\hat{y}$

kinetic energy, $T=\frac{1}{2}m(-g\sin(\theta)t+v_0)^2=...=T_0+mg(y_0-y)$

b) by conservation of energy (where constraint forces are assumed produces no work):



  • $\begingroup$ Could you elaborate why you think the kinetic energy has that form? $\endgroup$
    – Qmechanic
    Jun 12 '20 at 19:03
  • $\begingroup$ @Qmechanic: thanks for your interest in this question. Addendum added, with details of the two methods I used to find the expression of the kinetic energy. $\endgroup$ Jun 12 '20 at 19:32
  • $\begingroup$ @Qmechanic: However, even if this expression is not correct, another $T(x,y)$ will do, and the application of the Lagrange equations will have the same problem, partial derivatives respect $\dot{x}$ and $\dot{y}$ will be 0. $\endgroup$ Jun 12 '20 at 19:40

Lagrange's equations are based on generalized coordinates $q_i$.

A system of $N$ particles in a 2D system with $k$ holonomic constraints have $2N-k$ degrees of freedom and hence are described by $2N-k$ independent generalized coordinates.

Holonomic constraints are basically constraint equations of the form $f(\vec{r}_1,\vec{r}_2...,t)=0$.

In this problem, there is 1 holonomic constraint equation of the form $$y=Ax+C, $$where $A$ and $C$ are constants. This arises from requiring the mass to stay on the inclined plane.

Thus in this problem, there are $2-1=1$ degree of freedom and should be described by 1 independent generalized coordinate $q$.

You can choose either $x$ or $y$ as the independent generalized coordinate. It is incorrect to express the Lagrange equation in terms of both $x$ and $y$ as they are not independent coordinates. Hence you get the 'nonsense' equation like $0=0$ for your $x$ coodinate.

Furthermore, the derivation of Lagrange's equations has the kinetic energy $T$ specifically defined as $$T=\frac{1}{2}mv^2.$$ Following this definition, you should express $v$ in terms of the generalized coordinate you choose. For example if you choose $y$ as the generalized coordinate, you will get $v=v(\dot{y})$ and hence $T=T(\dot{y})$. This will resolve your error of getting $y(t)=0$.

Summing it up, you should always use generalized coordinates when using lagrange's equations and always express the kinetic energy as $T=\frac{1}{2}mv^2$.

  • $\begingroup$ Thanks for your answer. Using "y" as generalized coordinate, $T(y)=T_0+mg(y_0-y)$, but method still fails. Do you mean I must express $T$ as a function of only $\dot{y}$? That seems to limit the method. Moreover, Lagrange equations could then be simplified using $\frac{\partial L}{\partial y}=\frac{\partial V}{\partial y}$ and $\frac{\partial L}{\partial \dot{y}}=\frac{\partial T}{\partial \dot{y}}$ $\endgroup$ Jun 13 '20 at 18:11
  • $\begingroup$ @pasabaporaqui You did not express kinetic energy as $T={1\over 2}mv^2={1\over 2}m(\dot{x}^2+\dot{y}^2)$ $\endgroup$
    – TaeNyFan
    Jun 21 '20 at 1:06

The purpose of the Euler-Lagrange equations is to find the equation of motion of the system once we plug in the Lagrangian. Alternatively, one can use Newton's second law to find the equation of motion.

Your biggest mistake comes from the way in which you expressed the kinetic energy, because you had to solve the equation of motion to write it in that way (in part (a) of your addendum). In other words, you solved Newton's second law, and plugged the solution into the Euler Lagrange equations, so it makes sense that you didn't get a new equation. A simple analogy for what you did is the following: consider the equation $du/dt=1$, $u(0)=0$. The solution is $u(t)=t$, and putting this into our original equation we get $1=1$. Nothing new there.

The reason you also got a nonsense equation along with $0=0$ is that you actually solved for $y(t)$ and plugged that into $\dot{y}(t)$, but left $y$ as an independent variable. It's more subtle, but the nonsense equation comes from the same error.

Additional info

The Lagrangian is a functional of the particle's trajectory. You plugged a particular trajectory (the correct classical trajectory) into this function, and essentially this gives you a number (as a function of time). You can no longer plug this expression into the Euler-Lagrange equations, as those come from minimizing the action where the Lagrangian is expressed as a functional of possible trajectories. More explicitly, the equations of motion come from imposing $$ \frac{\delta}{\delta{\vec{q}}(t)} \int L(\vec{q},\dot{\vec{q}},t) dt = 0, $$ where the trajectory for $\vec{q}$ is not specified in the Lagrangian - it comes from solving the above equation. What you did is find $\vec{q}(t)$ and plug this into the kinetic energy, thereby changing the problem at hand; you can no longer vary the action with respect to possible trajectories $\vec{q}(t)$, when it's no longer a function of possible trajectories.


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