# How can we represent the motion of a particle in 2D space using Lagrange's equations?

Can we represent the motion of a particle in 2D space using Lagrange's equations? This is what I tried. Please tell me what is wrong?

Consider a particle on a plane have the co-ordinates $(x,y)$ with a velocity $v$ and mass $m$.

Now $v=\dot{x}+\dot{y}$ where $\dot{x}$ and $\dot{y}$ are the derivatives with respect to time.

According to Lagrange's equation,

$L=\frac{1}{2}m(\dot{x}+\dot{y})^2-mgy$

$\frac{{\partial}{L}}{\partial{y}}=-mg$ ---(1)

$\frac{{\partial}{L}}{\partial{\dot{y}}}=m(\dot{x}+\dot{y})$

Therefore $\frac {\mathrm d t}{\mathrm d t}(\frac{{\partial}{L}}{\partial{\dot{y}}})=\frac{{\partial}{L}}{\partial{\dot{y}}}=m(\ddot{x}+\ddot{y})$

Now $\ddot{x}$ and $\ddot{y}$ are the vertical & horizontal components of the acceleration $a$ of the particle. So let $\ddot{x}=a_x$ and $\ddot{y}=a_y$

Hence $\frac {\mathrm d t}{\mathrm d t}(\frac{{\partial}{L}}{\partial{\dot{y}}})=m(a_x+a_y)$ ---(2)

From Lagrange's equation, Eq (1)= Eq(2)

=> $-mg=m(a_x+a_y)$

$-g=a_x+a_y$

Does the above equation make any sense? I tried to apply it to a real problem and it gave me a wrong result. What is wrong with my working?

Thanks in advance.

## Update:

Thanks a lots guys. Yes, the expression for $v$ was wrong. When I calculated with the correct expression I got the result $a_y=-g$

What does it mean? Does it mean that the Y component of the acceleration is always equal to acceleration due to gravity?

And can someone explain how Lagrangian mechanics can be used to solve simple mechanical problems? When I tried applying it to problems all I got were few known results & other nonsensical ones (like the one in this question. Probably due to my erroneous application of the Lagrange.)

PS: Will someone explain why this question was downvoted?

• Your expression for v is wrong, it should be v^2=x_t^2+y_t^2 – Bernhard Jan 25 '12 at 6:43

## 3 Answers

What does it mean? Does it mean that the Y component of the acceleration is always equal to acceleration due to gravity?

Not always. But in this case, yes.

all I got were few known results & other nonsensical ones (like the one in this question. Probably due to my erroneous application of the Lagrange.)

You didn't get the results you got because of your own fault but because of virtue of Lagrangian formulation of mechanics. You got them because Lagrangian formulation of mechanics works! You took a simple system and got equations of motion for it which you already knew.

Do you know equations of motion for double pendulum? How will you find them(almost without using brain)?

Algorithm:

1. Find Lagrangian for double pendulum
2. Get equations of motion using Euler-Lagrange equations

The definition of your $L$ is wrong, IMHO it should be, $$L = \frac{1}{2}(mx'^{2} + my'^{2}) -mgy$$ There on if you apply variational principle you will find the correct answer.

PS: $x' = \dot{x}$

The problem is in the Lagrangian. The Lagrangian is of the independent variables x,y not of v. It should be L=(1/2)mx˙^2+(1/2)my'^2−mgy . By putting both in the parenthesis you gained the term mx'y' which caused the error.