# Point transformation for a constrained particle

• Lambda96
In summary, pasmith is not very familiar with the Lagrangian formalism and he is having difficulty solving task 1a. He has now proceeded by assuming that the potential energy is zero and has found the kinetic energy. He is Unsure about task b as he is not sure if the equation depends on x. He has then found the Euler-Lagrange equation.
Lambda96
Homework Statement
Lagrangian mechanics for a mass particle
Relevant Equations
none
Hi,

unfortunately, I'm not that fit concerning the Lagrangian formalism, so I'm not sure if I solved the problem 1a correctly.

I have now proceeded as follows

the Lagrangian is

$$L=T-U$$

Since there are no constraining or other forces acting on the point mass, I assume that the potential energy is 0 and thus the system has only kinetic energy, i.e.

$$T=\frac{1}{2}m*(\dot{x}^2+\dot{y}^2)$$

I would now represent the entire equation in the x coordinate only, so.

$$x=x$$
$$y=f(x)$$

Insert into the Lagrangian $$T=\frac{1}{2}m*(\dot{x}^2+\dot{f}^2(x))$$

Thus, I would be done with task 1a, or did I do something wrong?

That is correct, except that if $y = f(x)$ then $\dot y = f'(x)\dot x$, so $$T = \frac12 m (1 + f'(x)^2)\dot x^2$$.

PhDeezNutz
Thank you pasmith for your help , I had completely forgotten the chain rule.

Concerning task b, the Euler-Lagrange equation is as follows

$$\frac{\partial L}{\partial x}-\frac{d}{dt}\frac{\partial L}{\partial \dot{x}}=0$$

For $$\frac{\partial L}{\partial x}$$

I am not quite sure since f(x) depends on x, would the following then apply

$$\frac{\partial L}{\partial x}=m f''(x)$$

and

$$\frac{\partial L}{\partial \dot{x}}=m\dot{x}+m f'(x)^2\dot{x}$$

$$\frac{d}{dt}m\dot{x}+m f'(x)^2\dot{x}=m\ddot{x}+m f''(x)^2\ddot{x}\dot{x}$$

Then the Euler-Lagrange equation is

$$m f''(x)-m\ddot{x}-m f''(x)^2\ddot{x}\dot{x}=0$$

Your lagrangian is $L = \frac12 m g(x)\dot x^2$ where $g(x) = 1 + f'(x)^2$. Hence $$\begin{split} \frac{d}{dt}\left( \frac{\partial L}{\partial \dot x}\right) - \frac{\partial L}{\partial x} &= \frac{d}{dt}(mg(x)\dot x) - \tfrac12 m g'(x) \dot x^2 \\ &= mg(x) \ddot x + mg'(x) \dot x^2 - \tfrac12 m g'(x) \dot x^2 \\ &= mg(x) \ddot x + \tfrac12 m g'(x) \dot x^2. \end{split}$$ So what is $g'(x)$ using the chain rule?

Lambda96 said:
$$\frac{\partial L}{\partial \dot{x}}=m\dot{x}+m f'(x)^2\dot{x}$$

$$\frac{d}{dt}m\dot{x}+m f'(x)^2\dot{x}=m\ddot{x}+m f''(x)^2\ddot{x}\dot{x}$$
Don't forget the product rule.

Lambda96 said:
$$\frac{\partial L}{\partial \dot{x}}=m\dot{x}+m f'(x)^2\dot{x}$$

$$\frac{d}{dt}m\dot{x}+m f'(x)^2\dot{x}=m\ddot{x}+m f''(x)^2\ddot{x}\dot{x}$$
This is not good!

I suggest you take ##f(x) = \sin x## or something like that and check your differentiation against a definite function.

I don't want to be rude, but you are missing trivial points about elementary calculus, as chain rule and product rule.

May i ask, have you take those class? Don't try to skip steps in your learning process.

Thanks to all for the help, especially pasmith

I have now formed the derivative of ##g(x)=1+f'(x)^2## as follows.

The outer derivative is ##2f'(x)##and the inner derivative would be ##f''(x)##

Thus, the derivative of ##g'(x)## would be ##g'(x)=2f'(x)f''(x)##

The Euler-Lagrange equation is ##m\ddot{x}+mf'(x)^2\ddot{x}+mf'(x)f''(x)\dot{x}^2##Concerning task part c and d

##|l(x)|=\int_{x_0}^{x} \sqrt{1+f'(x)^2}dx##

I would now simply parameterize the curve ##l(x)=\left(\begin{array}{c} x \\ f(x) \end{array}\right)## of ##[x_0,x]## with ##l_1=x## and ##l_2=f(x)##

Then the Lagrange function would look like ##L=\frac{1}{2}m(\dot{l_1}^2+\dot{l_2}^2)##

## 1. What is a point transformation for a constrained particle?

A point transformation for a constrained particle is a mathematical technique used in mechanics to describe the motion of a particle that is subject to certain constraints or limitations. It involves transforming the coordinates of the particle from one reference frame to another in order to simplify the equations of motion.

## 2. How does a point transformation affect the equations of motion for a constrained particle?

A point transformation can simplify the equations of motion for a constrained particle by eliminating certain variables or constraints. This can make it easier to analyze and solve the equations, and can also reveal underlying symmetries or conserved quantities.

## 3. Are there any limitations to using point transformations for constrained particles?

Yes, there are some limitations to using point transformations for constrained particles. They may not be applicable to all types of constraints, and in some cases, the transformed equations may become more complex and difficult to solve.

## 4. How is a point transformation different from a coordinate transformation?

A point transformation and a coordinate transformation are similar in that they both involve changing the coordinates of a particle from one reference frame to another. However, a point transformation specifically refers to transformations that are used to simplify the equations of motion for constrained particles, while a coordinate transformation can be used for any type of transformation between reference frames.

## 5. Can point transformations be used for systems of multiple constrained particles?

Yes, point transformations can be used for systems of multiple constrained particles. However, the equations of motion become more complex and may require more advanced mathematical techniques to solve. In some cases, it may be necessary to use a combination of point transformations and other methods to fully describe the motion of the system.

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