# Finding the Lagrangian for an elastic collision

• eyenir
In summary: This means that the kinetic and potential energy are always equal at the point of collision.Assuming that the masses are connected by a spring of infinite stiffness, the total energy at the point of collision is conserved.
eyenir

## Homework Statement

a. Suppose two particles with mass $m$ and coordinates $x_1$, $x_2$ collides elastically, find the lagrangian and prove that the linear momentum is preserved.
b. Find new coordiantes (and lagrangian) s.t. the linear momentum is conjugate to the cyclical coordinate.

## The Attempt at a Solution

For (a) I thought that $L=m(\dot{x_1}^2+\dot{x_2}^2)/2$ but I don't know how to show the preservating of momentum.
For (b) I thought taking the new coordnates to be the center of mass and the distance of each particle from it but somehow I can't formulate it.

Can anybody assist with proving the momentum preservation and its conjugation in (b) for cyclical coordinates?

The question uses the word "preserved" in an odd context. It seems to mean that total linear momentum is conserved.

To include the collision you will need an interaction of some kind. I am hoping that the context of this question will provide you some guidance as to what is expected. An elastic collision means a conservative potential of some kind. This is "advanced physics homework" so it is probably not too much to expect you to know what that means with regard to a potential. So, what does it mean for a collision to be elastic, and what does a potential have to be such that it will produce an elastic collision?

Once you have that you show that total linear momentum is conserved by showing it is an invariant. And you do that by showing its bracket with the Hamiltonian is zero.

For part b) you need to remember what it means for the momentum to be conjugate. The question seems to be worded in slightly sloppy terms, possibly deliberately so. I think they are trying to get you to note that the conjugate variable and the momentum are not necessarily the same thing in different coordinate systems.

May can use: ## \int L\,dt = 0 ##

theodoros.mihos said:
May can use: ## \int L\,dt = 0 ##

Except that you forgot a bit. You forgot a ## \delta ## in front of that. Then you get ## \delta \int L\,dt = 0 ## which just gets you the Lagrange equation of motion. The integral you wrote is not usually zero.

theodoros.mihos
I would consider the two masses to be connected by a spring of infinite spring constant at the point of collision.

## 1. What is an elastic collision?

An elastic collision is a type of collision in which the total kinetic energy of the system is conserved. This means that the objects involved in the collision do not lose or gain any energy during the collision, and they bounce off each other without any deformation.

## 2. Why is it important to find the Lagrangian for an elastic collision?

The Lagrangian is a mathematical function that describes the dynamics of a system. In the case of an elastic collision, finding the Lagrangian allows us to analyze the behavior of the objects involved and predict their movements after the collision. It also helps us understand the conservation of energy and momentum in the system.

## 3. How do you determine the Lagrangian for an elastic collision?

The Lagrangian for an elastic collision can be determined by using the principle of least action, which states that the path taken by a system is the one that minimizes the action (the difference between the kinetic and potential energies). This involves setting up equations for the kinetic and potential energies of the system and finding the minimum value of the action.

## 4. Can the Lagrangian be used for non-elastic collisions?

Yes, the Lagrangian can be used for any type of collision. However, for non-elastic collisions, the total kinetic energy of the system is not conserved, so the equations for the Lagrangian will be different.

## 5. Are there any limitations to using the Lagrangian for an elastic collision?

One limitation is that the Lagrangian only applies to systems with conservative forces, meaning that there is no energy lost due to friction or other non-conservative forces. Additionally, the Lagrangian assumes that the objects involved in the collision are rigid and do not deform during the collision, which may not always be the case in real-world scenarios.

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