Checking if a stationary point is a minimum using Lagrangian Mechanics

In summary, the conversation discusses how to determine if a stationary point is a minimum using Euler's equation and second derivative tests. It also mentions the concept of 2nd variation in Lagrangian mechanics as an alternative method of solving the problem. The person is new to Lagrangian mechanics and is seeking clarification and advice on the topic.
  • #1
beans123
5
0
I'm having trouble understanding how to find out whether or not a stationary point is a minimum and I'm hoping for some clarification. In my class, we were shown that, using Euler's equation, the straight-line path:
Screenshot 2023-02-05 18.16.34.png

with constants a and b results in a stationary point of the integral:
Screenshot 2023-02-05 18.16.47.png

A certain practice question then asks to show that the stationary point corresponds to a minimum. My only attempt so far was performing a simple second derivative test on the function f(x') which turned out to be successful. However, I'm wondering if this is the only way to solve such a problem. I know that a minimum is satisfied if S(a) > S_actual, but can that same idea be mapped onto I(a), that is, is a minimum achieved if I(a) > I_actual (if that even makes sense)? I'm very new to Lagrangian mechanics and find it kind of overwhelming so forgive me if this is a silly question. It just seems that I took the calculus way of solving this when that may not be the ideal method for a class based on Lagrangian mechanics/. I appreciate any help/advice!
 
Physics news on Phys.org
  • #2
Try googling "2nd variation in Lagrangian mechanics". (This is analog of 2nd derivatives in ordinary calculus.)
 
  • Like
Likes vanhees71

FAQ: Checking if a stationary point is a minimum using Lagrangian Mechanics

1. What is a stationary point in the context of Lagrangian Mechanics?

A stationary point in Lagrangian Mechanics refers to a point where the first derivatives (partial derivatives) of the Lagrangian with respect to the generalized coordinates and velocities are zero. This indicates that the system is in a state of equilibrium or at an extremum of the action.

2. How can I determine if a stationary point is a minimum?

To determine if a stationary point is a minimum, you need to examine the second derivatives of the Lagrangian. Specifically, you form the Hessian matrix of second partial derivatives of the Lagrangian with respect to the generalized coordinates. If this Hessian matrix is positive definite at the stationary point, then the point is a local minimum.

3. What is the Hessian matrix and how is it used in this context?

The Hessian matrix is a square matrix of second-order partial derivatives of a scalar function, in this case, the Lagrangian. It is used to analyze the curvature of the Lagrangian function at the stationary point. For a function \( L(q_1, q_2, ..., q_n) \), the Hessian matrix \( H \) is composed of elements \( H_{ij} = \frac{\partial^2 L}{\partial q_i \partial q_j} \). If the Hessian is positive definite, the stationary point is a local minimum.

4. What does it mean for the Hessian matrix to be positive definite?

A Hessian matrix is positive definite if all its eigenvalues are positive. This implies that the quadratic form associated with the Hessian is always positive for any non-zero vector, indicating that the Lagrangian has a local minimum at the stationary point.

5. Are there any specific conditions or constraints that need to be considered when applying Lagrangian Mechanics to find minima?

Yes, when applying Lagrangian Mechanics, you need to consider any constraints present in the system. These constraints can be incorporated using Lagrange multipliers, which modify the Lagrangian to include terms representing the constraints. The modified Lagrangian is then analyzed to find stationary points and determine their nature (minimum, maximum, or saddle point).

Similar threads

Replies
23
Views
3K
Replies
2
Views
746
Replies
25
Views
2K
Replies
11
Views
2K
Replies
3
Views
1K
Replies
5
Views
1K
Replies
2
Views
758
Replies
2
Views
2K
Back
Top