I Action in Lagrangian Mechanics

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Lagrangian mechanics relies on the calculus of variations to determine the action, defined as an integral that needs to be minimized or maximized. The integral typically involves variables like time, potential energy, or distance, which are crucial for solving problems such as the brachistochrone or catenary problems. The Euler-Lagrange equation is then used to derive equations of motion from the defined Lagrangian, which is the difference between kinetic and potential energy. The discussion highlights the historical development of the principle of least action, emphasizing the foundational roles of Newton and Hamilton in formulating these concepts. Understanding the principle of least action is essential for grasping the underlying mechanics of motion.
Dario56
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Lagrangian mechanics is built upon calculus of variation. This means that we want to find out function which is a stationary point of particular function (functional) which in Lagrangian mechanics is called the action.

To know what this function is, action needs to be defined first. Action is defined via integral.

In problems which use calculus of variation such as brachistochrone problem, caternary problem or finding path of least distance between two points, appropriate integral is the integral between two points of question of appropriate variable (time, potential energy, distance etc.), that is the integral of variable which is usually needed to be minimized in the problem (can be maximized as well).

When integral is defined, function is known and with Euler - Lagrange equation we get the solution to the problem. For example that can be function which defines path of least time, distance or shape of the rope as a solution of caternary problem.

What I don't understand is how to define this integral in context of Lagrangian mechanics or in context of action? In another words, how did Lagrange found out that difference in kinetic and potential energy of the system (commonly known as Lagrangian) gives correct equations of motion when plugged in Euler - Lagrange equation?
 
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Dario56 said:
What I don't understand is how to define this integral in context of Lagrangian mechanics or in context of action? In another words, how did Lagrange found out that difference in kinetic and potential energy of the system (commonly known as Lagrangian) gives correct equations of motion when plugged in Euler - Lagrange equation?
Wikipedia has a page on the history and development of the principle of least action:

https://en.wikipedia.org/wiki/Stationary-action_principle
 
Probably similar to this

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You might try looking at Lanczos
https://www.amazon.com/dp/0486650677/?tag=pfamazon01-20
Rojo and Bloch
https://www.amazon.com/dp/0521869021/?tag=pfamazon01-20
Coopersmith
https://www.amazon.com/dp/0198743041/?tag=pfamazon01-20
 
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Likes Dario56 and vanhees71
You don't need lagrange, you need Newton+hamilton+idea of test particle. We, as physicists, use test particles to learn about the behavior and motion about the world around us.

Newton's first law is basically "test particles, when left alone, will move in a straight line at a constant speed".

Then a guy called Hamilton sets up a premise called "principle of least action" which uses Newton's ideal+geometry, in which you must notice that a straight line between two points can be thought of as "the shortest distance" or "least time" between two points. If you agree with this, then bam, you formulate the following integral formulation:
$$S =-mc \int^{\tau_{final}}_{\tau_{initial}} d \tau$$

Minimizing this for some $\tau$ will give you the trajectory of a free particle. There is more to this story, and I hand waved a little, but I believe your question isn't really with lagrange, but more so with hamilton, Newton, and the ideal of a test particle, and at the core, what IS a principle of least action?!

Now, I've never TRULY gotten classical mechanics because it never sat right with me, but if the above is what you're questioning, I'm sure Feynman can do a better (and more through) job than I can, and you can find more information here:
https://www.feynmanlectures.caltech.edu/II_19.html
 
Thread 'Why higher speeds need more power if backward force is the same?'
Power = Force v Speed Power of my horse = 104kgx9.81m/s^2 x 0.732m/s = 1HP =746W Force/tension in rope stay the same if horse run at 0.73m/s or at 15m/s, so why then horse need to be more powerfull to pull at higher speed even if backward force at him(rope tension) stay the same? I understand that if I increase weight, it is hrader for horse to pull at higher speed because now is backward force increased, but don't understand why is harder to pull at higher speed if weight(backward force)...

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