What does it mean to "minimise" an action ?

[Clueless]

This is with regards to the principle of least action.

I understand that an action is a functional (a function where functions have values assigned to them, I believe?)

When trying to figure this out, I understood minimising as finding the true path of the principle of least action. But that does not make complete sense - if the true path is the path where the action is minimised MOST, what about the other paths? They can be minimised as well, right? S (trial) = S (true ) + some terms l x(t)^2 l according to my lecturer in classical mechanics. So if S (trial) is S (true), than those terms on the far right need to disappear, right? But what if they can't disappear because we have a false path? How (or can we) minimize the action still?

As you can see, I am very confused about the meaning of "minimising" the action. Mathematical explanations would be helpful as much as conceptual ones.

 

[PWiz]

Think of it this way - out of all the paths that a system can take, there will always be one where the action change will be the lowest (stationary, if you wish) compared to all other paths.

Let's look at a simpler example first - the Euclidean geodesic problem (in 2D).
$ds^2= dx^2 + dy^2$
$$s=\int_{x_1}^{x_2} \sqrt {1+ ( \frac {dy}{dx})^2 } dx$$
Now this is an integral. Out of all the functions $f(x)=y$, there will be one that will serve as an extremizer to this integral (the shortest distance $s$ between two points). In this case, you can use Euler-Lagrange equations to calculate the function, which turns out to be of the form $Ax + B$ , which is nothing but a straight line (a minimizer).

The PLA states that for a system, you can construct a Lagrangian function $L$ which will depend upon time, the generalized coordinates in which the system is described, and the first order derivatives of each generalized coordinate with respect to time.
So the lagrangian is $L(t, q(t), \dot q (t) )$. PLA states that a system always takes a path for which $$ \int_{t_1}^{t_2} L (t, q(t), \dot q (t) ) dt = 0 $$
which is analogous to the example above (an integral is being minimized)

For conceptual clarity, remember that we're not taking a particular path and trying to make its action stationary - instead, we're basically "searching" in the "bag" of all paths (infinite paths, really) for one which has stationary action. There are various "tools" that we can use for searching, and in most cases, we get the result directly.

 

[Simon Bridge]

[PWiz beat me to it...]
The action is a box with an input and an output - it takes a trajectory or path as an input and spits out a number.
Different trajectories get you a different number.

You cannot minimize the action for a single trajectory - it's just one number so it is already a minimum.
Instead, you have to find the trajectory whose action changes the least for nearby paths.
You'll get it after you've done a few examples.

 

[Clueless]

Thanks a lot for the clarification!

I was just confused because a fellow student told me to "minimise" a false path - which just threw me off .

 

[PWiz]

Thanks a lot for the clarification!

I was just confused because a fellow student told me to "minimise" a false path - which just threw me off .

Actually, that is the mainstream approach that is taken when the Euler-Lagrange equations are derived - you take a path which is slightly different from the path which extremizes the integral of F:
$F=u(x) + ε η(x)$ where $ε$ is a small, real parameter and u(x) is the extremizer that we want to find. We then differentiate the integral with respect to the parameter and then set the parameter to 0 (some prefer using a limit, but it unnecessarily complicates matter), so that the derivative can have a turning point. η(x) must satisfy certain boundary conditions and the final result is obtained using the fundamental lemma of calculus of variation, but the math is not the main point here :)

 

[Clueless]

Actually, that is the mainstream approach that is taken when the Euler-Lagrange equations are derived - you take a path which is slightly different from the path which extremizes the integral of F:
$F=u(x) + ε η(x)$ where $ε$ is a small, real parameter and u(x) is the extremizer that we want to find. We then differentiate the integral with respect to the parameter and then set the parameter to 0 (some prefer using a limit, but it unnecessarily complicates matter), so that the derivative can have a turning point. η(x) must satisfy certain boundary conditions and the final result is obtained using the fundamental lemma of calculus of variation, but the math is not the main point here :)

Wow... I must be very lost

But thanks for the extra clarification! This is starting to make even more sense now!

 

[PWiz]

Wow... I must be very lost

But thanks for the extra clarification! This is starting to make even more sense now!

Hehe, just focus on my first post and wait for your teacher to start with the math. You'll get to it eventually, so fasten your seat belts until then and understand the concepts first (trust me, once you get used to Lagrangian mechanics, it will appear to be as easy and natural to you as Newton's laws)