# Does the action have any physical significance?

1. Jul 24, 2014

After studying the methods of Lagrange and Hamilton for a bit I still find myself uneasy about the action. I don't even know how to define it other than the integral of the Lagrangian with respect to time:
$$I=\int_{t_1}^{t_2}\mathrm{d}t\, L(q,\dot{q},t)$$
Does the action have any significance? I also still fail to see the connection between the action and its corresponding configuration / phase space (is it related to both?). What's really been getting to me is this: why does the action have to be a local min/max/stationary point? What if it wasn't? I'm planning to delve deeper into the topic soon (Hamilton-Jacobi, Canonical transformations, etc) but these gaps in my knowledge are quite irritating.

2. Jul 24, 2014

### Simon Bridge

You could think of it, conceptually, as a relationship between kinetic and potential energy.
That is not precise of course, but it should help give you a kinesthetic feel for it.

The significance, of course, is that Nature follows the path of least action.
So in the end it is a technique for calculating stuff.
I had a really good lecture on vid somewhere - I'll have a look for it.

3. Jul 24, 2014

### vanhees71

The action principle is at the heart of all contemporary physics. The reason is that it provides the mathematically most elegant way to use symmetry principles and (Lie) group theory in determining the fundamental laws of physics. These symmetry principles are often the only guiding principles we have to find the appropriate natural laws, and they are tightly related with observed conservation laws due to Noether's theorems.

The question, why the action is stationary for the trajectories of classical particles or why the classical field equations follow from corresponding action principles, is a very good one. The reason is that our world doesn't work as expected from classical physics but is following the rules of quantum theory. Using the path-integral formulation of quantum theory and functional methods, it turns out that the configurations around the stationary point of the action functional become most important, if the situation you are describing is well approximated by classical approximations to quantum theory. Formally it's the saddle-point expansion of the path integral for the propagators describing the time evolution of the system.

4. Jul 24, 2014

So the action is just a mathematical tool; I guess that's why I didn't expect it to have any physical meaning (like the Lagrangian).

5. Jul 25, 2014

### vanhees71

Yes, it's a mathematical tool, but mathematics is the best way to describe what's going on in Nature. The action principle is the single-most important concept in this most successful description of nature called (theoretical) physics! In this sense it has the most physical meaning of all concepts used in physics.

6. Jul 27, 2014

### Simon Bridge

The "mathematical tool" is the physical significance.
What counts as a physical object tends to be subject to philosophy though.

7. Jul 27, 2014

### random_user

If you refer to http://math.ucr.edu/home/baez/classical/texfiles/2005/book/classical.pdf you'll see that John Baez deduces the fact that granted the good old, F=ma it follows that there is some quantity whose critical points constitute the trajectory of a particle. And this quantity is what we define to be $$I=\int_{t_1}^{t_2}\mathrm{d}t\, L(q,\dot{q},t)$$Alternatively, you can opt for the opposite approach, which is to start with accepting the Principle of Least (or more, accurately, Stationary) Action and deriving F=ma from that.

I'd suggest you to continue reading Baez's notes to the point where he talks about the historical motivation behind taking this alternative view to Classical Mechanics.

8. Jul 27, 2014

### bolbteppa

When you think of L as T - V then T - V represents the excess of Kinetic over potential energy at a point. Thus integrating T - V over a path represents summing up the excess of one form of energy over another form, throughout the entire path. Therefore, minimizing the action just represents minimizing the excess of kinetic energy over potential energy over the motion of a particle. The principle of least action, in this form, is thus a kind of balancing act of forms of energy over the motion of a particle. There are all sorts of exceptions to this, but it's helpful to think like this to a first approximation, e.g. you can see the difference between forms of energy might be maximized instead of minimized in special situations etc...

That is a description of the action in terms of a Lagrangian, how do we get phase space from it? The Lagrangian is represented in terms of position and velocity, which we think of as point coordinates, and the point coordinates generate the surface L = L(x,v). The derivative of L w.r.t. velocity generates a tangent line p to the surface L, and intuitively we can generate the surface either with two point coordinates, x & v, over the whole surface, or using points x and tangent lines p to the surface at those points, x & p. The way to switch between these representations of the surface L is known as a Legendre transform. Doing this gives you the Hamiltonian form of the action, and extremizing it gives you Hamilton's equations. Geometrically the equivalence between the two approaches is completely obvious now, either way generates the same surface, so extremizing the action using either set of coordinates should give the same EOM. But while velocity is directly dependent on position, the tangent line to the point could feasibly point in any direction, so you have to count it as an independent coordinate. Therefore we go from n degrees of freedom (independent coordinates) to 2n d.o.f. and the set of 2n position and generalized momentum coordinates are known as phase space coordinates (while the x & v coordinates are configuration space coordinates).

9. Jul 27, 2014

Staff Emeritus
Why do you single out "action". What about "angular momentum"? How could it be that one has physical significance while the other does not?

10. Jul 27, 2014

### Delta Kilo

But there is a difference. Angular momentum appears in a whole bunch of formulas and interacts with other physical quantities in all sorts of interesting ways. This allows one to develop an intuitive understanding of the effect angular momentum is going to have in a particular physical situation. With action, while famous for its virtue of being stationary, the actual value does not seem to be used much by itself. This makes it difficult to relate observed behavoiur of the system to the value of action. It is not immediately clear for example, what physical significance would a large value of action have compared to a smaller value in a given physical setting.

11. Jul 27, 2014

### voko

It is similar (but not completely analogous) to the two ways of expressing a functional dependence. We can say $y = f(x)$, and we can say $F(x, y) = 0$. The second form, which is also known as implicit, is more general, even though it is less useful when you want, for example, to plot the graph of $y$ vs $x$. The action apparatus is akin to the second, more general form, hence why it is so important in fundamental theory, but may be too abstract in other ways.

12. Jul 27, 2014

### Delta Kilo

Yes, sure. But the question was whether the action by itself has a significance as a physical quantity.
I can imagine a whole bunch of situations where value of angular momentum is important (like if you are an astronaut on a space station spinning out of control you would be very much concerned with the exact amount of angular momentum in your CMGs). Can't think of any problem where the actual value of the action matters (as a numerical value expressed in J*s).

13. Jul 27, 2014

### Matterwave

The only place I can think of where the actual numerical value of the action might actually matter is perhaps in the action-angle formulation of Hamiltonian mechanics, and subsequently the Born-Sommerfeldt quantization procedure based on that. It's been a while since I've dealt with it though, so right now it's just a gut feeling more or less. You can perhaps look into it.

14. Jul 28, 2014

### voko

The original question did not have the "as a physical quantity" bit; the question as you put it is a very different question.

In classical non-relativistic mechanics, the numeric value of action had no significance. Only differences in the value would mean something. Total mechanical energy is exactly like that, too, in that same branch of physics (not co-incidentally as I see it). Would you find it acceptable to say that "total mechanical energy has no significance as a physical quantity"?

15. Jul 28, 2014

### Delta Kilo

Sorry I did not mean to change the subject question.At the end of the day action IS a quantity, measured in Js. How can it be significant if not as a quantity?

Can you please give an example where the value (and not just the sign) of the difference in action matters? Certainly the value of the difference in total energy matters quite a lot.

16. Jul 28, 2014

### voko

Just about any formulation of any physical theory deals with the extremization of action. Which means its first variation must be zero. "Variation" is the limit of difference in action.

17. Jul 28, 2014

### Delta Kilo

Which makes it an extremely valuable mathematical tool. But it does not necesserely make action itself physically significant in a sense energy or anguar momentum are.