Feynman tries to account for the relation between the principle of stationary action, which is a statement about the whole path of a particle, and Newton's second law, which is a statement about the instantaneous state of a particle. In the paragraphs, he means by the "differential" law, Newton's second law.

Now, I would like to explain why it is true that there are differential laws when there is a least action principle of this kind. The reason is the following: Consider the actual path in space and time. As before, let’s take only one dimension, so we can plot the graph of x as a function of t. Along the true path, S is a minimum. Let’s suppose that we have the true path and that it goes through some point a in space and time, and also through another nearby point b (Fig. 19–11). Now if the entire integral from t1 to t2 is a minimum, it is also necessary that the integral along the little section from a to b is also a minimum. It can’t be that the part from a to b is a little bit more. Otherwise you could just fiddle with just that piece of the path and make the whole integral a little lower.

“So every subsection of the path must also be a minimum. And this is true no matter how short the subsection. Therefore, the principle that the whole path gives a minimum can be stated also by saying that an infinitesimal section of path also has a curve such that it has a minimum action. Now if we take a short enough section of path—between two points a and b very close together—how the potential varies from one place to another far away is not the important thing, because you are staying almost in the same place over the whole little piece of the path. The only thing that you have to discuss is the first-order change in the potential. The answer can only depend on the derivative of the potential and not on the potential everywhere. So the statement about the gross property of the whole path becomes a statement of what happens for a short section of the path—a differential statement. And this differential statement only involves the derivatives of the potential, that is, the force at a point. That’s the qualitative explanation of the relation between the gross law and the differential law.

Given a path whose action is minimum, it follows that every arbitrarily small subsection of this path has minimum action as well. I accept this deduction.

Feynman then tells us, to consider such an infinitesimally small subsection of the whole path.

The part that I don't get: He says that the change in potential energy is not important, but rather the first order change in it, or in other words, the derivative of potential energy with respect to position, that is the force.

I know that force is given by the derivative of the potential energy, and that the first order change in the potential energy is given by the derivative. However I can't get why it's only the first order change that matters. Also, how an infinitesimally small subsection of the whole path having minimum action implies we should consider the first order change in the potential energy?


1 Answer 1


If you imagine drawing the potential as a function of position and pick a point and zoom in on it, then if you zoom in enough (and the potential is differentiable) then it will look like a line.

But the line might not be horizontal, when you assume the potential is spatially constant to first order you are assuming that straight line it looks like happens to be horizontal.

There is no reason to assume the potential has zero derivative. You might have thought it has to be zero derivative since there was talk about a minimum. But the minimum is of the action, and the action is a scalar associated with a path, sorta like how work is a scalar associated with a path, but this path is through time so it depends on how quickly you go as well as where and how far.

So all that talk about a minimum was just saying that the true minimum path had to get from end to end in the absolute most efficient way (by the action's judge of efficiency) so obviously bit can't make any wrong turns so each little of travel must also be as efficient as possible.

And as you zoom in to a short piece it is only looking at the potential in that region. But the potential might not be a minimum there, the action you get/make (it's not conserved so we don't really talk about giving or getting) as you go through the region is minimal. That just means you adjust your speed and direction as you go through it to go through efficiently.

Imagine that you have to pay a tax based on how fast you go, but you get paid based on long you spend in certain neighborhoods. Then you might drive slower through the regions that pay you more (less taxes. And more money, win win) but you are willing to drive faster through a region that pays you less because you do have a final destination you are heading for and you'd like to spend you time going slow through the neighborhood's that pay you more.

That is why you slow down in regions with higher potential and speed up in regions with lower potential.

This goal of having more money (pay less tax and get paid more) does have a best strategy (at least locally). But just because you found the best deal doesn't mean the potential is flat where you were. If the potential was flat then you'd just say the whole region is the same so you'd pick the optimal direction and optimal speed and just cruise though it. So you velocity doesn't change in the middle of a region where the potential is flats because your previous planning already figured out how to truck through it.

If all of a sudden you lost your map, then you might not want to keep going at the same speed if you were in a region of changing potential. If the potential is increasing then where you were was relatively low paying so you were blasting through it and now you should slow down. If the potential is decreasing then you aren't making as much money and you still have a destination to get to so you can't justify going as slow as you were you really dawdle in the regions that way well and truly went as slow as you could in that really really sweet spot. Thus new spot isn't quite as sweet so you can't go quite as slow. So you have to speed up.

So potentials can be increasing or decreasing and they will make you slow or speed up. That's fine.

  • $\begingroup$ I rephrased my question to make it clearer. Feel free to edit yours in response to that. $\endgroup$
    – Omar Nagib
    Commented Aug 1, 2015 at 16:56

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