# Does the integral in the action formula regarding the principle of stationary action represent an area or a length?

I am referring to the Feynman Lectures. The second volume has the "Principle of Least Action" as one of his lectures. (See after the 2nd paragraph below figure 19-6.) Although he does not explicitly say I read other sources that regard it as an area.

But I have a problem with this. It looks to me based on the dimensions of the variables that it represents a length where the action is stationary and an area for all the variations that need to be minimized.

Is it not similar to arc length in the sense that the dimension is 1 not a square and it represents a length not an area. Depending on how you treat the arc length integral will decide if it's a functional or a function for the arc length example.

• As the book is now available on-line (link to the chapter in question: feynmanlectures.caltech.edu/II_19.html), you should (a) link to it and (b) be explicit about what figures/equations you are referring to. – dmckee --- ex-moderator kitten Jul 27 '18 at 0:16
• 19 "Principle of Least Action ...after the 2nd paragraph below figure 19-6 . It is the standard action equation with the Lagrangian as the functional with respect to time. We don't have to use the Feynman lectures...the question is at the fundamental level . Does the integral used to define the action represent an area or a line? – user86411 Jul 27 '18 at 1:25

Here are some examples of how the action is connected to lengths and areas:

1. The action $S=\int \! dt ~L$ (being an integral) represents a signed area in a $(t,L)$ diagram.

2. The relativistic action for a point particle represents a length in spacetime, up to an overall dimensionful constant. The EL equations are the geodesic equations. The non-relativistic limit corresponds to $c\to \infty$.

3. The relativistic Nambu-Goto action for a string represents an area in spacetime, up to an overall dimensionful constant.

Let us for the rest of this answer for simplicity specialize to the case were the action $S$ has an interpretation as a length $s$ of a worldline, cf. Fig. 19-1 & Fig. 19-3. The important quantity is then the difference in lengths between two neighboring paths, not the area between them in a $(t,x)$-diagram, cf. Fig. 19-7 & Fig. 19-9.

• the shortest path of an object from a fixed point to a fixed point in a given length of time with a known kinetic and potential energy expressed in the form of a Lagrangian is represented by an area not a length ? I can see that with all the variational paths but isn't the actual path a length.? That is what is confusing me – user86411 Jul 27 '18 at 14:54
• I updated the answer. – Qmechanic Jul 27 '18 at 16:40
• After reading over the answers to my other questions on stationary action I believe I have isolated my confusion. The answer you posted and I accepted is addressing the issue I am having. In particular your item number 1 . Would it be possible for you to provide some trivial example so I can see that the units in this case will end up a square....since that will represent an area ? Thanx again. – user86411 Sep 2 '18 at 14:24
• Dimensional analysis: $[\text{action}] =[\text{Lagrangian}][\text{time}] =[\text{energy}][\text{time}] =ML^2/T$. – Qmechanic Sep 2 '18 at 14:57