# Physical meaning of line integral involving force

I'm curious about the physical meaning of the following equation: $$$$\oint \mathbf{F} \cdot d \mathbf{s} = 0$$$$

What does this physically mean? I think is has something to do with the chosen path of the particle but I'm not really sure. I have no experience with line integrals but looked up some information on them.

• It means the work done for that closed path. This is a well known definition. Obviously, if the force is a conservative force then the work is zero since the closed path has no boundary points.
– hft
Commented Feb 18 at 17:10
• (In case it is not obvious... the integral for a conservative force is just $U(\vec x_0) - U(\vec x_0) = 0$, for whatever point $\vec x_0$ you take as the starting (and ending) point of the closed path.)
– hft
Commented Feb 18 at 17:12

A conservative force is one for which the work done between any two points is path-independent.

In general, the work done by a force between two points depends on the path taken. Think about the work done by friction when sliding a box between two points, A and B, on the floor: (i) If you go on a straight line, $$\Delta \vec{r}_{AB}$$, between the points you get some amount of work done: $$W^{(i)}_{F_{fk}} = \vec{F}_{fk} \cdot \Delta \vec{r}_{AB}\,,$$ but, (ii) if you stop at some intermediate point, C, turn around and go back to A, and then go all the way to B, you will find a different value of work: $$W^{(ii)}_{F_{fk}} = \vec{F}_{fk} \cdot \Delta \vec{r}_{AC} + \vec{F}_{fk} \cdot \Delta \vec{r}_{CA} + \vec{F}_{fk} \cdot \Delta \vec{r}_{AB}\,.$$ Because the friction force always points opposite to the displacement, $$\Delta \vec{r}$$, each of these terms will be negative and you will get "more negative work" than for path (i).

But, if you had instead considered the work done by a spring force (a conservative force) between two points along a line, then any number of back-and-forths would cancel out in the calculation of work. All paths would give the same result as that of the direct path from A to B.

An immediate consequence of this definition of a conservative force is that the work done on any closed path is zero. It's just another way of saying that the works due to back-and-forths cancel. Mathematically, that is expressed as a line integral over a closed path $$C$$: $$\oint_C \vec{F}_\text{cons} \cdot d \vec{r} = 0 \quad (\forall \; C)$$ where the $$\oint$$ means "integral over a closed curve", and "$$C$$" is the path taken ("$$C$$" is for "curve" in math).

Where does the line integral come from? Let's start with the basic definition of work by a constant force along a straight-line path: $$W_F = \vec{F} \cdot \Delta\vec{r} = |\vec{F}| |\Delta \vec{r} | \cos \theta_{F,\Delta r}\,.$$ If we would like to calculate the work done along a path that is not straight, or over which the force has a changing magnitude or direction, then we must break up that path into small straight-line "chunks" over which the force is approximately constant. Let's say we break the path into some finite number $$N$$ (maybe $$N=20$$) chunks. Then we can approximate the value of the work as: $$W_F \approx \sum_{n=1}^N \vec{F}(\vec{r}_n) \cdot \Delta \vec{r}_n$$ where $$\vec{F}(\vec{r}_n)$$ is the average force vector over the $$n$$th chunk (at the position $$\vec{r}_n$$ along the path) and $$\Delta \vec{r}_n$$ is the straight-line displacement approximation to the $$n$$th chunk of the path.

Essentially, to obtain the integral form of the line integral, we take the limit of this approximate expression as $$N \rightarrow \infty$$. But the details are a little bit subtle. To make the above sum into a Riemann sum (for which the limit is the definition of the Riemannian integral), we define the (average) velocity at a chunk: $$\vec{v}_n = \frac{\Delta \vec{r}_n}{\Delta t}$$ and we assume that the chunks have been chosen at equal time intervals, $$\Delta t$$. Then the work sum becomes: $$W_F \approx \sum_{n=1}^N \vec{F}(\vec{r}_n) \cdot \Delta \vec{r}_n = \sum_{n=1}^N \vec{F}\left(\vec{r}(t_n)\right) \cdot \frac{\Delta \vec{r}_n}{\Delta t} \, \Delta t = \sum_{n=1}^N \vec{F}\left(\vec{r}(t_n)\right) \cdot \vec{v}(t_n) \, \Delta t \,.$$ Now this is a proper Riemann sum and we obtain the exact value of the work by taking the limit as $$N\rightarrow\infty$$: $$W_F = \lim_{N\rightarrow\infty} \sum_{n=1}^N \vec{F}\left(\vec{r}(t_n)\right) \cdot \vec{v}(t_n) \, \Delta t =: \int_{t_i}^{t_f} \vec{F}\left(\vec{r}(t)\right) \cdot \vec{v}(t) \, dt$$ where one must explicitly specify the parameterized path, $$\vec{r}(t)$$, for the chosen path, $$C$$ (with $$\vec{r}(t_i)$$ the initial point and $$\vec{r}(t_f)$$ the final point), and find its derivative $$\vec{v}(t) = \frac{d\vec{r}}{d t}$$. Note that the integral in the last equality is just symbolic notation for the limit of the Riemann sum. One further piece of notation is to write the integral over time as: $$\int_{t_i}^{t_f} \vec{F}\left(\vec{r}(t)\right) \cdot \vec{v}(t) \, dt = \int_{t_i}^{t_f} \vec{F}\left(\vec{r}(t)\right) \cdot \frac{d\vec{r}(t)}{d t} \, dt =: \int_C \vec{F} \cdot d\vec{r}$$ where the second expression is just another way of writing the first, and the last expression is nothing more than symbolic notation for the explicit integral over time.

So, we see that the work done by a force over a path $$C$$ can be written as: $$W_F^C = \int_C \vec{F} \cdot d\vec{r}$$ which we now know means the sum of the works over small chunks of the path. And the expression $$\oint \vec{F} \cdot d\vec{r} = 0$$ means that the work done by the force, $$F$$, over any closed path is zero, which means that force is conservative.

For a complete definition of the line integral (which is essentially the same as that presented here) see Stewart's Calculus textbook, Chapter 17 ("Vector Calculus").

Whatever closed path an object takes, the total work on it from a conservative force will be zero.

You are taking a closed path (i.e. a loop) and on each point of the path you are taking the dot product $$\vec{F}$$ and $${\rm d}\vec{s}$$ that is you are taking the projection of the force a small infinitesimal piece of path. Then you integrate i.e. you sum this projection on each piece of the path. A rough intuition you can have is that this quantity tells you how much a "marble" attached to the path would be moved around the path by the force (it would correspond to the work made by this force on the path). If you get zero, the bid would not move.

Without context, it doesn't mean much.

If the context is: $${\bf F}$$ is the force from a field on a fiducial (or small enough) particle that travels a closed path $${\bf s}$$, it means the field is conservative. In America (England) these are called maga (Tory) fields...no Sunday is pun-day.

Actually, wiki says it's a conservative vector field, which has two properties: your equation, and: it is the gradient of some scalar potential. Physics examples are Newtonian gravity and the electric field.

It would be true for a magnetic field (on a charged scalar particle), but that's because $${\bf F} \cdot d{\bf s} = 0$$ all the time, and there's a $$d{\bf s}/dt$$ dependences...but a magnetic field isn't really a vector field: it just looks like one, sometimes.

I wonder what we'd do with a arbitrary static magnetic field on a spinor (charge and magnetic moment) over closed path?

If you are familiar with dot products $$\vec{F} \cdot d\vec{\ell}$$ this is how much the force vector (magnitude and direction) acts in the direction of $$d\vec{\ell}$$ and for a line integral $$\oint F\cdot d\ell$$ the circle around the integral symbolizes a closed path i.e. the path ends up at the starting point again. What does it mean for $$\int_C\vec{F} \cdot d\vec{\ell}=0$$ for a curve $$C$$? If you think of a very simple path like $$(0,0,0) \to (5,0,0)$$ back down to $$\to (0,0,0)$$ we can set up the integrals $$\int_0^5 \vec{F}\cdot d\vec{x}\hat{i}+\int_5^0 \vec{F}\cdot d\vec{x}\hat{i}=\int_0^5 \vec{F}\cdot d\vec{x}\hat{i}-\int_0^5 \vec{F}\cdot d\vec{x}\hat{i}=0$$ the force isn't important in this case but let's just say $$\vec{F} = 2\hat{i}$$, you can see that the work done by the force is positive but then since the direction changes and we end back at 0, we haven't done any net work.