# Line integral calculates work, even though the force from the vector field doesn't cause the movement?

I'm afraid the title here was unclear, so I'll attempt to make things a bit more clear.

I am conceptually confused about the physical meaning of line integrals over a vector field, so I'll pick an arbitrary example here to illustrate my confusion.

Suppose we have the vector field $\mathbf{F} (0, x)$ and the curve parameterized by $C(t) = (\cos(t), \sin(t))$ for $0 \leq t \leq \pi/2$.

If we imagine a particle moving along this curve, I know that the work is $$Work = \int_C \mathbf{F} \cdot d\mathbf{s}$$

That is, the work is essentially the sum of the the component of the force in the direction of travel times the distance traveled, and you take the limit as the distance gets really small. The force here is not producing the movement, however, so I am confused how there is still work. That is, shouldn't the particle be moving straight up due to the vector field it is in? How can it be moving along the defined curve?

If we assume that the particle is somehow constrained to this curve, then the force from the vector field isn't really resulting in displacement, and thus there shouldn't there not be work from it? Hopefully I've explained my question clearly.

Can anyone clear this up, or perhaps correct some other conceptual error (abstract or otherwise) that is leading to this misunderstanding? Thanks.

P.S My searches didn't return any relevant posts, but I apologize if I somehow missed this question being asked already.

• The fact that the particle is limited to a path doesn't produce any work. If you take all the external forces away, moving the particle along that path can be accomplished with an arbitrary small amount of energy (even though that may require a very long time for a given energy budget). If you think about the additional force required to keep a particle on a certain path in the presence of an external force field, than that force will always be perpendicular to the path, i.e. the scalar product of that force and ds is zero. – CuriousOne Oct 17 '14 at 6:25

## 2 Answers

If the particle is moving along some constrained curve then there must be some external force, $\mathbf{F_e}$, acting to constrain it to the curve. So the net force on the particle is:

$$\mathbf{F_{net}} = \mathbf{F} - \mathbf{F_e}$$

and the work done on the particle is:

$$W_{net} = \int (\mathbf{F} - \mathbf{F_e}).d\mathbf{s} = \int \mathbf{F}.d\mathbf{s} - \int\mathbf{F_e}.d\mathbf{s}$$

The first term is the work done by the force field and that hasn't changed. What has changed is that the particle now does work of whatever is applying the external force, and that is given by the second term. Some of the work done by the force $\mathbf{F}$ ends up being done on the external system.

So the force field $\mathbf{F}$ does the same work regardless of whether the path is constrained or not. The only difference is where that work ends up.

• It is also very possible that $\textbf{F_e}$ will always be perpendicular relative to the tangent of the path and thus will only help curve it. For example a circular path, as mentioned by OP, can be achieved with a tether. – fibonatic Oct 17 '14 at 12:54

If we consider Newton's laws, then

$$m\ddot{ \vec x}=\sum_i \vec F_i,\tag1$$

where $\vec F_i$ is force from $i$th source.

When you calculate work of force $\vec F_i$, you have

$$W_i=\int_C \vec F_i d\vec s.\tag2$$

Actual motion, whatever you set it to, results from the sum of all the forces acting on the partice, and the work of any given force is given by the integral $(2)$. Even though it seems to you that the motion is pre-defined, the $i$th force, taking part in the equations of motion, does make its contribution to particle movement, thus does in general produce some work.

The sum of all the work done by each force is the total change of particle kinetic energy:

$$\Delta E=\sum_i W_i.$$