Assume $\mathbf F$ is a non-conservative force and it's the unique force acting on a particle, to simplify.

Elementary work done by $\mathbf F$ ($\delta W = \mathbf F \cdot \rm d \mathbf r $) is an inexact differential, as there's no function $U$ such that $\mathbf F \cdot {\rm d} \mathbf r = {\rm d} U = \frac{\partial U}{\partial x}{\rm d}x + \frac{\partial U}{\partial y}{\rm d}y + \frac{\partial U}{\partial z}{\rm d}z $.

But, thanks to work-energy theorem, $δW = {\rm d}(\frac{1}{2}mv^2) = {\rm d}K$, and we careless do: $\displaystyle \int_A^B \delta W = \int_A^B {\rm d}K = K_B - K_A$, but not $\displaystyle \int_A^B \delta W = W_B - W_A$.

Can we say that ${\rm d}K$ is also an inexact differential, although we can integrate it normally?


2 Answers 2


If $\mathbf F$ is the unique force: $\delta W = {m(\frac{\mathbf {dv}}{dt})\mathbf{⋅dr}} = {m(\mathbf {dv.}\frac{\mathbf dr}{dt})} = m(\mathbf v\mathbf {.dv}) = d(\frac{1}{2}mv^2)$

But $v = |\mathbf v| = \sqrt{v_x^2 + v_y^2 + v_z^2}$

So $$\frac{\partial v}{\partial v_x} = \frac{v_x}{v}$$

And $$\frac{\partial (\frac{1}{2}v^2)}{\partial v} = v$$

We can write: $$v_x = \frac{\partial (\frac{1}{2}v^2)}{\partial v}\frac{\partial v}{\partial v_x} = \frac{\partial (\frac{1}{2}v^2)}{\partial v_x}$$

And analogous for the other velocity components. Naming the scalar function $E_k = \frac{1}{2}mv^2$

$$dE_k = m\mathbf v\mathbf {.dv} = \frac{\partial E_k}{\partial v_x}dv_x + \frac{\partial E_k}{\partial v_y}dv_y + \frac{\partial E_k}{\partial v_z}dv_z$$

$\delta W$ is not an exact differential when $\mathbf F$ is not conservative, not a function of $\mathbf r$. But if $\mathbf F$ is the unique force acting on the body, it becomes an exact differential in the space of the velocities.


Can we say that $dK$ is also an inexact differential, although we can integrate it normally?

Little Base:

Writing the work-energy theorem $$\oint_{\mathbf{r}_a}^{\mathbf{r}_b}\mathbf{F}\cdot d\mathbf{r}=K_b-K_a$$

In words, It says that the work done by a force on a particle to take it from one point to another is equal to the difference in kinetic energy of the particle.

There are some points to be noted: First the integral on the left-hand side is a line integral and so it depends on the path. The other point I want to make is that the trajectory of the particle in configuration space is not unique in the sense that there can be two or more with different velocity. For instance :

Suppose a trajectory with $x=y=t$ and $x=y=t^2$ with boundary point $(x,y)=(0,0)$ and $(1,1)$. As you can see though the path are different but in configuration space (naivly $xy$-plane) they are same.

While the path in phase space is unique. It defines the velocity ( naive) and the position of the particle at any instant time.

Now On problem :

The fact is the line integral on the left-hand side is perform in a configuration space so that different paths have different values of the velocity on the boundary point. So that as you change your path the boundary velocities will be changed to maintain the same integral answer.

The inexact differential is naively defined as a quantity that depends on the path through which you have attained the state. kinetic energy integral is performed in space velocity so it needs only a boundary point. So it's an exact differential.

Edit: One point to note that as you change the path on the left-hand side the boundary value of velocity will change to maintain equality.


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