# Deep meaning of work integral formula [duplicate]

I want to understand very deeply the meaning of the work integral formula:

$$\int m\frac{d\bar{v}}{dt}d\bar{l} \, .$$

It is not enough for me to know that it was defined in this way, I want to know why it was defined in this way. My intuition says that is due to the conservation of energy, however I don't fully understand the meaning of this term.

To start, what is the physical meaning of $$m\frac{d\bar{v}}{dt}d\bar{l}$$, what we are looking for when we calculate it?

Work is defined as the line integral of the force because this makes it equal to the change in the kinetic energy.

\begin{align}W_{A\rightarrow B}&=\int_A^B\vec{F}\cdot d\vec{\ell}=\int_A^Bm\vec{a}\cdot d\vec{\ell}=\int_A^Bm\frac{d\vec{v}}{dt}\cdot d\vec{\ell}\\&=\int_A^Bm\,d\vec{v}\cdot\frac{d\vec{\ell}}{dt}=\int_A^Bm\vec{v}\cdot d\vec{v}=\int_A^Bd\left(\frac{1}{2}m \vec{v}^2\right)\\&=K_B-K_A\end{align}

As you can see, $$\vec{F}\cdot d\vec{\ell}$$ happens to be a total differential, so its integral is the difference in something. We call that something the kinetic energy.

Whether energy is conserved or not depends on whether the force is a conservative force. A conservative force is the negative gradient of some potential energy function $$U(\vec{r})$$. In this case the line integral does not depend on the path.

$$W_{A\rightarrow B}=\int_A^B\vec{F}\cdot d\vec{\ell}=-\int_A^B\vec{\nabla}U\cdot d\vec{\ell}=-(U_B-U_A)$$

Then there are two different expressions for the work, and equating them gives the conservation of the total (kinetic plus potential) energy,

$$K_B+U_B=K_A+U_A$$

• So are there dt and dl linked ? And does the integral$\int m\frac{d\bar{v}}{dt}d\bar{l} \,$ represents the sum of all the mdv / dt along a path ? Commented Apr 28, 2019 at 16:44
• $d\vec{\ell}$ and $dt$ are “linked” on the sense that given the trajectory $\vec{\ell}(t)$ you can compute the derivative $d\vec{\ell}/dt$ and thus have a relationship between $d\vec{\ell}$ and $dt$. Their ratio is simply the particle’s velocity $\vec{v}$. Commented Apr 28, 2019 at 18:04
• The path integral is the infinite sum of the infinitesimal work $\vec{F}\cdot d\vec{\ell}$ done along each infinitesimal path segment $d\vec{\ell}$. Commented Apr 28, 2019 at 18:04
• Do you know a book where i can find a deep and complete analysis of the work integral? Any book or website i find explains it very concisely, also the graduate books :( Commented Apr 28, 2019 at 18:05
• I have given a complete explanation. There is nothing more to it. Is there something about my explanation that you don’t understand? Commented Apr 28, 2019 at 18:06

$$K = \frac{p^2}{2m}$$

so

$$\frac{dK}{dp} = \frac{2p}{2m} = v$$

or

$$dK = dW = vdp = mvdv$$

Whether you approach it using momentum and rate of change of momentum or velocity and acceleration, the kinetic energy is quadratic in the variable, which means the differential is linear.

Adding up a bunch of differential elements is integration, so:

$$W = \int{Fdx}= \int{madx} = \int{m\dot v dx} = \int{\dot p dx}$$

The final twist is adding dimensions, so that the force only counts if it is in the direction of motion. Enter the dot product:

$$W = \int{\vec F \cdot d\vec l}$$

I think that is as deep as it gets in Newtonian mechanics.

The physical meaning of $$m\frac {\mathrm dv}{\mathrm dt}\cdot \mathrm dl$$ is the rate of change in momentum in the direction of $$\mathrm dl$$. Since the acceleration $$a=\frac{\mathrm dv}{\mathrm dt}$$ and $$F=ma$$, it is force times displacement in the direction of the force, or a differential amount of work $$\mathrm dW=ma\cdot\mathrm dl$$.

Hope this helps.

• What does mean "the rate of change in momentum in the direction of dl" ? Commented Apr 28, 2019 at 15:43
• Momentum is $mv$. If $m$ is constant and velocity is changing in time then momentum is changing in time. Since momentum is a vector, the dot product with $dl$, another vector is a scalor, which is work. Commented Apr 28, 2019 at 15:53
• In $m\frac{d\bar{v}}{dt}d\bar{l}$, is dt a function of dl ? Commented Apr 28, 2019 at 16:17
• No, it is not. But $\vec{\ell}$ is a function of $t$. The mass moves along the trajectory $\vec{\ell}(t)$. Commented Apr 28, 2019 at 16:18
• I thought dl and dt were linked, and that the integral $\int m\frac{d\bar{v}}{dt}d\bar{l} \,$ simply represented the sum of all the m*dv / dt along a path Commented Apr 28, 2019 at 16:42