# Work/Energy and Impulse/Momentum

1. Jun 29, 2015

### Bullwinckle

If we take F=ma and multiply both sides by dt, we get

Fdt = ma dt

And then:

Fdt = mdv

And then:

Impulse = change in momentum.

OK; I get that.
I get a similar process for Work/Energy multiplying F=ma by ds on both sides as follows

Fds = ma ds

And using a ds = v dv to get

Fds = m v dv

Work = change in kinetic energy.

Now I have been coming to learn that it is not wise to split the derivative
For example the form: ads = vdv is possible in 1D.
And even then, it is fairly contorted: one should not, in a pure sense, split the derivative.

(I have gotten wind of issues like force is a one form and that explains the ds... can we avoid that advanced stuff for now?)

Is it possible get to the core of work/energy and impulse/momentum without splitting the derivative?

2. Jun 29, 2015

### Lebesgue

The sppliting of derivatives is useful trick for intuitively get the concepts around. However, I do not find it rigorous and I prefer just doing the integral.
For the Work/Kinectic energy relation:
The work is a line integral and it must be calculated via a parametrization of the curve (the trajectory). Our parametrization is simply $\vec{r}(t)$. This vector line integral is calculated integrating over the domain of parameter $t \in [t_0,t_1]$ the function multiplied by the derivative/tangent vector (which happens to be the velocity).
$Work = \int_{Path} \vec{F}·d\vec{r}=\int_{t_0}^{t_1} \vec{F} · \vec{v} dt = \int_{t_0}^{t_1} m\vec{a} · \vec{v} dt=\int_{t_0}^{t_1} m \frac{d\vec{v}}{dt} · \vec{v} dt = \int_{t_0}^{t_1} \frac{d}{dt}[\frac{1}{2} m \vec{v} · \vec{v}] dt = \int_{t_0}^{t_1} \frac{d}{dt}[\frac{1}{2} m v^2] dt = \frac{1}{2} m v^2(t_1) - \frac{1}{2} m v^2(t_0) = \Delta E$
And this is valid for arbitrarily close values of $t_0,t_1$.

For the impulse I cannot help you since I have a very simplistic view of the concept. Since it is the change over time of the momentum (i.e. the derivative)
$\vec{I} = \frac{d\vec{p}}{dt} = m\frac{d\vec{v}}{dt} = m\vec{a} = \vec{F}$
(whenever tha mass is constant, which is pretty usual in classical mechanics)

3. Jun 29, 2015

### Bullwinckle

OK, so are you not also stumbling over this:

dr = v dt (to progress from the second to third term)

Is that not taking this: dr/dt = v

And multiplying by dt?

Now I feel I am back at square-1

Or, are your words "parametrization" the key. Is this allowed in a parametrization?
Is there something about BEGINNING with dr = vdt on which I should focus?

Last edited: Jun 29, 2015
4. Jun 29, 2015

### Lebesgue

The $d\vec{r}$ in the line integral is pure notation, a mere symbol. It does just means that the integral is a certain type integral: a line integral. Line integrals of scalar or vector functions/fields are mathematically completely different objects from typical integrals over subsets of $\mathbb{R}^n$ (they use a different measure).

https://en.wikipedia.org/wiki/Line_integral#Definition_2

I wouldn't matter if a chose a parametrization of the trajectory in which the particle travels the same path but at a different speed. As you can see, the definition(*) given by Wikipedia tells us that computing line integrals of vector fields requiere:
• A parametrization (physicist usually use the typical $\vec{r}(t)$).
• The derivative of that parametrization function (with our choice, it'll be just $\vec{v}(t)$). The splitting of derivatives gives you an intuitive mnemotecnic way of remembering this.
Then you just calculate the dot product of the $\vec{F}$ field and the tangent vector and integrate it over the parameter interval (in most of our cases will be the time interval). But we could have just used another parametrization of the same path (for example, instead of using time, we can use the arclength parameter).

This is mainly the theory that is behind the typical splitting of derivatives. It is valid to use but keep in mind it is just a way of remembering how to calculate line integrals. Mathematically, the splitting of derivatives makes no sense.