Passage not understood in a Physics formula

Question

I stumbled upon the demonstration of the energy problem and saw something I don't understand.

I thought mathematicians would be happier to solve his kind of problem

$$ \int_a^b \vec F \cdot d \vec s = \int_a^b m\vec a \cdot d\vec s = \int_a^b m {d \vec v \over dt} \cdot d\vec s = \int_a^b m d \vec v \cdot {d\vec s \over dt} = \cdots$$

Why can I do the last passage? I think it's a kind of scalar product property mixed to some differentials propery, but I can't figure out better, and on my book it's all taken for granted. I mean, it's not like I can take that $ dt$ and pass it below everything I want, right?

Sorry if the question is too simple but I want to know everything I go through, as I go through. Tell me to delete it and I will if it's a problem. Bye!

Answer 1

Yes you can. But if you think that is not "formal" enough, just look at it the following way:

1. Note that $$\frac{d}{dt}(\vec{v}\cdot\vec{v})=\frac{d}{dt}\left({\frac{d\vec{s}}{dt}\cdot\frac{d\vec{s}}{dt}}\right)=2\frac{d\vec{s}}{dt}\cdot\frac{d^2\vec{s}}{dt^2}= 2\vec{v}\cdot\frac{d\vec{v}}{dt}$$
2. Look at the definition of integration along a path $$\int \vec{F}\cdot d\vec{s} \equiv \int \vec{F}\cdot \frac{d\vec{s}}{dt} dt = \int \vec{F}\cdot \vec{v} dt$$

If you combine these two facts, you'll see that the manipulations that happened are just a shortcut for this.

Answer 2

Let me fix notation before arriving at the core of the answer

• Notation

The integral you are trying to understand it defined as follows:

$$\int_s F:=\int_a^b \langle F(s(t)),\frac{ds}{dt}\rangle dt=\sum_{i=1}^3 \int_a^b F_i(s(t))s_i(t) dt $$

i.e. the integral of the vector field $F=(F_1,F_2,F_3)$ along the curve $t\mapsto s(t)=(s_1(t),s_2(t),s_3(t))$, denoting by $\frac{ds}{dt}$ the tangent vector of the curve $s$ at $s(t)$. In your notation

$$ds=(s_1(t)dt,s_2(t)dt,s_3(t)dt)$$ $$\int_s F:=\int_a^b F\cdot ds, $$

and $F\cdot ds=\sum_{i=1}^3 F_i(s(t))s_i(t) dt$.

• Computation

We have

$$\int_a^b \langle F(s(t)),\frac{ds}{dt}\rangle dt= \int_a^b \frac{d}{dt}\left(\langle F(s(t)),s(t)\rangle\right)dt -\int_a^b \langle \frac{dF(s(t))}{dt},s(t)\rangle dt,~~(*) $$

which follows from the definition of the scalar product $\langle\cdot,\cdot\rangle$. In summary

$$\int_a^b \langle F(s(t)),\frac{ds}{dt}\rangle dt= \langle F(s(b)),s(b)\rangle-\langle F(s(a)),s(a)\rangle -\int_a^b \langle \frac{dF(s(t))}{dt},s(t)\rangle dt. $$

Note the presence of the boundary terms for $t=a$ and $t=b$.

• F=m\frac{d^2s}{dt^2}

In this specific case, we have

$$\int_s F=\int_a^b \langle F(s(t)),\frac{ds}{dt}\rangle dt= \int_a^b m \langle \frac{d^2s}{dt^2},\frac{ds}{dt}\rangle dt= \int_a^b m \sum_{i=1}^3 \frac{d^2s_i}{dt^2}\frac{ds_i}{dt}dt=(\text{first option})= \int_a^b m \sum_{i=1}^3 \frac{d^2s_i}{dt^2}ds_i=(\text{second option})= \int_a^b m \sum_{i=1}^3 dv_i\frac{ds_i}{dt},$$

denoting by $dv_i$ the expression

$$dv_i:=\frac{dv_i}{dt}dt.$$