Generalizing one motion equation

Question

So, in physics we have a pretty common equation of motion which is:

$$v_f^2 = v_0^2 + 2a(s_f-s_0)$$

which is valid for one-dimensional movement with constant acceleration. We can generalize it for any acceleration by the use of the chain rule (and derivative of the inverse):

$$\frac{dv}{ds} = \frac{dv}{dt} \cdot \frac{dt}{ds} =a \cdot \frac1{s'(t)} = \frac{a}v$$

therefore:

$$v dv = a ds \iff v_f^2 = v_0^2 + 2 \int_{s_0}^{s_f}a ds$$

valid for all kinds of one directional movement, which (I presume) is the equation that creates conservation of mechanical energy for a one-dimensional movement particle.

I wonder if the following generalisation for greater dimensions of this kinematic equation is true and, if it is so, how could one derive it:

$$|v_2|^2 = |v_1|^2 + 2 \int_{\gamma} \vec a \cdot d \gamma$$

Answer 1

In the general case $s'(t) = |v(t)|$, so you'd find something like $|v|dv =ads$. If you were in $n$-dimensions $v=(v_1,...,v_n)$, $a=(a_1,...,a_n)$ so you'd have $n$-equations like $$ \sqrt{v_1^2+...+v_n^2}dv_i = a_i ds $$.

In 2-d for example, you can always set $v_1 = |v|\cos(\omega)$, $v_2=|v|\sin(\omega)$ so that $$dv_1 = \cos(\omega)d|v| -|v|\sin(\omega)d\omega$$ $$dv_2=\sin(\omega)d|v| + |v| \cos(\omega) d\omega$$ and the LHS of the integeral becomes $$|v|\cos(\omega) d|v| - |v|^2\sin(\omega)d\omega = a_1 ds $$ $$|v|\sin(\omega)d|v| +|v|^2 \cos(\omega)d\omega = a_2 ds $$ You'll get some nicer equations $$|v| d|v| = [a_1\cos(\omega) + a_2\sin(\omega)]ds $$ $$ |v|^2 d\omega = [ a_2\cos(\omega) - a_1 \sin(\omega)] ds$$ The first equation is what you're looking for in regards to the magnitude of the velocities is: $$|v_f|^2 = |v_i|^2 + 2\int_{\gamma} a\cdot \frac{v}{|v|}ds $$ or $$|v_f|^2 = |v_i|^2 + 2\int_{\gamma} a\cdot d\gamma$$ if you set $v= \gamma'(t).$

The second equation for $d\omega$ just gives an equation for the angular part of the velocity.

Lets generalize to n-d now that we have the likely formula $v= |v|\hat{v}$ where $\hat{v}=\frac{v}{|v|}$ is the unit vector in the direction of $v$ so we have $|v|dv = a ds$ so $$ |v|\hat{v}\cdot dv = a \cdot \hat{v} ds.$$ Next since $\hat{v}\cdot v = |v|$ so $$d|v| = d\hat{v} \cdot v + \hat{v}\cdot dv= v\cdot \left(\frac{dv}{|v|}-v\frac{d|v|}{|v|^2} \right)+\frac{v}{|v|}\cdot dv = 2\hat{v}\cdot dv - d|v|$$ implying that $$d|v| = \hat{v}\cdot dv.$$ Plugging this into the one form, we find $$|v|d|v| = a\cdot\hat{v} ds$$ so that $$|v_f|^2 = |v_i|^2 + 2\int_{\gamma} a\cdot d\gamma $$ or $$|v_f|^2 = |v_i|^2 + 2\int_{\gamma} a\cdot \hat{v}ds$$ giving the desired equation.

Answer 2

If $\gamma$ is intended to be arc length, then yes, it is valid. It is derivable from plain old conservation of energy. In and of itself, it's not terribly useful, however. To use it to calculate $|v_2|^2$, you would have to know the path over which to do the integral. That is, you would already have to know the trajectory of the particle. In principle, I suppose you might know the trajectory, $x$, without really knowing at what time the particle actually got to any particular value of $x$. (If you did know that, then you could just go ahead and calculate $v$ and be done with it.) And then you could then calculate $a(x)$, and then get $v_2$. But I can't think of a realistic situation where you would have just that information.