# Why is the independence of orthogonal vector-quantities always implicit in books/lectures?

The "theorem" that I can "just" separately deal with orthogonal quantities (like horizontal and vertical force or velocity, etc), I never found explicitly mentioned, but just implicitly in the solution of exercises. Why?

I do not think that this is obvious.

For example projectile motion exercises are always solved such that vertical and horizontal motion is dealt with separately. One might think

when projectile has larger horiz velocity, then it hits more air particles per unit time, and since it moves vertically upwards, it hits them more often from below then from above and so with larger horiz velocity the vertical deceleration will be larger

Why is it always implicit in books, lectures, exercises, that the orthogonal (eg: verical and horizontal) components (of velocity/force/etc) are independent, and never explicitly mentioned as a theorem and explained why we can do that? Or is it in some?

• The example you just cite doesn't apply to the usual treatment in books on mechanics because friction is ignored. If it is not, it most certainly does depend on each component of the velocity (or rather, on the magnitude of the velocity), like you point out. Generally, books only treat quantities separately when it has first been justified that the degrees of freedom are really independent, of if it should be obvious why this is a reasonable assumption.
– Danu
Mar 13 '14 at 13:58

I think when you invoke Newton's law of force being proportional to acceleration as $\vec F = m \vec a$ you do indeed implicitly assume that both force $\vec F$ and acceleration $\vec a$ are 3D vectors. Once you do that the question regarding the componentwise independence does indeed follow as $F_x = m a_x, F_y = m a_y, F_z = m a_z$. Of course this formal independence ends when the force can be written as a position and/or velocity dependent field, such as $F_x = f_1(x,y,z,\dot x,\dot y,\dot z) = m \ddot x$, etc. Now the components have lost their independence both formally and substantially.

I think the answer to your question is more one of didactics than physics. These things are usually taught in the first semester - or even earlier in school. So it would be too complicated to start talking about separability of differential equations in the first lecture.

On the other hand, once you are a little bit more advanced as a physicist, and already know a little about separability of differential equations, this separability in those simple systems (and its limits as mentioned in the comments and other answers) is very obvious. So there is no need, again, to explicitly mention it.

In fact the example you give is one where the horizontal and vertical motion cannot be treated separately, and indeed the extra complexity this adds is the reason why there is no simple equation to describe the motion in the presence of aerodynamic drag. Let me come back to this, but for now let's take the simpler case where there is no aerodynamic drag and only gravity affects the motion.

Newton's law tells us that $F = ma$, or more precisely $\vec{F} = m\vec{a}$, where $\vec{F}$ is the vector $F_x \hat{x} + F_y \hat{y}$, and $\vec{a}$ is the vector $a_x \hat{x} + a_y \hat{y}$. The vectors $\hat{x}$ and $\hat{y}$ are the unit vectors in the $x$ and $y$ directions respectively. So:

\begin{align} F_x \hat{x} + F_y \hat{y} &= m \left( a_x \hat{x} + a_y \hat{y} \right) \\ &= m a_x \hat{x} + ma_y \hat{y} \end{align}

In the absence of drag the force is just the force due to gravity, $F = (0, -g)$, and our equation becomes:

$$-g \hat{y} = m a_x \hat{x} + m a_y \hat{y}$$

and this naturally splits into two equations, one just involving $x$ and one just involving $y$:

$$0 = m a_x \hat{x}$$

$$-g \hat{y} = m a_y \hat{y}$$

And that's why we can solve the differential equations for the horizontal and vertical motion separately.

I said at the outset that drag spoils this convenient separation. That's because the drag is (approximately) given by $F = A v^2$, or in a vector form $\vec{F} = A |v|^2 \hat{v}$, where $A$ is some constant that depends on the size and shape of the object. If we expand out the equation for the drag we get:

\begin{align} F_x \hat{x} + F_y \hat{y} &= A \left(v_x^2 + v_y^2 \right) \left(\tfrac{v_x}{\sqrt{v_x^2 + v_y^2}} \hat{x} + \tfrac{v_y}{\sqrt{v_x^2 + v_y^2}} \hat{y} \right) \\ &= A v_x \sqrt{v_x^2 + v_y^2} \hat{x} + A v_y \sqrt{v_x^2 + v_y^2} \hat{y} \end{align}

But there is no way to split up this equation into one part just involving $x$ and one part just involving $y$. That means motion in the $\hat{x}$ direction affects motion in the $\hat{y}$ direction and vice versa and we can't analyse them separately.

"Why is it always implicit that the orthogonal components are independent?"

The point is that the orthogonal components can be dependent on each other. Just take the vector $v=y\ \vec{e_x} + x\ \vec{e_y}$ as an example: the x component depends on the y component and vice versa.

When you write down a vector into his components, you're always (implicitly) choosing a basis for the vector in which you are representing him. Normally this basis is formed by the standard-orthonormal basis, in your case the orthogonal vectors $\vec{e_x}$ and $\vec{e_y}$. Those orthogonal vectors on the other hand are always independent from each other.

This is why you can write the horizontal and vertical components down in the first place.

So: - the orthogonal components can be dependent on each other BUT - the orthogonal vectors used to represent the vector in your basis are always independent from each other.

And for your question of "why orthogonal comp.": you don't have to choose the orthogonal basis. If you want you can create your own basis (as long as it's span is the same vector space) in which you represent your vector. But normally the orthogonal basis the simplest.

In fact, it is not always true that, say, horizontal and vertical components, or $x$ and $y$ components, are independent! Consider for example a skater riding a half-pipe! But it is certainly true in a simple dynamics exercise, where somebody told you that the only active force is a vertical one, of the type $\mathbf{F}=-mg\hat{\mathbf{y}}$. Here, of course, components do not mix, hence you can solve the problem considering components separately.

In principle, you can add whatever mechanism you want to mix up components and make them dependent; constrained motions, e.g. on an half-pipe, or air resistance in your example are mechanisms of this type.

• And yet, a half pipe you have two independent movements: along the surface of the pipe and jumping upwards. Mar 13 '14 at 14:12