How do you interpret the radian in physics? [duplicate]

Question

When calculating $\sin x$, $x$ needs to be radian to calculate it. so for example when solving Uniform Circular motion, $x(t)$, $y(t)$ can be expressed $$x(t)=R\cos(ωt) [m] $$ $$y(t)=R\sin(ωt) [m]$$ and when we differenciate it $$v_x(t)=-Rω\sin(ωt) [rad*m/s]$$ $$v_y(t)=Rω\cos(ωt) [rad*m/s]$$ how do you interpret rad in this way? furthermore $$a_x(t)=-Rω^2\cos(ωt) [rad^2*m/s^2]$$ $$ a_y(t)=-Rω^2\sin(ωt) [rad^2*m/s^2]$$ do we just ignore $$[rad],[rad^2],[rad^n] $$to see the unit as [m/s]? or is there a meaning in it when the unit is written with rad?

also for $e^{iωt}$ I heard ω needs to be $rad/s $ does that mean $e^{\text{blank}}$ the blank needs to be [rad] to have the meaning?

Answer 1

Radian is the ratio between arc and radius of a circle. So it has no dimension. In the case of the derivatives that you mentioned, the units are simply $ms^{-1}$ for speed and $ms^{-2}$ for acceleration.

Answer 2

Dimensional analysis is trickier than they sell it to be. Radians is one of those quirks. Consider this series:

$$ a = sin(\theta)$$ $$ b = \frac{da}{d\theta} = cos(\theta)$$ $$ c = \frac{d^2a}{d\theta^2} = -sin(\theta)$$

Just a stack of derivatives, right? Now let's invert the functions for $a$ and $c$: $$\theta = \text{sin}^{-1}(a)$$ $$\theta = -\text{sin}^{-1}(c)$$

There's nothing wrong with these two equations, but think about what that implies when we bring units in. What are the units for $a$ and $c$? Either they have to be the same, or $\text{sin}^{-1}$ needs to be an awfully specialized function which can somehow accept inputs in different units and produce an output that has the same units!

In reality, the units are only axiomized for some units and some operations. If you have something with units that are some combination of the base 7 SI units, and you do arithmetic operations on them, we have a pretty good sense of what should happen. However, bring in other operations like sin and cos, and it gets complicated fast. Instead of having hard and fast rules, we have soft ones.

Radians is how we handle those soft rules. Radians have no dimensionality, unlike meters (dimensionality: length) or miles per hour (dimensionality: length per time). They are actually a ratio of two lengths (length per length). We keep them around as a placeholder of sorts, reminding ourselves that they are an angle, but in fact they don't fit into the nice easy world of units.

Thus, when we do something like $sin(\theta)$, we may check the angular units, and convert degrees to radians if $\theta$ is in degrees, but otherwise we just silently drop them. Dimensional analysis just doesn't help with tricky functions.

Now this is the general rule. Most people drop "radians" silently. There are systems where you do not. The Boost library in C++ has a unit library where radians are a first class citizen. However, what you will find is that in any system which handles radians like this, there will be a need for games, multiplying by $1[rad]$ or $1[rad^{-1}]$ at times where you are doing something mathematically valid, but where the radians got in the way. For example, there is the small angle approximation of $sin\theta \approx \theta$ This works mathematically, but has to be mucked with to get the units right: $sin\theta \approx \theta\cdot 1[\text{rad}^{-1}]$ Its hard to justify that extra factor other than that it's the thing that made the units work.

Answer 3

Dimensionless quantities are not numbers, although many say so. But in a coherent system of units it happens that the unit $[A]$ of a dimensionless class of quantities $\{A\}$ does not depend on any basic unit. Therefore you have a natural bijective application $M$ (measure) between dimensionless quantities $A\in \{A\}$ and real numbers $x\in R$: $$ \{A\} \leftrightarrow R \qquad\qquad x=M(A)={A\over [A]}$$

To every ordinary function $ f:R \to R$, namely y = f (x), you can associate a compound function $ F=f\circ M:\{A\} \to R $ having an adimensional quantity as argument:

$$ F(A) = f[M(A)] = f\left({A\over [A]}\right) = f(x) \in R $$

e.g.:

$$ x = r\;Sin(\alpha) = r\;sin\left({\alpha \over [rad]}\right) $$

Unfortunately the compound function $ F: \{A\} \to R $ and the real function $ f: R \to R $ are usually denoted by the same symbol (e.g. $ x = r\;sin(\alpha)$), with consequent formal inconsistencies. The BIPM justifies such writings by saying that formally $[rad]=1$, a relationship which badly fits with the traditional definition of radian.

Answer 4

Consider the arc length formula $s = r \,\theta$ which has units

$$ \text{[len] = [rad]*[len]} $$

It is obvious then that $\text{[rad]}=[1]$ and considered as dimensionless.

Feel free to add it you units to make it clear you are talking about some form of rotation. For example rotational stiffness might be $$ k_\theta = 1000 \text{ Nm/rad}$$ to distinguish it from pure torque $$\tau = 200 \text{ Nm}$$

even though the units are identical in both cases.