Radian, an arbitrary unit too?

Question

Why is the radian defined as the angle subtended at the centre of a circle when the arc length equals the radius? Why not the angle subtended when the arc length is twice as long as the radius , or the radius is twice as long as the arc length? Isn't putting 2 $\pi$ radian similar to arbitrarily putting 360 degree in a revolution?

Answer 1

You are right when you say that the angle can be described in the ways you specified as well. The radian is arbitrary in that sense. The relationship between subtended arc length and angle is not arbitrary.

Radians are a convenient choice which you can see if you relate arc length to radius.

$$ s = r \cdot k \theta $$

If you want the $k$ to be a $1$ then you want to work in radians.

Physicists run into a similar issue when choosing units. If the speed of light is $c$ the distance travelled by a light ray is

$$ d = c t$$

The numerical value of $c$ depends on the units you are working in. If you choose to work in meters and seconds then $c=3.0\times 10^8 \frac{m}{s}$. If you choose to work in light-years and years then $c=1.0 \frac{\text{light-years}}{\text{year}}$.

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Update:

There is a comment requesting clarification as to what I mean by "The relationship between subtended arc length and angle is not arbitrary".

There are many ways you can assign numerical values to different angle measures however that is still what you might call a "human convention". The way in which we agree to assign numbers to angles doesn't change what an angle is. No matter how you do it there will still be a circular arc subtended by the angle which is unique to that angle. Therefore if you give me a scheme for assigning numbers to angles then it will always be possible for me take a given angle measure and tell you the arc length for that angle.

An example of this is in the way you proposed to define angles. We could say that the value of an angle is twice the arc length subtended by the angle. If we did this there would still be a 1-1 correspondence between any angle measurement and the arc-length. In our case if you told me you had an angle measured to be 2 meters in this convention I could telly you the subtended arc length is 1 meter.

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Update about a year later Please don't consider the following as a formal proof of anything. There are some loose ends involving infinite limits that I didn't completely tidy up. Nevertheless this is a sketch of how one might get at the inifinites series for sine using the idea of radian measure for angles and hopefully explains why this series specifically requires radians.

You asked in the comment below why the Taylor series for $\sin$ is only valid for angles measured in radians. To understand why this is we need to have some idea where such a series would come from in the context of geometry.

Remember we define the measure of an angle in radians in terms of the arc length subtended along a circle. Measuring the length of an arc is difficult to do geometrically. The standard approach involves approximating the circular arc with a arbitrarily large number of small line segments which become indefinitely small.

From this approach to measuring arc length we can conclude that if an angle is measured in radians then the following will be approximately true for very small angles,

$$ \sin( \delta \theta ) \approx \delta \theta \qquad \cos(\delta\theta) \approx 1 $$

The above "small angle" formulas are the critical point in this derivation where we use the notion of radian measure in particular. If we measured angles in terms of degrees we would have some conversion factors showing up in these formulas.

Now suppose we want to know the sine of some large angle $\theta$. For a large enough integer $N$ we can write $\theta$ as a multiple $N$ of some very small angle $\delta \theta$,

$$ N \delta \theta = \theta .$$

Using your favorite multiple angle identity it is possible to express $\sin(N \delta \theta)$ in terms of $\sin(\delta \theta)$ and $\cos(\delta \theta)$. In particular I'm going to use DeMoivre's Formula,

$$ \cos(N\delta\theta) + i \sin( N \delta \theta) = (\cos(\delta\theta) + i \sin(\delta\theta))^N$$

Applying the binomial theorem on the right allows us to express this as ,

$$ \cos(N\delta\theta) + i \sin( N \delta \theta) = \sum_{k=0}^N i^k \sin^k(\delta\theta) \cos^{N-k}(\delta \theta) \frac{N!}{k!(N-k)!}$$

Now if $\delta \theta $ is small enough we can use the small angle identities for sine and cosine,

$$ \cos(N\delta\theta) + i \sin( N \delta \theta) = \sum_{k=0}^N i^k (\delta\theta)^k \frac{N!}{k!(N-k)!}$$

Removing all explicit reference to $\delta \theta$ we get,

$$ \cos(\theta) + i \sin(\theta) = \sum_{k=0}^N i^k (\theta)^k \frac{ N!}{k! N^k (N-k)!}$$

Now if we take the limit as $N$ becomes infinite the term $\frac{ N!}{k! N^k (N-k)!}$ will go to $\frac{1}{k!}$. Since all of the approximations we made become exact for infinite $N$ we take the limit and get,

$$ \cos(\theta) + i \sin(\theta) = \sum_{k=0}^\infty i^k \theta^k \frac{1}{k!}$$

If we want the $\sin(\theta)$ then we need to take the imaginary part of this series which will give us only the odd terms in the series. We note that $Im(i) = 1$, $Im(i^3)=-1$, $Im(i^5)=1$, and so on.

$$\sin(\theta) = \theta - \theta^3 / 3! + \theta^5 / 5! -\theta^7/7!+\cdots$$

So you can see that the form of this series is all based on a hypothesis about how to measure angles in terms of arc length and the multiple angle formula for $\sin$.