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I've been staring at a formula for a while now which equates an angle with a solid angle. I can see how the equation is dimensionally consistent, which settles well with me, but I get tripped up at the idea of equating radians with steradians. To me these are two different units.

But apparently it's common for people to mix them all up? In other words, the area of a unit hemisphere is equal to the circumference of a unit circle? See how this doesn't seem to make sense?

Now that I think about it, I suppose a radian ought to be thought of as a ratio, instead of areas and circumferences on unit object. 2$\pi$ radians is the ratio of any sized circle's circumference to its radius. 2$\pi$ steradians is the ratio of any sized hemisphere's area to its radius squared.

Is this the correct way of thinking about it?

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  • $\begingroup$ In the SI, both radian and steradian are synonyms of 1. Therefore, from a purely dimensional point of view, they can be equated. Of course, the fact the an equation is dimensionally correct doesn't mean that it's physically correct: there is an infinity of quantities and just seven dimensions. To answer, it could be of help if you can post the equation and the context. $\endgroup$ Oct 18 '17 at 20:37
  • $\begingroup$ Consider a unit vector embedded in a rigid body, ${\bf e}{\tau}$. Consider that vector at some time, $t_1$. Now suppose that, at some time $t_2$, ${\bf e}(t_1) = {\bf e}(t_2)$. Let $\nu$ be the amount that the body has rotated about ${\bf e}{t_2}$ after the time interval. Let $\omega$ be the angular velocity vector. Then, using Gauss-Bonnet theorem, we have that $\nu = \int_{t_1}^{t_2} \omega \cdot {\bf e}(\tau) d \tau + A$, where $A$ is the solid angle traced out be ${\bf e}(\tau)$. $\endgroup$
    – Evan
    Oct 18 '17 at 20:55
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The unit steradian or solid angle are used for two-dimensional, unitless angles. Radians are used for one-dimensional, unitless angles. They are not the same thing.

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  • $\begingroup$ Then how can a formula equate the two? I am very certain that the formula is not wrong. $\endgroup$
    – Evan
    Oct 19 '17 at 20:46
  • $\begingroup$ To what formula/equation are you referring? It may not be wrong, but there may be something that is implied that you do not see. $\endgroup$ Oct 19 '17 at 20:54
  • $\begingroup$ See my response to massimo $\endgroup$
    – Evan
    Oct 19 '17 at 21:02
  • $\begingroup$ I have several questions for that comment. For instance, if $\mathbf{e}(t_{1}) = \mathbf{e}(t_{2})$, then does that not imply no rotation/change has occurred in $\mathbf{e}(\tau)$? Is $\omega$ the rate of rotation of $\mathbf{e}(\tau)$ about the center-of-mass (COM)? Is $\mathbf{e}(\tau)$ one of the eigenvectors for the moment of inertial tensor? $\endgroup$ Oct 19 '17 at 21:23
  • $\begingroup$ Consider a body in fixed axis rotation. You can take ${\bf e}(\tau)$ to be in the direction of the axis of rotation. Then ${\bf e}(\tau)$ is fixed for all time, yet the body has still rotated about ${\bf e}(t_1) = {\bf e}(t_2)$. $\endgroup$
    – Evan
    Oct 19 '17 at 21:27
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There is a relation between radian and steradian. $$2\pi\left(1-\cos\frac{Q}{2}\right) = \text{steradian}$$ where $Q$ is the radian measure. One can derive this from the volume of a sector of a sphere. Here, $Q$ ranges from $0$ to $2\pi$ radian. Angle Q is the plane angle subtended by a spherical cap at centre of a sphere. It can also be viewed in wekkipedia. Hope it helps.

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