How can I think of rotational diffusion inverse seconds?

AI Thread Summary
Rotational diffusion describes the motion of particles, like rods, rotating in a fluid, quantified in inverse seconds (1/s). This concept can be visualized by considering the angular displacement of the particle's ends over time, akin to tracing a path on a sphere. For instance, a rotational diffusion coefficient (D_r) of 10 s^-1 indicates a certain angular movement compared to a higher value of 100 s^-1, which suggests faster rotation. The discussion emphasizes understanding rotational diffusion in terms of angular displacement rather than linear distance. Overall, visualizing this motion involves imagining the distance traced by the particle's ends on a sphere over a given time.
Steve Drake
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When thinking of a spherical shaped particle moving about under Brownian motion, one describes its motion by Diffusion. The units being \frac{m^2}{s} I can understand this physically as a distance it will travel from a certain point in space averaged over x-y and z direction.
Now rotational diffusion on the other hand has units of inverse seconds \frac{1}{s} I cannot think of a way to visualize that? For e.g. a rod shaped particle, what does a
D_r = 10\, s^{-1} as opposed to say a D_r = 100\, s^{-1} mean? How can I visualize this like I can with a sphere moving an actual distance?
Thanks
 
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Think of rotational diffusion as having units radian2/s instead of 1/s.
 
Hi, Thanks

That makes a big more sense...

So in a geometrical sense... if a rod's center of mass with fixed in a liquid somehow, but it could still rotate around that point, does this mean the distance its ends would 'trace' out on a hypothetical sphere in 1 second would equal eg 40 radians?... I am still a bit confused thanks
 

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