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In a wikipedia article about rotating frames is written:

$$ \frac{d}{dt} \hat{u} = \Omega \times \hat{u}$$

What exactly is the intuition behind this equation? I seek a physical explanation of the above equation.

See under time derivatives in the two frames (here)

tryst with freedom
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2 Answers2

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The interpretation is simple. In a rotating frame, a fixed vector changes over a small time slice by an amount perpendicular to the vector and the rotation direction.

fig1

$$ {\rm d}\hat{u} = ( \vec{\Omega} {\rm d}t) \times \hat{u} $$

or

$$ \frac{{\rm d}\hat{u}}{{\rm d}t} = \vec{\Omega} \times \hat{u} $$

JAlex
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I will not be going into math here as you want physical intuition. Unit vectors express a direction. Loosely speaking, you can think of a unit vector to be synonymous to direction or rather a mathematical analogue of the word 'direction'. So we can make the following comment, if you do not change your direction, your unit vector do not change. But if you do change your direction during your motion, the unit vector change changes. If you agree with the last statement, then let us understand, why change of direction is connected to rotation.

We measure curvature of any curve in terms of radius of curvature. This radius is the radius of a circle that is tangent to the curve under consideration. If the curvature changes at different points, then we require multiple circles of different radius to define the radius of curvature and as a result the radius of curvature may vary from point to point depending on the curvature. Now clearly, if we move along a part of a circle, then we are experiencing rotation. So, when we move along some curve, we will experience rotation, due to the curvature of the curve. The rotation is more when the curvature is more or in other words, thus the radius of curvature is small. (curvature is inversely proportional to radius of curvature)

Now, a straight line also a curve. So this must also be a part of a circle. This is true but the radius of the circle is infinite. Thus the radius of curvature is infinite as well and the curvature of a straight line is zero. Since the radius is infinite, we do not experience any rotation while moving along a straight line and thus our unit vector do not change either.

As mentioned earlier, changing direction means change in unit vector. But if we change direction, then we are moving along a curve that is not a straight line and thus, the radius of curvature is finite. This means, changing direction means, we are moving along some part of a circle. Thus, changing direction must be related to rotation right?

You can think of this as a physical reasoning behind why change in unit vector is related to rotation.

Now, about the cross product:

I guess you know, that the quantity $\Omega \times \hat{u}$ is perpendicular to both $\Omega$ and $\hat{u}$. The direction this cross product implies, is along the tangent of the circle, of which the curve is a part of. Clearly, this tangent is the direction of the changing unit vector. This is why taking cross-product with $\Omega$ (the angular velocity vector) is justified.

Hope this clarifies things.

Samapan Bhadury
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  • "But if you do change your direction during your motion, the unit vector change changes."

    Aren't unit vector a property of co-ordinate system rather than of the body going through motion?

     – tryst with freedom Oct 14 '20 at 12:10
  • " if we move along a part of a circle, then we are experiencing rotation. " I agree intuitively but we need an axis to speak about rotation – tryst with freedom Oct 14 '20 at 12:10
  • I can't quite imagine the closing statement "The direction this cross product implies, is along the tangent of the circle, " – tryst with freedom Oct 14 '20 at 12:11
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    Okay, maybe my statement was not clear. I had something else on mind. Let me clarify, As I mentioned, $\Omega \times \hat{u}$ is perpendicular to both $\Omega$ and $\hat{u}$ simultaneously. To begin with, $\hat{u}$ is along the curve i.e. on the point the circle touches the curve (along tangent). $\Omega$ is perpendicular to the plane containing the curve. Thus the cross product of these two is directed towards to center of the circle. Now this is the direction of $\Omega\times\hat{u}$ vector. $\hat{u}$ changes along this. If you add the original $\hat{u}$ and this, you get the new unit vector – Samapan Bhadury Oct 14 '20 at 13:42
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    In a rotating frame, you have to deal with two set of co-ordinates. (i) The global one, which is fixed and about which the body (or you the observer) rotate, (ii) the local one, which is on the rotating body. Note that, the equation you have written, is for the unit vector of the rotating body and not the global one. The global one is still fixed and that is the frame of reference in this case. – Samapan Bhadury Oct 14 '20 at 13:46
  • Surely we need an axis for the rotation. What is the problem with that? The axis is along the direction of $\Omega$. – Samapan Bhadury Oct 14 '20 at 13:48
  • ". Now this is the direction of Ω×u^ vector. u^ changes along this. If you add the original u^ and this, you get the new unit vector " how did you conclude that it will still be unit even after this vector is added? – tryst with freedom Oct 14 '20 at 13:56
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    Well, there is a proof of this. You can find that in some books on vector analysis. I remember reading it in the book by Chatterjee and Sengupta "A treatise on general properties of matter". [link here]{https://books.google.co.in/books/about/A_Treatise_On_General_Properties_Of_Matt.html?id=qz8EjBtpZnsC&redir_esc=y} – Samapan Bhadury Oct 14 '20 at 15:17