A A x ds = v x dv or a • ds = v • dv?

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In 3D curvilinear motion, the correct formulation is a dot product, expressed as a • ds = v • dv, rather than a simple multiplication. This distinction is crucial because, in uniform circular motion, while the speed remains constant (making ds effectively zero), the direction of velocity changes, resulting in a non-zero dv. The left-hand side of the multiplicative equation would be zero, while the right-hand side would not, invalidating that form. The dot product maintains equality since both sides become zero when considering the perpendicular nature of acceleration and velocity. Thus, the dot product is essential for accurately describing the relationship in curvilinear motion.
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I have this question for 3D curvilinear motion of particles. I know that the relation ads = vdv is commonly used in rectilinear physics, but what about in 3D curvilinear motion?

Would it be a x ds = v x dv or a • ds = v • dv for describing curvilinear motion?
 
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It will be the latter - a dot product. Consider uniform circular motion, in which the acceleration is always perpendicular to the velocity. ds will be zero because the speed is constant, but dv will not, as the velocity is changing direction. So the simple multiplicative equality will not hold because the LHS is zero but the RHS is not. But the dot product equality will hold, as both sides will be zero (because dv is perpendicular to v).
 
andrewkirk said:
It will be the latter - a dot product. Consider uniform circular motion, in which the acceleration is always perpendicular to the velocity. ds will be zero because the speed is constant, but dv will not, as the velocity is changing direction. So the simple multiplicative equality will not hold because the LHS is zero but the RHS is not. But the dot product equality will hold, as both sides will be zero (because dv is perpendicular to v).
Hi sir, I was wondering why ds would be zero? Wouldn't a particle in circular motion have some sort of change in displacement over time? (e.g at 90° or so)
 
Iqminiclip said:
Hi sir, I was wondering why ds would be zero? Wouldn't a particle in circular motion have some sort of change in displacement over time? (e.g at 90° or so)
Ah, I was interpreting your s as meaning 'speed' (=|v|) but based on your response, I see you mean it to refer to displacement.

In that case the equation has to be a dot product, not a multiplication, as one cannot multiply vectors. We have an equation for each of the three dimensions:

\begin{align}
a_1\,ds_1 &= v_1\,dv_1\\
a_2\,ds_2 &= v_2\,dv_2\\
a_3\,ds_3 &= v_3\,dv_3
\end{align}

Adding them together, we can represent this as:
$$\vec a\cdot \vec {ds} = \vec v\cdot \vec {dv}$$
 
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