A x ds = v x dv or a • ds = v • dv?

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The discussion centers on the equations governing 3D curvilinear motion of particles, specifically addressing the relationship between acceleration, displacement, and velocity. The consensus is that the correct formulation is a dot product: a • ds = v • dv, rather than a simple multiplication. This is particularly relevant in uniform circular motion, where acceleration is perpendicular to velocity, leading to a zero displacement change (ds) despite a change in velocity direction (dv). The discussion emphasizes the necessity of using vector dot products in this context to accurately represent the relationships between these quantities.

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Iqminiclip
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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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