A x ds = v x dv or a • ds = v • dv?

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Discussion Overview

The discussion revolves around the appropriate mathematical representation of 3D curvilinear motion of particles, specifically whether the relationship should be expressed as a cross product or a dot product. Participants explore the implications of these formulations in the context of uniform circular motion and the nature of displacement and velocity changes.

Discussion Character

  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant suggests that the relationship for 3D curvilinear motion should be expressed as a cross product (a x ds = v x dv).
  • Another participant argues for the dot product (a • ds = v • dv), citing uniform circular motion where acceleration is perpendicular to velocity, leading to a zero displacement change (ds) while velocity (v) changes direction.
  • A participant questions the assertion that ds would be zero, proposing that a particle in circular motion experiences a change in displacement over time.
  • Further clarification is provided that the initial interpretation of 's' as speed was incorrect, and it should refer to displacement, reinforcing the need for the dot product formulation.
  • Equations for each dimension are presented to support the dot product approach, indicating that the relationship can be expressed as a vector equation involving the dot products of acceleration, displacement, and velocity.

Areas of Agreement / Disagreement

Participants express differing views on whether the relationship should be a cross product or a dot product, indicating that the discussion remains unresolved with multiple competing perspectives.

Contextual Notes

There is a lack of consensus on the interpretation of displacement in circular motion and the implications for the mathematical formulation. The discussion highlights the complexity of vector relationships in curvilinear motion.

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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