- #1

Chenkel

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Looking forward to your feedback, thank you!

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- Thread starter Chenkel
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- #1

Chenkel

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Looking forward to your feedback, thank you!

- #2

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Are you confusing

Looking forward to your feedback, thank you!

- #3

Chenkel

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If we designate the arc length (ds) traveled during a small amount of time (dt), then we have the displacement over time (ds/dt), which is the velocity in arc length per unit time. I'm wondering if (ds/dt) is equal to the magnitude of the velocity vector when dt is small.Are you confusinginstantaneousvelocity withaveragevelocity (over a finite time interval)?

- #4

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It's not when ##dt## is small, it's the limit as ##dt## tends to zero. Those are fundamentally different things mathematically.If we designate the arc length (ds) traveled during a small amount of time (dt), then we have the displacement over time (ds/dt), which is the velocity in arc length per unit time. I'm wondering if (ds/dt) is equal to the magnitude of the velocity vector when dt is small.

Technically, you ought to use ##\Delta s## and ##\Delta t## when you mean finite intervals (small or otherwise). And use ##dt## and ##ds## when you mean

- #5

Chenkel

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I'm wondering if ##\lim_{\Delta t \rightarrow 0} {|\vec V|} = \frac {ds} {dt} ## where ##\vec V## is the linear velocity of the point and ds/dt is the arc length per second where ds is the arc length traced out by the point

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- #6

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This doesn't make mathematical sense. What is ##\vec V## as a function of ##\Delta t##?I'm wondering if ##\lim_{\Delta t \rightarrow 0} {|\vec V|} = \frac {ds} {dt} ##

For uniform circular motion we have:where ##\vec V## is the linear velocity of the point and ds/dt is the arc length traced out by the point per second

$$\vec r = R\cos(\omega t) \hat x + R \sin(\omega t) \hat y$$ and everything can be calculated from that:

$$\vec v = \frac{d\vec r}{dt} = -R\omega \sin(\omega t) \hat x + R \omega \cos(\omega t) \hat y$$$$|\vec v| = R\omega$$$$\Delta s = R \Delta \theta = R\omega \Delta t$$And, we can see that:

$$|\vec v| = \frac{ds}{dt} = \lim_{\Delta t \to 0} \frac{\Delta s}{\Delta t}$$Note that this should hold for non-uniform motion in physical systems, although a mathematically pathological counterexample could be constructed (I think there was a thread about this recently).

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- #7

Chenkel

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I see! It's as if the linear velocity is a string we can wrap around any curve representing the path of the object. Hopefully I am seeing things clearly.This doesn't make mathematical sense. What is ##\vec V## as a function of ##\Delta t##?

For uniform circular motion we have:

$$\vec r = R\cos(\omega t) \hat x + R \sin(\omega t) \hat y$$ and everything can be calculated from that:

$$\vec v = \frac{d\vec x}{dt} = -R\omega \sin(\omega t) \hat x + R \omega \cos(\omega t) \hat y$$$$|\vec v| = R\omega$$$$\Delta s = R \Delta \theta = R\omega \Delta t$$And, we can see that:

$$|\vec v| = \frac{ds}{dt} = \lim_{\Delta t \to 0} \frac{\Delta s}{\Delta t}$$Note that this should hold for non-uniform motion in physical systems, although a mathematically pathological counterexample could be constructed (I think there was a thread about this recently).

Thanks for the reply!

- #8

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This is related to the definition of vectors related to a given trajectory ##\vec{x}(t)## in usual 3D Euclidean space. It defines specific (usually non-inertial) reference frame for an observer moving along this given curve.

You start with the normalized tangent vector,

$$\vec{T}=\frac{\dot{\vec{x}}}{|\dot{\vec{x}}|}.$$

For the following it is more convenient to introduce the arc-length parameter,

$$s=\int_{t_0}^t \mathrm{d} t' |\dot{\vec{x}}(t')|,$$

which is the arclength of the piece of curve given by ##t' \in [t_0,t]##.

Obviously

$$\dot{s} =|\dot{\vec{x}}|>0,$$

and thus you can parametrize the curve as well by ##s## rather than ##t##. Then the normalized tangent vector reads

$$\vec{T}=\frac{\dot{\vec{x}}}{\dot{s}}=\frac{\mathrm{d} \vec{x}}{\mathrm{d} s}.$$

Then, because of ##\vec{T}^2=1=\text{const}## you have

$$\vec{T} \cdot \mathrm{d}_s \vec{T}=0,$$

and thus you can define

$$\vec{N}=\frac{\mathrm{d}_s \vec{T}}{|\mathrm{d}_s \vec{T}|}$$

as one unit-normal vector to the curve and finally, as a third vector making a right-handed Cartesian coordinate system complete, the unit-binormal vector

$$\vec{B}=\vec{T} \times \vec{N}.$$

Further you have by definition of ##N##

$$\mathrm{d}_s \vec{T}=\kappa \vec{N},$$

where ##\kappa = |\mathrm{d}_s \vec{T}|##. Now from ##\vec{N}^2=1## you find that

$$\vec{N} \cdot \mathrm{d}_s \vec{N}=0 \; \Rightarrow \; \mathrm{d}_s \vec{N}=\alpha \vec{T} + \tau \vec{B}$$

and then

$$\mathrm{d}_s \vec{B} = (\mathrm{d}_s \vec{T}) \times \vec{N} + \vec{T} \times \mathrm{d}_s \vec{N} = \kappa \vec{N} \times \vec{N} + \vec{T} \times (\alpha \vec{T}+\tau \vec{B}) = \beta \vec{T} \times \vec{B} = \tau \vec{T} \times (\vec{T} \times \vec{N}) = \tau [ \vec{T} (\vec{T} \cdot \vec{N})-\vec{N} \vec{T}^2]=-\tau \vec{N}.$$

In matrix notation we can write

$$\mathrm{d}_s (\vec{T},\vec{N},\vec{B}) = (\vec{T},\vec{N},\vec{B}) \begin{pmatrix} 0 & \alpha & 0 \\

\kappa &0 & -\tau \\ 0 &\tau &0 \end{pmatrix}=(\vec{T},\vec{N},\vec{B}) \hat{M}. \qquad (*)$$

But now the matrix

$$\hat{O}=(\vec{T},\vec{N},\vec{B}),$$

built with the column vectors ##\vec{T}##, ##\vec{N}##, and ##\vec{B}## is an orthogonal matrix, because these three vectors form a right-handed Cartesian basis, i.e.,

$$\hat{O} \hat{O}^{\text{T}}=\hat{1}$$

and thus

$$\mathrm{d}_s (\hat{O} \hat{O}^{\text{T}})=0=(\mathrm{d}_s \hat{O}) \hat{O}^{\text{T}} + \hat{O} \mathrm{d}_s \hat{O}^{\text{T}},$$

which means that the matrix ##\mathrm{d}_s \hat{O}) \hat{O}^{\text{T}}## is antisymmetric. Now we can write (*) as

$$\mathrm{d}_s \hat{O} = \hat{O} \hat{M} \; \Rightarrow \; \hat{M}=\hat{O}^{\text{T}} \mathrm{d}_s \hat{O},$$

which implies that ##\hat{M}=-\hat{M}^{\text{T}}## and thus ##\alpha=-\kappa##, i.e., finally

$$\mathrm{d}_s \vec{T}=\kappa \vec{N}, \quad \mathrm{d}_s \vec{N} = -\kappa \vec{T} + \tau \vec{N}, \quad \mathrm{d}_s \vec{B}=-\tau \vec{N}.$$

These are the Frenet-Serret formulas, and ##\kappa## and ##\tau## are called curvature and torsion of the curve.

The geometrical meaning becomes clear from the following figure from Wikipedia:

https://commons.wikimedia.org/wiki/File:Frenet.svg#/media/File:Frenet.svg

You start with the normalized tangent vector,

$$\vec{T}=\frac{\dot{\vec{x}}}{|\dot{\vec{x}}|}.$$

For the following it is more convenient to introduce the arc-length parameter,

$$s=\int_{t_0}^t \mathrm{d} t' |\dot{\vec{x}}(t')|,$$

which is the arclength of the piece of curve given by ##t' \in [t_0,t]##.

Obviously

$$\dot{s} =|\dot{\vec{x}}|>0,$$

and thus you can parametrize the curve as well by ##s## rather than ##t##. Then the normalized tangent vector reads

$$\vec{T}=\frac{\dot{\vec{x}}}{\dot{s}}=\frac{\mathrm{d} \vec{x}}{\mathrm{d} s}.$$

Then, because of ##\vec{T}^2=1=\text{const}## you have

$$\vec{T} \cdot \mathrm{d}_s \vec{T}=0,$$

and thus you can define

$$\vec{N}=\frac{\mathrm{d}_s \vec{T}}{|\mathrm{d}_s \vec{T}|}$$

as one unit-normal vector to the curve and finally, as a third vector making a right-handed Cartesian coordinate system complete, the unit-binormal vector

$$\vec{B}=\vec{T} \times \vec{N}.$$

Further you have by definition of ##N##

$$\mathrm{d}_s \vec{T}=\kappa \vec{N},$$

where ##\kappa = |\mathrm{d}_s \vec{T}|##. Now from ##\vec{N}^2=1## you find that

$$\vec{N} \cdot \mathrm{d}_s \vec{N}=0 \; \Rightarrow \; \mathrm{d}_s \vec{N}=\alpha \vec{T} + \tau \vec{B}$$

and then

$$\mathrm{d}_s \vec{B} = (\mathrm{d}_s \vec{T}) \times \vec{N} + \vec{T} \times \mathrm{d}_s \vec{N} = \kappa \vec{N} \times \vec{N} + \vec{T} \times (\alpha \vec{T}+\tau \vec{B}) = \beta \vec{T} \times \vec{B} = \tau \vec{T} \times (\vec{T} \times \vec{N}) = \tau [ \vec{T} (\vec{T} \cdot \vec{N})-\vec{N} \vec{T}^2]=-\tau \vec{N}.$$

In matrix notation we can write

$$\mathrm{d}_s (\vec{T},\vec{N},\vec{B}) = (\vec{T},\vec{N},\vec{B}) \begin{pmatrix} 0 & \alpha & 0 \\

\kappa &0 & -\tau \\ 0 &\tau &0 \end{pmatrix}=(\vec{T},\vec{N},\vec{B}) \hat{M}. \qquad (*)$$

But now the matrix

$$\hat{O}=(\vec{T},\vec{N},\vec{B}),$$

built with the column vectors ##\vec{T}##, ##\vec{N}##, and ##\vec{B}## is an orthogonal matrix, because these three vectors form a right-handed Cartesian basis, i.e.,

$$\hat{O} \hat{O}^{\text{T}}=\hat{1}$$

and thus

$$\mathrm{d}_s (\hat{O} \hat{O}^{\text{T}})=0=(\mathrm{d}_s \hat{O}) \hat{O}^{\text{T}} + \hat{O} \mathrm{d}_s \hat{O}^{\text{T}},$$

which means that the matrix ##\mathrm{d}_s \hat{O}) \hat{O}^{\text{T}}## is antisymmetric. Now we can write (*) as

$$\mathrm{d}_s \hat{O} = \hat{O} \hat{M} \; \Rightarrow \; \hat{M}=\hat{O}^{\text{T}} \mathrm{d}_s \hat{O},$$

which implies that ##\hat{M}=-\hat{M}^{\text{T}}## and thus ##\alpha=-\kappa##, i.e., finally

$$\mathrm{d}_s \vec{T}=\kappa \vec{N}, \quad \mathrm{d}_s \vec{N} = -\kappa \vec{T} + \tau \vec{N}, \quad \mathrm{d}_s \vec{B}=-\tau \vec{N}.$$

These are the Frenet-Serret formulas, and ##\kappa## and ##\tau## are called curvature and torsion of the curve.

The geometrical meaning becomes clear from the following figure from Wikipedia:

https://commons.wikimedia.org/wiki/File:Frenet.svg#/media/File:Frenet.svg

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- #9

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- #10

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I guess you refer to the missing time derivative in the definition of ##s##. I've corrected it.

- #11

erobz

Gold Member

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Typo? You have replaced ##r## with ##x##. That should be confusing for the OP$$\vec v = \frac{d\vec x}{dt} = -R\omega \sin(\omega t) \hat x + R \omega \cos(\omega t) \hat y$$$$|\vec v| =

- #12

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Well spotted, thanks!Typo? You have replaced ##r## with ##x##.

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