Category: Differential Geometry

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202405091824 Pastime Exercise 011

About the graph below, tell some stories as probable as probable can be.

Roughwork.

Assuming linear (/rectilinear) motion in a single direction.

Assuming uniform acceleration, there are two cases: i. zero acceleration and ii. non-zero (constant) acceleration; in the former velocity being (a) constant and the latter (b) non-constant.

Assuming a flat spacetime metric of which the spatial part does not expands with the temporal part.

Imagine a man walking along a straight line from point (x_1,y_1) to point (x_2,y_2). From

\displaystyle{\textrm{Speed }(v)=\frac{\textrm{Distance }(s)}{\textrm{Time }(t)}}

as the path is fixed, i.e., \Delta s=\textrm{Const.}, we see speed v and time t

\begin{aligned} v\uparrow\enspace \Leftrightarrow \enspace& t\downarrow \\ v\downarrow\enspace\Leftrightarrow\enspace& t\uparrow \\ \end{aligned}

in an inverse relationship. As the man keeps his own fair pace and makes himself a good timekeeper, we can treat speed v as an independent variable, and time t a dependent variable.

\textrm{\scriptsize{CASE} \textbf{\texttt{(a)}}}: velocity being constant

Parameterize the Cartesian equation y=mx+c by the parameter t' (*as distinguished from the natural/unit-speed/arc-length parameter time t) so as to write a set of parametric equations:

\begin{aligned} & \begin{cases} s_x(t')=t' \\ s_y(t')=mt'+c \\ \end{cases} \\ & \\ & \begin{cases} v_x(t') =s_x'(t') = 1 \\ v_y(t') = s_y'(t') = m \\ \end{cases} \\ \end{aligned}

\begin{aligned} v^2(t') & = v_x^2(t')+v_y^2(t') \\ v(t') & = \displaystyle{\sqrt{\bigg(\frac{\mathrm{d}s_x(t')}{\mathrm{d}t'}\bigg)^2 +\bigg(\frac{\mathrm{d}s_y(t')}{\mathrm{d}t'}\bigg)^2}} \\ |\mathbf{v}(t')| & = \sqrt{1+m^2} \\ t & = \int_{0}^{t'}|\mathbf{v}(t)|\,\mathrm{d}t \\ & = (\sqrt{1+m^2})t'\\ \end{aligned}

\textrm{\scriptsize{CASE} \textbf{\texttt{(b)}}}: velocity being non-constant

The man begins with initial speed v_1 at start point (x_1,y_1) and ends with final speed v_2 at finish point (x_2,y_2). We have

\displaystyle{\textrm{Acceleration }(a)=\frac{\textrm{Change in speed }(v-u)}{\textrm{Time }(t)}}

Write by SUVAT equations of motion:

\begin{aligned} \mathbf{u} & = (u_x,u_y) \\ \mathbf{v} & = (v_x,v_y) \\ \mathbf{a} & = (a_x,a_y) \\ & = \bigg( \frac{v_x-u_x}{t}, \frac{v_y-u_y}{t}\bigg) \\ \mathbf{s} & = (s_x,s_y) \\ & = \bigg( u_xt+\frac{1}{2}a_xt^2, u_yt+\frac{1}{2}a_yt^2\bigg) \\ & = \bigg(\frac{1}{2}(u_x+v_x)t, \frac{1}{2}(u_y+v_y)t\bigg) \\ & = (x_2-x_1,y_2-y_1) \\ \end{aligned}

Is this time t also a natural parameter?

For a given parametric curve, the natural parametrization is unique up to a shift of parameter.

Wikipedia on Differentiable curve

(to be continued)

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201911190336 Homework 1, Differential Geometry

Let \boldsymbol{\alpha} :I\rightarrow \mathbb{R}^3 be a regular parametrized curve (not necessarily by arc length) and let \beta :J\rightarrow \mathbb{R}^3 be a reparametrization of \boldsymbol{\alpha} by the arc length s=s(t) measured from t_0\in I.

Let also t=t(s) be the inverse function of s and denote the derivative of \boldsymbol{\alpha} wrt t by \boldsymbol{\alpha}'. Prove that

i. \mathrm{d}t/ \mathrm{d}s=1/\|\boldsymbol{\alpha}'\| and \mathrm{d}^2t/ \mathrm{d}s^2=-\langle \boldsymbol{\alpha}',\boldsymbol{\alpha}''\rangle /\|\boldsymbol{\alpha}'\|^4;

ii. The curvature of \boldsymbol{\alpha} at t\in I is \kappa (t)=\displaystyle{\frac{\|\boldsymbol{\alpha}' \wedge\boldsymbol{\alpha}''\|}{\|\boldsymbol{\alpha}'\|^3}}; and

iii. The torsion of \boldsymbol{\alpha} at t\in I is \tau (t)=\displaystyle{-\frac{\langle \boldsymbol{\alpha}' \wedge \boldsymbol{\alpha}'',\boldsymbol{\alpha}'''\rangle }{\| \boldsymbol{\alpha}' \wedge\boldsymbol{\alpha}''\|^2}}.

Solution.

i. By definition \displaystyle{\frac{\mathrm{d}\boldsymbol{\alpha}}{\mathrm{d}t}}=\boldsymbol{\alpha}'.

Using chain rule,

\displaystyle{\frac{\mathrm{d}\boldsymbol{\alpha}}{\mathrm{d}t}}= \displaystyle{\frac{\mathrm{d}\boldsymbol{\alpha}}{\mathrm{d}s}}\displaystyle{\frac{\mathrm{d}s}{\mathrm{d}t}}.

After taking the norm, as a consequence of natural parametrization

(i.e., \bigg| \displaystyle{\frac{\mathrm{d}\boldsymbol{\alpha}}{\mathrm{d}s}}\bigg| =1),

we have

\| \boldsymbol{\alpha}'\|=\bigg| \displaystyle{\frac{\mathrm{d}\boldsymbol{\alpha}}{\mathrm{d}s}} \bigg| \bigg| \displaystyle{\frac{\mathrm{d}s}{\mathrm{d}t}} \bigg|= \bigg| \displaystyle{\frac{\mathrm{d}s}{\mathrm{d}t}}\bigg|.

Hence \mathrm{d}t/\mathrm{d}s=1/ \| \boldsymbol{\alpha}'\|.

That said,

\begin{aligned} \displaystyle{\frac{\mathrm{d}^2t}{\mathrm{d}s^2}} & =\bigg( \displaystyle{\frac{\mathrm{d}t}{\mathrm{d}s}} \bigg) \displaystyle{\frac{\mathrm{d}}{\mathrm{d}t}}\bigg( \displaystyle{\frac{\mathrm{d}t}{\mathrm{d}s}} \bigg) \\ & =\displaystyle{\frac{1}{\| \boldsymbol{\alpha}' \|}}\frac{\mathrm{d}}{\mathrm{d}t}\bigg(\displaystyle{\frac{1}{\|\boldsymbol{\alpha}' \|}}\bigg) \\ & =\displaystyle{\frac{1}{\| \boldsymbol{\alpha}' \|}}\frac{\mathrm{d}}{\mathrm{d}t}\bigg(\displaystyle{\frac{1}{\langle \boldsymbol{\alpha}',\boldsymbol{\alpha}' \rangle^{1/2}}}\bigg)\\ \end{aligned}.

But,

\begin{aligned} & \displaystyle{\frac{\mathrm{d}}{\mathrm{d}t}}\bigg(\displaystyle{\frac{1}{\langle \boldsymbol{\alpha}',\boldsymbol{\alpha}' \rangle^{1/2}}}\bigg)\\ & =-\displaystyle{\frac{1}{2}}\langle \boldsymbol{\alpha}',\boldsymbol{\alpha}' \rangle^{-3/2}\Big( \langle \boldsymbol{\alpha}',\boldsymbol{\alpha}'' \rangle +\langle \boldsymbol{\alpha}'',\boldsymbol{\alpha}' \rangle \Big) \\ & =-\langle \boldsymbol{\alpha}',\boldsymbol{\alpha}' \rangle^{-3/2}\langle \boldsymbol{\alpha}',\boldsymbol{\alpha}'' \rangle \\ \end{aligned}.

Thus,

\begin{aligned} \displaystyle{\frac{\mathrm{d}^2t}{\mathrm{d}s^2}} & =-\displaystyle{\frac{1}{\| \boldsymbol{\alpha}' \|}}\langle \boldsymbol{\alpha}',\boldsymbol{\alpha}' \rangle^{-3/2}\langle \boldsymbol{\alpha}',\boldsymbol{\alpha}'' \rangle \\ & =-\displaystyle{\frac{1}{\langle \boldsymbol{\alpha}',\boldsymbol{\alpha}' \rangle^{1/2}}}\langle \boldsymbol{\alpha}',\boldsymbol{\alpha}' \rangle^{-3/2}\langle \boldsymbol{\alpha}',\boldsymbol{\alpha}'' \rangle \\ & = \displaystyle{-\frac{\langle \boldsymbol{\alpha}',\boldsymbol{\alpha}'' \rangle}{\big( \langle \boldsymbol{\alpha}', \boldsymbol{\alpha}'\rangle^{1/2}\big)^4}} \\ & =-\langle \boldsymbol{\alpha}',\boldsymbol{\alpha}''\rangle /\|\boldsymbol{\alpha}'\|^4 \\ \end{aligned}.

ii.

\boldsymbol{\alpha}'= \displaystyle{\frac{\mathrm{d}\boldsymbol{\alpha}}{\mathrm{d}t}}=\displaystyle{\frac{\mathrm{d}\boldsymbol{\alpha}}{\mathrm{d}s}}\displaystyle{\frac{\mathrm{d}s}{\mathrm{d}t}}=\dot{\boldsymbol{\alpha}}s'.

Besides,

\boldsymbol{\alpha}''=\displaystyle{\frac{\mathrm{d}}{\mathrm{d}t}}(\dot{\boldsymbol{\alpha}}s')=\dot{\boldsymbol{\alpha}}\displaystyle{\frac{\mathrm{d}s'}{\mathrm{d}t}}+s'\displaystyle{\frac{\mathrm{d}\dot{\boldsymbol{\alpha}}}{\mathrm{d}t}}=\dot{\boldsymbol{\alpha}}s''+(s')^2\ddot{\boldsymbol{\alpha}}.

Then,

\langle \boldsymbol{\alpha}'\wedge \boldsymbol{\alpha}'' \rangle =\big( \dot{\boldsymbol{\alpha}}s' \big) \times \big( \dot{\boldsymbol{\alpha}}s''+(s')^2\ddot{\boldsymbol{\alpha}} \big) =(s')^3\langle\dot{\boldsymbol{\alpha}}\wedge \ddot{\boldsymbol{\alpha}} \rangle =\|\boldsymbol{\alpha}'\|^3 \langle \dot{\boldsymbol{\alpha}}\wedge \ddot{\boldsymbol{\alpha}} \rangle.

(For \mathrm{d}s/\mathrm{d}t=\| \boldsymbol{\alpha}'\| of part i. is used.)

It follows that

\langle \boldsymbol{\alpha}'\wedge \boldsymbol{\alpha}'' \rangle =\|\boldsymbol{\alpha}\|^3 \| \dot{\boldsymbol{\alpha}} \|\| \ddot{\boldsymbol{\alpha}}\|\sin \measuredangle (\dot{\boldsymbol{\alpha}},\ddot{\boldsymbol{\alpha}}).

Note that \dot{\boldsymbol{\alpha}}=\mathbf{t} and \ddot{\boldsymbol{\alpha}}=\dot{t} are orthogonal,

\| \boldsymbol{\alpha}\|=1 and \| \ddot{\boldsymbol{\alpha}}\|=\| \dot{\mathbf{t}}\|=\| \kappa\|.

We obtain

\kappa (t)=\displaystyle{\frac{\|\boldsymbol{\alpha}' \wedge\boldsymbol{\alpha}''\|}{\|\boldsymbol{\alpha}'\|^3}}.

iii.

First,

\dot{\boldsymbol{\alpha}}=\displaystyle{\frac{\mathrm{d}\boldsymbol{\alpha}}{\mathrm{d}s}}=\displaystyle{\frac{\mathrm{d}\boldsymbol{\alpha}}{\mathrm{d}t}}\displaystyle{\frac{\mathrm{d}t}{\mathrm{d}s}}=\boldsymbol{\alpha}'\dot{t};

secondly,

\ddot{\boldsymbol{\alpha}}=\displaystyle{\frac{\mathrm{d}}{\mathrm{d}s}}(\boldsymbol{\alpha}'\dot{t})=\boldsymbol{\alpha}'\ddot{t}+\bigg( \displaystyle{\frac{\mathrm{d}}{\mathrm{d}s}}\boldsymbol{\alpha}' \bigg)\dot{t}=\boldsymbol{\alpha}'\ddot{t}+\boldsymbol{\alpha}''\dot{t}^2;

thirdly,

\dddot{\boldsymbol{\alpha}}=\displaystyle{\frac{\mathrm{d}}{\mathrm{d}s}}\big( \boldsymbol{\alpha}'\ddot{t}+\boldsymbol{\alpha}''\dot{t}^2 \big) =\boldsymbol{\alpha}'\dddot{t}+\boldsymbol{\alpha}''\dot{t}\ddot{t}+\boldsymbol{\alpha}''2\dot{t}\ddot{t}+\boldsymbol{\alpha}'''\dot{t}^3=\boldsymbol{\alpha}'\dddot{t}+2\boldsymbol{\alpha}''\dot{t}\ddot{t}+\boldsymbol{\alpha}'''\dot{t}^3.

Compute \langle \dot{\boldsymbol{\alpha}}\wedge \ddot{\boldsymbol{\alpha}}, \dddot{\boldsymbol{\alpha}} \rangle as follows:

\begin{aligned} \langle \dot{\boldsymbol{\alpha}}\wedge \ddot{\boldsymbol{\alpha}}, \dddot{\boldsymbol{\alpha}} \rangle & = \big( \boldsymbol{\alpha}'\dot{t} \big) \wedge \big( \boldsymbol{\alpha}'\ddot{t}+\boldsymbol{\alpha}''\dot{t}^2 \big) \cdot \big( \boldsymbol{\alpha}'\dddot{t}+2\boldsymbol{\alpha}''\dot{t}\ddot{t}+\boldsymbol{\alpha}'''\dot{t}^3 \big) \\ & =\big( \boldsymbol{\alpha}'\dot{t}\wedge \boldsymbol{\alpha}''\dot{t}^2\big) \cdot \big( \boldsymbol{\alpha}'\dddot{t}+2\boldsymbol{\alpha}''\dot{t}\ddot{t}+\boldsymbol{\alpha}'''\dot{t}^3 \big)\\ \end{aligned}

Now that the cross product \boldsymbol{\alpha}'\wedge \boldsymbol{\alpha}'' is orthogonal to both \boldsymbol{\alpha}' and \boldsymbol{\alpha}'', we can ignore the dot product among them and what remains is

(\boldsymbol{\alpha}'\dot{t}\wedge \boldsymbol{\alpha}''\dot{t}^2)\cdot \boldsymbol{\alpha}'''\dot{t}^3, or,

\dot{t}^6\langle \boldsymbol{\alpha}'\wedge \boldsymbol{\alpha}'', \boldsymbol{\alpha}'''\rangle.

Using the result of part i., substitute 1/\| \boldsymbol{\alpha}'\| for \dot{t}, we obtain

\langle \dot{\boldsymbol{\alpha}}\wedge \ddot{\boldsymbol{\alpha}}, \dddot{\boldsymbol{\alpha}} \rangle=\displaystyle{\frac{\langle \boldsymbol{\alpha}'\wedge \boldsymbol{\alpha}'', \boldsymbol{\alpha}'''\rangle}{\|\boldsymbol{\alpha}'\|^6}}.

Using the formula for curvature in part ii. and put it into

\tau =\displaystyle{\frac{\langle \dot{\boldsymbol{\alpha}}\wedge \ddot{\boldsymbol{\alpha}},\dddot{\boldsymbol{\alpha}} \rangle}{\kappa^2}}=\displaystyle{\frac{\langle \boldsymbol{\alpha}'\wedge \boldsymbol{\alpha}'', \boldsymbol{\alpha}'''\rangle}{\kappa^2 \|\boldsymbol{\alpha}'\|^6}},

i.e.,

\tau (t)=\displaystyle{-\frac{\langle \boldsymbol{\alpha}' \wedge \boldsymbol{\alpha}'',\boldsymbol{\alpha}'''\rangle }{\| \boldsymbol{\alpha}' \wedge\boldsymbol{\alpha}''\|^2}}.

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201911181237 Example 1.1

Put the equations of the circular helix (3) in the form (8).

Background.

Eq. (3):

x=a\cos u, y=a\sin u, z=bu

is the parametric form of a circular helix.

Eq. (8):

x=x, y=F_2(x), z=F_3(x)

is another form of the equations of a curve.

Let x\stackrel{\mathrm{def}}{=} a\cos u,

or u=\displaystyle{\cos^{-1}\bigg( \frac{x}{a} \bigg)}.

\begin{aligned} x^2 & = a^2\cos^2 u \\ a^2 - x^2 & = a^2 - a^2 \cos^2 u = a^2\sin^2 u \\ \pm \sqrt{a^2-x^2} & = a\sin u \\ y & = \pm \sqrt{a^2-x^2} \end{aligned}

z=b\cos^{-1}\displaystyle{\bigg( \frac{x}{a} \bigg)}.

In sum, the circular helix is now expressed in the form

\begin{aligned} x & = a\cos u \\ y & = F_2(x)=\pm\sqrt{a^2-x^2} \\ z & =F_3(x)=b\cos^{-1}\bigg( \frac{x}{a} \bigg) \\ \end{aligned}

Remark. In this form the curve is really defined by the last two equations, or, if it be a plane curve in the xy-plane, its equation is in the customary form Eq. (9): y=f(x).

pg.3 , Chapter 1 Curves in Space