201902200829 Derivation 2.3

Prove that the shortest distance between two points in space is a straight line.

Solution.

This solution is not mine. It was found on the Internet some years ago, to whose author(s) I lost references.

Assume the path (of any curve $C$ ) connecting two points $(a,y(a))$ and $(b,y(b))$ is given by a function $C(x)=(x,y(x))$ , with $\displaystyle{\frac{\mathrm{d}}{\mathrm{d}x}}C(x)=(1,y')$ being the first derivative of the curve.

To minimise the path distance

$\displaystyle{\mathcal{L}=\int \| C'\| \,\mathrm{d}x=\int_a^b\sqrt{1+y'^2}\,\mathrm{d}x}$ ,

define now

$f(x,y,y')=\sqrt{1+y'^2}$ ,

having $\displaystyle{\frac{\mathrm{d}f}{\mathrm{d}y}}=0$ and $\displaystyle{\frac{\mathrm{d}f}{\mathrm{d}y'}}=\frac{y'}{\sqrt{1+y'^2}}$ .

From Euler-Lagrange (E-L) equation it follows that

$\displaystyle{\frac{\mathrm{d}}{\mathrm{d}x}}\bigg( \displaystyle{\frac{y'}{\sqrt{1+y'^2}}}\bigg) =0$ ,

i.e., $y'=\mathrm{Const.}$

In conclusion, the shortest distance between two points in space is a straight line.

Lemma. (Fundamental lemma of the calculus of variations)

If $\displaystyle{\int_{x_1}^{x_2}M(x)\eta (x)\,\mathrm{d}x=0}$ for any $\eta (x)$ continuous through second derivative, then $M(x)$ must identically vanish in the interval $x_1,x_2$ .

_{Text on pg.38, Goldstein}

M	T	W	T	F	S	S
				1	2	3
4	5	6	7	8	9	10
11	12	13	14	15	16	17
18	19	20	21	22	23	24
25	26	27	28

Leave a Reply Cancel reply