Irodov 1.343: Only the Height Is Contracted – Amit & Deepali Sinha Foundation

Problem Irodov 1.343

A right circular cone stands at rest in its own reference frame. Its taper angle, the angle between its axis and a generator of its lateral surface, is $\theta = 45^\circ$ , and the area of that lateral surface is $S_0 = 4.0$ m $^2$ . A second reference frame moves relative to the cone with velocity $v = (4/5)c$ directed along the cone’s axis; write $\beta = v/c$ for the ratio of that speed to the speed of light. As measured in the moving frame, find:

the taper angle $\theta'$ of the cone
the area $S$ of its lateral surface

Half the cone's axial section, in its rest frame: base radius R, height h along the axis, and the generator of slant l at taper angle θ. The right-angle tick sits at the base; the second frame moves along v. — Half the cone’s axial section, in its rest frame: base radius $R$ , height $h$ along the axis, and the generator of slant $l$ at taper angle $\theta$ . The right-angle tick sits at the base; the second frame moves along $\vect v$ .

Given

$\theta = 45^\circ$
$S_0 = 4.0$ m $^2$
$v = (4/5)c$
$\beta = v/c = 0.80$

Find

(a) the taper angle

\theta'

in the moving frame, (b) the lateral surface area

S

in that frame

Solution

Step 1Name the cone by lengths a boost can act on

A boost does different things along the motion and across it, so the cone must first be described by lengths that are purely one or purely the other. Let $R$ be the base radius, $h$ the height along the axis and $l$ the slant, that is, the length of a generator, all measured in the cone’s own rest frame. The half section drawn above is a right triangle in these three lengths, and the taper angle and the lateral area are read off it.

\tan\theta = \frac{R}{h}, \qquad l = \sqrt{R^2+h^2}, \qquad S_0 = \pi R\,l

Step 2What the boost does to the two independent lengths

The velocity is along the axis, so the height lies wholly along the boost and the base radius wholly across it. The rule is then applied to each separately, the along-motion length being multiplied by the contraction factor and the across-motion length being left alone.

h' = h\sqrt{1-\beta^2}, \qquad R' = R

(1)

Step 3The taper angle in the moving frame

The same half section after the boost. Dashed is the proper shape, of height h and slant l; solid is what the moving frame measures: base radius unchanged, height down to h', generator to l', taper angle opened to θ'. — The same half section after the boost. Dashed is the proper shape, of height $h$ and slant $l$ ; solid is what the moving frame measures: base radius unchanged, height down to $h'$ , generator to $l'$ , taper angle opened to $\theta'$ .

In the moving frame the body is still a right circular cone, of base radius $R$ and height $h'$ , so its taper angle obeys the same right-triangle relation as before with $h'$ in place of $h$ . Substituting (1),

\tan\theta' = \frac{R}{h'} = \frac{R}{h\,\sqrt{1-\beta^2}},

and the ratio $R/h$ in the numerator is the proper cone’s own $\tan\theta$ . The given speed fixes the contraction factor,

\sqrt{1-\beta^2} = \sqrt{1-\left(\frac{4}{5}\right)^{2}} = \sqrt{\frac{9}{25}} = \frac{3}{5},

and at the given taper angle the numerator is $\tan 45^\circ$ , which is $1$ , so the quotient is immediate.

\tan\theta' = \frac{\tan\theta}{\sqrt{1-\beta^2}} = \frac{5}{3}, \qquad \theta' = 59^\circ

(2)

Step 4The lateral area in the moving frame

The moving frame measures an ordinary cone of base radius $R$ and slant $l'$ , so its lateral area is $\pi R\,l'$ , and only the slant has to be worked out. Applying Pythagoras to the contracted half section and then using (1),

\begin{aligned} l'^2 &= R^2 + h'^2 \\[2pt] &= R^2 + h^2\left(1-\beta^2\right) \\[2pt] &= \left(R^2+h^2\right) - \beta^2 h^2 \\[2pt] &= l^2\left(1 - \beta^2\,\frac{h^2}{l^2}\right) \end{aligned}

The ratio $h/l$ is the cosine of the angle between the generator and the axis, which is the taper angle itself, so $h/l = \cos\theta$ and the bracket becomes a function of $\theta$ alone.

l' = l\,\sqrt{1-\beta^2\cos^2\theta}

(3)

Since the base radius is common to both frames, the areas are in the ratio of the two slants, and (3) supplies it.

\frac{S}{S_0} = \frac{\pi R\,l'}{\pi R\,l} = \sqrt{1-\beta^2\cos^2\theta}

With $\beta^2 = 0.640$ and $\cos^2 45^\circ = 0.500$ the radicand is $1-0.320 = 0.680$ , whose square root is $0.8246$ .

S = S_0\sqrt{1-\beta^2\cos^2\theta} = 3.3\ \text{m}^2

(4)

The area ratio S/S_0 against β, at taper angle θ = 45°. It falls from 1 slowly at first; the marked point P is the problem's β = 0.8, at 0.82. At β -> 1 the curve reaches sin θ = 0.71, not zero. — The area ratio $S/S_0$ against $\beta$ , at taper angle $\theta = 45^\circ$ . It falls from $1$ slowly at first; the marked point $P$ is the problem’s $\beta = 0.8$ , at $0.82$ . At $\beta \to 1$ the curve reaches $\mathrm{sin}\,\theta = 0.71$ , not zero.

Check

Two degenerate cones test the area formula. A needle along the axis has $\theta \to 0$ , so its lateral surface is a long thin tube lying along the motion and the formula gives $S \to S_0\sqrt{1-\beta^2}$ , the plain contraction of a length along $\vect v$ . A flat disc has $\theta \to 90^\circ$ , its whole surface across the motion, and the formula gives $S \to S_0$ , unchanged, as a transverse area must be. Setting $\beta = 0$ returns $\theta' = \theta$ and $S = S_0$ in both boxes. The taper angle also moves the right way: contraction shortens the height and leaves the radius, so the cone must open out, and $59^\circ$ is indeed larger than $45^\circ$ .

Remark

The area does not vanish in the limit $\beta \to 1$ , which is the surprise here: it tends to $S_0\,\mathrm{sin}\,\theta$ , and that number has a shape behind it. In the moving frame the cone is squashed onto its own base plane, so its lateral surface tends to the base disc, of area $\pi R^2$ . Writing that with the proper quantities, $S_0 = \pi R\,l$ and $\mathrm{sin}\,\theta = R/l$ , gives $\pi R^2 = S_0\,\mathrm{sin}\,\theta$ exactly, so the limit of the formula and the area of the limiting shape agree. The reason a surface can survive an infinite squash is that only one of its two dimensions lies along the motion: the base circle, entirely transverse, is never contracted at all, and it is what the collapsing cone folds down onto.