IPhO 2006 · Theory problem 2

Watching a Rod in Motion

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Enter all your answers into the Answer Script.

Figure 1

Physical situation A pinhole camera, with the pinhole at \(x=0\) and at distance \(D\) from the \(x\) axis, takes pictures of a rod, by opening the pinhole for a very short time. There are equidistant marks along the \(x\) axis by which the apparent length of the rod, as it is seen on the picture, can be determined from the pictures taken by the pinhole camera. On a picture of the rod at rest, its length is \(L\). However, the rod is not at rest, but is moving with constant velocity \(\upsilon\) along the \(x\) axis.

Basic relations A picture taken by the pinhole camera shows a tiny segment of the rod at position \(\tilde{x}\).

2.1 (0.6) What is the actual position \(x\) of this segment at the time when the picture is taken? State your answer in terms of \(\tilde{x}\), \(D\), \(L\), \(\upsilon\), and the speed of light \(c=3.00\times10^8\ \mathrm{m\,s^{-1}}\). Employ the quantities

\[ \beta=\frac{\upsilon}{c} \quad\text{and}\quad \gamma=\frac{1}{\sqrt{1-\beta^2}} \]

if they help to simplify your result.

2.2 (0.9) Find also the corresponding inverse relation, that is: express \(\tilde{x}\) in terms of \(x\), \(D\), \(L\), \(\upsilon\), and \(c\).

Note: The actual position is the position in the frame in which the camera is at rest

Apparent length of the rod The pinhole camera takes a picture at the instant when the actual position of the center of the rod is at some point \(x_0\).

2.3 (1.5) In terms of the given variables, determine the apparent length of the rod on this picture.

2.4 (1.5) Check one of the boxes in the Answer Script to indicate how the apparent length changes with time.

Symmetric picture One pinhole-camera picture shows both ends of the rod at the same distance from the pinhole.

2.5 (0.8) Determine the apparent length of the rod on this picture.

2.6 (1.0) What is the actual position of the middle of the rod at the time when this picture is taken?

2.7 (1.2) Where does the picture show the image of the middle of the rod?

Very early and very late pictures The pinhole camera took one picture very early, when the rod was very far away and approaching, and takes another picture very late, when the rod is very far away and receding. On one of the pictures the apparent length is \(1.00\ \mathrm{m}\), on the other picture it is \(3.00\ \mathrm{m}\).

2.8 (0.5) Check the box in the Answer Script to indicate which length is seen on which picture.

2.9 (1.0) Determine the velocity \(\upsilon\).

2.10 (0.6) Determine the length \(L\) of the rod at rest.

2.11 (0.4) Infer the apparent length on the symmetric picture.

2.4 Check one: The apparent length
☐ increases first, reaches a maximum value, then decreases.
☐ decreases first, reaches a minimum value, then increases.
☐ decreases all the time.
☐ increases all the time.		 1.5
Very early and very late pictures For Examiners Use Only
2.8 Check one:
☐ The apparent length is 1 m on the early picture and 3 m on the late picture.
☐ The apparent length is 3 m on the early picture and 1 m on the late picture.		 0.5

Official English text of the problem.

Answers

Use the abbreviations

\[ \beta=\frac{\upsilon}{c},\qquad \gamma=\frac{1}{\sqrt{1-\beta^2}},\qquad a=\frac{L}{2\gamma},\qquad R=\sqrt{D^2+a^2}. \]

  • 2.1 \[ \boxed{x=\tilde{x}+\beta\sqrt{D^2+\tilde{x}^{\,2}}.} \]

  • 2.2 \[ \boxed{\tilde{x}=\gamma^2\left(x-\beta\sqrt{x^2+\frac{D^2}{\gamma^2}}\right).} \]

  • 2.3 \[ \boxed{ \tilde{L}(x_0)=\gamma L-\beta\gamma^2 \left[ \sqrt{(x_0+a)^2+\frac{D^2}{\gamma^2}} - \sqrt{(x_0-a)^2+\frac{D^2}{\gamma^2}} \right]. } \]

  • 2.4 Select the third box: the apparent length decreases all the time.

  • 2.5 \[ \boxed{\tilde{L}_{\mathrm{sym}}=\frac{L}{\gamma}.} \]

  • 2.6 \[ \boxed{x_{0,\mathrm{sym}}=\beta R =\beta\sqrt{D^2+\frac{L^2}{4\gamma^2}}.} \]

  • 2.7 \[ \boxed{ \tilde{x}_{\mathrm{mid}} =\beta\gamma^2\left(R-\sqrt{D^2+\beta^2a^2}\right) =\frac{\beta a^2}{R+\sqrt{D^2+\beta^2a^2}}. } \] It lies to the right of the image’s geometric center, toward the leading end.

  • 2.8 Select the second box: \(3.00\ \mathrm{m}\) early, \(1.00\ \mathrm{m}\) late.

  • 2.9 \[ \boxed{\upsilon=0.500c=1.50\times10^8\ \mathrm{m\,s^{-1}}} \] in the positive \(x\) direction.

  • 2.10 \[ \boxed{L=\sqrt{3}\ \mathrm{m}=1.73\ \mathrm{m}.} \]

  • 2.11 \[ \boxed{\tilde{L}_{\mathrm{sym}}=1.50\ \mathrm{m}.} \]

Solution

Work throughout in the camera’s rest frame, with the rod moving in the positive \(x\) direction. The exposure is sufficiently short that all light forming one photograph arrives at the same time \(t\). Light propagates along straight lines at speed \(c\).

The stationary scale identifies the coordinate from which a ray was emitted. Thus \(\tilde{x}\) is the segment’s position at emission, whereas \(x\) is its position when the light reaches the pinhole.

2.1

Suppose the light was emitted at time \(t_{\mathrm e}\). Its distance to the pinhole is

\[ r=\sqrt{D^2+\tilde{x}^{\,2}}, \]

so its travel time is

\[ t-t_{\mathrm e}=\frac{r}{c}. \]

During this interval the segment moves through \(\upsilon(t-t_{\mathrm e})\). Therefore

\[ x=\tilde{x}+\upsilon(t-t_{\mathrm e}) =\tilde{x}+\frac{\upsilon}{c}\sqrt{D^2+\tilde{x}^{\,2}}, \]

giving

\[ \boxed{x=\tilde{x}+\beta\sqrt{D^2+\tilde{x}^{\,2}}.} \]

The relation concerns one segment and is independent of \(L\).

2.2

Rearranging and squaring the previous result gives

\[ (x-\tilde{x})^2=\beta^2(D^2+\tilde{x}^{\,2}), \]

or

\[ (1-\beta^2)\tilde{x}^{\,2} -2x\tilde{x} +x^2-\beta^2D^2=0. \]

The quadratic roots are

\[ \tilde{x} =\gamma^2\left(x\pm\beta \sqrt{x^2+\frac{D^2}{\gamma^2}}\right). \]

For positive velocity, the original, unsquared equation requires \(x-\tilde{x}>0\): the segment advances between emission and reception. This selects the minus sign. Hence define

\[ \boxed{ \tilde{x}=f(x) =\gamma^2\left(x-\beta \sqrt{x^2+\frac{D^2}{\gamma^2}}\right). } \]

2.3

The given \(L\) is the rod’s rest length. Special relativity therefore gives its simultaneous length in the camera frame as

\[ \ell=\frac{L}{\gamma}. \]

At the instant of the photograph, its endpoint positions are

\[ x_-=x_0-a,\qquad x_+=x_0+a, \qquad a=\frac{\ell}{2}=\frac{L}{2\gamma}. \]

Their apparent positions are \(f(x_-)\) and \(f(x_+)\). Since

\[ f'(x)=\gamma^2\left( 1-\frac{\beta x}{\sqrt{x^2+D^2/\gamma^2}} \right)>0, \]

the endpoint order is preserved, and the apparent length is

\[ \tilde{L}(x_0)=f(x_0+a)-f(x_0-a). \]

Substitution yields

\[ \boxed{ \tilde{L}(x_0)=\gamma L-\beta\gamma^2 \left[ \sqrt{(x_0+a)^2+\frac{D^2}{\gamma^2}} - \sqrt{(x_0-a)^2+\frac{D^2}{\gamma^2}} \right]. } \]

This includes both Lorentz contraction and the different light travel times from the endpoints.

2.4

Differentiate \(f'(x)\):

\[ f''(x) =-\frac{\beta D^2} {\left(x^2+D^2/\gamma^2\right)^{3/2}}<0. \]

Thus \(f'(x)\) strictly decreases with \(x\). Consequently,

\[ \frac{d\tilde{L}}{dx_0} =f'(x_0+a)-f'(x_0-a)<0. \]

Since \(dx_0/dt=\upsilon>0\),

\[ \frac{d\tilde{L}}{dt}<0. \]

Therefore the apparent length decreases all the time. Select the third box.

2.5

In the symmetric photograph the two apparent endpoint positions are \(-b\) and \(+b\), because their distances from the pinhole are equal:

\[ r_-=r_+=\sqrt{D^2+b^2}. \]

Equal travel distances and a common arrival time imply equal emission times. The endpoint separation at emission is therefore the simultaneous, Lorentz-contracted length:

\[ 2b=\frac{L}{\gamma}. \]

Thus \(b=a\), and

\[ \boxed{\tilde{L}_{\mathrm{sym}}=2a=\frac{L}{\gamma}.} \]

2.6

At the common endpoint emission time, the endpoints are at \(-a\) and \(+a\), so the rod’s actual middle is at \(x=0\).

Their light takes time

\[ \Delta t=\frac{R}{c}, \qquad R=\sqrt{D^2+a^2}, \]

to reach the pinhole. During this interval the middle advances through \(\upsilon\Delta t\). Its actual position when the photograph is taken is therefore

\[ \boxed{ x_{0,\mathrm{sym}} =\beta R =\beta\sqrt{D^2+\frac{L^2}{4\gamma^2}}. } \]

2.7

Apply the inverse relation to the actual position of the rod’s middle:

\[ \tilde{x}_{\mathrm{mid}} =f(\beta R) =\beta\gamma^2 \left[ R-\sqrt{\beta^2R^2+\frac{D^2}{\gamma^2}} \right]. \]

Using \(R^2=D^2+a^2\) and \(1/\gamma^2=1-\beta^2\),

\[ \beta^2R^2+\frac{D^2}{\gamma^2} =D^2+\beta^2a^2. \]

Hence

\[ \boxed{ \tilde{x}_{\mathrm{mid}} =\beta\gamma^2 \left[R-\sqrt{D^2+\beta^2a^2}\right]. } \]

Rationalizing gives the equivalent expression

\[ \boxed{ \tilde{x}_{\mathrm{mid}} =\frac{\beta a^2} {\sqrt{D^2+a^2}+\sqrt{D^2+\beta^2a^2}}. } \]

For the stated motion, \(0<\tilde{x}_{\mathrm{mid}}<a\). Thus the material midpoint appears between the image’s geometric center and its right endpoint.

Its light travels a shorter distance than the endpoint light, so it is emitted later. By that later emission time, the material midpoint has moved to positive \(x\).

2.8

For the very early and very late photographs, use the far-distance limit

\[ \sqrt{D^2+\tilde{x}^{\,2}}\simeq|\tilde{x}|. \]

In the early photograph the rod approaches from negative \(x\). Equation 2.1 becomes

\[ x\simeq(1-\beta)\tilde{x}. \]

Applying this to both endpoints gives the early apparent length

\[ A\equiv\tilde{L}_{\mathrm{early}} =\frac{\ell}{1-\beta} =\frac{L}{\gamma(1-\beta)}. \]

In the late photograph the rod recedes toward positive \(x\), so

\[ x\simeq(1+\beta)\tilde{x}, \]

and

\[ B\equiv\tilde{L}_{\mathrm{late}} =\frac{\ell}{1+\beta} =\frac{L}{\gamma(1+\beta)}. \]

Because \(\beta>0\), \(A>B\). Therefore

\[ \boxed{A=3.00\ \mathrm{m},\qquad B=1.00\ \mathrm{m}.} \]

Select the second box.

2.9

Taking the ratio eliminates the rod’s length:

\[ \frac{A}{B}=\frac{1+\beta}{1-\beta}. \]

Solving,

\[ \beta=\frac{A-B}{A+B} =\frac{3.00-1.00}{3.00+1.00} =0.500. \]

Thus

\[ \boxed{ \upsilon=\beta c =0.500(3.00\times10^8) =1.50\times10^8\ \mathrm{m\,s^{-1}}, } \]

directed along positive \(x\).

2.10

Multiplying the early and late apparent lengths gives

\[ AB =\frac{L^2}{\gamma^2(1-\beta)(1+\beta)} =\frac{L^2}{\gamma^2(1-\beta^2)} =L^2. \]

Therefore

\[ \boxed{ L=\sqrt{AB} =\sqrt{(3.00\ \mathrm{m})(1.00\ \mathrm{m})} =1.73\ \mathrm{m}. } \]

2.11

From part 2.5,

\[ \tilde{L}_{\mathrm{sym}}=\frac{L}{\gamma}. \]

With \(\beta=1/2\),

\[ \gamma=\frac{1}{\sqrt{1-1/4}}=\frac{2}{\sqrt{3}}. \]

Using the unrounded rest length \(L=\sqrt{3}\ \mathrm{m}\),

\[ \boxed{ \tilde{L}_{\mathrm{sym}} =\frac{\sqrt{3}}{2/\sqrt{3}}\ \mathrm{m} =1.50\ \mathrm{m}. } \]

The solution was checked against the official solution.