IPhO 1994 · Theory problem 1

Relativistic Particle

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In the theory of special relativity the relation between energy \(E\) and momentum \(P\) or a free particle with rest mass \(m_0\) is

\[ E=\sqrt{p^2c^2+m_0^2c^4}=mc^2 \]

When such a particle is subject to a conservative force, the total energy of the particle, which is the sum of \(\sqrt{p^2c^2+m_0^2c^4}\) and the potential energy, is conserved. If the energy of the particle is very high, the rest energy of the particle can be ignored (such a particle is called an ultra relativistic particle).

1) consider the one dimensional motion of a very high energy particle (in which rest energy can be neglected) subject to an attractive central force of constant magnitude \(f\). Suppose the particle is located at the centre of force with initial momentum \(p_0\) at time \(t=0\). Describe the motion of the particle by separately plotting, for at least one period of the motion: \(x\) against time \(t\), and momentum \(p\) against space coordinate \(x\). Specify the coordinates of the “turning points” in terms of given parameters \(p_0\) and \(f\). Indicate, with arrows, the direction of the progress of the mothon in the \((p,x)\) diagram. There may be short intervals of time during which the particle is not ultrarelativistic. However, these should be neglected.

Use Answer Sheet 1.

2) A meson is a particle made up of two quarks. The rest mass \(M\) of the meson is equal to the total energy of the two-quark system divided by \(c^2\).

Consider a one--dimensional model for a meson at rest, in which the two quarks are assumed to move along the \(x\)-axis and attract each other with a force of constant magnitude \(f\) It is assumed they can pass through each other freely. For analysis of the high energy motion of the quarks the rest mass of the quarks can be neglected. At time \(t=0\) the two quarks are both at \(x=0\). Show separately the motion of the two quarks graphically by a \((x,t)\) diagram and a \((p,x)\) diagram, specify the coordinates of the “turning points” in terms of \(M\) and \(f\), indicate the direction of the process in your \((p,x)\) diagram, and determine the maximum distance between the two quarks.

Use Answer Sheet 2.

3) The reference frame used in part 2 will be referred to as frame \(S\), the Lab frame, referred to as \(S\), moves in the negative \(x\)-direction with a constant velocity \(v=0.6c\). the coordinates in the two reference frames are so chosen that the point \(x=0\) in \(S\) coincides with the point \(x'=0\) in \(S''\) at time \(t=t'=0\). Plot the motion of the two quarks graphically in a \((x',t')\) diagram. Specify the coordinates of the turning points in terms of \(M\), \(f\) and \(c\), and determine the maximum distance between the two quarks observed in Lab frame \(S'\).

Use Answer Sheet 3.

The coordinates of particle observed in reference frames \(S\) and \(S''\) are related by the Lorentz transformation

\[ \begin{cases} x'=\gamma(x+\beta ct),\\[2pt] t'=\gamma\left(t+\beta\dfrac{x}{c}\right) \end{cases} \]

where \(\beta=v/c\), \(\gamma=1/\sqrt{1-\beta^2}\) and \(v\) is the velocity of frame \(S\) moving relative to the frame \(S''\).

4) For a meson with rest energy \(Mc^2=140\ \mathrm{MeV}\) and velocity \(0.60c\) relative to the Lab frame \(S''\), determine its energy \(E'\) in the Lab Frame \(S''\).

Official English text of the problem.

Answers

  1. Choose \(p_0>0\), and define \[ \tau_0=\frac{p_0}{f},\qquad A_0=\frac{cp_0}{f}. \] The position–time graph consists of straight segments through \[ (t,x):\quad (0,0)\to(\tau_0,A_0)\to(2\tau_0,0) \to(3\tau_0,-A_0)\to(4\tau_0,0). \] The turning points are \((\tau_0,A_0)\) and \((3\tau_0,-A_0)\), and \[ \boxed{T_1=\frac{4p_0}{f}}. \] With \(x\) horizontal and \(p\) vertical, the phase diagram is the clockwise diamond \[ (x,p):\quad (0,p_0)\to(A_0,0)\to(0,-p_0)\to(-A_0,0)\to(0,p_0). \]

  2. Define \[ p_*=\frac{Mc}{2},\qquad \tau=\frac{Mc}{2f},\qquad a=\frac{Mc^2}{2f}. \] Quark 1 follows the motion in part 1 with \(p_0=p_*\); quark 2 has \[ x_2=-x_1,\qquad p_2=-p_1. \] Their positions at the turning times are \[ t=\tau:\ (x_1,x_2)=(a,-a),\qquad t=3\tau:\ (x_1,x_2)=(-a,a). \] Both phase trajectories follow the same diamond clockwise, starting at opposite momentum vertices. Thus \[ \boxed{T=\frac{2Mc}{f}},\qquad \boxed{d_{\max}=\frac{Mc^2}{f}}. \]

  3. Here \(\gamma=5/4\). In coordinates \((t'/\tau,x'/a)\), join the following vertices by straight lines: \[ \begin{aligned} \text{Quark 1:}\quad &(0,0)\to(2,2)\to(3,1)\to(5,3),\\ \text{Quark 2:}\quad &(0,0)\to\left(\tfrac12,-\tfrac12\right) \to\left(\tfrac92,\tfrac72\right)\to(5,3). \end{aligned} \] The two interior vertices of each trajectory are its turning points. Subsequent cycles repeat after a translation \((\Delta t',\Delta x')=(5\tau,3a)\). The maximum simultaneous laboratory separation is \[ \boxed{d'_{\max}=a=\frac{Mc^2}{2f}}. \]

  4. The meson’s laboratory energy is \[ \boxed{E'=\frac{Mc^2}{\sqrt{1-0.60^2}}=175\ \mathrm{MeV}}. \]

Solution

Position–time plots below use time horizontally and position vertically. Phase diagrams use \(x\) horizontally and \(p\) vertically. The laboratory frame is denoted consistently by \(S'\).

1) Motion in the constant-magnitude central force

Choose the positive \(x\)-direction along the initial momentum, so \(p_0>0\). In the ultrarelativistic approximation,

\[ E_{\mathrm{particle}}\simeq c|p|, \qquad \dot x=\frac{pc^2}{E_{\mathrm{particle}}} \simeq c\,\operatorname{sgn}(p). \]

Thus the particle moves at speed \(c\) between reversals. The brief non-ultrarelativistic intervals at the turning points are neglected as instructed.

The attractive force and a corresponding potential are

\[ \dot p=-f\,\operatorname{sgn}(x), \qquad U(x)=f|x|. \]

Energy conservation gives

\[ c|p|+f|x|=cp_0. \]

At a spatial turning point, \(p=0\), so the maximum displacement is

\[ A_0=\frac{cp_0}{f}. \]

Initially \(x>0\), hence \(p=p_0-ft\). The first reversal occurs after

\[ \tau_0=\frac{p_0}{f}, \]

at \(x=c\tau_0=A_0\). The force continues to point toward the origin, so the particle returns with negative momentum. After crossing the origin, the force changes sign.

One complete cycle is:

\(t\) \(x\) \(p\)
\(0\) \(0\) \(p_0\)
\(\tau_0\) \(A_0\) \(0\)
\(2\tau_0\) \(0\) \(-p_0\)
\(3\tau_0\) \(-A_0\) \(0\)
\(4\tau_0\) \(0\) \(p_0\)

Join the position entries by straight lines. The slopes are \(+c\) up to \(\tau_0\), \(-c\) from \(\tau_0\) to \(3\tau_0\), and \(+c\) thereafter. The origin crossing at \(2\tau_0\) is not a reversal.

Consequently,

\[ \boxed{T_1=4\tau_0=\frac{4p_0}{f}}. \]

The phase curve follows directly from energy conservation:

\[ \boxed{\frac{|x|}{A_0}+\frac{|p|}{p_0}=1}. \]

It is a diamond with vertices \((0,p_0)\), \((A_0,0)\), \((0,-p_0)\), and \((-A_0,0)\). Starting at the upper vertex, the motion proceeds to the right vertex, then the lower vertex, then the left vertex, and back to the upper vertex: clockwise.

For an initially negative signed momentum, reflect both \(x\) and \(p\); the amplitude and period use \(|p_0|\).

2) The two quarks in the meson rest frame

Take the interaction potential to vanish when the quarks coincide. For separation \(r=|x_1-x_2|\),

\[ U(r)=fr. \]

Because the meson is at rest, the initial quark momenta are equal and opposite. Label the initially right-moving quark as quark 1:

\[ p_1(0)=p_*,\qquad p_2(0)=-p_*. \]

At \(t=0\), the potential energy is zero. The total energy therefore gives

\[ Mc^2=2cp_*, \qquad \boxed{p_*=\frac{Mc}{2}}. \]

Symmetry preserves

\[ x_2=-x_1,\qquad p_2=-p_1. \]

Each quark experiences a force of magnitude \(f\) directed toward the other quark, which is also toward the origin while they are separated. Therefore each follows the motion found in part 1, with

\[ \boxed{\tau=\frac{Mc}{2f}},\qquad \boxed{a=c\tau=\frac{Mc^2}{2f}}. \]

The complete energy equation confirms this:

\[ Mc^2 =2c|p_1|+f|x_1-x_2| =2\bigl(c|p_1|+f|x_1|\bigr). \]

The following table specifies both plots over one period:

\(t/\tau\) \(x_1/a\) \(p_1/p_*\) \(x_2/a\) \(p_2/p_*\)
\(0\) \(0\) \(1\) \(0\) \(-1\)
\(1\) \(1\) \(0\) \(-1\) \(0\)
\(2\) \(0\) \(-1\) \(0\) \(1\)
\(3\) \(-1\) \(0\) \(1\) \(0\)
\(4\) \(0\) \(1\) \(0\) \(-1\)

For the position–time diagrams, join each quark’s position entries by straight lines. The two triangular curves are reflections of each other about \(x=0\), and the quarks pass through one another at \(t=2\tau\).

For either phase diagram,

\[ \boxed{\frac{|x_i|}{a}+\frac{|p_i|}{p_*}=1}. \]

Quark 1 starts at \((x,p)=(0,p_*)\), while quark 2 starts at \((0,-p_*)\). Both traverse the diamond clockwise.

At the turning points, the quarks are at opposite extremes, so

\[ \boxed{d_{\max}=2a=\frac{Mc^2}{f}}. \]

Their individual period is

\[ \boxed{T=4\tau=\frac{2Mc}{f}}. \]

3) Motion and maximum separation in the laboratory frame

The given transformation has

\[ \beta=\frac35,\qquad \gamma=\frac54, \]

so

\[ \gamma(1+\beta)=2,\qquad \gamma(1-\beta)=\frac12. \]

A segment with velocity \(u=\pm c\) transforms to

\[ u'=\frac{u+\beta c}{1+\beta u/c}=\pm c. \]

Thus all laboratory trajectory segments also have slopes \(+c\) or \(-c\). We obtain their endpoints by transforming the turning events.

For example, quark 1’s first turn occurs at \((t,x)=(\tau,a)\), with \(a=c\tau\). Therefore

\[ t'=\gamma\tau(1+\beta)=2\tau, \qquad x'=\gamma a(1+\beta)=2a. \]

All four turning events in one cycle are:

Turning event \(t'\) \(x'\)
Quark 2, first turn \(\displaystyle \frac{\tau}{2}=\frac{Mc}{4f}\) \(\displaystyle -\frac a2=-\frac{Mc^2}{4f}\)
Quark 1, first turn \(\displaystyle 2\tau=\frac{Mc}{f}\) \(\displaystyle 2a=\frac{Mc^2}{f}\)
Quark 1, second turn \(\displaystyle 3\tau=\frac{3Mc}{2f}\) \(\displaystyle a=\frac{Mc^2}{2f}\)
Quark 2, second turn \(\displaystyle \frac{9\tau}{2}=\frac{9Mc}{4f}\) \(\displaystyle \frac{7a}{2}=\frac{7Mc^2}{4f}\)

Both trajectories begin at \((t',x')=(0,0)\). The rest-frame crossing at \((t,x)=(2\tau,0)\) becomes

\[ (t',x')=\left(\frac{5\tau}{2},\frac{3a}{2}\right). \]

The end of the rest-frame cycle, \((4\tau,0)\), becomes

\[ (t',x')=(5\tau,3a). \]

These points and the turning-point table completely specify the laboratory plots: connect the successive points for each quark by straight lines. Subsequent cycles repeat after translating both curves by

\[ \Delta t'=5\tau=\gamma T,\qquad \Delta x'=3a. \]

To find the observed separation, the two positions must be evaluated at the same laboratory time:

\[ d'(t')=|x'_1(t')-x'_2(t')|. \]

Initially,

\[ x'_1=ct',\qquad x'_2=-ct', \]

so the separation grows as \(2ct'\). Quark 2 reverses first, at \(t'=\tau/2\). At that instant,

\[ x'_1=\frac a2,\qquad x'_2=-\frac a2, \]

and hence \(d'=a\).

From \(t'=\tau/2\) to \(t'=2\tau\), both quarks move to the right at speed \(c\):

\[ x'_1=ct',\qquad x'_2=ct'-a. \]

Their separation therefore stays equal to \(a\). It then decreases to zero at \(5\tau/2\), increases back to \(a\) at \(3\tau\), stays at \(a\) until \(9\tau/2\), and decreases to zero at \(5\tau\). This pattern repeats.

Thus

\[ \boxed{d'_{\max}=a=\frac{Mc^2}{2f} =\frac12\,d_{\max}}. \]

The different laboratory reversal times are essential: the two simultaneous rest-frame turning events are not simultaneous in the laboratory.

4) Laboratory energy of the meson

For the complete meson, the total rest-frame momentum is zero and its total energy is \(Mc^2\). Transforming its energy gives

\[ E'=\gamma\bigl(E+\beta cP_x\bigr)=\gamma Mc^2. \]

With the supplied values,

\[ \gamma=\frac{1}{\sqrt{1-0.60^2}} =\frac{1}{0.80}=1.25, \]

and therefore

\[ \boxed{E'=1.25(140\ \mathrm{MeV})=175\ \mathrm{MeV}}. \]

The solution was checked against the official solution.