Irodov Solutions: Relativistic Mechanics

Q.340. A rod moves lengthwise with a constant velocity v relative to the inertial reference frame K. At what value of v will the length of the rod in this frame be η = 0.5% less than its proper length? 

Solution. 340. From the formula for length contraction

So,  

Q.341. In a triangle the proper length of each side equals α. Find the perimeter of this triangle in the reference frame moving relative to it with a constant velocity V along one of its
 (a) bisectors;
 (b) sides.

Solution. 341. (a) In the frame in which the triangle is at rest the space coordinates of the vertices are  all measured at the same time t. In the moving frame the corresponding coordinates at time t' are

The perimeter P is then

When a figure is observed from a frame moving relative to it, only the components of lengths parallel to the velocity are contracted by the factor √(1 - V²/c²), while perpendicular components are unchanged. Apply this rule to each side and sum the transformed side lengths to obtain the expression shown above.

(b) The coordinates in the first frame are shown at time t. The coordinates in the moving frame are,

The perimeter P is then

For part (b), one side lies along the direction of motion and is therefore contracted; the other two sides have components both parallel and perpendicular to V, so transform each component by the Lorentz rule and then compute the side lengths and sum them to get the displayed perimeter.

Q.342. Find the proper length of a rod if in the laboratory frame of reference its velocity is v = c/2, the length l = 1.00 m, and the angle between the rod and its direction of motion is θ = 45°. 

Solution. 342. In the rest frame, the coordinates of the ends of the rod in terms of proper length l₀

at time t. In the laboratory frame the coordinates at time t' are

To find the proper length l₀, resolve the rod into components parallel and perpendicular to the motion. The parallel component observed in the lab is contracted by √(1 - v²/c²), while the perpendicular component is unchanged. Combine these components (using Pythagoras) to recover l₀; substitute v = c/2 and θ = 45° into the displayed relations to get the numerical value.

Q.343.  A stationary upright cone has a taper angle θ =. 45°, and the area of the lateral surface S_o  = 4.0 m². Find: (a) its taper angle; (b) its lateral surface area, in the reference frame moving with a velocity v = (4/5)c along the axis of the cone. 

Solution. 343. In the frame K in which the cone is at rest the coordinates of A are (0,0,0) and of B are (h, h tan θ, 0) . In the frame K , which is moving with velocity v along the axis o f the cone, the coordinates of A and B at time t' are

Thus the taper angle in the frame K is 

and the lateral surface area is, 

Here S₀ - πh² secθ tanθ is the lateral surface area in the rest frame and 

When the cone is observed from a frame moving along its axis, lengths along the axis are Lorentz contracted while transverse dimensions are unchanged. Use these transformed dimensions to compute the new taper angle via tan θ' = (transverse)/(axial) and then recompute the lateral surface area using the transformed slant height and base parameters as indicated in the displayed formulae.

 Q.344. With what velocity (relative to the reference frame K) did the clock move, if during the time interval t = 5.0 s, measured by the clock of the frame K, it became slow by Δt = 0.10 s?

Solution. 344. Because of time dilation, a moving clock reads less time. We write,

Thus,  

or,
Interpret Δt = 0.10 s as the difference between the coordinate time t and the proper time carried by the moving clock; substitute t = 5.0 s and solve the displayed relation for v. The algebra uses γ = 1/√(1 - v²/c²), leading to the numerical value for v shown in the image expressions.

Q.345. A rod flies with constant velocity past a mark which is stationary in the reference frame K. In the frame K it takes Δt = 20 ns for the rod to fly past the mark. In the reference frame fixed to the rod the mark moves past the rod for Δt' = 25 ns. Find the proper length of the rod. 

Solution. 345. In the frame K the length l of the rod is related to the time of flight Δt by

l = v Δt

In the reference frame fixed to the rod (frame K')the proper length l₀ of the rod is given by

But    

Thus,    

So

and  
The two observed times are related by time dilation and the relative speed v. Use Δt' = γ Δt and l = v Δt (or the corresponding relations shown) to eliminate v and determine the proper length l₀. The displayed steps show this elimination explicitly; substitute the given numerical times to obtain l₀.

Q.346.  The proper lifetime of an unstable particle is equal to Δt₀  = 10 ns. Find the distance this particle will traverse till its decay in the laboratory fraine of reference, where its lifetime is equal to Δt = 20 ns. 

Solution. 346. The distance travelled in the laboratory frame of reference is vΔ t where v is the velocity of the particle. But by time dilation

Thus the distance traversed is

Since γ = Δt/Δt₀ = 20/10 = 2, one can find v from γ and then the distance vΔt directly. The displayed relations give the straightforward substitution and final distance.

Q.347. In the reference frame K a muon moving with a velocity v = 0.990c travelled a distance l = 3.0 km from its birthplace to the point where it decayed. Find:
 (a) the proper lifetime of this muon;
 (b) the distance travelled by the muon in the frame K "from the muon's standpoint". 

Solution. 347. (a) If τ₀ is the proper life time of the muon the life time in the moving frame is

Thus  
Use Δt = l/v to get the dilated lifetime in the laboratory and then obtain τ₀ = Δt/γ. (b) From the muon's own rest frame the distance to the decay point is contracted by the same γ so the spatial separation is l/γ; the displayed formulae above give these numerical relations when v = 0.990c is substituted.

Q.348. Two particles moving in a laboratory frame of reference along the same straight line with the same velocity v = (3/4)c strike against a stationary target with the time interval Δt = 50 ns. Find the proper distance between the particles prior to their hitting the target. 

Solution. 348. In the frame K in which the particles are at rest, their positions are A and B whose coordinates may be taken as,

In the frame K' with respect to which K is moving with a velocity v the coordinates of A and B at time t' in the moving frame are

Suppose B hits a stationary target in K' after time t'_B while A hits it after time t_B + Δt. Then,

To obtain the proper separation, apply the Lorentz transformation to convert the time interval between hits in the laboratory into the corresponding separation in the particles' rest frame; the displayed relations show this conversion step by step. Substitute v = 3/4 c and Δt = 50 ns to evaluate the proper distance.

Q.349. A rod moves along a ruler with a constant velocity. When the positions of both ends of the rod are marked simultaneously in the reference frame fixed to the ruler, the difference of readings on the ruler is equal to Δx₁  = 4.0 m. But when the positions of the rod's ends are marked simultaneously in the reference frame fixed to the rod, the difference of readings on the same ruler is equal to Δx₂ = 9.0 m. Find the proper length of the rod and its velocity relative to the ruler. 

Solution. 349. In the reference frame fixed to the ruler the rod is moving with a velocity v and suffers Lorentz contraction. If l₀ is the proper length of the rod, its measured length will be

In the reference frame fixed to the rod the ruler suffers Lorentz contraction and we must have

and  

From these two relations one eliminates γ (or v) and determines the proper length l₀ and the velocity v. Substitute the given Δx₁ and Δx₂ into the displayed formulae and solve algebraically to obtain the numerical answers.

Q.350. Two rods of the same proper length l₀  move toward each other parallel to a common horizontal axis. In the reference frame fixed to one of the rods the time interval between the moments, when the right and left ends of the rods coincide, is equal to Δt. What is the velocity of one rod relative to the other?

Solution. 350. The coordinates of the ends of the rods in the frame fixed to the left rod are shown. The points B and D coincides when

The points A and E coincide when

From this    
The algebra proceeds by writing the spatial positions of encounters, equating them at the appropriate times and solving for v given Δt and l₀; the displayed equations give this procedure explicitly.

Q.351. Two unstable particles move in the reference frame K along a straight line in the same direction with a velocity v = 0.990c. The distance between them in this reference frame is equal to l = 120 m. At a certain moment both particles decay simultaneously in the reference frame fixed to them. What time interval between the moments of decay of the two particles will be observed in the frame K? Which particle decays later in the frame K? 

Solution. 351. In K₀, the rest frame of the particles, the events corresponding to the decay of the particles are,

In the reference frame Kf the corresponding coordintes are by Lorentz transformation

Now  

by Lorentz Fitzgerald contraction formula. Thus the time lag of the decay time of B is 

B decays later (B is the forward particle in the direction of motion)
Because simultaneity is relative, two decays that are simultaneous in the particles' rest frame are separated in time in K. Apply the Lorentz time transformation to the pair of decay events; the expressions above show that the forward particle (in the direction of motion) decays later in K and the explicit time difference is given by the displayed formula when v = 0.990c and l = 120 m are substituted.

Q.352. A rod AB oriented along the x axis of the reference frame K moves in the positive direction of the x axis with a constant velocity v. The point A is the forward end of the rod, and the point B its rear end. Find:
(a) the proper length of the rod, if at the moment t_A the coordinate of the point A is equal to x_A, and at the moment t_B the coordinate of the point B is equal to x_B;
 (b) what time interval should separate the markings of coordinates of the rod's ends in the frame K for the difference of coordinates to become equal to the proper length of the rod. 

Solution. 352. (a) In the reference frame K with respect to which the rod is moving with velocity v, the coordinates of A and B are

The proper length l₀ is the spatial separation of the rod's ends measured simultaneously in the rod's rest frame. Use Lorentz transformations to convert the pair of lab measurements (x_A at t_A and x_B at t_B) into simultaneous coordinates in the rod's rest frame; the relations in the images show the precise algebra and lead to the formula for l₀. For (b), impose simultaneity in K (t_A = t_B + Δt) and solve for Δt such that the coordinate difference equals l₀.

Q.353. The rod A'B' moves with a constant velocity v relative to the rod AB (Fig. 1.91). Both rods have the same proper length l₀ and at the ends of each of them clocks are mounted, which are synchronized pairwise: A with B and A' with B'. Suppose the moment when the clock B' gets opposite the clock A is taken for the beginning of the time count in the reference frames fixed to each of the rods. Determine: 

(a) the readings of the clocks B and B' at the moment when they are opposite each other;
 (b) the same for the clocks A and A'. 

Solution. 353.  At the instant the picture is taken the coordintes of A, B, A', B' in the rest frame of AB are

In this frame the coordinates o f B ' at other times are B': (t, vt, 0, 0). So B ' is opposite to B at time  In the frame in which B', A' is at rest the time corresponding this is by Lorentz tranformation.

Similarly in the rest frame of A, B, te coordinates of A at other times are

A' is opposite to A at time  

The corresponding time in the frame in which A', B' are at rest is

The problem is solved by applying the Lorentz time transformation to the events of coincidence; the displayed expressions give the clock readings in both frames for parts (a) and (b). Note that relativity of simultaneity causes the clock readings to differ as shown.

Q.354. There are two groups of mutually synchronized clocks K and K' moving relative to each other with a velocity v as shown in Fig. 1.92. The moment when the clock A' gets opposite the clock A is taken for the beginning of the time count. Draw the approximate position of hands of all the clocks at this moment "in terms of the K clocks"; "in terms of the K' clocks".

Solution. 354. By Lorentz transformation

So at time  

 and we get the diagram given below "in terms o f th eif-clock ".

The situation in terms of the K' clock is reversed.

To construct the clock-face diagram, transform the event coordinates of each clock hand to the chosen frame using t' = γ(t - Vx/c²) and then place the hands accordingly. The transformations and the qualitative positions are given in the displayed expressions and figures; they reflect the relativity of simultaneity and show the opposite skewing of clock readings in the two frames.

Q.355. The reference frame K' moves in the positive direction of the x axis of the frame K with a relative velocity V. Suppose that at the moment when the origins of coordinates O and O' coincide, the clock readings at these points are equal to zero in both frames. Find the displacement velocity  of the point (in the frame K) at which the readings of the clocks of both reference frames will be permanent- ly identical. Demonstrate that 

Solution. 355. Suppose x (t) is the locus of points in the frame K at which the readings of the clocks of both reference system are permanently identical then by Lorentz transformation

So differentiatin  

Let  

 (tan h θ is a monotonically increasing function of θ).
The derivation shows that the locus of points where t = t' for all times is a line x(t) with velocity given by the displayed expression; differentiating the Lorentz relation yields the required constant displacement velocity and the identity written in the image is thereby confirmed.

Q.356. At two points of the reference frame K two events occurred separated by a time interval Δt. Demonstrate that if these events obey the cause-and-effect relationship in the frame K (e.g. a shot fired and a bullet hitting a target), they obey that relationship in any other inertial reference frame K'.

Solution. 356. We can take the coordinates of the two events to be

For B to be the effect and A to be cause we must have  

In the moving frame the coordinates of A and B become

Since

If the separation of events is time-like (so that a signal travelling at ≤ c can connect them), the sign of Δt' remains the same in all inertial frames. The displayed inequalities and Lorentz transformation demonstrate this invariance of causal order: a cause cannot become an effect in the past of its cause in any inertial frame.

Q. 357. The space-time diagram of Fig. 1.93 shows three events A, B, and C which occurred on the x axis of some inertial reference frame. Find:
 (a) the time interval between the events A and B in the reference frame where the two events occurred at the same point;
 (b) the distance between the points at which the events A and C occurred in the reference frame where these two events are simultaneous. 

Solution. 357. (a) The four-dimensional interval between A and B (assuming Δy = Δz = 0) is :

5² - 3² bs 16 units

Therefore the time interval between these two events in the reference frame in which the events occurred at the same place is

(b) The four dimensional interval between A and C is (assuming Δy = Δz = 0)

3² - 5² = - 16

So the distance between the two events in the fram in which they are simultaneous is 4 units = 4m.

Q. 358. The velocity components of a particle moving in the xy plane of the reference frame K are equal to v_x  and v_y. Find the velocity v' of this particle in the frame K' which moves with the velocity V relative to the frame K in the positive direction of its x axis. 

Solution. 358. By the velocity addition formula

and  

Q. 359. Two particles move toward each other with velocities v₁ = 0.50c and v₂ = 0.75c relative to a laboratory frame of reference.
 Find: 
 (a) the approach velocity of the particles in the laboratory frame of reference;
 (b) their relative velocity.

Solution. 359. (a) By definition the velocity of apporach is

in the reference frame K .The approach velocity in the lab is simply v₁ + v₂ = 1.25c classically, but relativistically the speed of approach as measured in the lab is their rate of decrease of separation and equals v₁ + v₂

(b) The relative velocity is obtained by the transformation law

their relative velocity (the velocity of one as seen from the other) is obtained via the relativistic addition formula v = (v₁ + v₂)/(1 + v₁v₂/c²), substituting the given numbers to get the explicit value less than c.

Q. 360. Two rods having the same proper length l₀  move lengthwise toward each other parallel to a common axis with the same velocity v relative to the laboratory frame of reference. What is the length of each rod in the reference frame fixed to the other rod? 

Solution. 360. The velocity of one of the rods in the reference frame fixed to the other rod is

The length of the moving rod in this frame is

Compute the relative speed using the relativistic velocity-addition formula v_rel = (2v)/(1 + v²/c²) and then apply length contraction l = l₀√(1 - v_rel²/c²). The displayed expressions give the direct substitution and final contracted length.

Q. 361. Two relativistic particles move at right angles to each other in a laboratory frame of reference, one 'with the velocity v₁ and the other with the velocity v₂. Find their relative velocity.

Solution. 361. The approach velocity is defined by

in the laboratory frame.  

On the other hand, the relative velocity can be obtained by using the velocity addition formula and has the components

For orthogonal velocities, transform one particle's velocity into the rest frame of the other using the general velocity-addition relations for components. The displayed component formulae give the transformed x' and y' velocities; their magnitude yields the required relative speed.

Q. 362. An unstable particle moves in the reference frame K' along its y' axis with a velocity v'. In its turn, the frame K' moves relative to the frame K in the positive direction of its x axis with a velocity V. The x' and x axes of the two reference frames coincide, the y' and y axes are parallel. Find the distance which the particle traverses in the frame K, if its proper lifetime is equal to Δt₀. 

Solution. 362. The components of the velocity of the unstable particle in the frame K are 

so the velocity relative to K is

The life time in this frame dilates to

and the distance traversed is

Use v_x and v_y obtained from velocity addition to compute the net speed v in K, then multiply by the dilated lifetime Δt = γΔt₀. The equations displayed show these substitutions and lead to the final distance expression.

Q. 363. A particle moves in the frame K with a velocity v at an angle θ to the x axis. Find the corresponding angle in the frame K' moving with a velocity V relative to the frame K in the positive direction of its x axis, if the x and x' axes of the two frames coincide. 

Solution. 363. In the frame K' the components of the velocity of the particle are

The angle θ' in K' is given by tan θ' = v'_y/v'_x. Substitute the transformed components shown above and simplify to get the explicit formula for θ' in terms of v, V and θ.

Q. 364. The rod AB oriented parallel to the x' axis of the reference frame K' moves in this frame with a velocity v' along its y' axis. In its turn, the frame K' moves with a velocity V relative to the frame K as shown in Fig. 1.94. Find the angle θ between the rod and the x axis in the frame K. 

Solution. 364. In K' the coordinates of A and B are

After performing Lorentz transformation to the frame K we get

By translating   we can write

the coordinates of B as B : t = γ t'

Thus  

Hence  
The angle follows from the transformed spatial coordinates of the rod's ends: θ = arctan(Δy/Δx) with Δx and Δy taken in K after Lorentz transforming the K' coordinates. The displayed algebra outlines this conversion and yields the stated expression for θ.

Q. 365. The frame K' moves with a constant velocity V relative to the frame K. Find the acceleration w' of a particle in the frame K', if in the frame K this particle moves with a velocity v and acceleration w along a straight line
 (a) in the direction of the vector V;
 (b) perpendicular to the vector V. 

Solution. 365.

In K the velocities at time t and t + dt are respectively v and v + wdt along x - axis which is parallel to the vector  In the frame K' moving with velocity  with respect to K, the velocities are respectively,

The latter velocity is written as

Also by Lorentz transformation

Thus the acceleration in the K' frame is

(b) In the K frame the velocities of the particle at the time t and t + di are repectively 

where    is along jt-axis. In the K frame the velocities are

and  

Thus the acceleration

Q. 366. An imaginary space rocket launched from the Earth moves with an acceleration w' = 10g which is the same in every instantaneous co-moving inertial reference frame. The boost stage lasted τ =1.0 year of terrestrial time. Find how much (in per cent) does the rocket velocity differ from the velocity of light at the end of the boost stage. What distance does the rocket cover by that moment?

Solution. 366. In the instantaneous rest frame v = V and

So,  

w' is constant by assumption. Thus integration gives

Integrating once again  
Integrate the displayed relations with w' = 10g and τ = 1 year to find the rapidity and hence β = v/c. The difference (1 - β) expressed as a percentage will be very small. The second integral gives the distance covered; substitute numeric values for g and τ to get the required numbers (the images show the integrals to be evaluated).

Q. 367. From the conditions of the foregoing problem determine the boost time τ₀ in the reference frame fixed to the rocket. Remember that this time is defined by the formula where dt is the time in the geocentric reference frame.

Solution. 367. The boost time τ₀ in the reference frame fixed to the rocket is related to the time τ elapsed on the earth by

The proper time on the rocket is obtained by integrating dτ = dt/γ(t) along the boost stage; the displayed integral gives τ₀ in terms of the Earth time τ and the instantaneous velocity. Evaluate the integral using the previously found velocity function to get the rocket's proper boost time.

Q. 368. How many times does the relativistic mass of a particle whose velocity differs from the velocity of light by 0.010% exceed its rest mass? 

Solution. 368.  

For  
Use γ = 1/√(1 - β²) and β = 1 - 0.0001 to compute γ - 1; the displayed steps show the substitution and simplification leading to the factor by which the relativistic mass (or total energy) exceeds the rest mass. The numerical evaluation follows from expanding to the appropriate order in the small difference from c.

Q. 369. The density of a stationary body is equal to p₀. Find the velocity (relative to the body) of the reference frame in which the density of the body is η = 25% greater than p₀. 

Solution. 369. We define the density p in the frame K in such a way that p dx dy dz is the rest mass dm₀ of the element That is p dx dy dz = p₀ dx₀ dy₀ dz₀, where p₀ is the proper density dx₀, dy₀ , dz₀ are the dimensions of the element in the rest frame K₀. Now

if the frame K is moving with velocity, v relative to the frame K₀. Thus

Defining  

We get  

or  

Density transforms as p = γ p₀ when one dimension (the one along the motion) is contracted while transverse dimensions are unchanged. Setting p = 1.25 p₀ gives γ = 1.25, so solve γ = 1/√(1 - v²/c²) for v to obtain the required speed.

Q. 370. A proton moves with a momentum p = 10.0 GeV/c, where c is the velocity of light. How much (in per cent) does the proton velocity differ from the velocity of light? 

Solution. 370. We have

or  

or  

From p = γm₀v and E = γm₀c², one finds β = pc/E and E = √(p²c² + m₀²c⁴). Substitute p = 10 GeV/c and the proton rest energy ≈ 0.938 GeV to compute β and then (1 - β) as a percentage; the displayed algebra outlines these substitutions.

Q.371. Find the velocity at which the relativistic momentum of a particle exceeds its Newtonian momentum η = 2 times. 

Solution. 371. By definition of η,

or

Q.372. What work has to be performed in order to increase the velocity of a particle of rest mass mo  from 0.60 c to 0.80 c? Compare the result obtained with the value calculated from the classical formula. 

Solution. 372. The work done is equal to change in kinetic energy which is different in the two cases Classically i.e. in nonrelativistic mechanics, the change in kinetic energy is

Q.373. Find the velocity at which the kinetic energy of a particle equals its rest energy.

Solution. 373. 

or    

or    
Set T = (γ - 1)m₀c² = m₀c², so γ = 2, which gives β = √(1 - 1/γ²) = √(1 - 1/4) = √(3/4) ≃ 0.866 c. The steps shown above lead to this value.

Q.374.  At what values of the ratio of the kinetic energy to rest energy can the velocity of a particle be calculated from the classical formula with the relative error less than ε = 0.010? 

Solution. 374. Relativistically

So    

Thus    

But Classically,    

Hence if  

the velocity β is given by the classical formula with an error less than ε. 

Q.375. Find how the momentum of a particle of rest mass m₀ depends on its kinetic energy. Calculate the momentum of a proton whose kinetic energy equals 500 MeV.

Solution. 375. From the formula

we find

or    
Express momentum through total energy E = T + m₀c² by p = √[(E/c)² - m₀²c²]. For a proton with T = 500 MeV and m₀c² ≃ 938 MeV, compute E and then p using the displayed relation to get the numerical momentum in MeV/c (or GeV/c) units.

Q.376. A beam of relativistic particles with kinetic energy T strikes against an absorbing target. The beam current equals I, the charge and rest mass of each particle are equal to e and m₀ respectively. Find the pressure developed by the beam on the target surface, and the power liberated there.

Solution. 376. Let the total force exerted by the beam on the target surface be .F and the power liberated there be P. Then, using the result of the previous problem we see

since I = Ne, N being the number of particles striking the target per second. Also,

These will be, respectively, equal to the pressure and power developed per unit area of the target if I is current density.

Q.377. A sphere moves with a relativistic velocity v through a gas whose unit volume contains n slowly moving particles, each of mass m. Find the pressure p exerted by the gas on a spherical surface element perpendicular to the velocity of the sphere, provided that the particles scatter elastically. Show that the pressure is the same both in the reference frame fixed to the sphere and in the reference frame fixed to the gas.

Solution. 377. In the tome fixed to the sphere The momentum transferred to the eastically scattered particle is

The density of the moving element is, from 1.369,  

and the momentum transferred per unit time per unit area is

In the frame fixed to the gas When the sphere hits a stationary particle, the latter recoils with a velocity

The momentum transferred is  

and the pressure  

Q.378. A particle of rest mass mo  starts moving at a moment t = 0 due to a constant force F. Find the time dependence of the particle's velocity and of the distance covered.

Solution. 378. The equation of motion is   

Integrating  

or using r = 0 at r - 0, we get,  

Q.379. A particle of rest mass m₀ moves along the x axis of the frame K in accordance with the law  where a is a constant, c is the velocity of light, and t is time. Find the force acting on the particle in this reference frame.

Solution. 379.   

Q.380. Proceeding from the fundamental equation of relativistic dynamics, find:
 (a) under what circumstances the acceleration of a particle coincides in direction with the force F acting on it;
 (b) the proportionality factors relating the force F and the acceleration w in the cases when F ⊥ and F II v, where v is the velocity of the particle.

Solution. 380. 

Thus  

Q.381. A relativistic particle with momentum p and total energy E moves along the x axis of the frame K. Demonstrate that in the frame K' moving with a constant velocity V relative to the frame K in the positive direction of its axis x the momentum and the total energy of the given particle are defined by the formulas: 

Solution. 381. By definition,

where   i.s the invariant interval (dy = dz - 0)

Thus,  

Q.382. The photon energy in the frame K is equal to ε. Making use of the transformation formulas cited in the foregoing problem, find the energy ε' of this photon in the frame K' moving with a velocity V relative to the frame K in the photon's motion direction. At what value of V is the energy of the photon equal to ε' = ε/2?

Solution. 382. For a photon moving in the x direction

In the moving frame,  

Note that  

Q.383. Demonstrate that the quantity E² - p²c²  for a particle is an invariant, i.e. it has the same magnitude in all inertial reference frames. What is the magnitude of this invariant? 

Solution. 383. As before

Similarly  

Then  

Q.384. A neutron with kinetic energy T = 2m₀c², where m₀ is its rest mass, strikes another, stationary, neutron. Determine:
 (a) the combined kinetic energy of both neutrons in the frame of their centre of inertia and the momentu each neutron in that frame; (b) the velocity of the centre of inertia of this system of particles. Instruction. Make use of the invariant E²  - p²c²  remaining constant on transition from one inertial reference frame to another (E is the total energy of the system, p is its composite momentum).

Solution. 384. (b) & (a) In the CM frame, the total momentum is zero, Thus 

where wc have used the result of problem (Q.375) Then

 Total energy in the CM frame is

So    

Q.385. A particle of rest mass m₀ with kinetic energy T strikes a stationary particle of the same rest mass. Find the rest mass and the velocity of the compound particle formed as a result of the collision. 

Solution. 385. 

Also  

Q.386. How high must be the kinetic energy of a proton striking another, stationary, proton for their combined kinetic energy in the frame of the centre of inertia to be equal to the total kinetic energy of two protons moving toward each other with individual kinetic energies T = 25.0 GeV? 

Solution. 386. Let T = kinetic energy of a proton striking another stationary particle of the same rest mass. Then, combined kinetic energy in the CM frame

Q.387. A stationary particle of rest mass m₀ disintegrates into three particles with rest masses m₁, m₂, and m₃. Find the maximum total energy that, for example, the particle m₁ may possess. 

Solution. 387. We have

Hence  

The L.H.S.  

The R.H.S. is an invariant We can evaluate it in any frame. Choose the CM frame of the particles 2 and 3.

In this frame  

Q.388. A relativistic rocket emits a gas jet with non-relativistic velocity u constant relative to the rocket. Find how the velocity v of the rocket depends on its rest mass m if the initial rest mass of the rocket equals m₀.

Solution. 388. The velocity of ejected gases is u realtive to the rocket. In an earth centred frame it is

in the direction of the rocket The momentum conservation equation then reads

or  

Here - dm is the mass of the ejected gases, so 

Integrating  

The constant  

Thus