Relativistic dynamics

In the worksheet SpecialRel.mws we introduced the phenomena of time dilation and length contraction, and motivated the transformation of position and time to relate two inertial frames in order to be consistent with the two postulates of relativity. Here we consider the equations of motion of classical particles, in particular, the question of how to make them consistent with the new transformation laws (Lorentz rather than Galilei transformation between reference frames).

We begin by rederiving the Lorentz transformation in a slightly more formal way. (1)

Then we introduce acceptable definitions for relativistic momentum and energy. (2)

Newton's equation with relativistic momentum is illustrated on the example of an oscillator (3)

How are energy and momentum transformed between inertial frames (4)

Relativistic billiard balls: transforming energy, considering momentum and energy conservation in elastic collisions (5,6)

Inelastic collisions: conversion of kinetic energy into mass (7)

The so-called invariant mass of a system of particles (8)

The special case of massless particles (photons) (9)

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1) Lorentz transformation for position, velocity, and acceleration

Let us consider again two reference frames with a boost with speed V between them along the x -axis. This choice can always be made. Let us further assume that a linear transformation mixing x and t with an as yet undetermined dimensionless parameter gamma determines xp = x ', the position in the boosted frame:

> restart; unprotect(gamma); gamma:='gamma': assume(V<c,V>-c); # assumption helps in sqrt's

> L1:=xp=gamma*(x-V*t);

This is a generalization of the Galilei transformation: for V /c small we need gamma=1 to reach this nonrelativistic limit. The inverse transformation has to look practically the same, if we wish the laws of physics to look the same in each reference frame. Of course the sign of the ( V t )-term better be different:

> L1i:=x=gamma*(xp+V*tp);

The two remaining coordinates are unchanged under the transformation ( yp=y , zp=z ) due to the choice of boost along the x -axis. According to Galilei we had tp=t . That this has changed with the generalization, and is only recovered for gamma=1 can be seen from the combination of the two transformations. We can solve L1i for xp and eliminate the variable from L1 :

> L1ip:=xp=solve(L1i,xp);

> L1p:=subs(xp=rhs(L1ip),L1);

> L1p:=tp=expand(solve(L1p,tp));

We see clearly how the Galilei concept of time is generalized for non-zero -values.

Now we determine gamma from the relativity postulates that the speed of light be the same in each reference frame, and that the laws of physics be the same in each inertial reference frame. Consider at t = tp = 0 a light flash originating at x = xp = 0. The wavefront of a spherical wave for observers in the unprimed and primed reference frames is specified by the light front conditions:

> L2:=x^2+y^2+z^2=c^2*t^2;

> L2p:=xp^2+yp^2+zp^2=c^2*tp^2;

The two statements have to transform into each other. We have the freedom of L1 and L1p to accomplish this. First we replace the unchanged coordinates in the primed coordinate system.

> L2p:=subs(yp=y,zp=z,L2p);

> L2p1:=subs(tp=rhs(L1p),L2p);

> L2p2:=subs(xp=rhs(L1),L2p1);

Now we compare coefficients between similar terms in L2p2 and L2 .

> L3:=L2-L2p2;

> L3:=expand(lhs(L3)-rhs(L3))=0;

> L3:=collect(L3,[x,y,z,t]);

Obviously, there are no y - or z -dependent terms left after the subtraction. Our task looks difficult: can we satisfy three conditions at the same time with a single choice of gamma?

> L3p1:=coeff(L3,x^2);

Error, unable to compute coeff

> L3p1:=op(1,op(1,L3));

> L3p1:=coeff(L3p1,x^2)=0;

We can't use the symbol gamma to solve L3p1 for it, as Maple still thinks that gamma is the Euler-Mascheroni constant. We need a change of name:

> L3p1:=subs(gamma=eta,L3p1);

> solve(L3p1,eta);

> simplify(1/sqrt(1-V^2/c^2));

The latter is identical to the fourth solution found above. Note that gamma=1 is also a solution. Now try the other conditions:

> L3p2:=op(2,op(1,L3));

> L3p2:=L3p2/(t*x)=0;

> L3p2:=subs(gamma=eta,L3p2);

> solve(L3p2,eta);

We are getting some good news: gamma=1 is excluded now, and we still have the same candidate(s) for gamma in the running to satisfy all terms.

> L3p3:=op(3,op(1,L3));

> L3p3:=coeff(L3p3,t^2)=0;

> L3p3:=subs(gamma=eta,L3p3);

> solve(L3p3,eta);

The magic has happened: all three conditions are satisfied by one constraint on gamma. We know how to pick out the choice for gamma that makes more sense: the sign of c^2-V^2 should be positive for a real-valued answer. The Lorentz transformation becomes:

> L1;

> L1p;

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> simplify(subs(gamma=1/sqrt(1-V^2/c^2),L1p));

One can convince oneself that this is equivalent to the compact form:

> L1p:=tp=gamma*(t-V*x/c^2);

The inverse transformations are given as:

> L1i;

and an equation for t that has to be the symmetric equivalent of L1p:

> L1pi:=t=gamma*(tp+V*xp/c^2);;

These are supplemented by the statements ( y = yp , and z = zp , and vice versa).

We can use the transformations to re-iterate previous findings.

Transformation of time intervals: take two events (x_a, t_0) and (x_b, t_1), and translate them into the frame of a moving observer. In particular look at the time interval in the primed frame:

> TI:=subs(x=x_b,t=t_1,rhs(L1p))-subs(x=x_a,t=t_0,rhs(L1p));

> expand(TI);

Not only the temporal separation is important, but also the spatial separation of the clocks that measured t in the rest frame.

Velocity transformation

The velocity transformation was also derived in the previous worksheet by forming differentials of L1 and L1p and the corresponding derivatives.

> Lv1:=up_x=(u_x-V)/(1-V*u_x/c^2);

> Lv2:=up_y=u_y/(gamma*(1-V*u_x/c^2));

> Lv3:=up_z=u_z/(gamma*(1-V*u_x/c^2));

The inverse transformation is given by symmetry (keep in mind the sign reversal on V ) by:

> Lvi1:=u_x=(up_x+V)/(1+V*up_x/c^2);

> Lvi2:=u_y=up_y/(gamma*(1+V*up_x/c^2));

> Lvi3:=u_z=up_z/(gamma*(1+V*up_x/c^2));

To understand relativistic dynamics we have to consider now the acceleration. In Galilean reference frames

We have to differentiate up_x with respect to tp .

> Lv1;

> subs(tp=dtp,t=dt,x=dx,L1p);

First we need to form the differentials in Lv1 . The appearance of u_x in the denominator complicates matters a little.

What happens when we form the derivative? V is a constant, i.e., in the numerator we obtain the diff(u_x, t) derivative; there is no x -dependence, and so apart from the gamma factor we expect no surprises from taking the derivative. However, the denominator also gets differentiated with respect to time, as u_x appears there as well. First we form the differential by taking a time derivative. For this purpose we have to introduce an explicit time dependence, and then remove it to recover the variables used normally.

> subs(u_x=u_x(t),rhs(Lv1));

> expand(diff(%,t));

> simplify(%);

> subs((c^2-V^2)=c^2/gamma^2,%);

> dup_x:=subs(diff(u_x(t),t)=du_x,u_x(t)=u_x,%);

Now we form the difference quotient:

> dup_x/expand(rhs(subs(tp=dtp,t=dt,x=dx,L1p)));

and remove the dx in the denominator by the trick used in the derivation of the velocity transformation:

> subs(dx=u_x*dt,%);

Now we can introduce the acceleration in the rest frame:

> simplify(subs(du_x=a_x*dt,%));

Apart from the c still present in the numerator, we generated the textbook result for the acceleration in the moving reference frame, ap_x . According to Galilei the acceleration is the same in two inertial reference frames. According to Lorentz it is not, unless if we take the limit ( gamma=1 and V /c approximately 0). We cannot expect Newton's law to survive in the simple form F = m a in the relativistic case.

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2) Relativistic momentum and kinetic energy

Relativistic momentum

We should first look at relativistic momentum to understand the situation when no forces act on a particle. Under these circumstances momentum is conserved, and we can ask how the nonrelativistic definition of momentum should be extended to maintain the conservation law.

Consider two observers with identical balls of mass m in the respective rest frame. The observers move with a boost velocity along the x -axis. They throw their balls in their respective y -directions (one up and one down), such that the balls collide and bounce elastically, i.e., reverse their velocities in the y -direction. From the velocity transformation discussed before we know that even the velocity components in the y -direction will transform between the two frames as a result of the relative boost in the x -direction. What will either observer find when comparing the magnitudes of the v_y velocity components for the two balls?

Suppose observer A is at rest. Initially, A's ball has no x -component and moves with speed u0 along the y -axis in the positive direction.

> vA_y:=u0;

> vA_x:=0;

In B's rest frame (denoted by primed coordinates) ball B runs downward along the y -axis:

> vB_yp:=-u0;

> vB_xp:=0;

In A's reference frame ball B has an x - and a y -component:

> vB_x:=V;

> gamma:=1/sqrt(1-beta^2);

> vB_y:=vB_yp/gamma/(1+V*vB_xp);

Here we used the velocity transformation law from the primed (moving) to the unprimed (rest) frame, which simplifies for the case of an object moving with the boosted frame in the direction of the boost ( vB_xp=0 ).

The result is that ball B from A's perspective moves with reduced speed, which is a consequence of the time dilation factor associated with the motion of B's reference frame along the x -axis. It takes more time for B to travel a certain distance along the y -direction in the unprimed frame, than the time it takes to travel the same distance in B's rest frame (primed). In A's rest frame (unprimed) the y -component of total momentum as defined according to Newton is not zero before the collision! Thus, the sum of mass times velocity for all particles is not conserved in A's rest frame. The same holds in B's frame. The classical momentum is only conserved in the non-relativistic limit.

How can we save the conservation law? The velocities were transformed according to the Lorentz transformation, so they should be correct. The only other freedom in converting the momentum to a relativistically acceptable quantity is to tamper with the mass. Let us call the speed of the ball from the opposite reference frame u to distinguish it from the own ball's speed u 0, and to account for the motion along the boost direction:

> u:=sqrt(V^2+vB_y^2);

We formulate the momentum conservation using a mass that depends on the speed, while using velocity reversal upon impact (the signs are equal as vB_y carries a minus sign relative to u 0):

> MC:=m(u0)*u0+m(u)*vB_y=-m(u0)*u0-m(u)*vB_y;

> MC:=simplify(lhs(MC)-rhs(MC))=0;

> solve(MC,m(u));

> abs(u0/vB_y);

> limit(u,u0=0);

For small speeds of the balls in their respective rest frames u is given by the boost speed and the mass ratio m ( u )/ m ( u 0) just describes the increase in mass from the mass at rest (speed u 0) to the mass m ( V ). This ratio is just given by the gamma factor. A stationary observer measures the mass of a moving object to be heavier than the rest mass. This causes the relativistic momentum of a material object to increase faster than just proportional to the particle speed.

This mass increase over the rest mass causes a problem when we attempt to accelerate particles: once particles reach a significant fraction of the speed of light c most of the additional energy deposited goes into increasing the mass rather than speeding up the particle.

Newton's law

We can salvage Newton's law by simply demanding that relativistic momentum be used.

The rate of change of relativistic momentum of a particle is proportional to the applied net force.

This will have to be verified by introducing energy: kinetic energy is defined by the work done by a net force while accelerating a particle from rest to some velocity.

In one dimension ( m denotes the rest mass):

> u:='u':

> Tkin:=Int(F,x=x0..x1)=Int(Diff(gamma*m*u,t),x=x0..x1);

The gamma (beta) factor refers to the actual speed u as attained at time t , or position x . The integration parameter x should be converted to speed u using u(t)=diff(x(t),t) . This results in the total differential D[m*gamma*u] to be integrated. The differential needs to be converted to d u which can be calculated by differentiating (chain rule):

> gamma:='gamma':

> subs(diff(u(t),t)=du,u(t)=u,simplify(diff(m*u(t)/sqrt(1-(u(t)/c)^2),t)));

To carry out the integration we need to remove the du, and multiply with the u -factor which arose during the shift of the integration variable from d x to d u .

> int(%*u/du,u=0..U);

> Tkin:=expand(%);

> subs((c^2-U^2)=c^2/gamma^2,Tkin);

> simplify(%);

> Tkin:=sort(%,[gamma,m,c]);

The relativistic kinetic energy is made up of a term that contains the energy of the rest mass, and then the rest mass energy is subtracted. It is of interest to Taylor expand this expression in order to see how the nonrelativistic kinetic energy is recovered. We have to substitute the gamma factor again, and note that we used the final speed U as an integration parameter. We return to use V in the form beta= V /c.

> subs(gamma=1/sqrt(1-beta^2),Tkin);

> taylor(%,beta=0,5);

> subs(beta=V/c,%);

We have arrived at the nonrelativistic result, and determined the lowest-order relativistic correction to the kinetic energy of a massive particle.

> TkNRC:=convert(%,polynom);

> TkNR:=1/2*m*V^2;

It is worthwhile to look at a graphical presentation of these results. We pick units where c is one, and the electron mass m =1, and graph the kinetic energy of an electron as a function of speed. To convert to SI, or particle physics units, we note that the mass of the electron is 0.511 MeV; this corresponds to one energy unit on our vertical scale.

> P1:=plot(subs(m=1,c=1,TkNR),V=0..1,color=blue):

> P2:=plot(subs(m=1,c=1,TkNRC),V=0..1,color=maroon):

> P3:=plot(subs(gamma=1/sqrt(1-V^2/c^2),m=1,c=1,Tkin),V=0..1,color=red):

> plots[display](P1,P2,P3,title="Kinetic energy as a function of speed: relativstic and nonrelativistic",view=[0..1,0..3]);

We notice from the graph that an electron moving at 90% of the speed of light has a kinetic energy that exceeds the rest mass energy. The general conclusion here is that the first relativistic correction may be adequate for particles moving at up to beta=0.6, but that for faster massive particles the full relativistic kinetic energy expression is required. The kinetic energy of a slow particle is negligible compared to its rest mass energy: an electron in the ground state of a hydrogen atom has a typical energy of 13.6 eV. This is small compared to the 511 keV rest energy, despite of the fact that the electron moves fast (on our life experience scale): it moves at about 1/137 times c (1/137 is the fine structure constant of electromagnetism).

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3) Relativistic trajectories

Relativistic trajectories: the relativistic harmonic oscillator (RHO)

Let us solve Newton's equation in the relativistic form to get some feeling for the effect of the increased mass. We consider a one-dimensional harmonic oscillator, e.g., for an electron attached to a spring (for quite some time approximations were made in atomic physics based on oscillator strengths). We will compare the analytically known solutions for the nonrelativistic case with numerical solutions based on the relativistic kinetic energy.

The units are chosen as before ( m =c=1) and the spring constant, and initial displacement is chosen to be large to make the particle acquire a substantial amount of kinetic energy. We need to understand the length scale in order to decide what we mean by this choice of parameters. Alternatively, we can just experiment around and observe the speed reached by the particle at x =0.

> m:=1; c:=1; k:=1; x0:=2;

For a nonrelativistic harmonic oscillator we have the solution:

> xoft:=x0*cos(sqrt(k/m)*t);

> voft:=diff(xoft,t);

The origin is reached at t = Pi/2, and the velocity exceeds the speed of light by a factor of 2. What can we find out about the relativistic motion in the same force law?

We solve the Newton equation in a way that uses both Maple's dsolve integrator. For the determination of the velocity from the momentum we have the nonlinear relationship:

> eqU:=U=P/m*sqrt(1-U^2);

> solve(eqU,U);

We use this expression for the velocity in terms of the momentum for the integration, and also for display purposes.

> NE1:=diff(p(t),t)=-k*x(t);

> NE2:=diff(x(t),t)=p(t)/sqrt(p(t)^2+1);

> IC:=p(0)=0,x(0)=x0;

> dt:=0.05; Nt:=trunc(4*Pi/dt);

> sol:=dsolve({NE1,NE2,IC},{x(t),p(t)},numeric,output=listprocedure);

> xt:=subs(sol,x(t)): pt:=subs(sol,p(t)):

> xt(0.05),pt(0.05);

> xL:=[[0,x0]]: vL:=[[0,0]]: for it from 1 to Nt do: ti:=it*dt; pti:=pt(ti); xti:=xt(ti); vti:=pti/sqrt(1+pti^2); vL:=[op(vL),[ti,vti]]: xL:=[op(xL),[ti,xti]]: od:

> plot([xL,vL],color=[red,blue],title=cat("RHO for x0= ",convert(x0,string)));

We observe that the motion is no longer linear. The particle reaches approximately beta=0.9 , and does not exceed the speed of light. The shape of the velocity as a function of time reminds one of the time dependence of the angle in the mathematical pendulum (not linearized!) for large initial angles, i.e., for cases when the pendulum starts near the top. Comparison with the nonrelativistic solution shows that the relativistic harmonic oscillator is no longer isochronous (as the pendulum): the period depends on the amount of energy. Let us veify the change in the period by a solution for a smaller initial stretch of the spring:

> x0:=0.2; IC:=p(0)=0,x(0)=x0;

> dt:=0.05; Nt:=trunc(4*Pi/dt);

> sol:=dsolve({NE1,NE2,IC},{x(t),p(t)},numeric,output=listprocedure);

> xt:=subs(sol,x(t)): pt:=subs(sol,p(t)):

> xt(0.05),pt(0.05);

> xL:=[[0,x0]]: vL:=[[0,0]]: for it from 1 to Nt do: ti:=it*dt; pti:=pt(ti); xti:=xt(ti); vti:=pti/sqrt(1+pti^2); vL:=[op(vL),[ti,vti]]: xL:=[op(xL),[ti,xti]]: od:

> plot([xL,vL],color=[red,blue],title=cat("RHO for x0= ",convert(x0,string)));

This solution is qualitatively comparable to the nonrelativistic case (harmonic oscillation with period depending only on k / m , no dependence on the intial stretch), even though a value of beta=0.2 is reached at the passage of the force-free point x = 0. The comparison of the qualitatively different plots helps to understand why the highly relativistic harmonic oscillator can no longer be isochronous: the slowdown of the mass due to its increase with velocity (energy is being dumped into increasing the relativistic mass), and the avoidance of the v =1 limit the mass is delayed and cannot execute oscillations on the same timescale as before, i.e., on a timescale just set by sqrt(k/m) . At the equilibrium point x =0 all energy resides in the form of kinetic energy. By increasing the initial stretch of the spring (i.e., x0) we can control the amount of kinetic energy of the mass while the spring is neither stretched nor compressed. It is important to realize that we can make the particle hardly go faster when doing so. The energy is dumped into the increase of the relativistic mass. Now that we appreciate this phenomenon, it is clear that the period of the RHO has to go down, as the mass increases substantially during the passages of x =0. Note that the term RHO is a misnomer: the generalization of the HO to the relativistic case leads to anharmonic motion when the speed of the mass becomes a significant fraction of c in the vicinity of the force-free point.

Exercise:

Find a start value x0 for which the period begins to change appreciably from the nonrelativistic harmonic oscillator case. Which maximum speed does the particle reach in this case?

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4) Transforming energy and momentum between reference frames

Conservation laws

An important aspect when studying dynamics is the investigation of conservation laws, which are obeyed by the solutions to the equations of motion. Conservation laws allow one to solve problems in a low number of dimensions, and to impose constraints on solutions in three dimensions without looking at the details of the interactions. Examples are provided by 'billiard-ball-type' scattering. In special relativity the question of which are the conserved quantities is tied closely to the properties of the Lorentz transformation.

Lorentz transformation of energy and momentum

Suppose we consider a particle moving with velocity vector u in a reference frame S, and observe it also in S' which is boosted compared to S along the x -axis with boost velocity V . We can write down the unprimed and primed energies and momentum components. We use g to denote gamma , and we do not assign it by the next statement, we write a (useless) equation to indicate its value:

> restart; assume(V<c,V>-c);

> g=1/sqrt(1-u^2/c^2);

> E:=g*m*c^2;

> p[x]:=g*m*u[x];

> p[y]:=g*m*u[y];

> p[z]:=g*m*u[z];

In S' gamma has a different value:

> gp=1/sqrt(1-up^2/c^2);

> Ep:=gp*m*c^2;

> pp[x] := gp*m*up[x]; pp[y] := gp*m*up[y]; pp[z] := gp*m*up[z];

To transform these four quantities we first express the primed gamma in terms of quantities measured in S. Of course, one could also proceed the other way around.

We need the components of u ' in terms of the components of u , as given by the LT. We use the previous notation for subscripts rather than making use of Maple's notation for table entries:

> Lv1:=up_x=(u_x-V)/(1-V*u_x/c^2);

> Lv2:=up_y=u_y/(1/sqrt(1-V^2/c^2)*(1-V*u_x/c^2));

> Lv3:=up_z=u_z/(1/sqrt(1-V^2/c^2)*(1-V*u_x/c^2));

> up_sq:=simplify(rhs(Lv1)^2+rhs(Lv2)^2+rhs(Lv3)^2);

> algsubs(u_x^2+u_y^2+u_z^2=u_sq,up_sq);

The expression for gp simplifies considerably:

> simplify(1/sqrt(1-up_sq/c^2),symbolic);

We recognize in the square root in the denominator a product: one part belongs to 1/gamma and the other to (1-u_sq/c^2) , i.e., to 1/g

> gp_S:=algsubs((u_x^2+u_y^2+u_z^2)=u_sq,%);

Therefore the kinetic energy of the particle in S' transforms into two contributions in S:

> Ep=gamma*(E-V*p_x);

To verify by comparison with our result which up to m*c^2 was the energy:

> subs(p_x=p[x],%);

Analogous derivations follow for the components of the momentum vector.

> subs(gp=gp_S,up[x]=rhs(Lv1),pp[x]);

> simplify(%);

> pp[x]=gamma*(p[x]-g*m*V);

The y -component:

> subs(gp=gp_S,up[y]=rhs(Lv2),pp[y]);

> simplify(%);

> pp[y]=p[y];

The same result follows obviously for the z -component. It is important to compare the transformation of the momentum vector to the transformation of the position vector. For the components perpendicular to the direction of the boost we find that they are unchanged in both cases. For the component along the boost direction a small manipulation reveals an interesting fact. One can introduce the energy into this expression:

> -m*gamma*V*g,E;

> E:='E':

> algsubs(g*m=E/c^2,-m*gamma*V*g);

> -gamma*V*E/(c^2);

Therefore we can write the momentum transformation as

> pp[x]=gamma*(p[x]-V*E/c^2);

The Lorentz transformation for the momentum vector components is now strikingly similar to the transformation for the position vector components, except that the role of time in the transformation has been taken by E/c^2 ! Now we should be curious how the energy transformation compares to the time transformation.

> Ep:='Ep':

> Ep=gamma*(E-V*p_x);

> tp=gamma*(t-V*x/c^2);

We observe that the role of x in the time transformation is taken uo by p_x*c^2 . Thus, it is fair to say that in the same way that position and time transform as one entity under Lorentz transformations, momentum and energy transform in an analogous way. The factors of c^2 are needed to make matters correct from the point of view of physical dimensions.

It is important to illustrate the derived results with some examples. (we pick example 2-3 from the book of Tipler and Llewellyn)

Example 1:

A micrometeroite of mass 10^(-9) kg moves past Earth at speed 0.01 c. An observer in a spaceship moving at V = 0.5 c in the same direction as the meteorite passes by the Earth. What values of energy and momentum will the moving observer detect for the meteorite?

Answer:

The total energy of the meteorite can be calculated in the Earth's frame by the non-relativistic approximation, or by the full relativisitic expression. We ignore the potential energy in the gravitational field, and use SI (MKSA) units throughout.

> m:=10^(-9):

> c:=3*10^8:

> u:=0.01*c:

> E_rel:=m*1/sqrt(1-u^2/c^2)*c^2;

> E_nr:=m*c^2+m/2*u^2;

> p_rel:=m*1/sqrt(1-u^2/c^2)*u;

> p_nr:=m*u;

Note that despite of the fact that the nonrelativistic approximation is highly accurate to describe the meteorite in the Earth's reference frame, the transformation to the spaceship frame should be carried out by a Lorentz transformation, since the spaceship moves at 0.5 c.

> V:=0.5*c;

> unprotect(gamma);

> gamma:=1/sqrt(1-V^2/c^2);

The kinetic energy of the meteorite in the spaceship frame:

> Ep:=gamma*(E_rel-V*p_rel);

The energy is 15 % larger in the spaceship frame. The effect is mostly due to the gamma-factor in this case.

> Ep/E_rel;

> pp:=gamma*(p_rel-V*E_rel/c^2);

> pp/p_rel;

Obviously the second term made the big contribution here. The momentum is in the opposite direction, and it is huge (Tipler and Llewellyn have a buboe in their solution: wrong numbers and a wrong conclusion by two orders of magnitude!).

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5) Relativistic billiard balls: calculating the energy in different frames

Energy conservation: relativistic billiard balls

Now that we have some familiarity as to how to transform the kinetic energy between inertial frames, as well as with the momentum conservation law (which we used to find an appropriate expression for relativistic momentum, we carry out an independent check: we consider the total energy of a system of two colliding classical particles, and investigate whether it is conserved.

First we return to the problem of elastic scattering of two objects of equal mass, which we considered to define momentum in special relativity.

We simply repeat the steps that led to the dependence of the mass on velocity.

We had formulated the velocities in the frame A where particle A has only a velocity y -component, and frame B in which particle B has only a velocity y -component, which opposes A's motion. The catch is that frame B moves along the x -axis with respect to A with a relativistic speed. This requires the use of the Lorentz transformation when calculating the momentum balance or the total energy. The balls are thrown such that they collide elastically. Due to the transformation of the velocity components nonrelativistic linear momentum is not conserved (when one transforms them according to Lorentz).

We should calculate the total energy before and after the collision.

> restart; unprotect(gamma); assume(V<c,V>-c);

> vA_y:=u0;

> vA_x:=0;

> vB_yp:=-u0;

> vB_xp:=0;

> vB_x:=V;

> gamma:=1/sqrt(1-beta^2);

> vB_y:=vB_yp/gamma/(1+V*vB_xp);

> u:=sqrt(V^2+vB_y^2);

> MC:=m(u0)*u0+m(u)*vB_y=-m(u0)*u0-m(u)*vB_y:

> MC:=simplify(lhs(MC)-rhs(MC))=0;

We have the velocity components of particle B in A's frame. How do we obtain the energy? We derived an expression for the relativistic energy using the definition of relativistic momentum, and by postulating Newton's law, and integrating free-particle motion to arrive at the kinetic energy as (gamma-1)*m*c^2 , where m is the rest mass, and the gamma-factor corrects this mass as a function of the particle speed.

We should obtain then for the total energy (simply ommit the subtraction of 1 in the round bracket:

> E_A:=m*c^2/sqrt(1-u0^2/c^2);

> E_B:=simplify(m*c^2/sqrt(1-subs(beta=V/c,u)^2/c^2));

> E_totA:=E_A+E_B;

To calculate matters from B's perspective one obtains the first term for B, and then hopefully for the second term the calculation gives the same answer after transforming A's velocity.

> vA_xp:=-V;

> vA_yp:=vA_y/gamma/(1-V*vA_x);

> w:=sqrt((-V)^2+vA_yp^2);

> E_Ap:=simplify(m*c^2/sqrt(1-subs(beta=-V/c,w)^2/c^2));

> E_Bp:=m*c^2/sqrt(1-u0^2/c^2);

The two particles have swapped their roles: this is consistent with the principle of relativity that the physics is the same in all intertial frames.

> E_totA-E_Ap-E_Bp;

The dependence on u 0 is quadratic, and therefore independent of the sign. The answer has to be the same after the collision.

Suppose we wish to calculate the energy in a frame that moves with respect to A and B along the x -axis. Given that B moves with respect to A with speed V , we can specify the frame C by stating that it moves with speed W with respect to A. To transform the energy of particle A is analogous to the the calculation in frame B, one just exchanges V for W in the gamma factor, etc.. The calculation of B's energy in frame C is trickier, as we do not know the relative speed between B and C! We should not take the difference between V and W , as we have rejected the additivity of velocities in special relativity. According to the velocity transformation we can calculate the energy of B in frame C via the rest frame of A. We use the gamma factor for the rest frame of A, and use the known velocity of particle B (namely V ) in the transformation law.

Let us write a procedure that calculates the required steps. We assume that A and B are not moving in the x -direction in their respective frame. The energy is calculated in frame C for which we have a gamma factor that relates it to the rest frame A.

> Eboost:=proc(W,V,vA_y,vB_yp) local gammaA,vA_xpp,vA_ypp,vB_xpp,vB_ypp,w_A,w_B,E_App,E_Bpp,vA_x,vB_xp,vB_x,vB_y; vA_x:=0; vB_xp:=0; vB_x:=V; vB_y:=vB_yp*sqrt(1-V^2/c^2)/(1+V*vB_xp);

> gammaA:=1/sqrt(1-W^2/c^2); vA_ypp:=vA_y/gammaA/(1-W*vA_x/c^2); vB_ypp:=vB_y/gammaA/(1-W*vB_x/c^2);

> vA_xpp:=-W; vB_xpp:=(vB_x-W)/(1-W*vB_x/c^2);

> w_A:=sqrt(vA_xpp^2+vA_ypp^2);

> w_B:=sqrt(vB_xpp^2+vB_ypp^2);

> E_App:=simplify(m*c^2/sqrt(1-w_A^2/c^2));

> E_Bpp:=simplify(m*c^2/sqrt(1-w_B^2/c^2));

> E_App+E_Bpp; end:

>

> Eboost(0,c/2,c/10,-c/10);

> evalf(subs(m=1,c=1,%));

> subs(u0=0.1,c=1,V=0.5,m=1,E_totA);

> Eboost(c/2,c/2,c/10,-c/10);

> evalf(subs(m=1,c=1,%));

> subs(u0=0.1,c=1,V=0.5,m=1,E_Ap+E_Bp);

> Eboost(c/4,c/2,c/10,-c/10);

> evalf(subs(m=1,c=1,%));

> plot(evalf(subs(m=1,c=1,Eboost(W,c/2,c/10,-c/10))),W=-0.9..0.9);

It appears as if the total energy is minimized in a reference frame that moves with half the speed ( W = V /2). This would be the result from the nonrelativistic kinetic energy expression (+ rest energy for two massive particles), where the sum of the squared velocities is minimized when the difference is split in half. A closer inspection performed with calculus (performed below) shows that the minimum is not exactly at the half-point. For larger boosts V the curve deviates more dramatically from the nonrelativistic expectation.

> solve(diff(Eboost(W,c/2,c/10,-c/10),W)=0,W);

> evalf(%);

In the nonrelativitic approximation we would calculate: rest energy + nonrelativistic kinetic energy. We add the statements to verify that the total energy is a frame-dependent quantity in the nonrelativistic case.

> E_nrA:=evalf(subs(u0=0.1,m=1,c=1,V=0.5,m*c^2+m/2*u0^2+m*c^2+m/2*(V^2+u0^2)));

> E_nrpp:=evalf(subs(u0=0.1,m=1,c=1,V=0.5,m*c^2+m/2*(u0^2+(V/2)^2)+m*c^2+m/2*((V/2)^2+u0^2)));

The increase in the total energy as the boost velocity approaches c or -c is obviously associated with the fact that the two particles in A and B acquire mass. This also explains the asymmetry in the curve shown above. For negative W the increase is larger than for positive W of equal magnitude due to the set-up of mass A being stationary, and mass B moving with c/2. For W >c/2 a substantial mass increase for particle B occurs only for large boosts. In the opposite direction the converse is true, but note that nothing too dramatic happens at W =-c/2, where one might argue naively that the relativistic mass for particle B should hit the roof. The relative speed between C and B reaches the speed of light in this regime only for W =-c.

We should be able to verify the result calculated in the double-primed frame (procedure Eboost ) by carrying out a Lorentz transformation of the total energy (e.g., from A's rest frame, i.e., from E_totA ). We boost from A with W , and realize that we need to know the momentum along the x -axis in frame A to carry out the boost. This momentum was calculated before when the momentum balance was formulated, but we need the explicit expression for the mass as a function of speed for the particles.

> u;

> px_totA:=m*0/sqrt(1-u0^2/c^2)+subs(beta=V/c,m*V/sqrt(1-u^2/c^2));

> gammapp:=1/sqrt(1-W^2/c^2);

> EboostLT:=unapply(simplify(gammapp*(E_totA-W*(px_totA))),W,V,u0);

> evalf(EboostLT(c/3,c/2,c/10));

> evalf(Eboost(c/3,c/2,c/10,-c/10));

> expand(simplify(EboostLT(c/3,c/2,c/10)));

> simplify(Eboost(c/3,c/2,c/10,-c/10));

The comparison verifies that the two calculations are consistent: in EboostLT we have Lorentz transformed the total energy for particles A and B as calculated in frame A to the reference frame C. In Eboost we have used Lorentz-transformed velocity components for particle B in A's restframe, and transformed the velocities to frame C to directly evaluate the relativistic mass increase as a function of the speed of particles A and B as seen in frame C. Note how we avoided the issue of the unknown relative velocity between frames B and C. We calculated the velocity of particle B as viewed by an observer in C by first transforming it into A (known relative velocity V), and then from A to C (known relative velocity W). The simple elastic billiard ball scattering example was not interesting from the energy conservation point of view, apart from the fact that energy conservation was ensured by the quadratic dependence on u 0.

More interesting cases involve unequal masses, and also inelastic scattering.

>

6) Billiard balls with unequal masses - elastic collisions

A non-trivial generalization would be to consider the case of different masses. Suppose particle B has twice the rest mass of A. Then we expect A to bounce off with more than - u 0, and B with less than u 0. Based on the equal-mass elastic scattering we figured out the dependence of relativistic energy on speed and rest mass of a particle. We should be able to formulate what happens. Let us simplify matters by going to one dimension only, and formulating momentum and energy conservation. Let us work in the rest frame of particle A.

> restart; assume(V<c,V>-c);

> v_A:=0;

> E_A:=m_A*c^2;

> v_B:=V;

> unprotect(gamma); gamma:=1/sqrt(1-V^2/c^2);

> E_B:=gamma*m_B*c^2;

> p_A:=m_A*v_A;

> p_B:=gamma*m_B*v_B;

Let us denote the velocities after the collision as primed. We can write for momentum and energy conservation:

> p_Ap:=m_A*v_Ap/sqrt(1-v_Ap^2/c^2); p_Bp:=m_B*v_Bp/sqrt(1-v_Bp^2/c^2);

> E_Ap:=1/sqrt(1-v_Ap^2/c^2)*m_A*c^2; E_Bp:=1/sqrt(1-v_Bp^2/c^2)*m_B*c^2;

> MC:=p_A+p_B=p_Ap+p_Bp;

> EC:=E_A+E_B=E_Ap+E_Bp;

> sol:=solve({MC,EC},{v_Ap,v_Bp});

So much for Maple magic...

Nevertheless, we should not despair. When we substitute numbers for the masses Maple does the work for us.

> sol:=solve({subs(m_A=1,m_B=1,V=c/2,MC),subs(m_A=1,m_B=1,V=c/2,EC)},{v_Ap,v_Bp});

> sol:=solve({subs(m_A=1,m_B=2,V=c/2,MC),subs(m_A=1,m_B=2,V=c/2,EC)},{v_Ap,v_Bp});

Note that there are always two solutions: one represents the initial condition, the other the final outcome.

We investigate various situations: a light particle with speed V hits a heavier one:

> sol:=solve({subs(m_A=10,m_B=1,MC),subs(m_A=10,m_B=1,EC)},{v_Ap,v_Bp});

> vA:=subs(sol[1],v_Ap): # we may need to pick by hand which one is the desired solution

> vB:=subs(sol[1],v_Bp): # and use sol[1] or sol[2] correspondingly.

> plot([subs(c=1,vA),subs(c=1,vB),-V],V=0..1,color=[red,blue,green],view=[0..1,-1..1]);

> limit(vA,V=c),limit(vB,V=c);

For small speeds we observe the known result from nonrelativistic energy and momentum conservation: the light impinging particle bounces back with appreciable speed and the hit particle runs off with a smaller speed. Note, however, that in the highly relativistic limit a crossover occurs!

The green line shows what would happen if one assumed naively that in the limit of the hit particle being very massive no energy transfer were to occur, and the impinging particle would rebound with its original speed reversed.

> plot([subs(c=1,vA),subs(c=1,-vB)],V=0..1,color=[red,blue],view=[0..1,0..1],numpoints=1500);

> fsolve(subs(c=1,vA+vB),V=0.9..1);

What can we say about the energies (in the rest frame of A)? Let us graph the kinetic energy of A and B after the collision:

> plot([subs(v_Ap=vA,m_A=10,c=1,E_Ap-m_A*c^2),subs(v_Bp=vB,m_B=1,c=1,E_Bp-m_B*c^2)],V=0..1,color=[red,blue],numpoints=1500,view=[0..1,0..5]);

As far as kinetic energy is concerned the crossover point is earlier: the massive hit particle A picks up more energy than the impinging light particle B retains in its rebound at an impact velocity of

> solve(subs(v_Ap=vA,m_A=10,c=1,E_Ap-m_A*c^2)=subs(v_Bp=vB,m_B=1,c=1,E_Bp-m_B*c^2),V);

> evalf(%);

These results are clearly incompatible with our intuition based on the nonrelativistic limit.

Exercise:

Explore how these crossover points move as the mass ratio is increased, i.e., as the hit particle is chosen to be more massive.

The other limit of a massive particle impinging on a light particle is solved less satisfactorily by Maple.

> sol:=solve({subs(m_A=1,m_B=10,c=1,MC),subs(m_A=1,m_B=10,c=1,EC)},{v_Ap,v_Bp});

> sol1:=allvalues(sol)[1];

The solution no longer recognizes the initial condition as a solution to the set of equations, even though they do satisfy the equations. A call to allvalues(sol) reveals a set of solutions, but no simplification is obvious.

> vA:=subs(sol1,v_Ap):

> vB:=subs(sol1,v_Bp):

> plot([vA,vB,V],V=0..1,color=[red,blue,green],view=[0..1,0..1],numpoints=1500);

> simplify(limit(vA,V=1)),simplify(limit(vB,V=1));

It appears as if there were fewer surprises in this case...?

The hit particle is 10 times lighter in this example, and one can see how its speed approaches c when the incident massive particle speed exceeds 0.8 c. For the kinetic energies this can imply a crossover where the hit particle acquires a big mass due to its speed.

> plot([subs(v_Ap=vA,m_A=1,c=1,E_Ap-m_A*c^2),subs(v_Bp=vB,m_B=10,c=1,E_Bp-m_B*c^2)],V=0..1,color=[red,blue],numpoints=1500,view=[0..1,0..15]);

> fsolve(subs(v_Ap=vA,m_A=1,c=1,E_Ap-m_A*c^2)=subs(v_Bp=vB,m_B=10,c=1,E_Bp-m_B*c^2),V=0.5..1);

We recognize from the comparison of the numerical results that the crossover in the energies occurs for the same incident speed in the cases of corresponding mass ratios for m_A/m_B and m_B/m_A respectively. Crossover in the present case of a heavy particle hitting a lighter one, which results in the heavy particle continuing its forward travel while the light particle zooms off, means that at this point the light particle has acquired more kinetic energy than the heavy one. These results refer to the rest frame of the particle initially at rest (laboratory frame).

>

7) Inelastic collisions in 1d: energy-mass conversion

Inelastic collision

We now proceed with an example of an inelastic collision. The inelasticity will play a different role than in nonrelativistic mechanics. Inelasticities in billiard ball scattering are associated with 'stickiness', a part of the mechanical energy is given up to heat. We say that it is dissipated and rely only on momentum conservation to calculate the dynamics of the one-dimensional motion. Complete inelasticity means that the two masses are stuck together, and proceed with a common velocity. The energy balance can be used to calculate the amount of mechanical energy that was dissipated in the system.

In the relativistic concept we do not calculate the mechanical energy only. The rest energy is included automatically in the energy expression. We know from chemical and nuclear reactions that the energy released from these is coming from the mass-to-energy conversion. The final products have a smaller mass than the initial ingredients to the reaction. In chemistry the mass differences which are converted into energy are small (but measurable); in nuclear physics they are considerably larger per reaction.

Thus, we will take the point of view that if we collide two objects in a completely inelastic way, such that they stick together, then the energy removed from the mechanical energy balance is not necessarily dissipated, but rather stored in the bound object. We can think of having compressed a spring to form the bond, and that at some point later (perhaps only induced by some reaction), the composite system can decay, and release the stored energy of the compressed spring. According to the energy-mass equivalence the rest mass of the composite object should account for the stored energy. We can use the energy conservation law to calculate this change in mass. Consistency of these ideas with special relativity will be demonstrated by performing the calculation in two different reference frames.

Center of mass frame

In the so-called center-of-mass frame the identical particles of mass m approach with equal speeds u 0 from left and right. After the collision they stick together at remain at rest in this frame (it is also called the zero-momentum frame as the total momentum vector vanishes at all times in this frame). If one looks at a worldline diiagram in this frame it looks like a Mercedes star (or a deformation thereof): for negative times the particles come in on worldlines with slopes 1/ u 0 and -1/ u 0 from left and right respectively. These worldlines merge at the origin (assuming that the collision takes place at c t = 0 in this frame, and emerge as a vertical worldline along the (c t )-axis for the newly formed object of mass M . The mass M is not simply 2 m , but has to be calculated to conserve energy. Let us denote this frame as S.

Laboratory frame

In the laboratory frame one of the two original masses m sits at rest (in the origin), while the other approaches with some speed u ' from the left. The initial speed u ' would be 2 u 0 in the nonrelativistic limit, but we anticipate a complication, since we cannot simply add velocities naively. The laboratory frame S' moves with respect to the center-of-mass frame with speed u 0 to the right, otherwise the mass that moves with - u 0 in S would not be at rest (it represents the target in the laboratory, while the other mass is a particle in the accelerator beam). Now that we have specified the boost velocity between S and S' as V = u 0, we have to be careful when determining the velocity of the impinging mass in S', i.e., we cannot simply say that it is 2 u 0. We will determine it from consistency with energy conservation or from the Lorentz transformation. After the sticky collision the composite object M dashes off to the right with some speed less than u ', as the impinging mass conserves momentum, i.e., the momentum that propelled m with speed u ' is the same that propels M with speed U .

Can we figure out the speed U without a calculation? Apparently yes, as we know the boost velocity between S and S' to be V = u 0, and M is at rest in S. Therefore, M has to move in S' with U = u 0 after the collision. The worldlines in S have to look as follows: the target mass m sits at the origin for c t ' < 0, while the beam particle mass m comes in with a slope 1/ u ' from the left.

At the origin c t ' = 0, x ' = 0 they form M which moves off for positive c t ' with a slope of 1/ U , i.e., 1/ u 0.

Let us figure out the incident particle speed u ' from momentum and energy conservation in S'. Before the collision in the lab frame all momentum resides in the beam particle.

> restart; assume(u0<c,u0>-c);

> pb_p:=m/sqrt(1-u_p^2/c^2)*u_p;

> Eb_p:=m*c^2+m/sqrt(1-u_p^2/c^2)*c^2;

After the collision we have

> pa_p:=M/sqrt(1-u0^2/c^2)*u0;

> Ea_p:=M/sqrt(1-u0^2/c^2)*c^2;

> sol:=solve({Eb_p=Ea_p,pb_p=pa_p},{M,u_p});

To look at the solution in detail we extract:

> M_sp:=simplify(subs(sol,M));

This is understood better by cancelling a factor of c in numerator and denominator.

> u_sp:=simplify(subs(sol,u_p));

> taylor(u_sp,u0=0);

We recover the expected nonrelativistic result in the limit of c being infinite (or u0/c being small).

Now let us perform the calculation in S, the center-of-mass, or zero-momentum frame:

The momentum is zero before and after the collision, which we do not formulate as an equation, but simply draw the conclusion that mass M moves with zero speed.

The energy conservation law reads:

> M:='M': Eb:=2*m/sqrt(1-u0^2/c^2)*c^2;

> Ea:=M*c^2;

> M:=simplify(solve(Ea=Eb,M));

We can verify that the energy expressions obtained in either frame are compatible with each other. Let us start in the CM frame and transform the energy to the lab frame: we carry out the calculation for the energy after the collision, Ea , first, and later perform the other calculation for Eb .

Energy after the collision:

To Lorentz transform to the lab frame we need to calculate (the object is not moving, i.e., has momentum 0):

> E_lab:=1/sqrt(1-u0^2/c^2)*(Ea+u0*0);

> E_lab:=simplify(E_lab);

> simplify(Ea_p);

Can we transform the other way too? Suppose we start with the lab frame result, we have the single mass M moving with u 0.

We are transforming to a frame that is boosted with - u 0 compared to this frame. Therefore,

> E_cm:=simplify(1/sqrt(1-u0^2/c^2)*(E_lab-u0*M/sqrt(1-u0^2/c^2)*u0));

> Ea;

This was simple. Now transform the energy before the collision. Let us go from the center of mass to the lab frame, as we know the velocities there.

Energy before the collision:

The lab frame is boosted with V = u 0 with respect to the CM frame.

> E_lab:=1/sqrt(1-u0^2/c^2)*((m*c^2/sqrt(1-u0^2/c^2)+u0*u0)+(m*c^2/sqrt(1-u0^2/c^2)-u0*u0));

> Eb_p;

> simplify(subs(u_p=2*u0/(1+u0^2/c^2),Eb_p));

> simplify(E_lab);

We notice that the results agree when we insert the speed of the beam particle (which we determined from momentum conservation). We could have used the Lorentz transformation of the energy from the CM (zero-momentum) frame to determine the beam particle speed:

> solve(Eb_p=E_lab,u_p);

For completeness one more check: the transformation from the lab frame to the CM frame. We begin with a definition of the beam particle speed u_p :

> u_p:=2*u0/(1+u0^2/c^2);

> E_cm:=1/sqrt(1-u0^2/c^2)*((m*c^2+u0*0)+(m*c^2/sqrt(1-u_p^2/c^2)-u0*m/sqrt(1-u_p^2/c^2)*u_p));

This is a formidable expression at first. Nevertheless, it simplifies to the desired result:

> E_cm:=simplify(E_cm);

> simplify(Eb);

We summarize the progress made in this section:

1) we understand now how to transform energy between inertial frames (such as lab vs cm frame) for systems of particles (demonstrated mostly in one dimension, but the generalization is straightforward);

2) we considered elastic and (completely) inelastic billiard ball scattering. For completely inelastic collisions in the center of mass frame we demonstrated how all kinetic energy was converted into an additional mass for a new particle of mass M in this zero-momentum frame. We can think of the inelasticity being effected by a virtual spring that was loaded and locked during the collision process; the converted energy that resulted in a mass increase can become available when the two particles break apart One concrete example of such a process would be the alpha-decay, or fissioning of nuclei, where after a tunneling process, or by collisional fragmentation the two daughter nuclei are pushed apart from each other by the Coulomb repulsion. The daughter products appear asymptotically will well-defined kinetic energies. As long as the spring is loaded (the system sticks together) the stored energy is noticeable via the increased mass (over 2 m ).

Usually bound systems in physics represent situations where the composite system has a lower energy (mass) than the sum of the masses of the constituents. This situation makes the composite system stable against break-up. The gain in energy by forming the composite system is the driving process for the compound system to be formed.

This implies that the idea of a rest mass for a composite system of fundamental particles may be tricky.

>

8) Invariant mass for systems of particles

Invariant mass

We noticed that the components of the momentum vector and energy transformed under LTs in analogy to the transformation of position and time. For the LT of a spacetime interval an invariant can be found:

Ds^2 = (c*Dt)^2 - (Dx^2 + Dy^2 + Dz^2)

(the mathematical reason for this quantity to be conserved is that it forms the inner product of a vector with itself in the four-dimensional spacetime. The LTs represent rotations in this space (that is why there is the peculiar sign chosen between the space and time contributions to ds), and leave the length of four-vectors invariant under the transformation.

It is of interest to consider the invariant quantity associated with the four-vector made up of the energy and the momentum components.

> restart; unprotect(gamma); assume(c>0,u0<c,u0>-c,m>0);

> gamma:=1/sqrt(1-u^2/c^2);

> px:=m*gamma*ux;

> py:=m*gamma*uy: pz:=m*gamma*uz:

> E:=gamma*m*c^2;

> E^2-c^2*(px^2+py^2+pz^2);

> algsubs(ux^2+uy^2+uz^2=u^2,%);

> simplify(%);

The invariant magnitude (under LT) of the fourvector made up of E and p is the rest energy of the mass m . We can state that any inertial-frame observer will measure the same value for the rest energy, and therefore the mass of an isolated system (since c is universal). There is a special reference frame where the particle is at rest, namely the frame specified by p =0. In this frame the total energy equals the rest energy.

The kinematic state of a system of particles is specified by the invariant spacetime interval

Ds^2 = (c*Dt)^2 - (Dx^2 + Dy^2 + Dz^2)

The dynamic state of a relativistic system of particles can be described by the following expression for the invariant total rest mass in terms of energy and momentum:

(m*c^2)^2 = E^2 - (p*c)^2

This relation allows to deduce the rest mass from an energy and momentum measurement. This is important, as one does not have to be in the rest frame of a particle to determine its rest mass - which is often encountered in particle physics.

Even if there are no interactions between the particles a system of moving particles does experience a peculiarity. The total energy and momentum of a system of particles was used to determine the invariant rest energy. We will now show that this invariant rest energy is not the sum of the rest energies of the individual particles.

A simple example of two freely moving particles suffices to make the point. We return to a set-up discussed in the context of inelastic collisions. Two particles of rest mass m with velocities + u 0 and - u 0 approach each other (CM frame). We can consider the invariant mass calculation also in different frames, such as the lab frame, where one of the particles is at rest.

For each particle we can calculate its energy

> E:=sqrt(m^2*c^4+p^2*c^2);

The total momentum in the CM frame is zero (it is also called the zero-momentum frame).

> P_tot:=0;

> E_tot:=2*E;

The invariant mass of the system of two free particles is now calculated to be:

> M:=simplify(sqrt((E_tot/c^2)^2-(P_tot/c)^2));

We notice that the rest mass of the system is greater than twice the rest mass of the particles! It does not make sense to associate the excess in the invariant rest mass of the system over twice the rest masses of the constituents with the motion of individual particles (even though the origin of this excess does lie in the non-vanishing momentum vectors). To illustrate this, we calculate the same invariant mass by transforming to another reference frame, and calculating the invariant mass of the entire system there (to check whether it really is a property of the system, i.e., independent of the reference frame of the observer).

Before we carry out this calculation we express the momentum of the particles in terms of their speeds. Given that the square of the momentum is required we do not need to distinguish the signs for the left-and right-moving particles:

> p:=m/sqrt(1-u0^2/c^2)*u0;

> simplify(eval(E_tot));

> M_cm:=simplify(eval(M));

We evaluate the expressions to make the use of the expression of p in terms of the mass m and speed u 0.

Suppose we pick a frame that moves with V = u 0 with respect to the center of mass, i.e., one of the masses is at rest, and the other (e.g., a beam particle) moves with an increased velocity vector (not quite - 2 u 0 according to a calculation done above, where the correct value of v_p was determined).

Now one particle is at rest and one moves with:

> u_p:=simplify(-2*u0/(1+u0^2/c^2));

We could repeat the calculation of the invariant mass from scratch:

> E_tot:=m*c^2+simplify(m/sqrt(1-u_p^2/c^2)*c^2);

> P_tot:=simplify(m*0-m/sqrt(1-u_p^2/c^2)*u_p);

> M_lab:=simplify(sqrt((E_tot/c^2)^2-(P_tot/c)^2));

> M_cm;

It is impressive that Maple 5 does recognize these two expressions to be the same once the restricition -c < u 0 < c is imposed in the assume statement (invoked after restart at the beginning of this section).

It is important to recognize that the same invariant mass was calculated for the system in both frames. This is remarkable as the relativistic mass calculation is different in the CM and Lab frames:

> RM_cm:=2*m/sqrt(1-u0^2/c^2);

> RM_lab:=simplify(m+m/sqrt(1-u_p^2/c^2));

> plot([subs(c=1,m=1,RM_cm),subs(c=1,m=1,RM_lab)],u0=0..1,color=[red,blue],view=[0..1,0..10]);

Clearly the invariant mass is not equal to the sum of the two relativistic masses of the particles! It has to be calculated according to the prescription for the dynamic state of the relativistic system.

Exercise:

Verify the invariant mass of the two-particle system by carrying out the Lorentz transformation of the energy from the CM to the Lab frame, and using the dynamic state of the system. Then illustrate the value of the invariant mass by considering the following values: m = 4 kg, u 0 = 0.6 c.

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9) Massless particles: photons,...?

Massless particles

It is possible to consider the dynamic state in the limit of zero rest mass. In principle, there could also be particles with negative rest mass (tachyons), but so far they have eluded observation if they indeed exist. For particles without a rest mass we find from the dynamic state the following relationship between energy and momentum:

> restart;

> E:=subs(m=0,sqrt(m^2*c^4+p^2*c^2));

We can determine the speed of the particle from the following observation. For massive particles the speed is related to energy and momentum as:

> p_m:=m/sqrt(1-u^2/c^2)*u;

> E_m:=m/sqrt(1-u^2/c^2)*c^2;

> p_m*c^2/E_m;

This relationship is valid irrespective of the mass of the particle, and presumably carries over to the massless case:

> simplify(p*c^2/E,symbolic);

Therefore a massless particle can only move with one speed, namely the speed of light.

What is the photon's momentum? We should get p = E /c according to our derivation above. Insertion of the speed u = c and mass m = 0 into the momentum expression results in an undefined answer (0/0). The appropriate step is to replace the m/sqrt(1-u^2/c^2) expression by E/c^2 .

The only particles known to be definitely massless are photons. Neutrinos are candidates for massless particles, although there is some evidence for a lower bound on the mass. Gravitons, the postulated qunatum carriers of gravity also should be massless (for photons and gravitons the masslessness is connected with the inverse squared force laws that appear in these theories).

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