mit relativity2

8/10/2019 MIT Relativity2

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Massachusetts Institute of TechnologyDepartment of Physics

Physics 8.022 Fall 2002

Special Relativity, Part II

Peter Fisher

Abstract

This note gives the derivation of relativistic momentum and energy,as well as the transformation of momentum and energy between iner-tial frames. The transformations are then used to derive the transfor-mations of forces between frames, which is necessary for understandinghow the electric and magnetic elds are related.

In the last handout, we worked out how to transform events from one inertialframe to another. In electricity and magnetism, we are most interested inthe forces (which arise from the elds) and F = d p/dt , so we must also learnhow momentum and energy work relativistically.

Momentum

If the maximum velocity attainable by a particle of mass m is c, then, usingthe p = m v means there is a maximum momentum p = mc and a maximumkinetic energy T = mc2 / 2. However, think of the following experiment: aparticle of mass m and charge q is accelerated by some means to a velocityclose to c and aimed through a set of parallel plates, Fig 1. Just as the


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particle reaches plate 1, a potential of strength V is turned on. Traversingthe plates gives the particle an additional energy T = qV . After leaving,the potential is turned off. Clearly, it is always possible to add energy toa particle, no matter what its velocity, so we need to come up with a newdenition of momentum. In coming up with a new denition, we must besure we recover the p = mv and T = mv2 / 2 when v


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Figure 1: Acceleration of a particle. a) A particle with velocity close to capproaches a set of parallel plates. b) When the particle passes plate 1, apotential V is turned on, creating a eld E which exerts a force F = qE onthe particle. c) When the particle leaves the plates, the potential is turnedoff. The particle now has qV more kinetic energy than before, regardless of its initial velocity.

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Figure 2: Scattering experiment. a. In frame S o, the particles A and Bcollide with equal and opposite momenta. b. In S A , particle A moves onlyalong the

y axis. c. In

S B, particle

B moves only along the

y axis. Frame

S B moves in the x direction with velocity v with respect to frame S A .4


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Energy

Now we want to know the energy of a particle moving with velocity v. Wework this out nding the work necessary to accelerate a particle from rest tovelocity v over some path. Then

E = F ds = d pdt ds= d pdt d sdt dt = d pdt vdt

where we have used the denition of velocity, v = d s/dt . The second lineresults from multiplying by 1 = dt/dt . Next, use the relation

ddt

( p v) = v d pdt + pdvdtto obtain

E = vo

o

ddt

( p v) p d vdt

dt

= m ov2o

vo

o p dv = m ov

2o

vo

o

mv dv 1 v2 /c 2

where we take the nal velocity of the particle to be vo. Since vdv = vdv =d(v

2

)/ 2, the second term is

vo

o

mv

1 v2 /c 2dv =

mc2

2 1 v2o /c 2

1

du u = mc

2 ( 1 v2o /c 2 1)where we have used the substitution u = 1 v

2 /c 2 . Then,

E = m ovo + mc2 1

o 1

= mc2 o 1 1 2o

+ 1 o 1

= omc2

mc2 .

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From the last line, a reasonable interpretation is that E = omc2 , so whena particle is at rest, o = 1 and we get the famous E = mc2 .1 We can nowalso prove other relations:

E 2 = p2 c2 + m2 c4

= pcE .

Given any two of m, p, v = c and E , we can nd the other two quantities.The relations are summarized in Fig. 3. You should be sure you know howto show all the relations in the triangle.

Now we have to make sure that we get the old expressions for kinetic energyand momentum when v


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Figure 3: The Relativity Triangle: knowing any two quantities, you can ndthe other two.

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y = y

(1 x )and

x = 1

1 2x = 1

1 x 1 x 2=

1 x (1 x )2 ( x )2

= 1 x

1 2 x + 2x 2 2x + 2 x 2

= 1 x (1 2x )(1 2 )

= (1 x ) x.

Now we just put in to nd the momentum and energy in the S frame:

px = m x c = m x (1

x )c (2)

= px E

c (3)

py = m yc = m x (1 x ) y

x(1 x )c = py (4)

E = mc2 = (1 x ) xmc2 (5)

= E p xc. (6)Notice energy and momentum transform in a way similar to time and thespatial coordinates.

Transformation of forces

The electric and magnetic elds create forces and transforming the forces iskey to transforming the elds. As usual, we start with a force F in S whichmoves relative to S with velocity u = c = ux and nd F . Starting rstwith F x

F x = dpxdt

dp xdE/cdt

c dx=

F x (dE/dt )/c1

v x

c.

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Remember dE = F ds, so dE/dt = F v and

F x = F x ( F v)/c

1 x.

For a particle at in S , F x = F x . For transverse forces (i.e. forces normal tothe relative velocity of the frames)

F y = dpydt

= dpx

dt c dx=

dpx /dt (1 vx)

= F y

(1 vx)and for a particle at rest in S , F y = F y/ .

Some examples

When physicists talk about relativistic kinematics, they usually express ener-gies in electron volts or eV . This is just the energy a particle of one electronof charge has when it sits at a potential of 1 Volt; 1 eV=4 .8 10

10 esu 1Volt / 300Volts / statvolt = 1 .6 10 12 ergs. Since E = mc2 , we can express

masses using units of eV/c2 = 1 .8 10 33 g. Thus, the electron, which has a

mass of 9.110 28 g would have a mass of me = 511, 000 eV/ c2 or 511keV/ c2 .

Similarly, momenta my be expressed using units of eV / c.

Pion decay

The lowest mass nuclear particle is the charged pion or . The is unstable,with a lifetime of 26 ns and decays to a muon () and a neutrino ( ), .The pion has a mass of m = 140MeV/ c2 , the muon has a mass m =106 MeV/ c2 and the neutrino is massless.

We start by asking: if a decays at rest, what are the energies and momentaof the decay products, the and the ? To nd the answer, use the fact thatenergy and momentum are conserved in the decay. Notice that since m

= 0,

E = p c. Referring to Fig. 4, p = p , so we take | p| = | p | = p. Energy9


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conservation says E = E + E and, since the pion is at rest, E = m c2 .Then, using E = p2 c2 + m2c4 ,

E = m c2

pc =

p2 c2 + m2c

4

p2

c2

2m pc3

+ m2 c

4

= p2

c2

+ m2c

4

(m 2 m2)c

4 = 2m 2 pc3

p = (m2 m

2)c

2m.

Putting the numbers for the masses in gives p = 30MeV/ c2 for the muonmomentum and E = p2 c2 + m2c4 = 110MeV for the muon energy.Next, we want to think about the decay of a pion moving with velocity v.When it decays, the momentum and energy of the muon and neutrino will

depend on the decay angle in the rest frame of the pion, Fig. 4b. We canuse relativity to nd out the energy and angle in the lab frame in which thepion moves with v = vx. In the pion rest frame, after the decay, the muonhas momentum

px = pcos py = psin

where p = 30MeV/ c. Using Eq. 3-6 to transform into a the lab frame gives

px = p cos + E py = psin E = E + p cos .

The range of possible energies is E p to E + p. For example, a pionwith kinetic energy of 500 MeV would have a total energy of E = 640 MeV.Then, = E /m = 4.6 and = 1 1/ 2 = 0 .98. Then, the minimummuon energy is 371 MeV and maximum is 641 MeV.

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Cutoff in cosmic rays

There is a very interesting effect in cosmic rays. Very high energy protonstravel vast distances without losing much energy. However, above a certain

energy, a new loss mechanism becomes important. A proton can absorba photon and make a particle, the next lightest nuclear particle. Thereaction is p + . The is unstable and quickly decays to a protonand a pion: p + .The universe is lled with photons emitted when electrons bound to protons300,000 years after the Big Bang. At the time, the photons had energiesof around 10eV. Since then, the expansion of the universe has caused themto cool to about 200 eV. These relic photons are hit by very high energyprotons to make s. If the mass of the is m = 1150MeV/ c2 , what energyproton is necessary to produce a ?

In order to make a , the total energy of the proton and photon must be atleast the m c2 : E p + E = m c2 . In addition, total momentum is conserved:E /c + p p = p . Remember, for a massless particle like a photon, E = pc.Inthe universe (unprimed) frame, the has and energy E and momentum p . Then

m c2 = E 2 p2 c

2

= E 2 + 2E E p + E 2

p p2 pc

2

2 p pcE E 2

p p =

2E E p (m2 m

2 p)c

4

2E .

Now that we have square both sides and solve for E p. Well use the notation 2 = ( m2 m

2 p).

p2 pc2 = E 2 p m

2 pc

4 = 4E 2 E

2 p 4E E p

2 c4 + 4 c8

4E 2 4E 2 m

2 pc

4 + 4 c8 = E E p2 c4

E p = 4E 2 m

2 p +

4 c8

4E 2 c4 . (7)

2 = m2

m2

p = 450, 000(MeV/ c2 )2 . Notice there are two terms in the

denominator of Eq. 7; the rst is 4 E 2 m2 p = 1.4 10

11 eV4 while the second

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is 4 = 2 1033 eV4 /c 4 , so for all practical purposes

E p = 2 c4

4E = 6 10

20 eV.

Once produced, the moves with essentially the proton energy and, after avery short time, decays to a proton and a pion, p + . The interestingquestion is than: what is the maximum energy the nal proton can have?In the rest (primed) frame, we have energy conservation: m c2 = E p + E and momentum conservation: p p = p . From momentum conservation

p = p pE 2 p m

2 pc

4 = E 2 m2 c

4

= ( m c2

E p)2

m2

c4

E p = m2 c4 m

2 c

4 + m2 pc4

2m c2 = 949MeV/ c2

which gives a proton momentum of p p = 144MeV/c . If the has an energyof 6 10

20 eV, then = 5.2 1011 (and of course = 1 to a very good

approximation). Then, the maximum energy the proton can have in the labframe is

E p = E p + p pc = 5.6 1020 eV.

Thus, the production and decay of the by very high energy protons resultsin a limit to the proton energy of 6 10

20 eV. Protons with higher energiesget knocked back down by the production and decay. Interestingly, someexperiments claim to have observed cosmic rays above this limit, which eitherthe originate nearby (from as yet unknown sources) or they are not protonsor nucleons. Several new experiments are attempting to search for theseunusual cosmics rays with higher sensitivity.

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Figure 4: Decay of a pion to a muon and neutrino: a) Decay in pion restframe. Since the total momentum is zero, the neutrino and muon have equaland opposite momenta. b) Pion decay in pion rest frame. The muon mo-mentum makes an angle with the x axis. c) Decay of a pion moving alongthe x axis. In this frame, the muon momentum makes an angle with thex axis.

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