Lecture 3. Relativistic Dynamics

Outline:

3-vectors vs. 4-vectors

Relativistic Momentum

Relativistic Kinetic Energy

Total Energy

Momentum and Energy in Relativistic Mechanics

Comment on 3- and 4-vectors

Galilean transformations do not affect the relations between vectors.

That makes vectors so useful in classical mechanics: if one can formulate a law that looks like “vector” = “vector” , it automatically means that this law is

invariant under G.Tr.

In particular, G.Tr. do not affect the length of a vector: the length of a vector is invariant under G.Tr.

t t

x x Vt

y y

z z

2 2 2

2 1 2 1 2 1

2 2 2

2 1 2 1 2 1 Invariant

length x x y y z z

x x y y z z

Comment on 3- and 4-vectors (cont’d)

Is there a combination of (x,y,z,t1) which remains invariant under L.Tr.? Indeed, such a combination does exist:

This is the square of the distance between two events (ict1x1,y1,z1) and (ict2,x2,y2,z2) in the 4-dimensional space. 1i

This quantity (a.k.a. the interval) is invariant under L.Tr. (please show this explicitly at home using L.Tr.), for two arbitrary events it might acquire any (zero, positive or negative) value, unlike the distance in 3D space (Appendix II).

L.Tr. correspond to a rotation of a 4-dimensional RF through a fixed angle; these rotations preserve the length of 4-vectors.

' cos sin

' cos sin

x x y

y y x

y

'x

'y

x

2 2 2 2 2 2 2 22 22 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1' ' ' ' ' ' ' ' =Invariantx x y y z z c t t x x y y z z c t t

x x ct

ct ct x

'

x x i

i x

2 2

2 2

1cos sin cos sin 1

1 1

ii

ict

tgV

i ic

Comment on 3- and 4-vectors (cont’d)

Note that both force and acceleration are 3-vectors. Thus, one should not expect that the 2nd Newton’s Law, being expressed in terms of force and acceleration, remains invariant under Lorentz Transformations. This is the reason why “force” and “acceleration” are not popular in relativistic mechanics. We’ll discuss relativistic dynamics in terms of the “momentum” and “total energy”.

Examples of 4-vectors:

0 1 2 3, , , , , ,ict x y z x x x x

, , ,x y zi A A A - the vector and scalar potentials in Electrodynamics

,A

, , ,x y z

Ei p p pc

- the momentum and total energy in relativistic mechanics

,p E

Thus, if one can express some physical law in a form : “4-vector A”=“4-vector B”, this would guarantee the invariance of this law under L.Tr!

Def.: A 4-vector is any set of four components which transforms in the same manner as the space-time vector under L.Tr.

, , ,x y zi k k kc

- the wave vector and angular frequency of a plane harmonic wave

,k

Relativistic Momentum

Newton’s 2nd Law:

p mv - the momentum of a particle in classical mechanics,

m is invariant (does not depend on the velocity)

dp dvF m ma

dt dt - expressed in terms of 3-vectors,

invariant under G.Tr. (but not L.Tr.!)

Relativistic form of the momentum (introduced by

Einstein):

2

21

mvp mv

vc

- definition of the momentum

in relativistic mechanics

p mv

/v c

Example: Calculate the momentum of an electron moving with a speed of 0.98c.

2

0.984.9

1 0.98

ee

m cp m c

1

By ignoring relativistic effects, one would get

0.98 ep m c

(m is still invariant, see comment below)

Comment

Einstein: “It is not good to introduce the concept of the mass m’ = m of a moving body for which no clear definition can be given. It is better to introduce no other mass concept than the “rest mass” m. Instead of introducing m’ it is better to mention the expression for the momentum and energy of a body in motion”.

“Relativistic” mass m’ = m – just another name for the momentum (energy).

Occam’s Resor: “Entities must not be multiplied beyond necessity”.

See also Okun’s paper on our Webpage.

Caution: some textbooks use the velocity-dependent mass m’ = m and the “rest” mass m.

Relativistic Kinetic Energy

Let’s calculate the kin. energy gained by an accelerated particle:

2 2 2 2 2

2 21 1 1 1 1

2 2

22 2

22 2 2 2 2 20

22 2 2

2 20

1 /

(integration by parts , )1 /

11 /

1 / 1 / 1 /

1 /1 /

f

f

v

v

dp dl mvK Fdl dl dp vd mv vd

dt dt v c

mvxdy xy ydx x v y

v c

mv vdv vdvm d v c

cv c v c v c

mvmc v c

v c

22 2 2 2

2 2

2 2 22 2 2 2 2

2 2 2 2

1 /1 /

/1 /

1 / 1 /

ff

f

ff

f f

mvmc v c mc

v c

v c mcmc v c mc mc

v c v c

1v = 0

2v = vf

2

2 2

2 21

1 /

mcK mc mc

v c

- kinetic energy of a particle of the mass

m moving with speed v

Total and Rest EnergiesWe expect this result to be reduced to the “non-relativistic” KE at low speed:

2 2 2

2 2

21 1 1

2 21

mc vK mc mc m

/v c

2

K

mc

1

Let’s rewrite the expression for K in the form:

22

2 21 /

mcK mc

v c

2E mc K the total energy

Limit ofsmall speed:

22

2

vE mc m

/v c

2

E

mc

1

1

The energy and momentum are conserved(the consequence of uniform and isotropic space).

For an isolated system of particles:

1

all particles

ii

E E const

1

all particles

ii

p p const

the rest energy

20E mc

- we must use Relativistic Mechanics when K and E0 become of the same order of magnitude.

Electron-Volt: convenient unit of energy

In relativistic mechanics, most of the time we deal with tiny particles like an electron or proton (after all, it’s hard to accelerate a macroscopic body to v~c). In this case, the most convenient unit of energy is an electron-volt, the kinetic energy acquired by an electron accelerated through a potential difference of 1Volt.

19 191.6 10 1 1.6 10 1K q V C V J eV

14

231 8 140 19

8.2 109.1 10 3 10 / 8.2 10 510,000 0.5

1.6 10 /e

JE kg m s J eV MeV

J eV

Thus, when the electrons are accelerated across a pot. difference ~ 10kV in a TV tube, they still can be considered as non-relativistic particles (K<<m0c2)

1V V

1819

16.2 10 /

1.6 10 /Energy eV Energy J Energy J eV J

J eV

For example, the rest energy of an electron:

The rest energy of a proton: 227 80 1.6726 10 3 10 / 938.3pE kg m s MeV

The rest energy of a neutron: 227 80 1.6749 10 3 10 / 939.6nE kg m s MeV

Problem

An electron whose speed relative to an observer in a lab RF is 0.8c is also being studied by an observer moving in the same direction as the electron at a speed of 0.5c relative to the lab RF. What is the kinetic energy (in MeV) of the electron to each observer?

VK

K’

v The electron speed v’ as seen by the moving observer K’:

2

0.8 0.5' 0.5

1 0.8 0.51

v V c cv c

vVc

In the lab IRF K:

2

2 2

1 11 0.5 1 0.34

1 / 1 0.8K mc MeV MeV

v c

In the moving IRF K’:

2

2 2

1 1' 1 0.5 1 0.08

1 '/ 1 0.5K mc MeV MeV

v c

Whereas the rest energy is RF-independent (m is invariant), both the total energy and kinetic energy do depend on our choice of the reference frame.

Rest Energy and Mass of a System of Particles

Classical mechanics: conservation of mass and energy

Relativistic mechanics: conservation of energy

Mass and energy are different aspects of the same “thing”, they become interchangeable: matter can be created or destroyed, but when it happens, an equivalent amount of energy vanishes or comes into being.

Important: for a system of many particles, the mass M includes the potential energy of all interactions between the particles.

The potential energy U for the particles that attract each other is negative (for repelling particles – positive). Thus, for the stability of a body, its mass should be smaller than the sum of masses of all particles that constitute the body.

2 2 2bind

i i ii i i

EE UM m m m

c c c

Mass Defect

Example: dynamite explosionWhen 1kg of the TNT explodes, the energy release is 5.4MJ. At the same time, the rest energy of 1kg is 22 8 161 3 10 / 9 10mc kg m s J

611

16

5.4 106 10

9 10

m J

m J

- it’s very difficult to detect this mass change

Mass defect: ii

M M m the mass of a composite body (system) the sum of masses of its constituents

For the stability of a body, its mass defect should be negative.

Binding energy: 2

bindE Mc We’ll consider the binding energy in detail when we consider nuclei and nuclear reactions.

Why was it difficult to notice in Cl. M.? Because in all processes of chemical transformations (the most “violent” processes of the pre-20-century physics) the energy release is tiny in comparison with the rest energy of reactants.

The mass loss for the Sun. The power of solar radiation P ~ 4 ·1026 J/s: the power per 1m2 on the Earth’s surface (~1400 W/m2) being multiplied by the area of a sphere with radius 1.5·1011 m (the Sun-Earth distance).

26

922 8

4 10 W4.4 10 kg/s

3 10 m

dm P

dt c

the mass loss per one second

ExamplesAn elementary particle (e.g. a free electron) cannot absorb/emit a photon. (Hint: use the

reference frame in which the electron was at rest after absorption/before emission).

Decay of a neutron: a free neutron is unstable (the lifetime ~ 15 min). It decays into a proton, an electron, and an electron anti-neutrino:

A composite particle can decay into two (or more) fragments if the mass of all fragments is less than the particle mass.

en p e

The masses involved:

2 938.3pm c MeV2 939.6nm c MeV2 0.5em c MeV

2 0e

m c

2

939.6 938.3 0.5

0.8

en p em m m m c

MeV

MeV

p e nm m m - otherwise, the neutron would be stable and most of the protons and electrons in the early Universe would have combined to form neutrons, leaving little hydrogen to fuel the stars.

per nuclonn n bind pm m m E m - the energy conservation prevents the neutron from decaying in nuclei.

Without neutrons we would not have the heavier elements needed for building complex systems such as life.

However, neutrons are stable in nuclei:

An important relationship between E and p

2 2 2

2 21 /

E m c

c v c

2 2

2

2 21 /

m vp

v c

2 2 2 222 2 2

2 2

1 /

1 /

m c v cEp m c Inv

c v c

this combination of E and p does not depend on the IRF!

(In fact, it is the length2 of a 4-vector formed by the components of p and i(E/c))

2 2 2E c p m c

2 2 2 2 4E p c m c

Ultra-relativistic and massless particles

In the ultra-relativistic case (K>>m0c2) E cp

This equation “works” for the particles with zero rest mass (photons). They must move with the speed of light: v=c (otherwise, both p and E are 0)

Massless particles: the photon (carrier of the electromagnetic interaction), the gluon (carrier of the strong interaction, never observed as a free particle), and, perhaps, the graviton (carrier of gravitational interaction, remains to be discovered).

External forces can bend the trajectories of massless particles (e.g., photons in gravitational fields), but cannot accelerate (decelerate) them (v is always =c).

2 2 2 2 4 0E p c m c

ph phE cp

Problems

1. What is the momentum of an electron with K =mc2?

2. How fast is a proton traveling if its kinetic energy is 2/3 of its total energy?

2 2 2 2 4E p c m c

22 22 2 2 2 2 2 2 24 3

E mc Kp m c m c m c m c mc

c c

2 2 22 22 3

3 3K E mc K K mc E mc

2

21 /

mcE

V c

2

2

1 1 83 1

9 31 /

VV c

cV c

Problem

An electron initially moving with momentum p=mc is passed through a retarding potential difference of V volts which slows it down; it ends up with its final momentum being mc/2. (a) Calculate V in volts. (b) What would V have to be in order to bring the electron to rest?

p = mc: 2 2 22 2 2 2 2 21 2E p c mc mc mc mc

Thus, the retarding potential difference

p = m0c/2: 2

22 2 22

1 5

2 2E mc mc mc

2 2 6 51 2

52 0.3 0.3 0.5 10 1.5 10

2E E E mc mc eV eV

51.5 10V V

(a)

(b) 2 2 2 5 51 22 2 1 2.1 10 2.1 10E mc E mc E mc eV V V

”Threshold” ProblemsExample: Find the minimum energy a proton must have to initiate the reaction

p p p p p p

(production of anti-protons (Berkeley Bevatron 1954)

The minimum energy – when the products of reaction are at rest in the “center of mass” reference frame: all the incoming energy is transformed into the rest energy. Take an advantage of the (E,p) invariant:

2 2( )sys sysE p c Inv

p

p

Now we know the magnitude of the invariant, let’s go back to the LAB RF in order to calculate the minimum energy of a colliding proton:

1 2 0p p p 2 21 2E K mc E mc

2 22 2 2 2 22 2 2 21 2 1 2 1 1 1 1 1

2 22 2 21

2

2 2 16

E E p p c E mc p c E E mc mc p c

mc E mc Inv mc

21 7E mc 2 2

1 6K E mc mc

22 2 2( ) 4sys sys pE p c m c

- no matter which IRF we use

In particular, to calculate this invariant for the “threshold” conditions (the products of reaction “stick” together), it is convenient to use the “center-of-mass” RF:

- a “fast” proton collides with another proton at rest

p

p

”Threshold” Problems (cont’d)

For “colliding beam” accelerators (e.g., LHC), the products of reaction are (more-or-less) at rest in the lab frame (the center-of-mass and lab RFs are the same):

Homework #2: Beiser Ch. 1, Problems 29, 30, 33, 34, 41, 42, 45, 49, 54, 56.

2 21 6K E mc mc

A “fast” proton collides with another proton at rest:

p

p

“Threshold” kinetic energy:

p

p

1 2p p p 21 2E E K mc

2 22 2 2 2 21 2 1 2 4 16E E p p c K mc mc

2K mc

22 2 2( ) 4sys sysE p c mc

2 22K mc mc -the “colliding beam” acceleratorsare much more efficient