Relativistic Mechanics
SR as a metatheory: laws of physics must be Lorentzinvariant.
Laws of classical mechanics are not: they are invariant under Galilean transformations, not Lorentz transformations.
=> we need a revised, relativistic, mechanics!
Starting point: Assume each particle has an invariant proper mass (or rest mass) m
0
Define 4momentum P = m0 U
Basic axiom: conservation of P in particle collisions: * Pn = 0
(in the sum, P of each particle going into the collision is counted positively while P of each particle coming out is counted negatively)
Since a sum of 4vectors is a 4vector, this equation is Lorentzinvariant.
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P = m0 U = m
0 (u) (u, c) ≡ (p, mc)
relativistic mass: m = (u) m0
relativistic momentum: p = m u
* p = 0 * m = 0Conservation of 4momentum yields both:
These are Newtonian conservation laws when c ∞
What about conservation of energy?
When v/c ≪ 1 :
Einstein proposed the equivalence of mass and energy: E = mc2
=> P = (p, E / c)
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Relativistic energies
rest energy: m0c2
kinetic energy: T = mc2 – m0c2 = ( 1) m
0c2
Note: Potential energy associated with position of particle in an external electromagnetic or gravitational field does not contribute to the relativistic mass of a particle. When particle's PE changes, it's actually the energy of the field that's changing.
H atom: energy of electron = 13.6 eV (binding energy)
=> matom
c2 = mp c2 + m
e c2 – 13.6 eV
=> atom has less mass than its constituents!
1 eV = 1.602 × 1012 ergs = 1.602 × 1019 J
mp = 1.67 × 1024 g ; m
e = 9.11 × 1028 g
=> m / m = 13.6 eV / (mp + m
e ) c2 ≈ 1.8 × 108
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Nuclear physicists use the atomic mass unit u, defined so that the massof the 12C atom is exactly 12 u.
1 u = 931.502 MeV / c2
mp = 1.00727647 u ; m
n = 1.00866501 u ; m
e = 5.485803 × 104 u
=> m / m ≈ 0.8% for the 12C nucleus
Binding energy ≈ 90 MeV (≈ 7.5 MeV per nucleon)
4
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P = m0 (u) (u, c) = (p, mc) = (p, E / c)
=> P2 = E2/c2 – p2
In particle's rest frame: P = (0, m0c) => P2 = m
02 c2
=> E2/c2 – p2 = m0
2 c2 => E2 = m0
2 c4 + p2c2
For a photon: m0 = 0 => E = pc , P = E/c (n, 1)
E = h = hc/ , n = a unit vector in dir of propagation
Since P is a 4vector, E and p transform as follows under a standard LT:
px' = (p
x E/c)
py' = p
y
pz' = p
z
E' = (E pxc)
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Applications
1. Compton scattering: photon strikes stationary electron
Before:E e
E'
e
After:
7
.
h / mec is called the
“Compton wavelength”of the electron.
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SP 4.1
2. Inverse Compton scattering: photon scatters off a highly energetic charged particle
BeforeE
m0
After
E2
E1
As before:
=>
Assume E ≫ m0c2 ≫ E
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(E ≫ m0c2)
(from previous slide)
=>
Maximum energy is transferred to photon when cos = 1:
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Cosmic Microwave Background (CMB): 3 K blackbody spectrum
=> typical photon energy ~ kT = 3 × 104 eV
Very high energy cosmic ray proton: 1020 eV ; mp c2 = 938 MeV
=> CMB photon can be upscattered to ~ 1019 eV (source of gamma rays)
3. Particle accelerators: “available energy” for creating new particles is the energy in the CM frame
Consider 2 arrangments:
a) a 30 GeV proton collides with a proton at rest
b) two 15 GeV protons collide headon
(lab and CM frames are identical => 30 GeV of available energy)
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a) lab frame: Pi = [ m
pc, (+1) m
pc ] CM frame: P = (0, E/c)
To reach an available energy of 30 GeV in arrangement (a), we would need a 480 GeV proton!
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4. Cosmic ray cutoff: highenergy nucleons can undergo the reaction + N N +
Reactions with CMB photons (E ~ 3 × 104 eV) probably produces a cutoff in the cosmic ray spectrum: no CRs with energy above the threshold energy for the reaction.
Assuming a headon collision, what's the threshold energy?
We need E = (mN + m
) c2 in CM frame => P
f = (0, m
Nc + m
c)
Lab frame, before collision:
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=> EN ≫ m
Nc2 => LHS 2 E
N => E
N ≈ 3 × 1014 MeV
(mNc2 ≈ 940 MeV , m
c2 ≈ 140 MeV)
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SP 4.28
What about forces?
Define 4force F = dP/d = d(m0U) / d = m
0A + (dm
0/d) U
Limited version of Newton's 3rd Law for contact collisions between 2 particles: during contact, is the same for both particles.
F1 + F
2 = d(P
1 + P
2) / d = 0 =>
F
2 = F
1
Relativistic 3force f = dp/dt = d(mu)/dt
f becomes Newtonian force when 0
4th component of F is proportional to the power absorbed by the particle.
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=> F⋅U = proper rate at which particle's internal energy increases
Also:
Equating above 2 expressions for F⋅U => rest masspreserving forces are characterized by:
F⋅U = 0
f ⋅u = dE/dt => f ⋅dr = dE F = (u) (f, f ⋅u /c)
(as in Newtonian theory; the added energy is entirely kinetic)
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4force transformation: (Q ≡ dE/dt)
Note: for rest masspreserving force, f is invariant among IFs with relative velocities parallel to force.
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SP 4.9
For a rest masspreserving force:
a is not generally parallel to f
A moving particle offers different inertial resistances to the same force, depending on whether it is applied longitudinally or transverse.
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Also for a rest masspreserving force:
F = m0A + dm
0/d U = m
0 d2R/d2 = (u) (f, c1 dE/dt)
=> (u) f = m0 d2r/d2
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