The Lorentz Velocity Transformations defining velocities as: ux = dx/dt, uy = dy/dt, u’x = dx’/dt’, etc. it is easily shown that: With similar relations for uy and uz:

The Lorentz Velocity Transformations In addition to the previous relations, the Lorentz velocity transformations for u’x, u’y , and u’z can be obtained by switching primed and unprimed and changing v to –v:

The Lorentz Velocity Transformations: an object moves with the speed of light u’x = c (light or, if neutrinos are massless, they must travel at the speed of light)

Spacetime diagrams For two events Spacetime interval lightlike lightlines Spacetime interval lightlike spacelike Causality: cause-and-effect relations Invariant: timelike Courtesy: Wikimedia Commons and John Walker

Doppler effect λ=(c-v)t/n=λ0(1-v/c) f=1/T=1/(λ/c)=f0/(1-v/c) source observer Light wave proper period seen in the source frame T0 (f0=1/ T0) v period in the observer’s frame T=T0 light, x=ct=n source, x=vt x=(c-v)t=nλ’ Astronomy: redshift

Another example with Doppler effect: laser cooling of atoms Example: DOPPLER EFFECT IN Fast ion beam precision laser spectroscopy IN collinear and anti-collinear Geometries referenced TO a frequency comb Another example with Doppler effect: laser cooling of atoms

Exclusion of relativistic frequency shifts by combining collinear and anticollinear measurements Frequency of light perceived by a moving ion   Laser tuned to resonance; the perceived frequency equals the transition frequency Thus, the resonance frequencies for collinear geometry and for anticollinear geometry To obtain the transition frequency we take the product and this is an exact relativistic formula!

2.6 #31

2.6 #32 Solution:

Twin Paradox (Courtesy: Wikimedia Commons)

2.7: Experimental Verification Time Dilation and Muon Decay Figure 2.18: The number of muons detected with speeds near 0.98c is much different (a) on top of a mountain than (b) at sea level, because of the muon’s decay. The experimental result agrees with our time dilation equation.

v=0.98c

Atomic Clock Measurement Figure 2.20: Two airplanes took off (at different times) from Washington, D.C., where the U.S. Naval Observatory is located. The airplanes traveled east and west around Earth as it rotated. Atomic clocks on the airplanes were compared with similar clocks kept at the observatory to show that the moving clocks in the airplanes ran differently.

Flight time (41.2 h) (48.6 h) The time is changing in the moving frame, but the calculations must also take into account corrections due to general relativity (Einstein). Analysis shows that the special theory of relativity is verified within the experimental uncertainties.

2.11: Relativistic Momentum Because physicists believe that the conservation of momentum is fundamental, we begin by considering collisions where there do not exist external forces and dP/dt = Fext = 0

Relativistic Momentum Frank (fixed or stationary system) is at rest in system K holding a ball of mass m. Mary (moving system) holds a similar ball in system K that is moving in the x direction with velocity v with respect to system K. Add paragraph from figure 2.29

Relativistic Momentum If we use the definition of momentum, the momentum of the ball thrown by Frank is entirely in the y direction: pFy = mu0 The change of momentum as observed by Frank is ΔpF = ΔpFy = −2mu0

According to Mary (the Moving frame) Mary measures the initial velocity of her own ball to be u’Mx = 0 and u’My = −u0. In order to determine the velocity of Mary’s ball as measured by Frank we use the velocity transformation equations: (we used velocity summation formula with ux=0)

Relativistic Momentum Before the collision, the momentum of Mary’s ball as measured by Frank (the Fixed frame) becomes Before For a perfectly elastic collision, the momentum after the collision is After The change in momentum of Mary’s ball according to Frank is (2.42) Spaced a bit more evenly (2.43) (2.44)

Relativistic Momentum (con’t) The conservation of linear momentum requires the total change in momentum of the collision, ΔpF + ΔpM, to be zero. The addition of Equations (2.40) and (2.44) clearly does not give zero. Linear momentum is not conserved if we use the conventions for momentum from classical physics even if we use the velocity transformation equations from the special theory of relativity. There is no problem with the x direction, but there is a problem with the y direction along the direction the ball is thrown in each system. ΔpF = ΔpFy = −2mu0

Relativistic Momentum Rather than abandon the conservation of linear momentum, let us look for a modification of the definition of linear momentum that preserves both it and Newton’s second law. To do so requires reexamining mass to conclude that: Relativistic momentum (2.48)

With modified (relativistic) momentum Now ΔpF + ΔpM =0 and momentum conserved!

Relativistic Momentum: two points of view physicists like to refer to the mass in Equation (2.48) as the rest mass m0 and call the term m = γm0 the relativistic mass. In this manner the classical form of momentum, p=mu, is retained. The mass is then imagined to increase at high speeds. other physicists prefer to keep the concept of mass as an invariant, intrinsic property of an object. We adopt this latter approach and will use the term mass exclusively to mean rest mass.

Behavior of relativistic momentum and classical momentum for v/c->1

2.11 #60 (a)

Homework (will be not graded): Problems 2.5. #20,21,23,27

Homework 2.6 #31, #32 2.7 #37 2.11 #60

Thank you for your attention!