PHYS344 Lecture 6 Homework #1 Due in class Wednesday, Sept 9 th Read Chapters 1 and 2 of Krane, Modern Physics Problems: Chapter 2: 3, 5, 7, 8, 10, 14, 16, 17, 19, 20

Addition of Velocities Taking differentials of the Lorentz transformation [here between the rest frame (K) and the space ship frame (K ’ )], we can compute the shuttle velocity in the rest frame ( u x = dx/dt ): Suppose a shuttle takes off quickly from a space ship already traveling very fast (both in the x direction). Imagine that the space ship’s speed is v, and the shuttle’s speed relative to the space ship is u ’. What will the shuttle’s velocity ( u ) be in the rest frame? v

The Lorentz Velocity Transformations Defining velocities as: u x = dx/dt, u y = dy/dt, u ’ x = dx ’ /dt ’, etc., we find: with similar relations for u y and u z : Note the  ’s in u y and u z.

The Inverse Lorentz Velocity Transformations If we know the shuttle’s velocity in the rest frame, we can calculate it with respect to the space ship. This is the Lorentz velocity transformation for u ’ x, u ’ y, and u ’ z. This is done by switching primed and unprimed and changing v to –v :

Relativistic velocity addition Speed, u ’ 0.25c Speed, u 0.50c0.75c v = 0.75c 1.0c 0.9c 0.8c 1.1c Galilean velocity addition Relativistic velocity addition 0

v Rg = velocity of Romulans relative to galaxy v tR = velocity of torpedo relative to Romulans v Eg = velocity of Enterprise relative to galaxy v Rg = 1/2c v Eg = 3/4cv tR = 1/3c Romulans Enterprise torpedo Example: Lorentz velocity transformation Capt. Kirk decides to escape from a hostile Romulan ship at 3/4c, but the Romulans follow at 1/2c, firing a matter torpedo, whose speed relative to the Romulan ship is 1/3c. Question: does the Enterprise survive?

We need to compute the torpedo's velocity relative to the galaxy and compare that with the Enterprise's velocity relative to the galaxy. Using the Galilean transformation, we simply add the torpedo’s velocity to that of the Romulan ship: Galileo’s addition of velocities

Einstein’s addition of velocities Due to the high speeds involved, we really must relativistically add the Romulan ship’s and torpedo’s velocities: The Enterprise survives to seek out new worlds and go where no one has gone before…

Example: Addition of velocities We can use the addition formulas even when one of the velocities involved is that of light. At CERN, neutral pions (  0 ), traveling at % c, decay, emitting  rays in opposite directions. Since  rays are light, they travel at the speed of light in the pion rest frame. What will the velocities of the  rays be in our rest frame? (Simply adding speeds yields 0 and 2c !) Parallel velocities: Anti-parallel velocities:

Experimental Verification of Time Dilation Cosmic Ray Muons: Muons are produced in the upper atmosphere in collisions between ultra-high energy particles and air-molecule nuclei. But they decay (lifetime = 1.52  s) on their way to the earth’s surface: No relativistic correction With relativistic correction Top of the atmosphere Now time dilation says that muons will live longer in the earth’s frame, that is,  will increase if v is large. Their average velocity is 0.98c!

Detecting muons to see time dilation It takes 6.8 ms for the 2000-m path at 0.98c, about 4.5 times the muon lifetime. So, without time dilation, of 1000 muons, we expect only 1000 x = 45 muons at sea level. In fact, we see 542, in agreement with relativity! And how does it look to the muon? Lorentz contraction shortens the distance! Since 0.98c yields  = 5, instead of moving 600 m on average, they travel 3000 m in the Earth’s frame.

v K’K’ v K’K’ Relativistic Momentum Because physicists believe that the conservation of momentum is fundamental, we begin by considering collisions without external forces: Frank is at rest in K and throws a ball of mass m in the - y -direction. Mary (in the moving system) similarly throws a ball in system K ’ that’s moving in the x direction with velocity v with respect to system K. dp/dt = F ext = 0 K x z y u

Relativistic Momentum If we use the classical definition of momentum, the momentum of the ball thrown by Frank is entirely in the y direction: p Fy =  m u The change of y -momentum as observed by Frank is:  p Fy = 2 m u Mary measures the initial velocity of her own ball to be: u ’ Mx = 0 and u ’ My = u. In order to determine the velocity of Mary’s ball, as measured by Frank, we use the relativistic velocity transformation equations: v K’K’ K x z y

Relativistic Momentum Before the collision, the momentum of Mary’s ball, as measured by Frank, becomes: Before For a perfectly elastic collision, the momentum after the collision is: After The change in y -momentum of Mary’s ball according to Frank is: v K’K’ K x z y whose magnitude is different from that of his ball:  p Fy = 2 m u

The conservation of linear momentum requires the total change in momentum of the collision, Δp F + Δp M, to be zero. The addition of these y -momenta is clearly not 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 special relativity. There is no problem with the x direction, but there is a problem with along the direction the ball is thrown in each system, the y direction. Relativistic Momentum v K’K’ K

Rather than abandon the conservation of linear momentum, we can make a modification of the definition of linear momentum that preserves both it and Newton’s second law. Relativistic Momentum v K’K’ K where: Important: note that we’re using  in this formula, but the v in  is really the velocity of the object, not necessarily that of its frame. To do so requires re-examining momentum to conclude that:

Does this modification work? v K’K’ K The initial y-momentum of Mary’s ball is now: where u M is the speed of Mary’s ball in K: The initial y-momentum of Frank’s ball is now: so: from the relativistic velocity addition equations after some simplification which perfectly cancels the y- momentum of Frank’s ball

Relativistic momentum

Some physicists like to refer to the mass as the rest mass m 0 and call the term m =  m 0 the relativistic mass. In this manner the classical form of momentum, m, is retained. The mass is then imagined to increase at high speeds. Most 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. Although we may use the terms mass and rest mass synonymously, we will not use the term relativistic mass. At high velocity, does the mass increase or just the momentum?