4-vectors: Four components that Lorentz transform like ct, x, y, z. Answer c Time is not invariant. Proper time is.

The line is a worldline for a particle that is (in this frame) A] stationary B] in constant acceleration Ans a.

The magnitude of the 4-velocity of this particle (in this frame) is A] zero B] positive C] negative

The 4-velocity in this frame is (c, 0, 0, 0) The magnitude of this (its “length”) is c.

In a different frame, this particle is moving. In the primed frame shown, is the particle moving in the [A] +x’ or [B]-x’ direction?

In a different frame, this particle is moving. In the primed frame shown, is the particle moving in the The frame is moving toward +x; IN THIS FRAME, the particle is moving toward -x.

In this primed frame, what is the magnitude (i. e In this primed frame, what is the magnitude (i.e. length) of the particle’s 4-velocity? A] larger than c, because now it has an x-component B] smaller than c, because now it has an x-component C] c Answer c. Invariant.

The magnitude of a particle’s 4-velocity is invariant The magnitude of a particle’s 4-velocity is invariant. Moving to a different frame “mixes up the components”, just as a cartesian rotation mixes up components.

In the primed frame, the size of the x-component of the 4-velocity increased. What happens to the size of the ct component (compared to the particle’s rest frame)? A] it decreases to “compensate” B] it increases to “compensate” C] it is unchanged

In the primed frame, the size of the x-component of the 4-velocity increased. What happens to the size of the ct component (compared to the particle’s rest frame)? B] it increases to “compensate” Because of the minus sign in our length formula, d(ct’)/d must be larger than c. In other words, t’ is larger than  … no surprise, really, since the proper time is the slowest clock.

Just as we sometimes draw velocity vectors on a plot of object positions, we can draw 4-velocities on a spacetime diagram. What is the direction of the 4-velocity vector for the particle we have been discussing, on the diagram shown? A] straight down B] straight up C] parallel to the ct axis for the unprimed frame, but parallel to ct’ for the primed frame

“A worldline is a worldline is a worldline” - Gertrude Stein (maybe?) It’s the same worldline, so a small piece of it, divided by an invariant number, is the same 4-vector for all observers: straight up in this diagram.

Since the length of a 4-vector is invariant, the length of your 4-velocity vector is the same regardless of how fast you move! (That makes sense… there is always a frame where you are stationary!) If you are moving, both the time and the space components increase.

Which 4-vector is 4-longest? Or choose E] they are the same length

All have same 4-length

4-momentum or the “energy-momentum 4-vector” is the rest mass x the 4-velocity. The gammas come from dt/d

4-momentum We showed that mvx is x-momentum. The “time” component of the energy-momentum 4-vector (when multiplied by c) gives a constant plus the kinetic energy. Since momentum is conserved, Pt must also be conserved.

Remember: “invariant” = same for all observers “conserved” = unchanging over time What is true? Any conserved quantity is also invariant Any invariant quantity is also conserved Neither is necessarily true

Neither is true. These are different ideas. ========== An object at rest has an energy-momentum 4-vector that looks like this: (mc, 0, 0, 0) Where m is its mass. If we collide two lumps of clay moving in opposite directions, the final lump has energy-momentum 4-vector of (2imc, 0, 0, 0)

If we collide two lumps of clay moving in opposite directions, the final lump has energy-momentum 4-vector of (2imc, 0, 0, 0) This is the energy-momentum 4-vector of a stationary object of mass 2im > 2m. We have CREATED NEW MASS! MASS IS NOT CONSERVED IN RELATIVITY. (But Pt is!)

The amount of new mass is just the kinetic energy that “disappeared” when the clay lumps stuck together (divided by c2) The kinetic energy did not disappear, of course, it went into heat, i.e. internal energy of the lump. So the internal energy of a system gives it inertia, i.e. mass. What if there is negative internal energy? Bound particles have negative energy, compared to free particles. In that case, the mass is LESS than the mass of the free particles. In nuclei, this is called the mass defect… the difference in mass between the nucleus and the masses of the separated neutrons and protons.

Which has more mass, a hydrogen atom or an electron plus a proton? Electron + proton This isn’t a nucleus, so they have exactly the same mass

Which has more mass, a hydrogen atom or an electron plus a proton? Electron + proton This isn’t a nucleus, so they have exactly the same mass Answer b. More energy. (Binding energy is negative!)

Hydrogen is lighter than a proton + an electron, but the difference is very small! 13.7 eV/c2 is tiny. What do 4-vectors look like in a different frame? Their lengths4 are invariant, but their components get mixed up via the Lorentz transformation. All 4-vectors Lorentz transform. You can just substitute Pt for ct, Px for x, Py for y, Pz for z.

Recall: all these have the same length4 Recall: all these have the same length4. (They are displacement 4-vectors. But I could relabel the vertical axis Pt and the horizontal axis Px and they are still the same length4.)

What is the length4 of the energy-momentum 4-vector What is the length4 of the energy-momentum 4-vector? A) 0 B) mc C) mc2 D) it depends on the reference frame

What is the length4 of the energy-momentum 4-vector What is the length4 of the energy-momentum 4-vector? A) 0 B) mc C) mc2 D) it depends on the reference frame Answer mc

Length4 = mc. That’s the length in the particle’s rest frame, so that’s the length in all frames. If the particle is moving, both the first and second components increase.

Which graph(s) could correctly show the energy-momentum 4-vectors for a particle with the worldline shown? Or choose (e) more than one could be correct

Answer A.

(It’s sort of a shell in the same way the light cone is a cone.) The tip of the energy-momentum 4-vector for a given particle lies on a hyperbola, which is called the “mass shell” (It’s sort of a shell in the same way the light cone is a cone.) Pythagorean triples. The “long side” is always Pt = mc = E/c Another is 20,21,29

The invariant length4 gives us a universal equation relating energy, mass, and momentum: (mc2)2 + (cP)2 = E2 If a particle has zero mass, cP = E. If cP = E, then the world line is at 45° - massless particles MUST travel at speed c.

Pythagorean triples let you solve some relativity problems VERY EASILY. In one reference frame, two events are separated by 5 seconds in time and 3 light-seconds in space. In the frame where these events occur in the same place, how much time separates them? 3 seconds 4 seconds 5 seconds 8 seconds Answer 4 seconds

If particle p1 moves west, what direction does p2 move? The invariant length4 gives us a universal equation relating energy, mass, and momentum: (mc2)2 + (cP)2 = E2 In relativity, it is easiest to measure mass and momentum in units of energy. Mass is MeV/c2, momentum is MeV/c. Some clickers: Unstable particle p0 at rest decays into to identical particles p1 and p2. If particle p1 moves west, what direction does p2 move? Up b) East c) South d) west e) not enough info

Unstable particle p0 at rest decays into to identical particles p1 and p2. If particle p1 moves west, what direction does p2 move? Up b) East c) South d) west e) not enough info East. Momentum is conserved. If particle p1 moves with speed v, what is the speed of p2? a) 0 b) v c) 2v d) c-v e) not enough info

Unstable particle p0 at rest decays into to identical particles p1 and p2. If particle p1 moves with speed v, what is the speed of p2? 0 b) v c) 2v d) c-v e) not enough info Answer b) speed = v. Momentum is conserved. If particle p1 has momentum 3 MeV/c, and total energy 5 MeV, what is its rest mass? a) 4 MeV/c2 b) 8 MeV/c2 c) sqrt(34) MeV/c2

Unstable particle p0 at rest decays into to identical particles p1 and p2. If particle p1 has momentum 3 MeV/c, and total energy 5 MeV, what is its rest mass? 4 MeV/c2 b) 8 MeV/c2 c) sqrt(34) MeV/c2 Answer a. What is the rest mass of particle p0? a) 6 MeV/c2 b) 7 MeV/c2 c) 8 MeV/c2 d) 10 MeV/c2

Unstable particle p0 at rest decays into to identical particles p1 and p2. If particle p1 has momentum 4 MeV/c, and total energy 5 MeV, what is its rest mass? 3 MeV/c2 b) 9 MeV/c2 c) sqrt(41) MeV/c2 Answer a. What is the rest mass of particle p0? 6 MeV/c2 b) 7 MeV/c2 c) 8 MeV/c2 d) 10 MeV/c2 Answer d. Mass-energy is conserved (even though mass is not!)

***** Pythagorean triples ( 3 , 4 , 5 ) ( 5, 12, 13) ( 7, 24, 25) ( 8, 15, 17) ( 9, 40, 41) (11, 60, 61) (12, 35, 37) (13, 84, 85) (16, 63, 65) (20, 21, 29) (28, 45, 53) (33, 56, 65) (36, 77, 85) (39, 80, 89) (48, 55, 73) (65, 72, 97) Unstable particle p0 at rest decays into to identical particles p1 and p2. If particle p1 has rest mass 20 MeV/c2, and momentum 21 MeV/c, what is its total energy? 1 MeV b) √41 MeV c) 29 MeV

What is the rest mass of p0? a) 40 MeV/c2 b) 42 MeV/c2 c) 58 MeV/c2 ***** Pythagorean triples ( 3 , 4 , 5 ) ( 5, 12, 13) ( 7, 24, 25) ( 8, 15, 17) ( 9, 40, 41) (11, 60, 61) (12, 35, 37) (13, 84, 85) (16, 63, 65) (20, 21, 29) (28, 45, 53) (33, 56, 65) (36, 77, 85) (39, 80, 89) (48, 55, 73) (65, 72, 97) Unstable particle p0 at rest decays into to identical particles p1 and p2. If particle p1 has rest mass 20 MeV/c2, and momentum 21 MeV/c, what is its total energy? 1 MeV b) √41 MeV c) 29 MeV Answer c. E = 29 MeV. What is the rest mass of p0? a) 40 MeV/c2 b) 42 MeV/c2 c) 58 MeV/c2

Unstable particle p0 at rest decays into to identical particles p1 and p2. If particle p1 has rest mass 20 MeV/c2, and momentum 21 MeV/c, what is its total energy? 1 MeV b) √41 MeV c) 29 MeV Answer c. E = 29 MeV. What is the rest mass of p0? a) 40 MeV/c2 b) 42 MeV/c2 c) 58 MeV/c2 Answer c. Mass-energy is conserved.

A different unstable particle p0 at rest decays into to identical particles p1 and p2. If particle p1 has momentum 3 MeV/c, and total energy 5 MeV, what is its speed/c ? a) 1/5 b) 3/5 c) 9/25 d) 5/3

Unstable particle p0 at rest decays into to identical particles p1 and p2. If particle p1 has momentum 3 MeV/c, and total energy 5 MeV, what is its speed/c ? 1/5 b) 3/5 c) 9/25 d) 5/3 Answer: b. What is the energy and momentum of particle 2 in a frame where particle 1 is at rest? You can do this by velocity addition.

The two particles are each moving at speed 3/5 c in opposite directions. What is the speed of particle 1 seen by particle 2? (as a fraction of c?) a) 3/5 b) 16/25 c) 30/34 d) 6/5

The two particles are each moving at speed 3/5 c in opposite directions. What is the speed of particle 1 seen by particle 2? (as a fraction of c?) Use v = (v1 + v2)/(1 + v1*v2) to find v = 30/34.