Topic 11 — relativity - energy and momentum — Use the fundamental relations between the mass, velocity, energy and momentum of a particle and conservation of energy and momentum to solve problems in relativistic kinematics and to simplify calculations involving space and time. Use 4-vectors and the invariant product. Explain and use the connection between the energy-momentum 4-vector and the space-time interval 4-vector. Understand and use the work-energy relation in relativistic systems. For a massive particle with velocity ~v (in sensible units)

E = mγ ~p = mγ ~v

q 2 Nutty and comes from a nutty Lagrangian −m 1 − ~r˙ — we won’t use this — But this is what is conserved! Tested quadrillions of times. Put the cs back in and see what it looks like for small v

E = mγ ~p = mγ ~v

q q E = mc2/ 1 − v2/c2 ~p = m~v/ 1 − v2/c2

 −1/2  −1/2 E = mc2 1 − v2/c2 ~p = m~v 1 − v2/c2

1 E ≈ mc2 + mv2 + O(mv4/c2) m~v + O(mv3/c2) 2

Newton returns as v/c → 0 imagine that the particle has a clock on it and consider the space-time interval between two ticks of the particle’s clock — (∆t, ∆~r ) we know because of time dilation

(∆t, ∆~r ) = γ (∆τ,~v ∆τ) = γ ∆τ (1,~v )

But this looks suspiciously like

(E, ~p ) = (mγ, mγ~v ) = mγ (1,~v ) or m (E, ~p ) = (∆t, ∆~r ) ∆τ m and ∆τ don’t depend on the frame, so this is true in any frame. That means the (E, ~p ) is a 4-vector, because it transforms just like the interval 4-vector when I go from one frame to another. If (E, ~p ) is a 4-vector, we should probably think about its invariant product with itself.

(E, ~p ) · (E, ~p ) = E2 − ~p 2 = (mγ)2 − (mγ~v )2 = m2γ2(1 − v2) = m2

Ahah! This is the significance of the mass - it is the invariant combination of E and ~p. Important relation #1

E2 − ~p 2 = m2

In dumb units this is E2 − ~p 2c2 = m2c4 Every fundamental object we have ever seen carries energy and momentum that bundled into something that satisfies this for some particular real m (so that m2 ≥ 0) that is characteristic of the particle. All electrons have the mass of the electron. All protons have the mass of a proton. Etc. The nice thing about this relation is that it makes sense even if m = 0. This is important. There is another relation that make sense when m− > 0.

~p mγ~v = = ~v E mγ or in dumn units ~pc2 ~v = E This is the other fundamental between energy, momentum mass and velocity. These two together are the key to understanding relativistic particles. E2 − ~p 2 = m2 ~v = ~p/E We can go backwards — if m 6= 0 you can solve for E and ~p. E2 − ~p 2 = m2 ~v = ~p/E What about m = 0? E2 − ~p 2 = m2 ~v = ~p/E

What happens to work-energy — suppose we apply a force F~ so that (just because there is a Lagrangian) d~p = F~ dt because the mass of a particle doesn’t change

d   d   dE d~p E2 − ~p 2 = m2 = 0 = 2E − 2~p · dt dt dt dt dE ~p d~p ⇒ = · = ~v · F~ dt E dt Power fed in is the rate of change of the energy — work-energy holds! This means for example that if you apply a force to a massless particle in the direction it is moving (what ever this means — it is not so easy to understand how to do this) the energy increases even though it continues to move at the speed of light. E2 − ~p 2 = m2 ~v = ~p/E Massless particles travel at the speed of light. What about massive particles?

q ~p E = ~p 2 + m2 ~v = √ < 1 ~p 2 + m2 Even if you keep feeding energy into a massive particle, you can never get it moving faster than the speed of light! E2 − ~p 2 = m2 ~v = ~p/E Massless particles travel at the speed of light. What about massive particles?

q ~p E = ~p 2 + m2 ~v = √ < 1 ~p 2 + m2

Example: Particle at rest at the origin at t = 0 subject to a constant force F~ — ~p = F~ t — how fast is it going at time t?

 F~ t  F~ t/m for small t  m/|F~ | ~v = q → F~ 2t2 + m2  Fˆ for large t  |m/|F~ | E2 − ~p 2 = m2 ~v = ~p/E Massless particles travel at the speed of light. What about massive particles?

q ~p E = ~p 2 + m2 ~v = √ < 1 ~p 2 + m2

Example: Particle at rest at the origin at t = 0 subject to a constant force F~ — ~p = F~ t — how far has it gone at time t — use work/energy — ∆E = E − m = F d where F = |F~ | and d is the distance traveled

√  F E − m F 2t2 + m2 − m  t2 for small t  m/|F~ | d = = → 2m F F  t for large t  |m/|F~ | The energy-momentum 4-vector is conserved. Elastic scattering — means something different in relativity In small v physics this means that kinetic energy is conserved. In large v physics this means that the particles are the same before and after the collision — so mass is conserved. Compare elastic scattering of a moving particle from a particle of the same mass m at rest. Initial ~v and final ~v1 and ~v2 m 2 m 2 m 2 small v: m~v = m~v1 + m~v2 and 2 ~v = 2 ~v1 + 2 ~v2 ⇒ ~v1 · ~v2 large v: mγ~v = mγ1~v1 + mγ2~v2 and energy conservation looks like this mγ + m = mγ1 + mγ2 m 2 In this case, we could define KE as E − m which goes to ≈ 2 v for small v, and we could write conservation as

(mγ − m) = (mγ1 − m) + (mγ2 − m) but this is really stupid Now there are much better ways to do the calculations using 4-vectors. Inelastic at large v means that new particles are created or destroyed! Higgs production — p + p → p + p + H