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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 p~ = m~v / 1 − v 2 /c2 E = mc2 / 1 − v 2 /c2 E = mc2 1 − v 2 /c2 −1/2 1 E ≈ mc2 + mv 2 + O(mv 4 /c2 ) 2 Newton returns as v/c → 0 q p~ = m~v 1 − v 2 /c2 −1/2 m~v + O(mv 3 /c2 ) 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 (∆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. (E, p~ ) = If (E, p~ ) is a 4-vector, we should probably think about its invariant product with itself. (E, p~ ) · (E, p~ ) = E 2 − p~ 2 = (mγ)2 − (mγ~v )2 = m2 γ 2 (1 − v 2 ) = m2 Ahah! This is the significance of the mass - it is the invariant combination of E and p~. Important relation #1 E 2 − p~ 2 = m2 In dumb units this is E 2 − p~ 2 c2 = m2 c4 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. mγ~v p~ = = ~v E mγ or in dumn units ~v = p~ c2 E This is the other fundamental between energy, momentum mass and velocity. These two together are the key to understanding relativistic particles. E 2 − p~ 2 = m2 ~v = p~/E We can go backwards — if m 6= 0 you can solve for E and p~. E 2 − p~ 2 = m2 What about m = 0? ~v = p~/E E 2 − 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 2 dE d 2 d~p E − p~ 2 = m = 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. E 2 − p~ 2 = m2 ~v = p~/E Massless particles travel at the speed of light. What about massive particles? E= q p~ 2 + m2 ~v = √ p~ <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! E 2 − p~ 2 = m2 ~v = p~/E Massless particles travel at the speed of light. What about massive particles? E= q p~ 2 + m2 ~v = √ p~ <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/m for small t m/|F~ | ~v = q → F̂ for large t |m/|F ~| F~ 2 t2 + m2 F~ t E 2 − p~ 2 = m2 ~v = p~/E Massless particles travel at the speed of light. What about massive particles? E= q p~ 2 + m2 ~v = √ p~ <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 E−m d= = F √ F t2 for small t m/|F ~| F 2 t2 + m2 − m → 2m 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 small v: m~v = m~v1 + m~v2 and m 2 ~v 2 = m 2 ~v 2 1 + m 2 ~v 2 2 ⇒ ~v1 · ~v2 large v: mγ~v = mγ1~v1 + mγ2~v2 and energy conservation looks like this mγ + m = mγ1 + mγ2 In this case, we could define KE as E − m which goes to ≈ we could write conservation as m 2 v 2 for small v, and (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