Download Topic 11 — relativity - energy and momentum — Use the

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