20 Sept Relativity Lecture
 We started with the position 4vector
Xμ=(ct,x,y,z)=(ct,X)=(ct,xi)
and recall
dt=γdτ so the operator (d/dτ)=γ(d/dt) where τ is the proper time
and the proper time derivative of position, the 4velocity becomes
(d/dτ)Xμ=Vμ=(γc,γvi)     where vi=dxi/dt is the usual 3velocity.

We showed (remember a minus sign turns Vμ into Vμ)

VμVμ = |V|² = (γc,γvi)(γc,-γvi) = γ²c²-γ²c²=γ²(c²-v²)=c²     since γ²=c²/(c²-v²)
But wait, here we learn a great relativity trick: Whenever you evaluate AμAμ use the v=0 γ=1 frame... the answer is (must be) the same in any frame but the calculation is trivial! Then we have
VμVμ = (c,0)(c,-0)=c².     Howzat!

The 4momentum is just mass times the 4velocity

Pμ=mVμ = (mγc,mγvi) = (E/c, pi)
where we have defined the relativistic energy E=mγc² and momentum pi=mγvi Note that
p/E=mγvi/mγc²=v/c² so p=(v/c)(E/c) and for photons p=E/c=hν/c.
Squaring and adding gives E²=p²c²+m²c4
but it is more fun to use the 4vectors:
PμPμ = (mγc,mγvi)(mγc,-mγvi) = m²c² (use the trick),
= (E/c, pi)(E/c, -pi) = E²/c²-p² = m²c²     QED.

And another (d/dτ) gives the Minkowski force

Fμ=(d/dτ)Pμ = γ(d/dt)(mγc, mγvi) = (mγc dγ/dt, γdpi/dt) = (mγc dγ/dt, γf i)
where the spacelike component looks like the classical F=dp/dt.

Now for a shock. Recall VμVμ = c² a constant, so Vμ(d/dτ)Vμ = (c,0)(d/dτ)(c,-0) = 0
so...
VμFμ = 0, V and F are orthogonal!

Now consider the rate at which work is done, the Minkowski scalar power P=FV:

P = FμVμ = (mγc dγ/dt, -γdpi/dt) (γc,γvi) = 0
mγ²c² dγ/dt - γ² dpi/dt vi = 0
mc² dγ/dt = dpi/dt vi = Fi dot vi = power
Now integrate from t1=0 to t to get work done = kinetic energy = T:
mc²∫(dγ/dt)dt = mc²(γ2-γ1) → mc²(γ - 1)
so the relativistic energy E = mγc² = mc² + T where, expanding γ, we see
mγc² = mc² + ½mv² + (3/8)mv4/c² + (15/48)mv6/c4 + ...
or if K=½mv² the classical kinetic energy
mγc² = mc² + K + (3/4)β²K + (15/24)β4K + ...
Next we see what setting momentum before = momentum after gets us.