Four Vectors

PHYS2100
Four Velocity, Momentum and Energy
How can we express a velocity in Special Relativity (SR)?
If you try V^μ=DX^μ/Dt = (c,V_x,V_y,V_z) you find the norm² of V (the norm is the square root of V^μV_μ) is not invariant under a Lorentz boost. No need to work an example here, I will let you in on a great secret often exploited in SR: Since the square of a four vector (F_µF^µ) must be invariant, evaluate the thing in the rest frame! In this case we would have V_x=V_y=V_z=v=vⁱ=0 and we see V²=c² which is certainly pretty strange ... that would suggest that all velocities, even "zero" have a norm equal to the speed of light! But worse than that is the fact that if V_x is nonzero then V²=c²-V_x² so that cannot be a four vector. It does not transform right.

When we defined the velocity above, we took dX/dt where dt refers to our "stationary frame" clock. Perhaps we should refer to the moving clock, dt'=dt/γ (moving clocks run slow) then

V = V^μ=dX^μ/dt' = γdX^μ/dt = (γc,γVx,γVy,γVz) = (γc,γv)
and V^µV_µ=|V|²=c² as before in the frame with v=0 (γ=1) but in any frame we get

V^µV_μ= (gc,γv)(γc,-γv) = γ²(c²-v²) = c²γ²(1-β²) = c²

Think about it... now every frame should get the same result because all use the same clock ... the clock attached to the object. The time atached to the object is called the proper time and is usually denoted τ instead of t'. This is why you get the same value for the norm².

Now evaluate V² in the rest frame where v=vⁱ=0 and γ=1 and we get ... c² again! But now if we apply a Lorentz X-boost of B you should get ß'=(ß+B)/(1+ßB) where ß=speed in S and B is speed of S w.r.t. S'....See what you get if you give a particle two boosts.

The Four Momentum
A similar argument to the one above for velocity is applied to momentum. In fact, we just multiply the four-velocity by mass ("rest mass") to get the four-momentum, or energy momentum four-vector. In short,

P^μ = (mγc,mγv) where as usual v=vⁱ = (v_x,v_y,v_z) is the "usual" 3-velocity.

The spacelike part mγv reduces to the "usual" definition of momentum

p = pⁱ = mv = mvⁱ
when v² is small. But we define momentum to be p=mγv. If you like, call this p the relativistic momentum but never confuse this with the momentum 4-vector or 4-momentum or energy-momentum four-vector. Remember, you are supposed to know what's happening!

Next, define the timelike part mγc=E/c that is, define the relativistic energy E=mγc².

Recall γ=(1-β²)^½~1+½β²+... for small β²=v²/c².
Then the relativistic energy becomes E=mc²+½mv²+...

The ½mv² you recognise as the kinetic energy K in nonrelativistic mechanics. We should really define K=E-mc².

The mc² you recognise as Einstein's E=mc², and it is! mc² is the rest mass energy.

Energy-momentum relations

We have E=mγc² and p=mγv (p and v are magnitudes)

the ratio p/E=v/c² or p=Ev/c². For photons v=c and p=E/c.

Also, E²-p²c² = m²γ²c⁴-m²γ²v²c² = m²c⁴γ²(1-v²/c²) = m²c⁴.

E² = p²c² + m²c⁴