The momentum of a particle is defined as the vector |p> = ∂L/∂|v> or <p| = <∇|L. In 3d, don't worry too much about a vector being written as a bra or as a ket. Focus on how each component is calculated: for example, pₓ=∂L/∂x.
For a free particle, since L=-mc²·√, we get
|p> = m|v> / √.
For small velocities, √ → 1 and |p> ≃ m|v>, the classical result. For v → c, the √ → 0 and momentum tends to infinity, suggesting that we cannot a finite mass travelling at this speed.
The force suffered by a particle is the total time derivative of its momentum. Suppose, as a particular case, that the velocity of the particle only changes in its direction, which happens when the suffered force is perpendicular to its velocity. Then, the total time derivative does not act upon √, since only the magnitude appears there. We can write for this case
|ṗ> = (m/√)·|v̇>.
The opposite particular case is when velocity only changes in magnitude, which occurs when the force acts parallel to the velocity. In this case, the time derivative only acts upon the magnitude v. We can write then
|ṗ> = (m/√)·(v̇ + (v²/c²)/√)·|^v^>.
The energy ℰ of a particle is defined as ℰ = <p|v> - L. Here, momentum shows its role as a form instead of a vector.
For a free particle, and using the momentum we got for the same free particle, we obtain
ℰ = m·c² / √ .
For v=0 we don't get zero energy, as in classical mechanics. Instead, we get ℰ = m·c². This is called the rest energy of the particle.
For small velocities, √ ≃ 1 + v²/(2c²), so ℰ ≃ mc² + m·v²/2.
These formulas can be applied not only to elementary particles, but also to any composite body made of many particles. We just focus all its mass as m and consider |v> the velocity of its centre of mass. We, then, ignore any rotation effects.
In classical mechanics, the energy of a body is free to be redefined by adding or subtracting any constant, so it can be either positive or negative. However, in relativistic mechanics we don't have this freedom. Here, the energy of a body is always a positive quantity, and it is completely defined, without allowance for adding or subtracting constants.
If a body consists in many particles and if the body as a whole is at rest, then the total rest energy of the body is ℰ = mc², but this is NOT equal to ∑mₐc² (mₐ being the mass of each component). So the rest energy of the body is not equal to the sum of rest energies of the constituent particles. To recover the rest energy of the whole body we must take into account the kinetic energy of each individual particle (they can move even though the body can be at rest as a whole) and also the interaction energies between them.
As we consider mass as the rest energy (up to a constant c²), from this discussion we see that in relativistic mechanics the mass is not conserved. The mass of a composite body is not equal to the masses of the constituent parts.
The conservation of total energy is still valid.
For a free particle, if we square our expressions for momentum and energy, we arrive to the important relation
ℰ² = (p·c)² + (m·c²)².
We can isolate the energy and it will be the energy with momentum as variable: this is what we call the Hamiltonian function ℋ :
ℋ = ( (p·c)² + (m·c²)² )¹⸍² = m·c²·( 1 + (p/(mc))² )¹⸍².
For low velocities (p << mc) we can use the latter expression to expand
ℋ ≃ mc² · (1 + p²/(2m²c²) = mc² + p²/(2m).
We identify here the classical term p²/(2m), while mc² is a constant not usually considered in classical mechanics (since there we can consider constants as arbitrary and ignore them).
Comparing again the expressions for the momentum and the energy of a free particle, we can arrive to another interesting expression,
c|p> = ℰ |v/c> .
It is interesting because here the mass does not appear. In the individual relations of momentum and energy, mass appears, so that for v = c we get infinite momentum and infinite energy, telling us that a particle with finite mass cannot travel at the speed of light. However, in the latter expression, mass is not explicit there, and this expression, if we make the limit v → c, we get c|p> = ℰ |^v^>, which is a perfectly valid expression. Squaring this expression we get (p·c)² = ℰ², which means that m=0.
Conclusion: in relativistic mechanics, particles with zero mass can travel at the speed of light, with a relation p·c=ℰ.
There is another limit that we can sometimes take. It is called the *ultrarelativistic* limit, and it consists in considering the rest mass of the particle small wrt the particle's energy ℰ. We can see that this is equivalent to approximate its mass by 0, and then we also get ℰ ≃ p·c, although here the expression is not exact as before.
Up to now we have been using 3d vectors and classical notation. Let's move now to energy and momentum in 4d.
The principle of least action says that δS = 0, so for a free particle
δS = -m·c·δ∫ds {s;a;b} = 0.
Recall that ds = (dxᵢdxⁱ)¹⸍². Now become convinced that we can write this:
δds = δ(dxᵢdxⁱ)¹⸍² = δ(dxᵢdxⁱ) / (2ds) = (dxⁱδdxᵢ+ dxᵢδdxⁱ) / (2ds) = = (dxⁱδdxᵢ)/ds = (dxᵢδdxⁱ)/dx.
This, added to dxᵢ/ds = uᵢ, and that δdxⁱ=dδxⁱ, makes
δS = -mc∫uᵢdδxⁱ {δxⁱ;a;b}
We integrate by parts, taking U=uᵢ and dV=dδxⁱ, so that dU=duᵢ=(duᵢ/ds)·ds and V=δxⁱ, and we apply U·dV = d(U·V)-V·dU, so
δS = -mcuᵢδxⁱ|{a;b} + mc∫δxⁱ(duᵢ/ds)·ds{a;b}.
About the notation here: | means that we have integrated and we still need to apply the values from a to b. For the integral, we also need to apply these limits.
By definition of our protocol, δxⁱ is 0 at a and b, so the first term on the RHS vanishes. We write then what remains and apply that δS=0,
δS = mc∫δxⁱ(duᵢ/ds)·ds{a;b} = 0.
This is true only for duᵢ/ds = 0. This means that our path consists in keeping the four velocity constant, as expected for a free particle.
Here we can perform a trick: to leave the initial point a fixed as before but allowing the final point b to be variable. Then we don't completely kill the term -mcuᵢδxⁱ|{a;b} but only kill the "a" part and we leave the b part as simply -mcuᵢδxⁱ, since now point b is a generic end point.
We force δS=0 again and obtain
δS = -mcuᵢδxⁱ + mc∫δxⁱ(duᵢ/ds)·ds{a;b} = 0.
But the real trick comes now: we only accept classical true trajectories, so that the 2nd term on the right, the integral, has to be 0. We then get
δS = -mcuᵢδxⁱ.
It seems a nasty trick but it is completely valid: the trajectory must have δS=0, so the 2nd term of the right must be zero. But later, we say, OK, if we only allow valid trajectories, we can play by freeing the final point and see which value of δS corresponds to that.
Now we see that mcuᵢ = (mc/√, m<v|/√) = (ℰ/c,<p|) = pᵢ, so we write
δS = -pᵢδxⁱ.
We can substitute δ by ∂ and get
∂S/∂xⁱ = -pᵢ.
Although I prefer
pᵢ = -∂ᵢS.
We call pᵢ the momentum four-vector. From mechanics, we can split this object in spatial and temporal part as
<p| = -<∇|S ℰ = -∂ₜS,
where ∂ₜ=∂/∂t and where
pᵢ = (ℰ/c, -<p|).
For contravariant components, pⁱ = (ℰ/c, |p>) = mcuⁱ. Use contravariant components if you want to place the positive version of the spatial part.
Conclusion: energy and momentum are the components of a single 4-v in relativistic mechanics. And, if they are a 4-v, their Lorentz transf are trivial. For a boost u in x, we have p₂=p'₂ and p₃=p'₃, and (being b=u/c),
p₁ = (p'₁ + b·p'₀)/√ p₀ = (p'₀ + b·p'₁)/√
where <p| = (p₁,p₂,p₃) and ℰ/c = p₀.
The invariant magnitude of this 4-v is clearly pᵢpⁱ=(m·c)².
We can *define* a four-vector for the *force* fⁱ as
fⁱ = dpⁱ/ds = m·c·duⁱ/ds.
We see how this 4-v is "perpendicular" to the four-velocity, since gᵢuⁱ = mc(duᵢ/ds)uⁱ=0. Remember that uⁱ and its derivative duᵢ/ds are "perpendicular".
From classical mechanics we know |f>=|ṗ>, so we can write the four-force in more familiar terms as
fⁱ = ( <f|v> /(c²√) , |f> / (c√) ).
The 0 (time) component is not as familiar as the spatial part. But remember that in classical mechanics, the infinitesimal work done by a force <f| is dW = <f|dr> = <f|v>dt. Then, the power developed is P = dW/dt = <f|v>. In our 0 component of fⁱ, we also have a term that is related to the power developed by the force.
From classical mechanics, we know that the Hamilton-Jacobi (HJ) equation is obtained by replacing the momentum in the Hamiltonian by its equivalent ∂S/∂xⁱ. So by going from pᵢ=∂ᵢS, we go from the 1st expression to the second:
pᵢ pⁱ = (m·c)² ∂ᵢS∂ⁱS = (m·c)² = gⁱᵏ∂ᵢS∂ₖS.
We can develop the last expression with the explicit four components as
(∂S/(c∂t))² - (∂S/∂x)² - (∂S/∂y)² - (∂S/∂z)² = m²c².