Four-momentum - Aetherwisp

The four-momentum, or energy-momentum four-vector, of an object of mass $m$ is a four-vector that extends the concept of linear momentum to special relativity:

p^{\mu}=m\eta^{\mu}=\left(\frac{E}{c},\;\mathbf{p}\right)=(\gamma mc,\gamma m\mathbf{v})

where $\eta$ is the proper four-velocity, $E$ is the relativistic energy and $\mathbf{p}$ is the relativistic momentum, which differs from the ordinary momentum in that it uses proper velocity instead of ordinary velocity: $\mathbf{p}=m\boldsymbol{\eta}$ .

The total four-momentum of a system is a conserved quantity. This does not mean it is a relativistic invariant (it's not even a Scalar); conservation has nothing to do with frames of reference, it just means that the quantity remains the same before and after a process.

The norm $p_{\mu}p^{\mu}=-m^{2}c^{2}$ is called the center-of-mass energy. This name comes from nuclear and particle physics where it is used in the context of an $N$ -body system.

The four-momentum is related to relativistic energy by

E^{2}-p^{2}c^{2}=m^{2}c^{4}

Transverse momentum#

Four-momentum secretly contains a second relativistic invariant. To see it, consider a particle $p$ in spherical coordinates. Its components are

\begin{cases} p_{x}=p\sin\theta\cos\phi \\ p_{y}=p\sin\theta\sin\phi \\ p_{z}=p\cos\theta \end{cases}

Now, do a Lorentz transformation on the $z$ axis:

\pmatrix{\frac{E'}{c} \\ p_{x}' \\ p_{y}' \\ p_{z}'}=\pmatrix{\gamma & 0 & 0 & \beta\gamma \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ \beta\gamma & 0 & 0 & \gamma}\pmatrix{\frac{E}{c} \\ p_{x} \\ p_{y} \\ p_{z}}

which can be equivalently written as

\begin{cases} E'=\gamma E+\beta\gamma p\cos\theta \\ p'\sin\theta'\cos\phi'=p\sin\theta\cos\phi \\ p'\sin\theta'\sin\phi'=p\sin\theta\sin\phi \\ p'\cos\theta=\gamma p\cos\theta+\beta\gamma E \end{cases}

Square the middle two and sum them to find

p'^{2}\sin^{2}\theta'(\cos^{2}\phi'+\sin^{2}\phi')=p^{2}\sin^{2}\theta(\cos^{2}\phi+\sin^{2}\phi)

and using the fundamental trigonometric identity $\sin ^{2}\theta+\cos ^{2}\theta=1$ we see

p'\sin\theta'=p\sin\theta

This quantity is called the transverse momentum $p_{T}$ . As you can see, it does not change under a Lorentz transformation on the $z$ axis. This makes it another relativistic invariant. In fact, there's a third one hidden here. From the same equation, while it's trivial, we can state

\cos ^{2}\phi'+\sin ^{2}\phi'=\cos ^{2}\phi+\sin ^{2}\phi

which implies $\phi'=\phi$ . Thus, the azimuthal angle is also a relativistic invariant.

Examples#

> Of course, energy is also conserved, so $E_\text{bef}=E_\text{aft}$. But remember that the collision is inelastic: the particles get "stuck" together, they don't bounce[^2]. You'd expect the total mass $M$ of this object to be the the sum of individual masses, so $M=2m$, but it's not! Since the velocity of this new object is *zero* (momenta cancel, leading to standstill), the only energy it has left is rest energy $E_{0}=Mc^{2}$, which means > $$E_\text{bef}=E_\text{end}\quad\Rightarrow \quad \frac{5}{2}mc^{2}=Mc^{2}\quad\Rightarrow \quad M=\frac{5}{2}m>2m

Evidently, we just "created" mass by transferring kinetic energy to mass energy. This is how we can make heavier particles from smaller, more stable ones like electrons and protons: we just need high enough kinetic energies and a strong enough collision.

Massless particles

Some particles have no mass, yet they still possess some energy when moving. In classical mechanics this is nonsense, as you wouldn't be able to apply a force ( $\mathbf{F}=m\mathbf{a}$ ) to a massless particle ( $m=0$ ), thus making it inconsequential in physical theory. But massless particles certainly do exist and they can only be appreciated through a quantum viewpoint, although that's not really the point of this example. We just want to see what happens when $m=0$ . Of course, the most intuitive thing to say is that $p^{\mu}=0$ . After all, how can a massless particle have momentum? Well, you'd be wrong, for massless particles do have momentum, and as de Broglie showed it depends on the wavelength of the particle. So, what now? de Broglie "says" that $p>0$ but Einstein "says" $p=0$ . The issue here is that we haven't talked about the speed of the particle. Mathematically speaking, it is possible for $p$ to be nonzero even when $m=0$ : we just need some term that goes to infinity just as fast as $m$ goes to zero, so that their asymptotic behaviors cancel out and leave us with a finite value. We don't have much to play with, really. The only other non-constant in $\mathbf{p}$ is the speed of the particle:

> For $p>0$, we must simultaneously have $m\to0$ and $v\to c$. In this case — and only this case — does $p$ come out to be nonzero and finite. In other words, *all* massless particles *must* move at the speed of light (in the vacuum, at least). This is why "light moves at the speed of light", so to speak. I suppose you can't know this if you haven't studied quantum mechanics, but light is a [[Photon]] beam, photons are massless particles, and so photons move (and must move!) at $c$. In fact, I suppose it would be more correct to call $c$ the "speed of massless particles", of which light happens to be made of, but scientists did not know this when they first coined the term "speed of light". [^1]: The actual experiments are obviously more complicated, usually testing the validity of numerous conservation laws for [[Numero quantico|quantum numbers]]. See for instance, [[Particle scattering]]. [^2]: In practice, what happens is that a random particle process occurs, reassigning quantum numbers around while respecting conservation laws. Either way, the system ends with a different set of objects compared to what it started with.