In Newtonian mechanics, the total momentum of a system is a conserved quantity. Further, as we’ve seen, the law of conservation of momentum, $P_{\text{f}}=P_{\text{i}}$, is true in all inertial reference frames if the particle velocities in different reference frames are related by the Galilean velocity transformations.

The difficulty, of course, is that the Galilean transformations are not consistent with the principle of relativity. It is a reasonable approximation when all velocities are very much less than c, but the Galilean transformations fail dramatically as velocities approach c. We’ll leave it as a homework problem to show that $P_{\text{f}}^{\prime }\ne P_{\text{i}}^{\prime }$ if the particle velocities in frame $\text{S}\prime$ are related to the particle velocities in frame S by the Lorentz transformations.

There are two possibilities:

1. The so-called law of conservation of momentum is not really a law of physics. It is approximately true at low velocities but fails as velocities approach the speed of light.

2. The law of conservation of momentum really is a law of physics, but the expression $p=mu$ is not the correct way to calculate momentum when the particle velocity u becomes a significant fraction of c.

Momentum conservation is such a central and important feature of mechanics that it seems unlikely to fail in relativity.

The classical momentum, for one-dimensional motion, is $p=mu=m(\text{Δ}x/\text{Δ}t)$. $\text{Δ}t$ is the time to move distance $\text{Δ}x$. That seemed clear enough within a Newtonian framework, but now we’ve learned that experimenters in different reference frames disagree about the amount of time needed. So whose $\text{Δ}t$ should we use?

One possibility is to use the time measured by the particle. This is the proper time $\text{Δ}\tau$ because the particle is at rest in its own reference frame and needs only one clock. With this in mind, let’s redefine the momentum of a particle of mass m moving with velocity $u=\text{Δ}x/\text{Δ}t$ to be

We can relate this new expression for p to the familiar Newtonian expression by using the time-dilation result $\text{Δ}\tau ={(1-u^2/c^2)}^{1/2}\text{Δ}t$ to relate the proper time interval measured by the particle to the more practical time interval $\text{Δ}t$ measured by experimenters in frame S. With this substitution, Equation 36.32 becomes

You can see that Equation 36.33 reduces to the classical expression $p=mu$ when the particle’s speed $u\ll c$. That is an important requirement, but whether this is the “correct” expression for p depends on whether the total momentum P is conserved when the velocities of a system of particles are transformed with the Lorentz velocity transformation equations. The proof is rather long and tedious, so we will assert, without actual proof, that the momentum defined in Equation 36.33 does, indeed, transform correctly. The law of conservation of momentum is still valid in all inertial reference frames if the momentum of each particle is calculated with Equation 36.33.

The factor that multiplies mu in Equation 36.33 looks much like the factor $\gamma$ in the Lorentz transformation equations for x and t, but there’s one very important difference. The v in the Lorentz transformation equations is the velocity of a reference frame. The u in Equation 36.33 is the velocity of a particle moving in a reference frame.

With this distinction in mind, let’s define the quantity

where the subscript p indicates that this is $\gamma$ for a particle, not for a reference frame. In frame $\text{S}\prime$, where the particle moves with velocity $u\prime$, the corresponding expression would be called ${\gamma }_{\text{P}}^{\prime }$. With this definition of ${\gamma }_{\text{p}}$, the momentum of a particle is