Proper velocity and momentum YouTube

Proper velocity

Figure 5.2 - World-line of a particle

Figure 5.2 - World-line of a particle

Consider the world-line of a particle through space-time, with coordinates $X^\mu$.

If two points, $P,Q$, are close to each other on the curve, then we can define a small displacement four-vector, $\Delta X^\mu$ between them.

Thus, if $P$ has coordinates $X^\mu$ then $Q$ has coordinates $X^\mu + \Delta X^\mu$.

Since world-lines are time-like curves - they have to be to be the paths of massive objects in space-time - then so small displacements along the curve must also be time-like.

Hence, the proper length of $\Delta X^\mu$ is in fact the proper time, $$\Delta \tau = \sqrt{\Delta X_\mu \Delta X^\mu}$$

Since $\Delta \tau$ is invariant, the set of quantities $$\frac{\Delta X^\mu}{\Delta \tau}$$ is a four-vector, and we define the proper velocity, $U^\mu$, to be the limit of these quantities where $\Delta \tau \rightarrow 0$. That is, $$U^\mu = \frac{\mathrm{d} X^\mu}{\mathrm{d} \tau}$$

We can calculate the proper velocity in terms of Newtonian velocity, $$\frac{\mathrm{d} \vec{x}}{\mathrm{d} t} = \left(\frac{\mathrm{d} x}{\mathrm{d} t}, \frac{\mathrm{d} y}{\mathrm{d} t}, \frac{\mathrm{d} z}{\mathrm{d} t}\right) = \left(v_x, v_y, v_z\right)$$

Then, for some path $X^\mu \rightarrow (t,x,y,z)$, we can write $$U^\mu = \frac{\mathrm{d} X^\mu}{\mathrm{d} \tau} = \frac{\mathrm{d} X^\mu}{\mathrm{d} t} \frac{\mathrm{d} t}{\mathrm{d} \tau} \rightarrow \frac{\mathrm{d} t}{\mathrm{d} \tau} (1, v_x, v_y, v_z)$$

Now, we have already seen that $$\mathrm{d} \tau^2 = \mathrm{d} t^2 - \mathrm{d} x^2 - \mathrm{d} y^2 - \mathrm{d} z^2$$ so that $$\left(\frac{\mathrm{d} \tau}{\mathrm{d} t}\right)^2 = 1 - v_x^2 - v_y^2 - v_z^2 = 1 - v^2$$

Hence, $$\frac{\mathrm{d} t}{\mathrm{d} \tau} = \frac{1}{\sqrt{1 - v^2}}$$

Thus, $$U^\mu = \frac{\mathrm{d} X^\mu}{\mathrm{d} \tau} \rightarrow \frac{(1, v_x, v_y, v_z)}{\sqrt{1 - v^2}}$$

If we set $c \neq 1$, so that $X^\mu \rightarrow (ct, x, y, z)$, then $$U^\mu \rightarrow \frac{(c, v_x, v_y, v_z)}{\sqrt{1 - \tfrac{v^2}{c^2}}}$$ and so if $v$ is much smaller than $c$ then the spatial components $$U^m \rightarrow (v_x, v_y, v_z)$$ correspond directly to the Newtonian velocity.

Unit tangent to the world-line

If we go back to $c \equiv 1$, then from the definition of the proper velocity, $$U^\mu = \frac{\mathrm{d} X^\mu}{\mathrm{d} \tau}$$ and from figure 5.2, it is clear that, in the $U^\mu$ is tangent to the world-line at $X^\mu$.

Also, $$U_\mu U^\mu = \frac{\mathrm{d} X_\mu}{\mathrm{d} \tau} \frac{\mathrm{d} X^\mu}{\mathrm{d} \tau} = \frac{\mathrm{d} X_\mu \mathrm{d} X^\mu}{\mathrm{d} \tau^2} = \frac{\mathrm{d} \tau^2}{\mathrm{d} \tau^2} = 1$$

Hence, the proper velocity, $$U^\mu = \frac{\mathrm{d} X^\mu}{\mathrm{d} \tau}$$ is also the unit tangent to the world-line $X^\mu$.

Nb. This is analogous to the spatial unit tangent $$\vec{u} = \frac{\mathrm{d} \vec{x}}{\mathrm{d} s} = \frac{\mathrm{d} \vec{x}}{\mathrm{d} t} \frac{\mathrm{d} t}{\mathrm{d} s}= \frac{1}{v} \frac{\mathrm{d} \vec{x}}{\mathrm{d} t}$$ to the curve $\vec{x} = \vec{x}(s)$ where $s$ is arclength and $v$ is the speed along the curve.

Proper momentum

Analogous to three-momentum, $\vec{p} = m \vec{v}$, we define the proper four-momentum of a particle to be mass times velocity, $$P^\mu = mU^\mu$$ where the mass, $m$ is an scalar invariant associated with a particle.

Nb. The modern view of the mass of an object is that it doesn't depend on the velocity - it's what used to be referred to as the rest mass of an object.

As for velocity, we can calculate the components of four-momentum in a stationary frame, $$P^\mu \rightarrow \frac{(m, m v_x, m v_y, m v_z)}{\sqrt{1 - v^2}}$$

If we set $c \neq 1$ then $$P^\mu \rightarrow \frac{(m c, m v_x, m v_y, m v_z)}{\sqrt{1 - \frac{v^2}{c^2}}}$$

The time component becomes $$P^0 = \frac{mc}{\sqrt{1 - \tfrac{v^2}{c^2}}}$$ and if we multiply both sides by $c$, we get a quantity with the same dimensions as energy, $$\left[P^0 c\right] = \left[mc^2\right] = \text{mass x speed}^2$$

We can express this more clearly using the binomial theorem to write $$\frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} = 1 + \frac{1}{2}\frac{v^2}{c^2} + O\left(\frac{v^4}{c^4}\right)$$ (where $O\left(\frac{v^4}{c^4}\right)$ just means terms of the series involving powers of $\frac{v^2}{c^2}$ of two and higher) and then $$P^0 c = \frac{mc^2}{\sqrt{1 - \tfrac{v^2}{c^2}}} \approx mc^2\left(1 + \frac{1}{2}\frac{v^2}{c^2}\right) = mc^2 + \frac{1}{2}mv^2$$ which, apart from a constant, is the kinetic energy associated with a particle.

Thus, the time component of the proper momentum, multiplied by the speed of light is energy, and in particular, the constant, $mc^2$ is called the rest energy associated with the particle.

Setting $v \ll c$ means the spatial components of the momentum, $P^m$ become their Newtonian counterparts.