Relativity Series Part 5 - Relativistic Dynamics

In the previous part, we discussed vectors and tensors. In this part, we will discuss relativistic dynamics.

In Newtonian mechanics, we express the motion of a particle in three dimensions as a function of time ($\vec{r}(t)$). For Minkowski spacetime, however, we must replace these old formulas with new ones. With what we have done so far, we have already generalized the expression for a particle's position in Newtonian mechanics ($x^\mu$). Our next task in special relativity is to reformulate the concept of velocity.

Four-Velocity Vector (4-Velocity)

We called the curve traced by a particle through spacetime its worldline. The tangent vector to this curve, obtained by parametrizing the particle's position vector with $\tau$, namely $u^\mu\equiv dx^\mu/d\tau$, is called the four-velocity vector. Let us examine the components of this vector one by one.

$$u^0=\frac{dx^0}{d\tau}=\frac{cdt}{d\tau}=\frac{c}{\sqrt{1- \frac{v^2}{c^2}}}=\gamma c$$

Here, $dt=\gamma d\tau$ has been used. Rather than writing out the values $\mu=1,2,3$ separately, we will use the index $i$ for these coordinates.

$$\frac{dx^i}{d\tau}=\frac{dx^i}{dt}\frac{dt}{d\tau}=\gamma \frac{dx^i}{dt}= \gamma \mathbf{v}$$

Here, $\mathbf{v}$ is the three-dimensional velocity vector familiar from Newtonian mechanics. From these results, we can write the vector $u^\mu$ as

$$u^\mu=(\gamma c, \gamma \mathbf{v})$$

Let us calculate the magnitude of this velocity vector.

$$||\mathbf{u}||^2=\mathbf{u}\cdot \mathbf{u}=\eta_{\mu\nu}u^\mu u^\nu=u_\nu u^\nu$$

Recalling that the metric $\eta$ is a diagonal matrix, the components $\eta_{\mu\nu}$ for which $\mu\ne\nu$ are equal to zero.

$$||\mathbf{u}||^2=\eta_{\mu\nu}u^\mu u^\nu=\eta_{00}u^{0}u^{0}+\eta_{11}u^{1}u^1{}+\eta_{22}u^{2}u^{2}+\eta_{33}u^{3}u^{3} = - \gamma^2c^2+\gamma^2||\mathbf{v}||^2 =-c^2$$

Four-Acceleration Vector (4-Acceleration)

As you might expect, the acceleration vector in four dimensions is defined in a manner similar to the vector concept in Newtonian mechanics.

$$a^\mu\equiv \frac{du^\mu}{d\tau}=\frac{d^2x^\mu}{d\tau^2}$$

Let us write out the components of the acceleration vector explicitly.

$$a^\mu=\frac{du^\mu}{d\tau}=\gamma\frac{du^\mu}{dt}=\gamma \frac{d}{dt}(\gamma c,\gamma \mathbf{v})=\gamma (c \frac{d\gamma}{dt}, \frac{d\gamma}{dt}\mathbf{v}+\gamma \frac{d\mathbf{v}}{dt})$$

For the particle under consideration, suppose that we make our observations from the particle's inertial reference frame. In this case, $\mathbf{v}=0$ and $\gamma=1$. Using these, let us write the components of our velocity and acceleration vectors. Because we will write the equations in the particle's own reference frame, we will use $\doteq$ rather than $=$ to indicate this.

$$u^\mu\doteq(c,\mathbf{0})\ \ \ \ \ \ a^\mu\doteq(0,\frac{d\mathbf{v}}{dt})$$

Thus, in the particle's frame of observation, the spatial three-dimensional acceleration vector must be zero for $a^\mu$ to be zero. The velocity vector, however, can never be zero. Another important property of the acceleration vector is that it is orthogonal to the velocity vector.

$$a_\mu u ^\mu=0$$

To prove this, let us differentiate the equation $||\mathbf{u}||^2=-c^2$, which we obtained earlier, with respect to $\tau$.

$$0=\frac{d}{d\tau}(-c^2)=\frac{d}{d\tau}(u_\mu u ^\mu)=\frac{d}{d\tau}(\eta_{\mu\nu}u^\mu u^\nu) =\frac{d}{d\tau}(\eta_{\mu\nu})u^\mu u^\nu+\eta_{\mu\nu}\frac{d}{d\tau}(u^\mu u^\nu) \\=\eta_{\mu\nu}u^\nu\frac{d}{d\tau}(u^\mu)+\eta_{\mu\nu}u^\mu\frac{d}{d\tau}(u^\nu) =u_\mu\frac{d}{d\tau}(u^\mu)+u_\nu\frac{d}{d\tau}(u^\nu)=2a^\mu u_\mu$$

In the final step, whether we call the index $\mu$ or $\nu$ makes no difference, since the index label itself is immaterial. We have also used the equality $d\eta_{\mu\nu}/d\tau=0$ during the calculation.

Four-Momentum Vector (4-Momentum)

By analogy with Newtonian mechanics, we define the four-momentum vector as follows.

$$p^\mu=m u^\mu$$

Here, $m$ represents the mass of the particle. This quantity, commonly taught as the rest mass, is constant in every reference frame. The components of this momentum vector are written as

$$p^\mu=\gamma m(c,\mathbf{v})=(\gamma mc, \gamma m\mathbf{v})=(\gamma mc,\mathbf{p})$$

It is also possible to define the zeroth, or temporal, coordinate of the momentum vector in another way. When defining the temporal coordinate of the position vector, for example, we arrived at a result using what we knew together with dimensional analysis. There is only one variable on which the position of a particle depends: time. Moreover, the fact that the speed of light has the same constant value for all observers imposed an important constraint on the position vector we were constructing. In light of this information and through dimensional analysis, we defined the zeroth coordinate of the position vector as $ct$. We can now make a similar inference for the momentum vector. As we know from classical physics, the momentum and energy of a particle are related by $E=p^2/2m$. Rather than taking this relation directly as our basis, we can consider the two parameters that matter to us ($E$ and $c$) and make the very simple definition $p^0=E/c$.

$$p^\mu \equiv \left(\frac{E}{c},\mathbf{p}\right)$$

With this definition, just as we defined time as a component paired with spatial position, we can regard energy as a quantity paired with three-dimensional spatial momentum.
Using what we have obtained, let us now examine the product $p^\mu p_\mu$.

$$p^\mu p_\mu=\eta_{\mu\nu}p^\mu p^\nu=m^2\eta_{\mu\nu}u^\mu u^\nu=m^2u^\mu u_\mu=-m^2c^2$$

At the same time, this product is also equal to

$$p^\mu p_\mu=\eta_{\mu\nu}p^\mu p^\nu=-\frac{E^2}{c^2}+||p||^2$$

and consequently we obtain

$$E^2=||p||^2c^2+m^2c^4$$

In the particle's own inertial reference frame ($\mathbf{v}=0$), we arrive at the famous equation

$$E\doteq mc^2$$

!

And with that, my part in the Relativity Series comes to an end. In the next part, Furkan will explain the energy–momentum tensor.

References

[1] F. Rahaman, (2014). The Special Theory of Relativity. (India, Springer)
[2] V. Faraoni, (2013). Special Relativity. (Switzerland, Springer)
[3] L.F. Landau, E.M. Lifshitz, (1980). The Classical Theory of Fields: Volume 2. (Butterworth-Heinemann)
[4] S.M. Carroll, (2003). Spacetime and Geometry: An Introduction to General Relativity. (Addison-Wesley)