Here we’re going to discuss the equations of motion for a charged particle in a curved spacetime with an electromagnetic tensor, $F_{ab}$, show why massive particles have a contracted $4-$velocity, $u^a$, that is constant along a charged-particle path, and why massive particles move slower than the speed of light.

In flat Euclidean space, the theory of electromagnetism describes the force on a charged particle by the Lorentz force;

\begin{equation} \mathbf{F} = m \frac{d\mathbf{v}}{dt} = q(\mathbf{E}+\mathbf{B}\times\mathbf{v})\end{equation}

However, there is nothing stopping us from discussing electromagnetism in a curved spacetime setting. In fact, Maxwell’s equations are fully formulated to be applicable to curved spacetime. We first write the Lorentz force, in flat spacetime, in terms of the electromagnetic tensor given by

\begin{equation} \frac{d^2 x^a}{d \tau^2 } = \frac{q}{m} g_{bc}F^{ab}\frac{dx^c}{d\tau}.\end{equation}

Now, the left hand side of the equation can be expressed as $u^a\partial_a u^b$ by writing

\begin{equation} \frac{d^2 x^a}{ d\tau^2 } =\frac{d u^a(x(\tau))}{ d\tau }= \frac{dx^a}{d\tau}\frac{\partial u^b}{\partial x^a} = u^a\partial_a u^b. \end{equation}

Now, we can finally discuss curved spacetime by taking the $\partial_a $ to $\nabla_a$ which encodes the curvature of the spacetime inside the Christoffel connection, $\Gamma_{ab}^c.$ So the full equation will be given by

\begin{equation} u^a\partial_a u^b + \Gamma^b_{ac}u^au^c = \frac{q}{m} g_{bc}F^{ab}u^c\end{equation}

So what does this mean physically? Well a charged particle moving in curved spacetime is represented by the geodesic equation on the left hand side. The geodesic equation describes the motion of the particle in the curvature of the surrounding space. The right hand side describes the particles behavior with respect to the electric and magnetic fields generated by the particle which are then included inside the electromagnetic tensor.

If you are wondering, why the geodesic equation is not equal to 0, or why we have a term on the right hand side in the first place if we are in a coordinate system that is not accelerating $-$ in fact, those two questions are exactly the same $-$ realize that, since the particle is charged, it would be accelerating locally in accordance to the Lorentz Force. After all, there may be electric and magnetic fields permeating the surrounding spacetime that influence the particle’s motion.

Now do you want to see something really cool? We will show why the particles have to move at speeds less than the speed of causality, $c$, or in other words why $g_{ab}u^au^b = -1$ for massive particles. The last statement amounts to saying that $g_{ab}u^au^b$ is constant along a particle’s path, and one way we can show that something is constant is to take a derivative of that object so

\begin{equation}\nabla_c[g_{ab}u^au^b] = \nabla_c(g_{ab}) u^au^b +g_{ab}(\nabla_c u)^a u^b +g_{ab}u^a(\nabla_c u)^b. \end{equation}

Behind the scenes, we choose our connection such that $g_{ab}$ vanishes when operated on by the covariant derivative $\nabla$, so the first term disappears. Next, we have to be careful about the $4-$velocity terms that the covariant derivative is operating on, because in general $\nabla_c u^a \ne (\nabla_c u)^a.$

First of all, the reason that we cannot assert the above equality is because when the covariant derivative operates on the four velocity component we would have a real number. Why? Well the covariant derivative is a one-form, it lives in the dual vector space, while the $4-$velocity itself is a tangent vector or an element of the tangent space. So upon operation, the one-form and the vector map their product back into the real numbers, $\mathbb{R}$, and then the contraction between $g_{ab}$ and $u^{b}$ wouldn’t make much sense. The source of all of this trouble is due to the fact that we’re in curved space, so effectively the Christoffel connection muck everything up. That is unless we go to $\textbf{Riemann Normal Coordinates}.$

Riemann normal coordinates is a way to cheat and recover the nice properties of Euclidean (flat spacetime) while still pretending that you are in curved space. The method effectively boils down to considering a point $p$ of the total trajectory of the particle. At that point, if you institute a normal coordinate basis, then the Christoffel terms vanish, i.e. $\Gamma_{ac}^b=0.$ So at a point, this is true, but the kicker is that this defines a locally inertial reference frame, which is precisely what we are interested in and what we have been doing for the last ~$17$ paragraphs.

So if we switch to normal coordinates, we not only have $\Gamma_{ac}^b=0.$, we get (for free) $\nabla_c u^a = (\nabla_c u)^a.$ And with this the previous equation becomes

\begin{equation}\nabla_c[g_{ab}u^au^b] = g_{ab}(\nabla_c u^a) u^b +g_{ab}u^a(\nabla_c u^b). \end{equation}

Then from $(2)$, we have that $\nabla_c u^b=(q/m)F^b_c$. So replacing everything, we have

\begin{align*}\nabla_c[g_{ab}u^au^b] &= \frac{q}{m}\left(g_{ab}F^a_c u^b +g_{ab}u^a F^b_c \right). \\ &= 2\frac{q}{m} g_{ab}(F^a_c) u^b \\ &= 0,\end{align*}

because the product of a symmetric tensor $g_{ab}$ and an anti-symmetric tensor $F^{ab}$ is $0.$ So where does the part with the particles moving less than the speed of light come from? That’s from having three different classes of particles; massive, massless, and negative mass $-$ the last of which we will ignore. Now, if we were to integrate the left hand side of the above equation, we would obtain something of the form $g_{ab}u^au^b = k,$ where k is some real number. In nature, however, we find particles with mass (you, me, the electrons whizzing around underneath this keyboard) and particles without mass, for which we only have one contender: light. Since massless particles move at a set speed, lets call that speed $c$, and notice that since light has no charge, $q=0$, and also that the acceleration $u^a\partial_au^b=0$, so (4) becomes $\Gamma^b_{ac}u^au^c = 0 $ or $g^l_{ac}u^au^c=0$ since $\Gamma^b_{ac}$ is not zero over the entire trajectory.

Now, let’s look at massive particles. These particles have some charge, and they have some mass, therefore (4) doesn’t change and we can instead write it as

\begin{equation} u^a\partial_a u^b = \frac{q}{m} g_{bc}F^{ab}u^c – \Gamma^b_{ac}u^au^c .\end{equation}

This just means that as you increase the acceleration on the left hand side, which is saying that you increase the Lorentz force on the right hand side, the Christoffel symbols (with the curvature encoded into them) push you back. So, for massive particles, $g^m_{ab}u^au^b < g^l_{ac}u^a u^c = 0 \implies g_{ab}u^au^b = -1$ or some other arbitrary negative number. Which is to say that no matter how much force you might give your system, the underlying spacetime will always slow you down $-$ and that’s a pretty good life lesson.

Thanks to L. Lancaster, and A. Ballin for notational and mathematical clarifications.

You made the Christwafuls less so

That’s exactly what I call them too when I have to calculate them.