Are geodesics on a flat spacetime with a background electromagnetic field still straight lines?

Question

In Minkowski, the metric covariant derivative is just $\nabla = \partial$, and therefore geodesics (solutions $\gamma(t)$ to the geodesic equation $\nabla_{\dot\gamma}\dot\gamma = 0$) are just straight lines. However, when considering a background electromagnetic field, the covariant derivative is given by the connection over the U(1) bundle so that $D_\mu = \partial_\mu + ieA_\mu$, with $A_\mu$ the electromagnetic vector potential.

I am having trouble in seeing the behavior of geodesics in this space. Are they still straight lines? Can one use the geodesic equation $$D_{\dot \gamma} \dot \gamma =0 ?$$

Having an $i$ in the $D_\mu$ makes everything even weirder, as it feels like this 'geodesic' will take complex values.

My guess from my very basic understanding of bundles is that this geodesic should be studied not in my flat spacetime $\mathbb{R}^4$, but rather on the total space $U(1)\times \mathbb{R}^4$, and somehow what I should be interested in is the projection of this geodesic onto my manifold, or something like that.

Context of the question:

In QFT in curved spacetimes (QFTCS) the vacuum expectation value of observables such as $\phi(x)^2$ is defined through a point splitting prescription, i.e. $$ \left<\phi(x)^2\right> := \lim_{x'\to x} \langle0\rvert\phi(x)\phi(x')\lvert 0\rangle. $$ This prescription can be extended to the stress-energy tensor by parallel transporting the $T_{ab}$ along the (unique) geodesic that joins x' and x.

In my case, I am studying the charge current density $\left<j_\mu(x)\right> $ on flat spacetime with a background EM following the same procedure. I want to know if this geodesic changes due to having to do this parallel transport, and along which path it is done.

I also don't really understand what exactly parallel transport in that limit means.

Answer 1

If the spacetime is flat, it doesn't matter what else you have in the spacetime. Geodesics will be straight lines.

You are mixing up two different covariant derivatives. A covariant derivative is defined with respect to a connection. When the spacetime is not flat, there is a non-trivial Levi-Civita connection, and you use that connection to define the appropriate covariant derivative for finding geodesics. The covariant derivative with respect to the $U(1)$ connection doesn't tell you anything about the geometry of the spacetime because the $U(1)$ connection is not related to the metric.

This means also that parallel transport is trivial in a flat spacetime; the $U(1)$ connection does not factor into it.

Answer 2

Let $(M,g)$ be a smooth manifold equipped with a semi-Riemannian metric $g$. Then its tangent bundle $TM$ is equipped with a natural affine connection $\nabla$ which is the Levi-Civita connection of the metric $g$ (the unique torsion-free, metric compatible connection). The autoparallel curves $\gamma:(a,b)\to M$ defined by

$$\nabla_{\gamma'}\gamma'=0$$

are the geodesics of the metric $g$.

On the other hand, suppose $G$ is a Lie group and you have a principal $G$-bundle $G\dashrightarrow P\to M$ over $M$ equipped with some principal connection. If $\rho:G\to {\rm GL}(V)$ is a representation of $G$ you can construct the associated bundle $\pi_V: P_V\to M$. This is a vector bundle that carries one affine connection $D$ which is induced by the principal connection on the principal bundle.

Now observe that the connection $\nabla$ acts on sections of $TM$ whereas $D$ acts on sections of $P_V$. These are differential operators acting on different spaces of functions. The connection $\nabla$ allows you to define the parallel transportation of objects of $TM$ whereas $D$ allows you to define the parallel transportation of objects of $P_V$. To make the distinction clear, you should understand that parallel transportation is really about finding the so-called horizontal lift of a curve.

So if $\gamma: (a,b)\to M$ is a curve on the base manifold, we look for a curve $\Gamma:(a,b)\to E$ where $E$ is the total space of whatever vector bundle you have defined over $M$ that satisfies a few properties: (1) it projects down to $\gamma$ under the projection $E\to M$, (2) it is horizontal. The horizontal part can be defined using the connection one-form on the principal bundle (check Isham's Differential Geometry for Physicists), but it can be rewritten in terms of the associated affine connection on the vector bundle as an equation that takes the form $\mathscr{D}_{\gamma'}\Gamma=0$.

When we prescribe one initial condition by choosing some $\xi\in E$ and solving the problem

$$\mathscr{D}_{\gamma'}\Gamma=0,\quad \Gamma(a)=\xi,\quad \Gamma:(a,b)\to E$$

we say that we are computing the parallel transportation of $\xi$ along $\gamma$.

So we see that parallel transportation is the motion of an initial object of a vector bundle along a curve according to a specified connection. In the present scenario we have two different vector bundles, which are associated bundles to two different principal bundles: one is the tangent bundle $TM$ which allows to parallely transport tangent vectors to the manifold, whereas the other is the bundle $P_V$ which allows to parallely transport elements of $P_V$.

These are different things! In particular, a geodesic is defined to be a curve $\gamma$ such that its tangent vector is parallely transported along itself. This is something happening in $TM$, it has absolutely nothing to do with another vector bundle like $P_V$.

So there are different parallel transportations happening on different bundles and you should not conflate them.