# Attraction near a straight cosmic string

I'm having an interpretational issue with the particles behavior around a straight cosmic string.

Consider the infinite cylindrical universe described by the following metric: $$\tag{1} ds^2 = dt^2 - dr^2 - (1 - 4 G \mu)^2 \, r^2 \, d\varphi^2 - dz^2.$$ To simplify things, I'll write $$\tag{2} \lambda = 1 - 4 G \mu \equiv 1 - \frac{\alpha}{2 \pi} \le 1,$$ and consider motions in the orthogonal plane only : $$z = 0$$ and $$\dot{z} = 0$$. From metric (1), it can be shown that the full Riemann curvature tensor is 0 everywhere around the string. So this spacetime is locally Minkowskian, but it has a conical topology with deficit angle $$\alpha = 8 \pi G \mu$$ (which gives $$\lambda < 1$$). I was expecting free particles to move in straight lines in this spacetime, since there's no curvature at all, and it is well known that the straight cosmic string doesn't apply any gravitational force on test-particles. Because of the deficit angle, I had a doubt so I made a numerical resolution of the geodesics equations using Mathematica. The particle's lagrangian is simply this, where the dots represent the derivative with respect to proper time $$\sigma$$ or an arbitrary parametrization in the case of light (when $$\mathcal{L} = 0$$): $$\tag{3} \mathcal{L} = \dot{t}^2 - \dot{r}^2 - \lambda^2 \, r^2 \, \dot{\varphi}^2.$$ Applying the Euler-Lagrange equations (or geodesics equation) give the following equations: \begin{align} \dot{t} \equiv \frac{d t}{d\sigma} &= \mathcal{K} = \text{constant,} \tag{4} \\[1ex] \frac{d^2 r}{d\sigma^2} &= \lambda^2 \, r \, \dot{\varphi}^2, \tag{5} \\[1ex] \dot{\varphi} \equiv \frac{d\varphi}{d\sigma} &= \frac{\mathcal{J}}{r^2}. \tag{6} \end{align} Using (6) in (5) gives $$\tag{7} \frac{d^2 r}{d\sigma^2} = \frac{\lambda^2 \mathcal{J}^2}{r^3}.$$ The constant of motion $$\mathcal{J}$$ is interpreted as the angular momentum of the particle, per unit mass. I then use cartesian coordinates for the numerical simulation: $$x = r \cos \varphi$$ and $$y = r \sin \varphi$$. This give the following differential equations (with $$r = \sqrt{x^2 + y^2}$$): \begin{align} \frac{d^2 x}{d\sigma^2} &= -\, \frac{(1 - \lambda^2) \mathcal{J}^2}{r^4} \, x, \tag{8} \\[1ex] \frac{d^2 y}{d\sigma^2} &= -\, \frac{(1 - \lambda^2) \mathcal{J}^2}{r^4} \, y. \tag{9} \end{align} So according to these equations, there's a "force" acting on the particle. Of course, it is 0 when $$\lambda = 1$$ (case of no deficit angle = global flat Minkowskian world). Running these equations in Mathematica gives typically the following kind of trajectories in the $$x \, y$$ plane (the gray disk is the cosmic string. Its radius is the unit of time and lenght. The three sliders are the deficit angle $$\alpha$$, the impact parameter $$b$$ and initial velocity $$v$$):

The force (8)-(9) is 0 when the particle is initially at rest (angular momentum $$\mathcal{J} = 0$$). This is consistent with the idea that the string doesn't apply a gravitationnal force. But from (8)-(9), there's a velocity dependant force acting on the particle, and the pictures clearly show that the string is attractive. (for some parameters, the particle could be captured by the string, and then be expulsed ...)

So how can this be, when there's no spacetime curvature at all, everywhere around the string?

Can someone point me to a freely available paper (arXiv ?) that describe these weird features of the straight cosmic string?

• Interesting problem, just a thought: the $xy$ plane you plot from the polar coordinates $r$, $\varphi$ does not have the usual flat metric, since a circle's circumference does not equal $2 \pi r$; so seeing the trajectories "curve" does not necessarily mean that they do. Have you tried plotting them on a cone? Commented Jan 4, 2021 at 20:32
• @JacopoTissino, I didn't made a plot on a cone. I'm not sure this is worth the effort. About the factor $\lambda^2$ in front of the angular part $r^2 \, d\varphi^2$, it could be absorbed in a redefinition of the radial coordinate $r$ (and would reappear in front of the radial part $dr^2$, in the metric). With the current coordinates, the circle is just missing an angle ("deficit angle" of the conical world)
– Cham
Commented Jan 4, 2021 at 20:43
• Well, that is my point: you can redefine $r$ or $\varphi$, but in any case you will not be able to plot your manifold isometrically on the whole plane, because of the missing angle. Your graphs are precisely an orthogonal projection of the cone onto a plane perpendicular to its axis: as such, they distort the trajectories and make it seem like they are curving, even though they are not. Commented Jan 4, 2021 at 21:52
• I think I get it now. If we take the extreme case of a straight cylinder, we could throw random free particles on its surface. Many of them will move on an helicoidal path, with various inclinations and velocities. Squashing the cylinder on a plane will show spiral-like curves encircling a central circular hole (like on my pictures above). Yet, it is cleat that the particles aren't deflected by some force on the 2D surface (the cylinder is locally flat). The cone has a similar property.
– Cham
Commented Jan 5, 2021 at 0:07
• I suggest you made a mistake in your numerical implementation. I can't reproduce your plots with the given equation. Commented Jan 5, 2021 at 10:54

Instead of using those coordinates, you could identify the space around the string with a wedge of Minkowski space, with the usual Minkowski coordinates and metric, and the edges of the wedge identified. Geodesics that don't cross the edges are straight lines.

If $$α\ge π$$ then you can always arrange for any particular geodesic to lie entirely in the wedge, and it clearly can't circle the string.

If $$α<π$$ then you can put $$\lceil π/α \rceil$$ copies of the space next to each other in a pie-slice arrangement, and draw the geodesic so that it passes through those copies. The portion of the geodesic that lies within each slice is one "circling" of the string. When you plot the geodesic in your original coordinates, it looks curved because you are, in effect, putting standard Minkowski cylindrical coordinates on the wedge and then scaling the angular coordinate by $$1/λ$$.

• Note that if the OP solves the equation via my method, they'll get the equation $1/r = A \cos(\lambda (\phi - \phi_0))$, where $A$ and $\phi_0$ are constants. This is exactly the equation of a straight line with the $\phi$ coordinate rescaled. Commented Jan 4, 2021 at 23:23

Too long for a comment:

Note that the geodesic equations are exactly soluble in this case. Specifically, we can derive an analog of the Binet equation. Defining $$u = 1/r$$, we have $$\frac{dr}{d\sigma} = \frac{d\phi}{d\sigma}\frac{dr}{d\phi} = \frac{\mathcal{J}}{r^2} \frac{d(1/u)}{d\phi} = - \frac{\mathcal{J}}{r^2u^2 } \frac{du}{d\phi} = - \mathcal{J} \frac{du}{d\phi}$$ and by similar logic $$\frac{d^2 r}{d\sigma^2} = \frac{d\phi}{d\sigma} \frac{d}{d\phi} \left( - \mathcal{J} \frac{du}{d\phi} \right) = -\mathcal{J}^2 u^2 \frac{d^2 u}{d \phi^2}.$$ Your Eq. (7) then becomes $$-\mathcal{J}^2 u^2 \frac{d^2 u}{d \phi^2} = \lambda^2 \mathcal{J}^2 u^3 \quad \Rightarrow \quad \frac{d^2 u}{d \phi^2} = -\lambda^2 u$$ which I trust that you can solve.

• Very nice! I'll try to use this to get back the curves of Mathematica.
– Cham
Commented Jan 4, 2021 at 21:25

I've fixed all of my issues (I think). Here's the $$x y$$ plane with a typical curve (the starting point is on the right):

And here's a view of the cone on which the geometry is defined, with conical deficit angle $$\alpha$$. There are three typical curves drawn on it, with different impact parameter $$b$$. I'm a bit surprised that the initial velocity $$v$$ doesn't make any difference, but since there's no force acting on the particle, it is not so much surprising after all:

The gray cylinder is the cosmic string.

What is going on is pretty clear, I think. There's no gravitational force, but the cone still have a geometrical effect even if its curvature is 0.

What is still not clear, is the diffusion effect on a bunch of particles moving on the cone. Is there diffusion or not?

EDIT: What is weird here, is that you have a flat spacetime around the cosmic string, and yet you may get back the ball that you throw away, without any force applied to the ball! (see for example the red curve, on the second picture above).

• Great! As for the diffusion (although perhaps that should be a different question?), geodesic deviation is directly proportional to the Riemann tensor, so it should be zero here - the distance between parallel geodesics remains constant. Commented Jan 5, 2021 at 19:59
• @JacopoTissino, yes, this is right. My code is showing that there's no diffusion. But then there are deflections without force and curvature. You may even throw away a ball and catch it back after a while, without any force applied to it. This is weird!
– Cham
Commented Jan 5, 2021 at 22:27