I'm having a difficulty in finalizing a resolution of the Einstein equation for a static cosmic string. I start with the following metric ansatz, for a static straight string oriented along the $z$ direction. Using cylindrical coordinates: $$\tag{1} ds^2 = \mathcal{P}^2(r) \, dt^2 - dr^2 - \mathcal{Q}^2(r)\, d\vartheta^2 - dz^2. $$ Plugging this metric into the Einstein equation gives the following equations, after a few pages of calculations (I'm using the $\eta = (1, -1, -1, -1)$ convention and $G_{ab} = -\, \kappa T_{ab}$. Indices are associated to the "flat" local inertial frames): \begin{align} G_{00} &= \frac{\mathcal{Q}^{\prime \prime}}{\mathcal{Q}} = -\, \kappa T_{00} = -\, \kappa \rho, \tag{2} \\[1ex] G_{11} &= -\, \frac{\mathcal{P}^{\prime} \mathcal{Q}^{\prime}}{\mathcal{P} \mathcal{Q}} = -\, \kappa T_{11} = 0, \tag{3} \\[1ex] G_{22} &= -\, \frac{\mathcal{P}^{\prime \prime}}{\mathcal{P}} = -\, \kappa T_{22} = 0, \tag{4} \\[1ex] G_{33} &= -\, \frac{\mathcal{P}^{\prime \prime}}{\mathcal{P}} - \frac{\mathcal{P}^{\prime} \mathcal{Q}^{\prime}}{\mathcal{P} \mathcal{Q}} - \frac{\mathcal{Q}^{\prime \prime}}{\mathcal{Q}} = -\, \kappa T_{33} = +\, \kappa \tau. \tag{5} \end{align} Here, $\rho > 0$ and $\sigma = -\, \tau < 0$ are the string's energy density and tension, respectively. Now, (2) implies $\mathcal{Q}^{\prime} \ne 0$, so (3) and (4) give $\mathcal{P}^{\prime} = 0$ and $\mathcal{P}^{\prime \prime} = 0$. Then (5) implies $\tau = \rho$, which is good for a relativistic string. Then, the trouble begins when I attempt to solve (2). I was expecting a thin string, so a Dirac's delta for the density $\rho$. But (2) suggest another solution and I wasn't expecting this. Assuming a constant density $\rho = \rho_0$ for $r < R$ (the string's radius) and $\rho = 0$ for $r > R$, I get $$\tag{6} \mathcal{Q}^{\prime \prime} + \kappa \rho_0 \, \mathcal{Q} = 0. $$ Then this is an harmonic equation which have the general solution (I write $\lambda \equiv \sqrt{\kappa \rho_0}$ to simplify things) $$\tag{7} \mathcal{Q}(r) = \alpha \sin(\lambda r) + \beta \cos(\lambda r). $$ For the exterior metric: $\rho = 0$, (2) reduces to $\mathcal{Q}^{\prime \prime} = 0$, which have solution $\mathcal{Q}(r) = a r + b$. Matching the solution at $r = R$ gives $$\tag{8} \mathcal{Q}(R) = \alpha \sin(\lambda R) + \beta \cos{\lambda R} = a R + b. $$ So my problem is to find the constants $\alpha$, $\beta$, $a$ and $b$ (four constants!), from regularity and junction conditions.

EDIT: From A.V.S and Michael answers below, I should apply the regularity at $r = 0$: \begin{align} \mathcal{Q}_{\text{int}}(0) &= 0, \tag{9} \\[1ex] \mathcal{Q}_{\text{int}}^{\prime}(0) &= 1. \tag{10} \end{align} This gives $\alpha = \lambda ^{-1}$ and $\beta = 0$, so $$\tag{11} \mathcal{Q}_{\text{int}}(r) = \frac{1}{\lambda} \, \sin(\lambda r). $$ This is fine. But then I imposes the junction at the string's surface. Apparently, there's a subtlety that I don't understand here. According to some obscure papers I've found, the radial coordinate $r$ isn't the same on the interior side and on the exterior of the string, so $R_{\text{int}} \ne R_{\text{ext}}$: \begin{align} \mathcal{Q}_{\text{int}}(R_{\text{int}}) &= \mathcal{Q}_{\text{ext}}(R_{\text{ext}}), \tag{12} \\[1ex] \mathcal{Q}_{\text{int}}^{\prime}(R_{\text{int}}) &= \mathcal{Q}_{\text{ext}}^{\prime}(R_{\text{ext}}). \tag{13} \end{align} This gives the following junction conditions: \begin{align} \frac{1}{\lambda} \, \sin(\lambda R_{\text{int}}) &= a R_{\text{ext}} + b, \tag{13} \\[1ex] \cos(\lambda R_{\text{int}}) &= a. \tag{14} \end{align} I can't solve this system of equations without an extra constraint (??). I get the relation of some paper if I impose $b = 0$ so $$\tag{15} R_{\text{ext}} = \frac{1}{\lambda} \, \tan(\lambda R_{\text{int}}). $$ Defining the energy per unit length $\mu = \rho_0 A_{\text{int}}$, I get $a = 1 - 4 G \mu$, which relates to the angle deficit.

But how can I justify that $b = 0$ and that the radial coordinate isn't the same on both side of the string's surface? Why can't I just use $R_{\text{int}} = R_{\text{ext}} = R$, and then find $a$ and $b \ne 0$ ?

An "obscure" paper that shows some details (with pesky weird notation!), without explaining the different radial coordinate and why $b$ should be 0. See expressions (6), (7), (8), on page 2:


See also pages 3 and 4 of this paper:


  • $\begingroup$ I don’t have time to write a proper answer. But answer should follow from imposing regularity at $r=0$ and junction conditions at $r=R$. $\endgroup$
    – TimRias
    Dec 21, 2020 at 19:51
  • 1
    $\begingroup$ It may not be a homework problem in a class you're taking right now, but it's still a pretty standard exercise. I assigned it in the undergraduate GR class I taught last spring. $\endgroup$ Dec 21, 2020 at 20:21
  • $\begingroup$ @MichaelSeifert, well, I'm not even taking a "class". I'm studying this subject on my own, to add it to my notes for a future class! ;-) $\endgroup$
    – Cham
    Dec 21, 2020 at 20:29
  • $\begingroup$ your ansatz does not seem correct. the term $\mathcal{Q}^2(r)\, \vartheta^2$ does not have a differential. Also you say that you use polar coordinates, but you probably mean cylindrical coordinates. $\endgroup$
    – magma
    Dec 22, 2020 at 9:00
  • $\begingroup$ Concerning your most recent edits, you should probably ask a new question about them and link to the papers making those claims (or include excerpts from them, if they're truly obscure.) $\endgroup$ Dec 22, 2020 at 17:53

2 Answers 2


As noted by @mmeent in the comments, the metric must be regular at the origin. This means, in particular, that $$ \mathcal{Q}(r) \approx r $$ in the limit $r \to 0$. Equivalently, we must have $\mathcal{Q}(0) = 0$ and $\mathcal{Q}'(0) = 1$. In addition, the metric components and their first derivatives must be continuous at the boundary $r = R$.

We thus have four equations (two from regularity at the origin, two from continuity at the boundary) in the four unknowns $\alpha$, $\beta$, $a$, and $b$. Take it from there.

  • $\begingroup$ Your junction conditions doesn't give the right deficit angle (unless I'm missing something). It give $$\mathcal{Q}_{\text{int}}(r) = \frac{1}{\lambda} \, \sin(\lambda r),$$ which appears right, and $$\mathcal{Q}_{\text{ext}}(r) = (r - R) \cos(\lambda R) + \frac{1}{\lambda} \, \sin(\lambda R),$$ which I think is wrong. It's not the "usual" $$\mathcal{Q}_{\text{ext}}(r) = (1 - 4 G \mu) \, r$$ that I could find in many papers (I do have $\cos(\lambda R) = 1 - 4 G \mu$ from the thick string cross section area). Do I have to make the limit $R \rightarrow 0$ ? $\endgroup$
    – Cham
    Dec 21, 2020 at 21:37
  • $\begingroup$ Please, do you confirm the expressions in my previous message? Should I take a limit to get back the deficit angle presented in many papers? $\endgroup$
    – Cham
    Dec 22, 2020 at 0:43
  • $\begingroup$ @Cham: I believe that the standard solution assumes that $\lambda R \ll 1$, yes. $\endgroup$ Dec 22, 2020 at 12:48
  • $\begingroup$ It is weird that the thick string solution involves a trigonometric - oscillating - sin function. On the interior, when $r$ increases, the function $\mathcal{Q}(r)$ is oscillating (unless we impose the restriction $\lambda R < \dfrac{\pi}{2}$). $\endgroup$
    – Cham
    Dec 22, 2020 at 16:00
  • $\begingroup$ @Cham: If $\mathcal{Q} \to 0$, then you have a coordinate singularity (and possibly a real one, I'm not sure.) The same sort of thing happens in the Schwarzschild solution at the event horizon. $\endgroup$ Dec 22, 2020 at 16:02

First, to me it seems that the ansatz $(1)$ for this metric of thick cosmic string is wrong, since for a general $\mathcal{P}(r)$ the metric lacks the $SO(1,1)$ invariance under boosts along the string direction, i.e. Lorentz transformations in $(t,z)$ plane.

My suggestion: $$ \tag{1*} ds^2 = \mathcal{P}^2(r) \,( dt^2 - dz^2) - dr^2 - \mathcal{Q}^2(r)\,d \vartheta^2 . $$

Both OP's and my metrics reduce to the same thing if condition $\mathcal{P}\equiv 1 $ is imposed (which happens if stresses $T_{rr}$ and $T_{\theta\theta}$ are zero). However, if $\mathcal{P}$ varies with $r$ in OP's version this boost symmetry disappears. So OP's ansatz would not work for a thick string that is (for example) a solution of Einstein–Abelian Higgs system .

Second. The correct choice of constants is $\alpha=1/\lambda$, $\beta=0$ (while $b\ne 0$). This follows from

  1. the interpretation of $r$ as distance from symmetry axis (which would be $r=0$) along the radial direction (so at $r=0$ metric must have coordinate singularity, so $\mathcal{Q}(0)=0$);

  2. near the symmetry axis there must be no additional angle deficit. If $\alpha\ne 1/\lambda$ there would be a thin string inside the thick string!

The remaining constants $a$ and $b$ are obtained from the conditions of continuity of $\mathcal{Q}$ and $\mathcal{Q}'$ across the boundary $r=R$. If $\mathcal{Q}'$ jumps then this would correspond to nonzero surface stress–energy tensor of the cylinder.

As an interesting variation on the thick string I suggest the metric of a “relativistic solenoid”: the inside metric is Melvin spacetime (with magnetic field along the $z$–axis) and with stress energy tensor depending on $r$ like this: $T^{\mu}_{\nu}=\rho(r)\,\mathop{\mathrm{diag}}(1,1,-1,-1)$, while outside it is flat spacetime with angle deficit. On the surface of a cylinder then there would be a current and distributional surface energy density and stresses. For such metric the function $\mathcal{P}(r)$ would not be constant.

  • $\begingroup$ Let us continue this discussion in chat. $\endgroup$
    – A.V.S.
    Dec 22, 2020 at 18:47
  • $\begingroup$ After thinking more about it, I still don't understand your Lorentz invariant ansatz. The string proper reference frame defines a privileged frame that may destroy the Lorentz invariance of the metric, so I don't see any problem in the ansatz (1) of my question (while at the end we're left with the same thing since $\mathcal{P} = 1$). $\endgroup$
    – Cham
    Dec 23, 2020 at 15:07
  • $\begingroup$ This is not Lorentz invariant but SO(1,1) (rather actually ISO(1,1)) invariant, since only $(z,t)$ plane is transformed. I don't see any problem in the ansatz (1) of my question The problem is that it is not most general. More general must have $\mathcal{B}(r)^2 dz^2$ rather than $dz^2$ and you cannot justify $\mathcal{B}=1$ for a general matter of the thick string. $\endgroup$
    – A.V.S.
    Dec 23, 2020 at 15:21
  • $\begingroup$ Well evidently, the metric $ds^2 = \mathcal{P}^2(r) dt^2 - dr^2 - \mathcal{Q}^2(r) d\vartheta^2 - \mathcal{B}^2(r) dz^2$ is more general, but it also makes the calculations much more complicated. By definition, an ansatz is a kind of "guess" to find solutions. I don't see any problem in restricting ourselves to $\mathcal{B}(r) = 1$ just to simplify things. $\endgroup$
    – Cham
    Dec 23, 2020 at 15:28
  • $\begingroup$ The function $\mathcal{P}^2(r)$ in front of $dt^2$ is natural since we would expect that time is slowing down close to the string. It is less obvious why there should be $\mathcal{B}^2(r)$ in front of $dz^2$, $\endgroup$
    – Cham
    Dec 23, 2020 at 15:35

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