A point particle near a relativistic string does not experience any gravitational force. Why? I would like to understand the following comment mentioned in section 7.6 of the book 'A First Course in String Theory' by Barton Zwiebach,

"... we begin by discussing the gravitational effects of a straight, infinitely long relativistic string. Suppose you place a massive particle some distance away from such a string. Since the string has rest energy one may imagine that the particle would experience gravitational attraction. This is not the case; the particle would experience exactly zero force. This is a result in general relativity and would not hold in Newtonian gravitation where only the effective mass density $\mu_0$ of the string contributes to the gravitational attraction. In general relativity the tension of the string gives an additional contribution, in fact, a gravitational repulsion. The total attractive force is proportional to $(\mu_0 - \frac{T_0}{c^2})$, a combination that precisely vanishes for relativistic strings."

How can one derive the above statement?
 A: If the energy-momentum tensor for an infinitely long string lying on the z-axis is $\tilde T_{\mu\nu}$, then the energy-momentum tensor obtained by "zooming out" and integrating over the cross-section (the x- and y-directions) is given by
$$
T_{\mu\nu}=\delta(x)\delta(y)\int\mathrm dx\ \mathrm dy\ \tilde T_{\mu\nu}
$$
For a uniform, static string (a Nambu-Goto string) of total mass per length $\mu$, the EM tensor is clearly invariant under Lorentz boosts in the tz-plane, which forces $T_{00}=T_{33}$. The only other possible non-zero components can be $T_{11}$ and $T_{22}$, but these must be zero since
$$
0=\int\mathrm dx\ \mathrm dy\ \partial_i\tilde T^{ij} x^k\qquad\text{(current conservation)}
\\=\int\mathrm dx\ \mathrm dy\ \tilde T_{ik}\qquad\text{(integrating by parts for }i,k\in\{1,2\})
$$
For a more physical argument, the pressure along the x and y directions in this diagonal energy-momentum tensor vanishes since we are averaging over a string whose internal structure cylindrically symmetric but otherwise irrelevant.
Therefore $T_{\mu\nu}=\delta(x)\delta(y)\mathrm{diag}(\mu, 0, 0, \mu)$. It is essentially because of the pressure in the z-direction in a relativistic setting (unlike for a classical string) that in the Newtonian limit, $\nabla^2\phi=4\pi G(T_{00}-T_{11}-T_{22}-T_{33})$ is identically zero, and hence a test mass will not experience any force due to the string.
For a full general relativistic analysis, one may consult Vilenkin, Phys. Rev. D23 (1981) 852. The solution to the metric (in a cylindrical coordinate system) in such a case is given by
$$
\mathrm ds^2=-\mathrm dt^2+d\rho^2+(1-4G\mu)^2\rho^2\mathrm d\phi^2+\mathrm dz^2
$$
Through a coordinate transformation $\phi\to\phi(1-4G\mu)$, we recover locally flat spacetime in cylindrical coordinates
$$
\mathrm ds^2=-\mathrm dt^2+d\rho^2+\rho^2\mathrm d\phi^2+\mathrm dz^2
$$
with one quirk: $\phi$ now runs from $0$ to $2\pi(1-4G\mu)$, which creates a conical defect in the spacetime. However, for any test particle anywhere away from the string, the geodesics are clearly straight and hence the gravitational attraction produced by the string vanishes. This is very surprising, since it is in stark contrast to the Newtonian prediction that an infinite string generates a non-zero potential of $\vec\phi=2\lambda\ln(\rho/\rho_0)\hat\rho$!
