Съдържание

Условие Introduction Part A: Gravitational Field of a Cosmic String (22 т.)Part B: Cosmic string as a photon gas (17 т.)Part C: Gravitational Lensing from cosmic Strings (16 т.)Отговори Решение

IOAA 2021, Теория — Задача 15. Cosmic String (55 points).

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Theory — Q15 · Cosmic String (55 points) · English (Official) · 55 т.

Условие

Introduction

According to our current understanding, just after the Big Bang, when the Universe was extremely hot, the electromagnetic force, the strong nuclear force as well as the weak nuclear force were unified as one Grand Unified (GUT) force.

When the Universe cooled down to $T_{GUT} = 10^{29}\ K$ , the strong nuclear force decoupled from the electroweak force. Later, when the temperature reduced to $T_{EW} = 10^{15}\ K$ , the weak force decoupled from the electromagnetic force. These transitions happened in a rapid succession within a small fraction of a second after the Big Bang. It is thought that these phase transitions produced a variety of peculiar objects, called vacuum defects, which may still be observed today.

This question will discuss properties of one such possible type of defect called cosmic strings and their observational effects.

Note 1. Unless otherwise stated use the laws of Newtonian Mechanics

Note 2. You will use the following constants:

Stefan Boltzmann Constant

\sigma = \frac{2\pi^5 {k_B}^4}{15 h^3 c^2} = \frac{\pi^2 {k_B}^4}{60 \hbar^3 c^2}

The reduced Planck constant

\hbar = \frac{h}{2\pi}

Universal Radiation Constant

a = \frac{4\sigma}{c} = 7.5657 \times 10^{-16} J\ m^{-3} K^{-4}

Planck Temperature

T_{pl} = \sqrt{\frac{\hbar c^5}{G {k_B}^2}} = 1.416784 \times 10^{32} K

Note 3. Recall that the gravitational field $\vec{g}$ satisfies the Gauss theorem:

\vec{g} \cdot \vec{A} = -4\pi G M_{in}

Where $M_{in}$ is the mass enclosed by the surface A.

Part A: Gravitational Field of a Cosmic String (22 т.)

As a first approximation, let us consider a cosmic string as an infinitely long cylinder of radius $r_0$ and mass per unit length $\mu$ .

An infinitely long vertical cylinder of radius r0 with a dashed axis line through its centre.

A.1 Write an expression in terms of the constants $G$ , $\mu$ and $r_0$ for the gravitational field produced by the string, $\vec{g}(r)$ .
Consider the cases $r_0 < r$ and $r_0 > r$ independently [6 т.]

A.2 Write an expression in terms of the constants $G$ , $\mu$ and $r_0$ for $g_0 \equiv |\vec{g}(r_0)|$ . [1 т.]

A.3 Let $g$ be defined as $\vec{g}(r) \cdot \hat{r}$ . Draw a rough sketch of $g$ vs. $r$ in the figure given in the answer sheet

Answer-sheet coordinate grid for the gravitational field sketch in A.3.

[3 т.]

A.4 It is possible to define a stable orbit around a Cosmic String. For circular orbits of radius $R > r_0$ and period $\tau$ , the following relation is attained

R = A \tau^{\alpha}

where $A$ and $\alpha$ are constants. Find $A$ and $\alpha$ in terms of $G$ and $\mu$ [4 т.]

A.5 The following three questions refers to a classical newtonian particle moving with speed $v$ when at a distance $r > r_0$ from the string. You will need to use the result below:

\int_{x_0}^{x} \frac{dx}{x} = \ln\left(\frac{x}{x_0}\right)

Show that the gravitational potential energy of the particle is

U = G m \mu\ \ln\left(\frac{r}{b}\right)

where $b$ is any fixed distance. [3 т.]

A.6 What is the maximum distance, $R_{\max}$ , from the string, that the particle can reach? [4 т.]

A.7 Is it possible for the particle to escape the gravitational field? Write YES/NO in the answer sheet. [1 т.]

Part B: Cosmic string as a photon gas (17 т.)

Consider now a cosmic string as a photon gas inside a very long cylinder of radius $r_0$ with adiabatic walls, and in thermal equilibrium at temperature $T$ .

B.1 What is the energy density $\rho$ of the string in terms of $T$ , $\hbar$ , $k_B$ and $c$ ? [2 т.]

B.2 The radius $r_0$ is related to the temperature $T$ vía

r_0 = \frac{\hbar^{n_1}\ c^{n_2}}{k_B T},

where $\hbar$ is the reduced Planck constant, and $c$ is the speed of light in vacuum, $k_B$ is the Boltzmann constant, and $n_1$ and $n_2$ are integer numbers. Determine $n_1$ and $n_2$ [4 т.]

B.3 What is the mass per unit length, $\mu$ , of the string in terms of $\rho$ and $r_0$ ? [2 т.]

B.4 Express the inequality for the weak field condition, defined as

\frac{2G\mu}{c^2} \ll 1,

only in terms of $T$ and $T_{pl}$ . [5 т.]

B.5 Calculate $\frac{2G\mu}{c^2}$ for

$T = T_{EW}$
$T = T_{GUT}$ [3 т.]

B.6 Does the weak field condition hold for $T_{EW}$ ? Answer YES or NOT.
Does the weak field condition hold for $T_{GUT}$ ? Answer YES or NOT. [1 т.]

Part C: Gravitational Lensing from cosmic Strings (16 т.)

So far, in part $A$ and $B$ , we have neglected the internal pressure of the photon gas inside the string. If we include it in our analysis, we need to consider the General Theory of Relativity.

After solving the Einstein field equations, one finds that the spacetime around a cosmic string is conical as if a narrow wedge were removed from a flat sheet and the edges connected, as shown below.

Illustration of a conical spacetime: a circle with a narrow wedge removed (left) and the resulting cone with two light rays converging (right). — http://www.ctc.cam.ac.uk/outreach/origins/cosmic_structures_five.php

A remarkable result of this model is light deflection by a cosmic string, which leads to the possibility of detection through gravitational lensing.

The angle of deflection (in radians) of a light ray coming from a distant quasar (O in the figure below), as the light passes close to a cosmic string (S in the figure below) and eventually reaching an observer on the Earth, (E in the figure below), is

\delta \phi = \frac{4\pi G\mu}{c^2}

and is independent of the parameter, $p$ , as shown in the figure below:

In the figure $E$ and $O$ are in a plane perpendicular to the string. The distance between the observer and the string is $D_{ES}$ and the distance between the observer and the source is $D_{OE}$

Points E and S on a horizontal line; O lies above the rightmost point at perpendicular distance p.

C.1 Although the angle of deflection does not depend on parameter $p$ , an Earth-based observer will be able to see more than one image only if the value of $p$ is within a certain range. Find a condition on the value of the parameter $p$ in terms of $D_{ES}$ , $D_{OE}$ , and temperature $T$ , for an Earth-based observer to see more than one image of the object $O$ [6 т.]

C.2 In case the observer sees more than one image, what is the angular separation between each pair? Find an expression in terms of $D_{ES}$ , $D_{OE}$ and $\delta \phi$ [6 т.]

C.3 If $D_{OE} = 2D_{ES}$ , determine the minimum size of an optical telescope needed to resolve this lensing event produced by GUT string. [4 т.]

Отговори

Решение

Part A

A.1 Due to cylindrical symmetry,

\vec g=\vec g(r)=g(r)\hat r=g\hat r.

[1 point]

By axial symmetry we can apply Gauss's theorem. [1 point]

\vec g\cdot\vec A=-4\pi GM_{in},

where $M_{in}$ is the mass inside the surface $A$ , as shown in the figure.

Cylindrical Gaussian surface around the cosmic string.

For a cylinder of length $L$ and radius $r$ :

2\pi rLg=-4\pi GM_{in},\qquad g=-\frac{2GM_{in}}{rL}.

[1 point]

There are two regions:

$r>r_0$ : $M_{in}=\mu L$ [0.5 points], so $g=-2G\mu/r$ [1 point].
$r<r_0$ : $M=(r^2/r_0^2)\mu L$ [0.5 points], so $g=-(2G\mu/r_0)(r/r_0)$ [1 point].

A.2 Writing $\vec g(r_0)$ in terms of $G$ , $\mu$ and $r_0$ :

\vec g(r_0)=-\frac{2G\mu}{r_0}\hat r,\qquad g_0\equiv|\vec g(r_0)|=\frac{2G\mu}{r_0}.

[1 point]

A.3 The following points should be included in the sketch:

$(0,0)$ [0.5 points]
$(1,-1)$ [0.5 points]

Additionally,

the sketch is linear between $(0,0)$ and $(1,-1)$ [0.8 points];
the sketch is concave for $r>r_0$ , $(\sim1/r)$ [0.8 points];
the sketch is drawn below the x-axis [0.4 points].

A.4 The centripetal acceleration equals the gravitational acceleration, thus

\frac{v^2}{R}=\frac{2G\mu}{R}.

[1 point]

\frac{2\pi R}{\tau}=\sqrt{2G\mu}.

[1 point]

R=\frac{\sqrt{2G\mu}}{2\pi}\tau,

A=\frac{\sqrt{2G\mu}}{2\pi}.

[1 point]

\alpha=1.

[1 point]

A.5 The potential energy of the particle is given by

U(r)=-\int_b^r\vec F(r)\cdot d\vec r=m\int_b^r g(r)\,dr.

[1 point]

=2Gm\mu\int_b^r\frac1r\,dr

[1 point]

=2Gm\mu\ln\left(\frac rb\right).

[1 point]

Where $b$ is a constant that sets the 0 of $U$ .

Note 1: Usually, $U=0$ is set at $b=\infty$ . For a cosmic string this is not possible. Instead, we can choose $U=0$ , for example, at the string's surface $b=r_0$ (this choice is not relevant!). Thus

U=2Gm\mu\ln\left(\frac r{r_0}\right).

Note 2: For completeness, notice that the potential inside the string, $r<r_0$ is

U=2Gm\mu\left(\frac r{r_0}\right)^2-U_0,

where

U_0=2Gm\mu\left(\frac r{r_0}\right)^2-2Gm\mu\ln\left(\frac r{r_0}\right)

is a constant (again, not relevant!) that can be chosen such that $U$ is continuous at the surface of the string. However, for the solution of the problem, it is not necessary to show this.

A.6 The total energy of the particle is conserved, thus:

\frac12mv^2+2Gm\mu\ln\left(\frac rb\right)=2Gm\mu\ln\left(\frac{R_{max}}b\right).

[2 points]

Solving for $R_{max}$ :

R_{max}=R e^{v^2/(4G\mu)}.

[2 points]

Notice that the answer does not depend on $b$ .

A.7 NO. [1 point]

From the previous result,

R_{max}=R e^{v^2/(4G\mu)},

we see that it is not possible to escape the gravitational field since for any speed, there is always a maximum distance $R_{max}<\infty$ .

Part B

B.1 The energy density is given by

\rho=aT^4.

[1 point]

Equivalently:

\rho=\frac{4\sigma}{c}T^4=\frac{\pi^2k_B^4}{15\hbar^3c^3}T^4.

[1 point]

Note 3: This result arrives from the integration of the spectral energy density given by the Planck law over all frequencies. However, it is not required to show this integral.

B.2

r_0=\frac{\hbar^{n_1}c^{n_2}}{k_BT}.

$r_0$ has dimensions of length: $[l]$ .
$\hbar$ has dimensions of energy × time: $[e\ t]$ .
$c$ has dimensions of speed: $[l/t]$ .
$k_BT$ has dimensions of energy: $[e]$ .

Then:

[l]=\frac{[e\ t]^{n_1}[l/t]^{n_2}}{[e]}.

Since the LHS and the RHS should have the same dimensions, we get:

[l]:1=n_2,\qquad[e]:0=n_1-1,\qquad[t]:n_1=n_2.

[2 points]

The solution is

n_1=1

[1 point]

n_2=1.

[1 point]

B.3 The energy of a piece of the string of length $L$ is

E=\rho\pi r_0^2L=Mc^2.

[1 point]

where $M=\mu L$ is the mass of the piece. Solving for $\mu$ :

\mu=\frac{\rho\pi r_0^2}{c^2}.

[1 point]

B.4 The weak field condition is

\frac{2G\mu}{c^2}\ll1,

2G\frac{\rho\pi r_0^2}{c^2}\ll1

[1 point]

2G\frac\pi{c^2}(aT^4)\left(\frac{\hbar c}{k_BT}\right)^2\ll1

[1 point]

\frac{2Ga\hbar^2\pi}{c^2k_B^2}T^2\ll1.

[2 points]

Where we used:

\mu=\frac{\rho\pi r_0^2}{c^2},\qquad\rho=aT^4,\qquad r_0=\frac{\hbar c}{k_BT}.

\frac{2Ga\hbar^2\pi}{c^2k_B^2}T^2\ll1,

\frac{2G\hbar^2\pi}{c^2k_B^2}\left(\frac{\pi^2k_B^4}{15\hbar^3c^3}\right)T^2\ll1,

\frac{2\pi^3}{15}\left(\frac{Gk_B^2}{\hbar c^5}\right)T^2\ll1,

\frac{2\pi^3}{15}\frac{T^2}{T_{Pl}^2}\ll1.

[2 points]

where $T_{Pl}=1.416784\times10^{32}\ K$ (known as the Planck Temperature).

Note 4: the numerical factor $2\pi^3/15\sim4.13$ (a 5% error tolerance is accepted).

Note 5: From A.4 we get that

2G\mu=v^2,

where $v$ is the speed with which a particle would orbit a string. Thus the weak field is rewritten as

\frac{v^2}{c^2}\ll1,

which is equivalent to a non-relativistic condition.

B.5 The weak field condition is

4.13\left(\frac T{T_{Pl}}\right)^2\ll1,

equivalently

\sim2\frac T{T_{Pl}}\ll1.

[1 point]

$2T_{EW}/T_{Pl}\sim2\times10^{15}/10^{32}\approx1.4\times10^{-17}\ll1$ [1 point].
$2T_{GUT}/T_{Pl}\sim2\times10^{29}/10^{32}\approx1.4\times10^{-3}\ll1$ [1 point].

Note 6: Using the following expression

4.13\left(\frac T{T_{Pl}}\right)^2\ll1

one gets

$4.13(T_{EW}/T_{Pl})^2\sim2.1\times10^{-34}\ll1$ .
$4.13(T_{GUT}/T_{Pl})^2\sim2.1\times10^{-6}\ll1$ .

And are also acceptable answers (a 5% error tolerance is accepted).

B.6

Yes [0.5 points].
Yes [0.5 points].

Part C

C.1 From the figure, the asymptotic condition for the observer to see a second image is that the light ray travelling from O directly to S should bend and travel along SE. This is the maximum angle of bending.

As all angles are small, we can safely use

\sin x\approx\tan x\approx x.

Now,

\frac p{SP}<\delta\phi

[2 points]

<\frac{4\pi G\mu}{c^2}<2\pi\left(\frac{2G\mu}{c^2}\right)<2\pi\left(\frac{2\pi^3}{15}\right)\left(\frac{T^2}{T_{Pl}^2}\right).

[2 points]

And $SP\approx(D_{OE}-D_{ES})$ ,

p<\frac{4\pi^4}{15T_{Pl}^2}T^2(D_{OE}-D_{ES}).

[2 points]

Two light paths and their apparent source positions.

C.2 In the figure, the blue lines represent the bending of two light rays corresponding to the images O1 and O2. Note that $D_{ES}\approx D_{ES1}\approx D_{ES2}$ and $D_{OS}\approx D_{OS1}\approx D_{OS2}$ . Thus the angle $S1EO\approx S2EO\equiv\alpha$ and $S1OE\approx S2OE\equiv\beta$ . $2\alpha$ is the angular separation we are looking for. [2 points]

Further, notice

\alpha+\beta=\delta\phi.

[1 point]

Using law of sines in the triangle EOS1

\frac\alpha{D_{S1O}}=\frac\beta{D_{ES1}}.

[1 point]

we get

2\alpha=2\delta\phi\left(\frac{D_{OE}-D_{ES}}{D_{OE}}\right).

[2 points]

C.3 If $D_{OE}=2D_{ES}$ ,

2\alpha=\delta\phi=2\pi\left(\frac{2\pi^3}{15}\right)\left(\frac{T_{GUT}^2}{T_{Pl}^2}\right)\approx1.29\times10^{-5}.

[2 points]

and

\delta\phi=1.22\frac\lambda D.

Thus, using $\lambda\in[3\times10^{-7}\ m,8\times10^{-7}\ m]$ ,

D=1.22\frac\lambda{\delta\phi}\in[3.75\times10^{-2}\ m,7.51\times10^{-2}\ m].

[2 points]

Any answer within this interval is valid.

A remarkable result of this model is light deflection by a cosmic string, which leads to the possibility of detection through gravitational lensing. For instance, cosmic string trings moving across the line of sight with respect to an Earth-based observer will cause line-like discontinuities as shown in the figure below (taken from [1]).

Three examples of galaxies lensed by cosmic strings.

Figure 6. (a) This picture represents the lensed image of a spiral galaxy. The direction of a string is almost perpendicular to galaxy plane. Duplicated details are clearly visible. (b) In this case, the string is inclined with respect to the galaxy plane. As a result sharp edge appears. (c) Also in this case the direction of string is inclined with respect to galactic plane.

[1] M. V. Sazhin, O. S. Khovanskaya, M. Capaccioli, G. Longo, M. Paolillo, G. Covone, N. A. Grogin, E. J. Schreier, Gravitational lensing by cosmic strings: what we learn from the CSL-1 case, Monthly Notices of the Royal Astronomical Society, Volume 376, Issue 4, March 2007, Pages 1731–1739, https://doi.org/10.1111/j.1365-2966.2007.11543.x.

Official source discrepancies

Question A.5 prints U = G m mu ln(r/b), while the official key prints a factor2. The key also prints a positive m integral of g after the negative force integral, an interior potential and U0 expression as reproduced, and uppercase R in the A.6 result despite lowercase r in its energy equation. In B.4 the intermediate c powers and printed marking (1+1+2+2, against5 question points) are retained. B.5 accepts both its transformed weak-field quantities and their squared equivalents; the printed approximations are retained. C.1 uses chained strict inequalities; C.3 gives wavelength endpoints3 and8 times10^-7 m but diameter endpoints3.75 and7.51 times10^-2 m. These source inconsistencies are not silently corrected.

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