OPhO 2023, Теория — Задача 3. T3: General Relativity

2023 Online Physics Olympiad: Invitational Contest — Theoretical Examination · 5 август 2023 г.

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Instructions for Theoretical Exam

The theoretical examination consists of 3 long answer questions and 160 points over 2 full days from August 5, 0:01 am GMT.

  • The team leader should submit their final solution document in this google form.
  • If you wish to request a clarification, please use this form. To see all clarifications, view this document.
  • Participants are given a google form where they are allowed to submit up-to 100 megabytes of data for their solutions. It is recommended that participants write their solutions in LaTeX. However, handwritten solutions (or a combination of both) are accepted too. If participants have more than one photo of a handwritten solution (jpg, png, etc), it is required to organize them in the correct order in a pdf before submitting. If you wish a premade LaTeX template, we have made one for you here.
  • Since each question is a long answer response, participants will be judged on the quality of your work. To receive full points, participants need to show their work, including deriving equations. As a general rule of thumb, any common equations (such as the ones in the IPhO formula sheet) can be cited without proof.
  • Remember to state any approximations made and which system of equations were solved after every step. Explicitly showing every step of algebra is not necessary. Participants may leave all final answers in symbolic form (in terms of variables) unless otherwise specified. Be sure to state all assumptions.

Problems

  • T1: Booster
  • T2: The Complex Potential
  • T3: General Relativity
List of Constants
Proton massmp=1.67⋅10−27 kgm_p = 1.67 \cdot 10^{-27}\ \mathrm{kg}
Neutron massmn=1.67⋅10−27 kgm_n = 1.67 \cdot 10^{-27}\ \mathrm{kg}
Electron massme=9.11⋅10−31 kgm_e = 9.11 \cdot 10^{-31}\ \mathrm{kg}
Avogadro’s constantN0=6.02⋅1023 mol−1N_0 = 6.02 \cdot 10^{23}\ \mathrm{mol}^{-1}
Universal gas constantR=8.31 J/(mol⋅K)R = 8.31\ \mathrm{J/(mol \cdot K)}
Boltzmann’s constantkB=1.38⋅10−23 J/Kk_B = 1.38 \cdot 10^{-23}\ \mathrm{J/K}
Electron charge magnitudee=1.60⋅10−19e = 1.60 \cdot 10^{-19}
1 electron volt1 eV=1.60⋅10−19 J1\ \mathrm{eV} = 1.60 \cdot 10^{-19}\ \mathrm{J}
Speed of lightc=3.00⋅108 m/sc = 3.00 \cdot 10^{8}\ \mathrm{m/s}
Universal Gravitational constantG=6.67⋅10−11 (N⋅m2)/kg2G = 6.67 \cdot 10^{-11}\ (\mathrm{N \cdot m^2})/\mathrm{kg}^2
Acceleration due to gravityg=9.81 m/s2g = 9.81\ \mathrm{m/s^2}
1 unified atomic mass unit1 u=1.66⋅10−27 kg=931 MeV/c21\ \mathrm{u} = 1.66 \cdot 10^{-27}\ \mathrm{kg} = 931\ \mathrm{MeV/c^2}
Planck’s constanth=6.63⋅10−34 J⋅s=4.41⋅10−15 eV⋅sh = 6.63 \cdot 10^{-34}\ \mathrm{J \cdot s} = 4.41 \cdot 10^{-15}\ \mathrm{eV \cdot s}
Permittivity of free spaceϵ0=8.85⋅10−12 C2/(N⋅m2)\epsilon_0 = 8.85 \cdot 10^{-12}\ \mathrm{C^2/(N \cdot m^2)}
Coulomb’s law constantk=14πϵ0=8.99⋅109 (N⋅m2)/C2k = \frac{1}{4\pi\epsilon_0} = 8.99 \cdot 10^{9}\ (\mathrm{N \cdot m^2})/\mathrm{C}^2
Permeability of free spaceμ0=4π⋅10−7 T⋅m/A\mu_0 = 4\pi \cdot 10^{-7}\ \mathrm{T \cdot m/A}
Stefan-Boltzmann constantσ=5.67⋅10−8 W/m2/K4\sigma = 5.67 \cdot 10^{-8}\ \mathrm{W/m^2/K^4}

Условие

The Alphabet of Spacetime Metrics

In multivariable calculus, we study the generalization of line elements to various coordinate systems. A coordinate system in R3\mathbb{R}^3 can be described by a function ψ\psi that takes a point (x,y,z)∈R3(x, y, z) \in \mathbb{R}^3 and outputs the three coordinates describing the same point in the coordinate system. For example, the ψ\psi describing spherical coordinates would be defined as ψ(x,y,z)=(r,θ,ϕ)\psi(x, y, z) = (r, \theta, \phi). As there must be a unique representation of each point in R3\mathbb{R}^3 in the ψ\psi coordinates (ex. r=3,θ=π2,ϕ=πr = 3, \theta = \frac{\pi}{2}, \phi = \pi cannot correspond to two points on the 3D space) there must be a one-to-one correspondence between points in R3\mathbb{R}^3 and points described in ψ\psi coordinates. Hence, there’s a well defined inverse function ψ−1\psi^{-1}. In spherical coordinates, for instance, ψ−1(r,θ,ϕ)=(x,y,z)\psi^{-1}(r, \theta, \phi) = (x, y, z).

Suppose we have a coordinate system ψ(x,y,z)=(q1,q2,q3)\psi(x, y, z) = (q_1, q_2, q_3). We ask a very important question: if we make an increment of dqidq_i in each qiq_i in the ψ\psi coordinates, what’s the physical distance between the starting and the finishing points in R3\mathbb{R}^3? In other words, what’s the distance between the points ψ−1(q1+dq1,q2+dq2,q3+dq3)\psi^{-1}(q_1 + dq_1, q_2 + dq_2, q_3 + dq_3) and ψ−1(q1,q2,q3)\psi^{-1}(q_1, q_2, q_3)? This distance is known as the line element generated by a coordinate system ψ\psi. If the coordinate system is orthogonal, meaning the basis vectors q1^\hat{q_1}, q2^\hat{q_2}, q3^\hat{q_3} are all mutually perpendicular, we know that the physical displacements in taking qi→qi+dqiq_i \to q_i + dq_i are in perpendicular directions in R3\mathbb{R}^3, so we can find the line element by Pythagoras’s theorem.

(a) Show that for an orthogonal coordinate system ψ\psi in an arbitrary Rn\mathbb{R}^n, the square of the line element ds2ds^2 has the form ∑i=1ngi(q1,...,qn)dqi2\sum_{i=1}^n g_i(q_1, ..., q_n)dq_i^2 where each gig_i is a function of q1,...,qnq_1, ..., q_n. Find an expression for gig_i in terms of the derivatives of ψ\psi or ψ−1\psi^{-1}.

We can think of physical observers as coordinate systems. If an arbitrary observer moving in space-time measures an event (t,x,y,z)(t, x, y, z), the coordinates seen by the observer can be expressed as ψ(t,x,y,z)\psi(t, x, y, z) by some function ψ\psi. It turns out that special relativity can be described with the math we’ve developed so far. In particular, we can consider measuring space-time lengths by integrating the line element along paths in space-time. You are given that for any observer with the coordinate system ψ≡(t,x,y,z)\psi \equiv (t, x, y, z), the line element is given by ds2=−c2dt2+dx2+dy2+dz2ds^2 = -c^2dt^2 + dx^2 + dy^2 + dz^2

(b) It’s a well-known result of special relativity that space-time lengths are measured to be the same regardless of the frame. Resolve the twin paradox by integrating the line element along the spaceship twin’s world-line.

It turns out that physics in weak gravitational fields can also be described with the math we’ve developed so far. Consider a 1 dimensional system with a time-independent gravitational potential Φ(x)\Phi(x) everywhere.

(c) An important postulate of general relativity states that observers that move solely under the influence of gravity take the path with the longest space-time length. Suppose an observer of this system takes a path parameterized by a variable σ∈[0,σf]\sigma \in [0, \sigma_f]: (t(σ),x(σ),y(σ),z(σ))(t(\sigma), x(\sigma), y(\sigma), z(\sigma)). Show that the line element ds2=−(1+2Φ(x))dt2+(1−2Φ(x))dx2ds^2 = -(1 + 2\Phi(x))dt^2 + (1 - 2\Phi(x))dx^2 generates equations of motion that are consistent with Newton’s laws (neglecting special relativity).

Hint: Refer to the Euler-Lagrange equation, but you do NOT have to perform the derivation to solve this problem.

(d) Derive a statement of time dilation at a point (x,y,z)(x, y, z).

The Spacetime Metric of a Black Hole

As we saw in part 1, the Minkowski metric describes geometry of spacetime in special relativity and is given as

ds2=−c2dt2+dx2+dy2+dz2.ds^2 = -c^2dt^2 + dx^2 + dy^2 + dz^2.

In general relativity, the result is more complex, and we describe spacetime via the Schwarzchild metric, a solution of Einstein’s equations under spherical symmetry. The spacetime interval in spherical coordinates can be expressed in terms of proper time τ\tau and the speed of light cc as

ds2=−c2dτ2=−(1−rsr)c2dt2+dr21−rsr+r2(dθ2+sin⁡2θdϕ2)ds^2 = -c^2d\tau^2 = -\left(1 - \frac{r_s}{r}\right)c^2dt^2 + \frac{dr^2}{1 - \frac{r_s}{r}} + r^2(d\theta^2 + \sin^2\theta d\phi^2)

where tt is the coordinate time, rr is the radial coordinate, θ\theta is the polar angle, ϕ\phi is the azithumal coordinate, and rsr_s is the Schwarzchild radius of a massive body. The Schwarzchild radius is related to the body’s mass MM by rs=2GMc2r_s = \frac{2GM}{c^2} where GG is the universal gravitational constant.

General relativity describes gravity as the consequence of curving spacetime. Geodesics generalize straight lines to curved spacetime and are very useful as a result. In the following problem, the radial geodesic equation may be helpful:

d2rdτ2+Γμνrdμdτdνdτ=0\frac{\mathrm{d}^2r}{\mathrm{d}\tau^2} + \Gamma^r_{\mu\nu}\frac{\mathrm{d}\mu}{\mathrm{d}\tau}\frac{\mathrm{d}\nu}{\mathrm{d}\tau} = 0

where Γ\Gamma represents a Christoffel symbol. The relevant terms of Γ\Gamma are Γttr\Gamma^r_{tt} , Γrrr\Gamma^r_{rr} , Γθθr\Gamma^r_{\theta\theta} , and Γϕϕr\Gamma^r_{\phi\phi}. Define B=1−rsrB = 1 - \frac{r_s}{r}, then we can get:

Γttr=c2BdBdr2Γrrr=−B−1dBdr2Γθθr=−rBΓϕϕr=−Brsin⁡2θ\begin{aligned}\Gamma^r_{tt} &= \frac{c^2B\frac{\mathrm{d}B}{\mathrm{d}r}}{2}\\ \Gamma^r_{rr} &= -\frac{B^{-1}\frac{\mathrm{d}B}{\mathrm{d}r}}{2}\\ \Gamma^r_{\theta\theta} &= -rB\\ \Gamma^r_{\phi\phi} &= -Br\sin^2\theta\end{aligned}

The Photon Sphere

In 2019, the first image of a black hole was taken in 2019 by the Event Horizon Telescope (EHT) at the center of the galaxy M87. The historic image of the black hole showcased a glowing ring of light surrounding the dark abyss known as the ”photon ring.” In this region, light rays follow closed circular paths due to the strong gravitational pull of the black hole. When a photon follows a closed circular path on the photon ring, it can effectively orbit the black hole multiple times before either escaping to infinity or being captured by the event horizon. This phenomenon is known as the ”photon sphere.”

You may recall the massive outburst of media to the images of M87. Many were already talking about it before the images came out, like Veritasium. Take 10 minutes to watch the video.

In this problem, we will be analyzing the optical effects for an observer next to a black hole. Assume for an idealized case that the black hole has a mass M=15M⊙M = 15M_\odot and is uncharged and not rotating.

The Event Horizon Telescope image of the M87 black hole: a glowing orange ring, brighter at the bottom, around a dark centre on a black background.
Figure 2: Event Horizon Telescope Collaboration

(e) Prove that the minimum radius of the black hole’s photon ring follows the relationship rp=32rsr_p = \frac{3}{2}r_s.

(f) Bill is near the black hole a distance rpr_p away. Assume Bill’s width is w=2δw = 2\delta where δ=0.25\delta = 0.25 m. Sketch light ray trajectories showing how Bill could see his own back. Determine the range for the angle of view α\alpha enabling this.

(g) Approximately what is the orbital period (in miliseconds) of a singular photon in coordinate time tt around the black hole. At what mass of a black hole would this time be greater than the visual reaction time of a human tr=150t_r = 150 ms (meaning if you closed your eyes, it would take some time for you to visualize your back again)? Is this mass possible?

(h) Propose a method (realistic or not) for Bill to safely get close to the black hole without getting ripped apart by gravity. If you believe there is no possible method, then give your reasoning behind why. You might want to consult various sources for information. To simplify the process and avoid teams searching for data such as ”shear stress of a human”, assume that Bill is a cylindrical object made of a material of your choosing that has a uniform density, a radius δ\delta, and a length of L=1.75L = 1.75 m. Bill possesses exceptional engineering skills, enabling him to construct virtually anything, provided it doesn’t contravene the principles of physics.

(i) Let’s assume that the galaxy has NstarN_{\mathrm{star}} stars, each emitting light isotropically with the same total power LL. Assume the stars are distributed uniformly throughout a spherical volume of radius RR. For an ideal case where the path of the light from the stars is not obstructed by any other celestial objects, estimate the expected number of photons from starlight that would contribute to the photon ring or fall in the event horizon per unit time.

Отговори

Решение


Оригинал в Архива: theory.pdf · официални решения: theory_sol.pdf