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# 6CPP3630

## 相对论代考

Consider the metric
$$d s^{2}=d t^{2}-d r^{2}-r^{2} d \phi^{2} .$$
Find expressions for the covariant derivative $\nabla_{\mu} V_{\nu}$ and the divergence $\nabla_{\mu} V^{\mu}$.

Using the given metric $d s^{2}=d t^{2}-d r^{2}-r^{2} d \phi^{2}$, we find that
$$g_{t t}=g^{t t}=1, \quad g_{r r}=g^{r r}=-1, \quad g_{\phi \phi}=\frac{1}{g^{\phi \phi}}=-r^{2},$$
which only give rise to one nonzero ordinary derivative, i.e., $\partial_{r} g_{\phi \phi}=-2 r$. Using the general expression for the Christoffel symbols, we obtain the following nonzero Christoffel symbols
$$\Gamma_{r \phi}^{\phi}=\Gamma_{\phi r}^{\phi}=\frac{1}{r}, \quad \Gamma_{\phi \phi}^{r}=-r$$
Using the definition of the covariant derivative $\nabla_{\mu} V_{\nu}=\partial_{\mu} V_{\nu}-\Gamma_{\mu \nu}^{\lambda} V_{\lambda}$, we obtain the components of the covariant derivative

The Christoffel symbols for the flat Euclidean metric in $\mathbb{R}^{3}$ vanish. Compute the Christoffel symbols in the spherical coordinates $(r, \theta, \varphi)$.

a) We can express $\theta$ and $\phi$ in terms of $x$ and $y$ as
$$\theta=2 \arctan \left(\frac{r}{2 R}\right), \quad \phi=\arctan \left(\frac{y}{x}\right),$$
where $r^{2}=x^{2}+y^{2}$. We can now use the relations
$$d \theta=\frac{\partial \theta}{\partial x} d x+\frac{\partial \theta}{\partial y} d y, \quad d \phi=\frac{\partial \phi}{\partial x} d x+\frac{\partial \phi}{\partial y} d y$$
to obtain the following expressions for $d \theta^{2}$ and $d \phi^{2}$ :
$d \theta^{2}=\frac{1}{r^{2} R^{2}} \frac{1}{\left(1+\frac{r^{2}}{4 R^{2}}\right)^{2}}\left(x^{2} d x^{2}+y^{2} d y^{2}+2 x y d x d y\right)$ $$d \phi^{2}=\frac{1}{r^{4}}\left(y^{2} d x^{2}+x^{2} d y^{2}-2 x y d x d y\right)$$ We can also express $\sin \theta$ in terms of $r$ as $$\sin \theta=2 \cos \frac{\theta}{2} \sin \frac{\theta}{2}=\frac{2 \tan \frac{\theta}{2}}{1+\tan ^{2} \frac{\theta}{2}}=\frac{4 R r}{4 R^{2}+r^{2}}$$
Inserting into the original metric, we obtain
$$d s^{2}=\left(\frac{1}{1+\frac{r^{2}}{4 R^{2}}}\right)^{2}\left(d x^{2}+d y^{2}\right)$$
b) The Christoffel symbols can be computed from
$$\Gamma_{v \lambda}^{\mu}=\frac{1}{2} g^{\mu \omega}\left(\partial_{\nu} g_{\lambda \omega}+\partial_{\lambda} g_{\nu \omega}-\partial_{\omega} g_{v \lambda}\right)$$
Using the fact that this is symmetric under $v \leftrightarrow \lambda$ as well as that the metric is symmetric under $x \leftrightarrow y$, we only need to compute three of the Christoffel symbols:
\begin{aligned} &\Gamma_{x x}^{x}=-\frac{x}{2\left(R^{2}+r^{2} / 4\right)} \\ &\Gamma_{x y}^{x}=-\frac{y}{2\left(R^{2}+r^{2} / 4\right)} \\ &\Gamma_{y y}^{x}=-\Gamma_{x x}^{x} \end{aligned}
c) The stereographic projection can be visualized according to Figure

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# 广义相对论写作有小技巧吗? Viking Essay教你如何在物理dissertation中得高分!

1

## The Einstein Equations

It is now time to do some physics. The force of gravity is mediated by a gravitational field. The glory of general relativity is that this field is identified with a metric $g_{\mu \nu}(x)$ on a 4d Lorentzian manifold that we call spacetime.

This metric is not something fixed; it is, like all other fields in Nature, a dynamical object. This means that there are rules which govern how this field evolves in time. The purpose of this section is to explore these rules and some of their consequences.

All our fundamental theories of physics are described by action principles. Gravity is no different. Furthermore, the straight-jacket of differential geometry places enormous restrictions on the kind of actions that we can write down. These restrictions ensure that the action is something intrinsic to the metric itself, rather than depending on our choice of coordinates.

Spacetime is a manifold $M$, equipped with a metric of Lorentzian signature. An action is an integral over $M$. We know from Section 2.4.4 that we need a volume-form to integrate over a manifold. Happily, as we have seen, the metric provides a canonical volume form, which we can then multiply by any scalar function. Given that we only have the metric to play with, the simplest such (non-trivial) function is the Ricci scalar $R$. This motivates us to consider the wonderfully concise action
$$S=\int d^{4} x \sqrt{-g} R$$
This is the Einstein-Hilbert action. Note that the minus sign under the square-root arises because we are in a Lorentzian spacetime: the metric has a single negative eigenvalue and so its determinant, $g=\operatorname{det} g_{\mu \nu}$, is negative.

As a quick sanity check, recall that the Ricci tensor takes the schematic form (3.39) $R \sim \partial \Gamma+\Gamma \Gamma$ while the Levi-Civita connection itself is $\Gamma \sim \partial g .$ This means that the Einstein-Hilbert action is second order in derivatives, just like most other actions we consider in physics.

## When Gravity is Weak

The elegance of the Einstein field equations ensures that they hold a special place in the hearts of many physicists. However, any fondness you may feel for these equations will be severely tested if you ever try to solve them. The Einstein equations comprise ten, coupled partial differential equations. While a number of important solutions which exhibit large amounts symmetry are known, the general solution remains a formidable challenge.

We can make progress by considering situations in which the metric is almost flat. We work with $\Lambda=0$ and consider metrics which, in so-called almost-inertial coordinates $x^{\mu}$, takes the form
$$g_{\mu \nu}=\eta_{\mu \nu}+h_{\mu \nu}$$
Here $\eta_{\mu \nu}=\operatorname{diag}(-1,+1,+1,+1)$ is the Minkowski metric. The components $h_{\mu \nu}$ are assumed to be small perturbation of this metric: $h_{\mu \nu} \ll 1$.

Our strategy is to expand the Einstein equations to linear order in the small perturbation $h_{\mu \nu}$. At this order, we can think of gravity as a symmetric “spin 2 ” field $h_{\mu \nu}$ propagating in flat Minkowski space $\eta_{\mu \nu}$. To this end, all indices will now be raised and lowered with $\eta_{\mu \nu}$ rather than $g_{\mu \nu}$. For example, we have
$$h^{\mu \nu}=\eta^{\mu \rho} \eta^{\nu \sigma} h_{\rho \sigma}$$
Our theory will exhibit a Lorentz invariance, under which $x^{\mu} \rightarrow \Lambda_{\nu}^{\mu} x^{\nu}$ and the gravitational field transforms as
$$h^{\mu \nu}(x) \rightarrow \Lambda^{\mu}{ }_{\rho} \Lambda_{\sigma}^{\nu} h^{\rho \sigma}\left(\Lambda^{-1} x\right)$$
In this way, we construct a theory around flat space that starts to look very much like the other field theories that we meet in physics.

## Black Holes

The Schwarzschild Solution

We have already met the simplest black hole solution back in Section 1.3: this is the Schwarzschild solution, with metric
$$d s^{2}=-\left(1-\frac{2 G M}{r}\right) d t^{2}+\left(1-\frac{2 G M}{r}\right)^{-1} d r^{2}+r^{2}\left(d \theta^{2}+\sin ^{2} \theta d \phi^{2}\right)$$
It is not hard to show that this solves the vacuum Einstein equations $R_{\mu \nu}=0$. Indeed, the calculations can be found in Section $4.2$ where we first met de Sitter space. The Schwarzschild solution is a special case of the more general metric $(4.9)$ with $f(r)^{2}=$ $1-2 G M / r$ and it’s simple to check that this obeys the Einstein equation which, as we’ve seen, reduces to the simple differential equations $(4.10)$ and $(4.11)$.
$\mathrm{M}$ is for Mass
The Schwarzschild solution depends on a single parameter, $M$, which should be thought of as the mass of the black hole. This interpretation already follows from the relation to Newtonian gravity that we first discussed way back in Section $1.2$ where we anticipated that the $g_{00}$ component of the metric should be $(1.26)$
$$g_{00}=-(1+2 \Phi)$$
with $\Phi$ the Newtonian potential. We made this intuition more precise in Section 5.1.2 where we discussed the Newtonian limit. For the Schwarzschild metric, we clearly have
$$\Phi=-\frac{G M}{r}$$
which is indeed the Newtonian potential for a point mass $M$ at the origin.
The black hole also provides an opportunity to roadtest the technology of Komar integrals developed in Section 4.3.3. The Schwarzschild spacetime admits a timelike Killing vector $K=\partial_{t}$. The dual one-form is then
$$K=g_{00} d t=-\left(1-\frac{2 G M}{r}\right) d t$$

Following the steps described in Section 4.3.3, we can then construct the 2 -form
$$F=d K=-\frac{2 G M}{r^{2}} d r \wedge d t$$
which takes a form similar to that of an electric field, with the characteristic $1 / r^{2}$ fall-off. The Komar integral instructs us to compute the mass by integrating
$$M_{\text {Komar }}=-\frac{1}{8 \pi G} \int_{\mathbf{S}^{2}} \star F$$
where $\mathbf{S}^{2}$ is any sphere with radius larger than the horizon $r=2 G M$. It doesn’t matter which radius we choose; they all give the same answer, just like all Gaussian surfaces outside a charge distribution give the same answer in electromagnetism. Since the area of a sphere at radius $r$ is $4 \pi r^{2}$, the integral gives
$$M_{\text {Komar }}=M$$
for the Schwarzschild black hole.
There’s something a little strange about the Komar mass integral. As we saw in Section 4.3.3, the 2 -form $F=d K$ obeys something very similar to the Maxwell equations, $d \star F=0$. But these are the vacuum Maxwell equations in the absence of any current, so we would expect any “electric charge” to vanish. Yet this “electric charge” is precisely the mass $M_{\text {Komar }}$ which, as we have seen, is distinctly not zero. What’s happening is that, for the black hole, the mass is all localised at the origin $r=0$, where the field strength $F$ diverges.

We might expect that the Schwarzschild solution only describes something physically sensible when $M \geq 0$. (The $M=0$ Schwarzschild solution is simply Minkowski spacetime.) However, the metric (6.1) is a solution of the Einstein equations for all values of $M$. As we proceed, we’ll see that the $M<0$ solution does indeed have some rather screwy features that make it unphysical.

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# 固体物理论文写作有小技巧吗? Viking Essay教你如何在物理dissertation中得高分!

0

## Scattering and the Correlation Function

We ended the last lecture with a brief discussion of the connection between scattering experiments and measurements of the correlation function $S(\vec{q}, \omega)$. In this lecture we will discuss scattering in more depth in terms of two concrete examples (electron and neutron scattering). After that, we will look at some more general properties of response functions.
1.1 Scattering
The picture we have is of some blob of material, with a plane wave $\left|\vec{k}_{i}\right\rangle$ coming in, and a different plane wave $\left|\vec{k}_{f}\right\rangle$ coming out. We define the momentum and energy transfer to the sample
\begin{aligned} \vec{Q} &=\vec{k}_{i}-\vec{k}_{f} \\ \omega &=E_{\vec{k}_{i}}-E_{\vec{k}_{f}} \end{aligned}
Let $\vec{R}$ be the coordinate of the scattering particle. Recall from last time that application of Fermi’s Golden Rule and the $1^{\text {st }}$ order Born Approximation leads to the differential rate
\begin{aligned} W_{i \rightarrow[f]} d^{3} k_{f} &=2 \pi \sum_{n}\left|\sum_{\vec{q}} v_{\vec{q}}\left\langle n\left|\hat{\rho}_{\vec{q}}^{\dagger}\right| \phi_{0}\right\rangle \int d \vec{R} e^{i\left(\vec{k}_{f}-\vec{k}_{i}\right) \cdot \vec{R}} e^{-i \vec{q} \cdot \vec{R}}\right|^{2} \delta\left(\omega-\left(E_{n}-E_{0}\right)\right) d^{3} k_{f}(1.3) \\ &=\left|v_{\vec{Q}}\right|^{2} 2 \pi \sum_{n}\left|\left\langle n\left|\hat{\rho}_{\vec{Q}}^{\dagger}\right| \phi_{0}\right\rangle\right|^{2} \delta\left(E_{f}-E_{i}\right) d^{3} k_{f} \\ &=\left|v_{\vec{Q}}\right|^{2} S(\vec{Q}, \omega) d^{3} k_{f} \\ P(\vec{Q}, \omega) &=\left|v_{\vec{Q}}\right|^{2} S(\vec{Q}, \omega) \end{aligned}
for scattering into a final state with momentum somewhere in a volume element $d^{3} k_{f}$ of momentum space centered on $k_{f} .$ Here, $v_{\vec{Q}}$ is the Fourier Transform of the interaction potential. The key result here is that the rate of scattering with momentum transfer $\vec{Q}$ and energy loss $\omega$ is directly proportional to the correlation function $S(\vec{Q}, \omega)$.

## Application: Electron Energy Loss Spectroscopy (EELS)

The experiment we imagine here is that of shooting high energy electrons ( $100 \mathrm{keV})$ at a thin film of material, and collecting them as they emerge with an energy-resolved detector. For this case, the interaction potential is just the Coulomb interaction between the electron and the sample’s charge density, so
$$\left|v_{\vec{q}}\right|=\frac{4 \pi e^{2}}{q^{2}}$$
Recall the definition
\begin{aligned} \frac{1}{\epsilon(\vec{q}, \omega)} &=\frac{U_{T o t}}{U_{E x t}} \\ &=1+\frac{U_{s c r}}{U_{E x t}} \end{aligned}
Remembering that $U_{s c r}(\vec{q})=\frac{4 \pi e^{2}}{q^{2}} \delta n(\vec{q})$, where $n(\vec{q})$ are the Fourier components of the density fluctuations,
$$\frac{1}{\epsilon(\vec{q}, \omega)}=1+\frac{4 \pi e^{2}}{q^{2}} \frac{\delta n(\vec{q}, \omega)}{U_{E x t}(\vec{q}, \omega)}$$
As defined in the previous lecture, the (linear) density response function $\chi(\vec{q}, \omega)$ is defined by the ratio
$$\chi(\vec{q}, \omega)=\frac{\delta n(\vec{q}, \omega)}{U_{E x t}(\vec{q}, \omega)}$$
Substituting this into the relation for $\frac{1}{\epsilon(\vec{q}, \omega)}$, we get
$$\frac{1}{\epsilon(\vec{q}, \omega)}=1+\frac{4 \pi e^{2}}{q^{2}} \chi(\vec{q}, \omega)$$
With $\chi^{\prime \prime}(\vec{q}, \omega)$ defined as the imaginary part of $\chi$, the relation
$$S(\vec{q}, \omega)=-2 \chi^{\prime \prime}(\vec{q}, \omega)$$
combined with equation (1.7) for the scattering rate into momentum space volume $d^{3} k_{f}$ gives the following relation for the scattering rate in terms of the dielectric function:
$$P(\vec{q}, \omega)=\frac{8 \pi e^{2}}{q^{2}}\left(-\operatorname{Im}\left[\frac{1}{\epsilon(\vec{q}, \omega)}\right]\right)$$
What useful information can we get out of this? For one, we are able to investigate the dielectric constant at finite values of $\vec{q}\left(0\right.$ to $k_{F}$ ). In optical experiments, the vanishingly small photon momentum in comparison with typical electron/nucleus momenta means that we are only able to investigate the $\vec{q} \approx 0$ regime with photons.

On the downside, the best energy resolution we can achieve today is around $0.1 \mathrm{eV}$, which is far too coarse to obtain much useful information. This energy resolution is already $1: 10^{6}$ when compared with the total electron energy of around $100 \mathrm{keV}$. To get around this, one might

consider trying lower energy experiments. However, the problem with low energy experiments is that the probability of multiple scattering events within the sample becomes significant, leading to complicated and messy results.

With EELS, we can also look at high energy excitations of the electrons in a metal. Recall that there is a high energy collective mode of the sample electrons at a frequency equal to the plasma frequency $\omega_{p l}$. The plasma frequency is defined in terms of the zero of the dielectric function
$$\epsilon\left(\vec{q}, \omega_{p l}\right)=0$$
The situation where the dielectric function becomes zero is interesting, because it represents a singularity in the system’s response to an external perturbation:
$$\frac{1}{\epsilon\left(\vec{q}, \omega_{p l}\right)}=\frac{U_{T o t}}{U_{E x t}}$$
Thus even a tiny perturbation at the plasma frequency results in a large response of the system.

## Application: Neutron Scattering

Since neutrons are uncharged, they do not see the electrons as they fly through a piece of material $^{1}$. The dominant scattering mechanism is through a contact potential with the nuclei of the sample
$$V(\vec{r})=\frac{2 \pi b}{M_{n}} \delta(\vec{r})$$
where $b$ is the scattering length and $M_{n}$ is the mass of the neutron. Since the Fourier transform of a delta function in space has no $\vec{q}$ dependence, the Fourier components of the interaction potential are all simply
$$v_{\vec{q}}=\frac{2 \pi b}{M_{n}}$$
Inserting this into equation (1.7) for the scattering rate, we get
$$P(\vec{Q}, \omega)=\left(\frac{2 \pi b}{M_{n}}\right)^{2} S(\vec{Q}, \omega)$$
Here, $S(\vec{Q}, \omega)$ is the correlation for the nuclear positions (density)
$$S(\vec{Q}, \omega)=\int d t e^{i \omega t}\left\langle\hat{\rho}_{\vec{Q}}(t) \hat{\rho}_{-\vec{Q}}(0)\right\rangle_{T}$$
with
$$\hat{\rho}_{\vec{Q}}=\sum_{i} e^{i \vec{Q} \cdot \vec{R}_{i}(t)}$$

where $\left\{\vec{R}_{i}(t)\right\}$ are the coordinates of the nuclei at time $t$. Now we can substitute this in to the expression for $S(\vec{Q}, \omega)$
$$S(\vec{Q}, \omega)=\int d t e^{i \omega t} \sum_{j, \ell}\left\langle e^{-i \vec{Q} \cdot \vec{R}_{j}(t)} e^{i \vec{Q} \cdot \vec{R}_{\ell}(0)}\right\rangle_{T}$$
To make progress, we must put in a specific form for $\vec{R}_{j}(t)$. We consider the case of small distortions from a Bravais lattice:
$$\vec{R}_{j}=\vec{R}_{j}^{0}+\vec{u}_{j}$$
where $\left\{\vec{R}_{j}^{0}\right\}$ are the Bravais lattice sites, and $\left\{\vec{u}_{j}\right\}$ are small displacements. The $\left\{\vec{u}_{j}\right\}$ can be expanded in phonon coordinates, yielding
$$\vec{u}_{j}=\sum_{\alpha} \sum_{\vec{q}} \vec{\lambda}_{\alpha} \frac{1}{\sqrt{2 N M \omega_{\vec{q}}}}\left(\hat{a}_{\vec{q}} e^{i\left(\overrightarrow{(} \cdot \vec{R}-\omega_{q}(t)\right)}+\hat{a}_{\vec{q}}^{\dagger} e^{-i\left(\vec{q} \cdot \vec{R}-\omega_{q}(t)\right)}\right)$$
where the sum over $\alpha$ is a sum over all phonon polarizations, $\vec{\lambda}_{\alpha}$ is the polarization of the $\alpha^{t h}$ mode.
After some algebra (see problem set), it can be shown that this decomposition yields
\begin{aligned} &S(\vec{Q}, \omega) \propto e^{-2 W}\left[\sum_{\vec{Q}} \delta(\vec{Q}-\vec{G}) \delta(\omega)+\sum_{\vec{q}} \frac{Q^{2}}{2 N M \omega_{\vec{q}}}\left\{\left(n_{\vec{q}}+1\right) \sum_{\vec{G}} \delta(\vec{Q}-\vec{q}-\vec{G}) \delta\left(\omega-\omega_{q}\right) 1.26\right)\right. \\ &\left.\left.\quad+n_{\vec{q}} \sum_{\vec{G}} \delta(\vec{Q}+\vec{q}-\vec{G}) \delta\left(\omega+\omega_{\vec{q}}\right)\right\}\right] \end{aligned}
where $W$ is the Debye-Waller factor, and $n_{\vec{q}}$ is the Bose statistical occupation factor.
There are several interesting features about this expression for the correlation function. The first term corresponds to simple elastic Bragg scattering through a momentum transfer $\vec{Q}$. Even in the presence of fluctuations, this term is still a sum of delta function peak. Thus the effect of fluctuations on the Bragg peaks is only to decrease their amplitude via $e^{-2 W}$, and not to induce any broadening.

The $2^{\text {nd }}$ and $3^{\text {rd }}$ terms give rise to peaks at $\pm \hbar \omega_{\vec{q}}$ arising from the emission/absorption of a phonon with wave vector $\vec{q}$. Note that each of these terms is multipled by a prefactor $Q^{2}$. Because of this prefactor, it is possible to experimentally achieve enhancement of the phonon emission/absorption peaks by looking at large $\vec{Q}$ scattering. Because the crystal momentum is conserved only up to a reciprocal lattice vector $\vec{G}, \vec{Q}$ is allowed to run outside of the first Brillouin Zone. Thus very large values of $\vec{Q}$ are possible. However, there is a dependance on $\vec{Q}$ hidden the Debye-Waller factor, which kills this enhancement for large $Q^{2}$
\begin{aligned} 2 W &=\frac{1}{3} Q^{2}\left\langle u_{j}^{2}\right\rangle \\ &=\frac{1}{3} \frac{Q^{2}}{2 N M} \sum_{\alpha, \vec{q}} \frac{2 n_{\vec{q}}+1}{\omega_{\vec{q}}} \end{aligned}
The expectation value $\left\langle u_{j}^{2}\right\rangle$ in this expression represents the mean square fluctuations of the nuclei from their ideal Bravais lattice positions. These fluctuations result in the overall

suppression of both elastic and inelastic scattering peaks. Furthermore, as noted above, the Bragg peak delta functions are not smeared out by thermal fluctuations.

In the low temperature limit, we can employ the Debye model $^{2}$ to evaluate the sum in equation (1.28). This gives
$$2 W \rightarrow \frac{3}{4} \frac{Q^{2}}{M \omega_{D}} \quad \text { as } T \rightarrow 0$$
which is the damping due to zero-point fluctuations.
For $k_{B} T \gg \hbar \omega_{D}$, the Bose factors $n_{\vec{q}} \rightarrow \frac{k_{B} T}{\hbar \omega_{D}} .$ In this case
$$2 W=\frac{Q^{2}}{2 M \omega_{D}^{2}} k_{B} T \quad \text { for } k_{B} T \gg \hbar \omega_{D}$$
which comes from the fact that at high temperatures, the mean square fluctuations are proportional to $k_{B} T$ according to the equipartition theorem.

In two dimensions, we get an interesting result. Using the fact that (for an “infinite” sample) there are phonon modes of arbitrarily small frequency, we can approximate the numerator of equation $(1.28)$ with $k_{B} T$. Using the Debye relation $\omega_{\vec{q}}=v|\vec{q}|$
\begin{aligned} \sum_{\vec{q}} \frac{2 n_{\vec{q}}+1}{\omega_{\vec{q}}} & \approx \sum_{\vec{q}} \frac{k_{B} T}{\omega_{\vec{q}}^{2}} \\ & \approx k_{B} T \int_{0}^{k_{B} T / \hbar v} d^{2} q \frac{1}{v^{2} q^{2}} \rightarrow \ln (0) \end{aligned}
which is logarithmically divergent. Thus $2 W$ is infinite for a $2 \mathrm{D}$ crystal. Although this would seem to imply the complete disappearance of the Bragg peaks, a more careful calculation reveals that the Bragg delta peaks are actually broadened to a power law.

What is the reason for this strange behavior? The answer is that in two dimensions, thermal fluctuations are sufficiently influential that they can destroy the long-range order of a crystal. If you imagine nailing down a single nucleus to be used as the origin of a Bravais lattice, then at large distances the mean positions of the nuclei will not be described by lattice vectors for a $2 \mathrm{D}$ crystal with thermal fluctuations. Because of this, some authors claim that there is no such thing as a $2 \mathrm{D}$ crystal.

However, we may ask a different question about our material to judge its crystallinity. Is orientational order preserved at long distances? Imagine nailing down two adjacent nuclei at their equilibrium separation, with the line connecting the two nuclei oriented along a particular direction. Far away from these two nuclei, are similar bonds still parallel to this one? The answer is yes, bond orientation is preserved over large distances for a $2 \mathrm{D}$ crystal $^{3} .$ In this sense, it still does make sense to speak of a two dimensional crystal.

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