Then the asymmetry parameter $$\eta = \frac{\rho_c \Omega_\mathrm{B}}{\langle m \rangle n_\gamma}.$$
Sakharov conditions
Assume $B=0$ in (very) early universe; $B>0$
later.
In 1967 Andrei Sakharov (implicitly) wrote down the
necessary conditions for baryogenesis:
Baryon number $B$ violation
$C$ and $CP$ violation
Departure from thermal equilibrium
These specify only what is needed,
not how it works.
About the Sakharov conditions: $C$
Note that if we had $B$ violation without $C$
violation, then $\overline{B}$ violation
would occur at the same rate:
$$ \Gamma(X \to Y + B) = \Gamma(\overline{X} \to
\overline{Y} + \overline{B}) $$
Thus over time $B=0$ still, unless we have $C$
violation too:
$$ \frac{\mathrm{d} B}{\mathrm{d} t} \propto \Gamma(\overline{X} \to
\overline{Y} + \overline{B}) - \Gamma(X \to Y + B).$$
About the Sakharov conditions: $CP$
In fact, also need $CP$ violation
Consider $B$-violating $X \to
q_\mathrm{L} q_\mathrm{L}$ process making left handed
baryons
$CP$ symmetry turns this equation into
$ \bar{X} \to \bar{q}_\mathrm{R}
\bar{q}_\mathrm{R}$
Applying lattice methods to extensions of the Standard Model
Can add extra fields (a singlet $\phi$, a triplet $\Sigma$, etc.) to Standard Model.
Choose suitable parameters, can get a first-order phase transition again.
Can thus 'regain' baryogenesis, postulate a dark matter candidate, etc.
Many studies of such models use perturbation theory. Can we test perturbation theory with the same style of lattice simulations as used for the Standard Model?
Lattice Monte Carlo benchmarks
e.g. $\Sigma$SM, with extra triplet scalar field arXiv:2005.11332
Need benchmarks: perturbation theory out by $\gtrsim 10\%$
Intro to EPWT - conclusion
SM is a crossover
Many simple extensions with first order phase
transitions
Will take a next-generation collider (or GW
detection/exclusion!) to rule out most models
And need new Monte Carlo simulations to benchmark perturbation theory studies
Why do we need simulations? A toy model example.
Toy model of a real scalar field
Let us explore how all this machinery works for the scalar field $\phi$, with Lagrangian
$$ \begin{align} \mathcal{L} & = \frac{1}{2} (\partial_\mu \phi)^2 - V(\phi) \\
V(\phi) & = \sigma \phi + \frac{1}{2}m^2 \phi^2 + \frac{1}{3!}g \phi^3 + \frac{1}{4!} \lambda \phi^4 \end{align}$$
and neglect couplings to other fields for simplicity.
(If we coupled it to the Standard Model, could be a DM candidate for example)
Infrared problem
At low temperatures the perturbative expansion parameter is $\lambda$...
...so at high temperatures it will be of the form $\frac{\lambda T}{m}$, with $m$ the mass of some bosonic excitation.
But the field undergoing the phase transition is often light compared to the temperature $T$.
Thus $T/m \gg 1$ and the expansion is poorly behaved.
Aside: pure gauge theory
Furthermore gauge bosons are perturbatively massless at high temperatures (in their symmetric phase)
If we want to study (de)confinement phase transitions, then lattice methods are essential
We studied bubble nucleation for the first time recently in $\mathrm{SU}(8)$, see arXiv:2603.22088
.
3D effective field theory
As mentioned above, the field becomes light close to the phase transition, due to cancellations between tree-level and thermal contributions.
Then the expansion parameter $\lambda T/m$ can become large, and perturbation theory breaks down.
The solution is high-temperature dimensional reduction: one writes down a 3D effective field theory with only the light fields.
3D model
Let us consider the following theory in three Euclidean dimensions (which happens to be the most general theory with the same symmetries and number of fields):
$$ \begin{align} \mathcal{L}_3 & = \frac{1}{2} (\partial_i \phi_3)^2 - V_3(\phi_3) \\
V_3(\phi_3) & = \sigma_3 \phi_3 + \frac{1}{2}m_3^2 \phi_3^2 + \frac{1}{3!}g_3 \phi^3 + \frac{1}{4!} \lambda_3 \phi_3^4 \end{align}.$$
This is interesting in itself, but is also a high-temperature effective field theory for our 4D model.
Dimensional reduction
How do we match the 4D model (possibly the whole Standard Model with extra field content) to the 3D effective field theory?
Having written down the most general 3D model with all the light degrees of freedom, we:
Compute correlation functions in the 4D and 3D theories
Compare the results of the two theories for momenta $p\sim \sqrt{\lambda} T$
Use these results determine the relationship between the 4D and 3D couplings
We can discretise our three-dimensional effective field theory with Euclidean Lagrangian $$ \mathcal{L}_{3,\text{lat}}(\mathbf{x})
= \frac{1}{2a^2} \sum_{i} \left[\phi_3(\mathbf{x} + \hat{\imath}) - \phi_3(\mathbf{x})\right]^2
+ V_{3,\text{lat}}[\phi_3(\mathbf{x})] $$
where
$$ \begin{multline} V_{3,\text{lat}}[\phi_3(\mathbf{x})] \\
= \sigma_{3,\text{lat}} \phi_3(\mathbf{x}) + \frac{1}{2} m_{3,\text{lat}}^2 \phi_3(\mathbf{x})^2 + \frac{1}{3!} g_3 \phi_3(\mathbf{x})^3 + \frac{1}{4!} \lambda_3 \phi_3 (\mathbf(x))^4 \end{multline} $$
(soon you will play with a code that implements this)
As well as taking $a \to 0$ we need to extrapolate to infinite volume $V \to \infty$, with $L_x, L_y, L_z \to \infty$.
Ideally:
Fix $a$ and extrapolate $V\to \infty$.
Then vary $a$ and extrapolate again.
Then extrapolate $a \to 0$.
Often it is enough to vary $a$ at some fixed volume $V$ to check the dependence on lattice spacing is not too strong.
Periodic boundary conditions
Physics is translation-invariant, so usually impose periodic boundary conditions
Lattice-continuum relations and continuum limit
How do we match the 3D continuum to the 3D lattice model?
Compute the effective potential in continuum and on the lattice.
Equate the two, in $\overline{MS}$ for continuum, and for finite lattice spacing $a$ on the lattice.
Ensures that the lattice theory converges to the continuum theory as $a\to 0$.
We then need to extrapolate simulation results to the continuum ($a\to 0$) as well.
What parameters do we choose?
Having picked 4D parameters, we match them to our 3D EFT parameters, and then relate them to the lattice parameters with the lattice-continuum relation.
The 4D to 3D mapping is not one-to-one because we integrate out a lot of field content, can thus sometimes recycle simulations
Then need to extrapolate to zero lattice spacing $a\to 0$, and infinite volume $V \to \infty$
Monte Carlo - importance sampling
We want to measure quantities like $\langle \phi \rangle$ on the lattice, i.e. evaluate
$$ \langle \phi \rangle = \frac{1}{Z} \int \mathcal{D} \phi \; \phi \, e^{-S_\text{lat}}; \quad S_\text{lat} = \sum_\mathbf{x} \mathcal{L}_{3,\text{lat}}(\mathbf{x}) $$
Very high-dimensional integral, $L_x \times L_y \times L_z$ points, each with its own value of $\phi$.
But this integral is peaked around the extremum of the exponent - use importance sampling.
Monte Carlo - measurements
Generate lattice configurations $i$ with probability weighted by $e^{-S_\text{lat}}/Z$.
Then $ \langle \phi \rangle $ is just a sum over measurements with configuration, $ \langle \phi \rangle = \sum_i \phi_i $.
$Z$ cannot be measured directly, so we only know relative probability of configurations.
Metropolis algorithm
Markov Chain: generate next configuration from current configuration, no memory (e.g. vary $\phi$ at one site by a random amount)
Calculate change in Euclidean action $\Delta S_\text{lat}$.
Nicholas Metropolis, Arianna W. Rosenbluth, Marshall Rosenbluth, Augusta H. Teller and Edward Teller (1953):
accept next configuration with probability
$$ W = \text{min} (1, e^{-\Delta S_\text{lat}}).$$
What do the histograms look like?
At the critical coupling, the probability density between the phases falls exponentially with volume:
If we simulate with one set of parameters we can reweight our results to 'nearby' parameters, so long as we keep track of the corresponding term in the action:
Suppose we want to reweight in $m_3^2$ to $m_3'^2$.
It enters into the action as $$ S \supset \frac{1}{2}m_3^2 \sum_\mathbf{x} \phi_3(\mathbf{x})^2 \equiv \frac{1}{2}m_3^2 \theta.$$
We need to record $\theta$ during our simulation, as well as what we want to observe $\mathcal{O}$.
Once we have determined the critical coupling we can use the machinery we discussed yesterday to get the critical temperature:
Lattice-continuum relations give us the continuum 3D critical couplings
Matching relations give us 4D critical couplings
Matching can be non-unique, but if we fix the 4D couplings and vary the temperature, we can usually match to the 3D theory
Relating lattice results to 4D physics
Often good to determine the 'trajectory' in 3D parameter space as a function of $T$ beforehand.
Fix $4D$ parameters $\sigma, m^2, g, \lambda$.
Determine matching relations for continuum 3D parameters $\sigma_3, m_3^2, g_3, \lambda_3$ as a function of $T$.
Also useful to have the partial derivatives with respect to $T$ (as we will see):
$$ m_3^2(T) = f(T | \sigma, m^2, g, \lambda); \quad \frac{\partial m_3^2}{\partial T} = \frac{\partial f}{\partial T}. $$
May need to reweight lattice results in more than one parameter to vary $T$.
For concreteness:
Recalling:
$$ V_3(\phi_3) = \sigma_3 \phi_3 + \frac{1}{2}m_3^2 \phi_3^2 + \frac{1}{3!}g_3 \phi^3 + \frac{1}{4!} \lambda_3 \phi_3^4 $$
the matching relations at leading order in our theory are
$$\begin{align}
\sigma_3(T) & = \frac{\sigma}{\sqrt{T}} + \frac{g T^{3/2}}{24} \\
m_3^2 (T) & = m^2 + \frac{\lambda T^2}{24} \\
g_3(T) & = \sqrt{T}g \\
\lambda_3 (T) & = T \lambda \\
\end{align} $$
We associate the probability $P(\theta)$ of observing a given order parameter $\theta$ with a constrained free energy $F(\theta)$, such that $P(\theta) \propto e^{-F(\theta)}$.
We can't measure $Z$, since we are doing importance sampling; only know relative probabilities $P(\theta)$.
From our histogram ($\equiv$ estimate of the probability distribution of $\theta$), can read off free energy differences: $$F(\theta) = \log P(\theta) + \text{constant}$$ (personally think this defines constrained free energy)
Another justification of the constrained free energy
This comes from James Langer's paper, Metastable statesPhysica73 (1974) 61-72.
$$ e^{-F(\theta)} = \sum_{\substack{\text{(constrained microscopic } \\ \text{variables with $\theta$ fixed)}}} e^{-H}$$ where $H$ is a Hamiltonian describing the short-range physics.
See arXiv:2108.04377 for a modern discussion in the context of bubble nucleation.
What do we want?
We need to know the following to study the out-of-equilibrium dynamics of the phase transition (including gravitational waves):
Critical temperature [done]
Surface tension [next]
Latent heat/phase transition strength
Nucleation rate, and hence nucleation temperature
Wall velocity [later on]
Thin wall approximation
Consider a cubic lattice for simplicity.
Suppose our lattice contains a mixed configuration of two phases.
Assume the phase boundary is well-localised compared with the lattice size.
Energy per unit area of the phase boundary is the surface tension $\sigma$.
For any given value of the order parameter $\theta$, the favoured mixed configuration will be that which minimises the free energy.
$\Delta p$ is the pressure difference between the two phases.
We treat the two phases as having separate subsystem partition functions so we can measure $\Delta \log Z$.
$\Delta F \equiv - T \Delta \log Z$ is the free energy difference between the phases. We infer $\Delta F/V = \Delta p $ at coexistence.
And $\Delta P$ is the difference in probabilities under the two peaks.
There is another way!
Starting from the free energy density $\Delta f$, can write down another version of the Clapeyron equation
$$ L = - \frac{\partial \Delta f}{\partial T}; \qquad \Delta f \equiv T \Delta \varepsilon_\text{vac} $$
where $\Delta \varepsilon_\text{vac}$ is the 3D vacuum energy.
Derivatives with respect to $T$ only enter this through the couplings in the potential energy, so
$$ \frac{L}{T} = \ldots + \frac{1}{2} \frac{\partial m_3^2}{\partial \log T}\Delta \langle \phi_3^2 \rangle + \ldots. $$
When the universe drops below the critical temperature,
broken phase is the new global minimum.
Quantum (or thermal) fluctuations will excite the
field across the potential barrier to the new
minimum.
Consider a single scalar field $\phi$ with
Lagrangian
$$ \mathcal{L} = \frac{1}{2} \partial_\mu \phi
\partial^\mu \phi - U(\phi) $$
and equation of motion
$$ \frac{\partial^2 \phi}{\partial t^2} + \nabla^2
\phi = U'(\phi).$$
More nucleation basics
Want to calculate probability for field $\phi$ to
tunnel from false vacuum $\phi_+$ to true vacuum
$\phi_-$.
Like quantum mechanical tunnelling.
Solve for trajectory that 'bounces' from $\phi_+$
to $\phi_-$ and back again in a localised region
Obtain exponential factor $B$ in
nucleation rate $\Gamma/V \propto Ae^{-B}$.
A bit more about Langer's theory
Langer's theory puts this on a rigorous footing.
The rate is the integral of the probability flux $J$ over the transition surface in free energy, $\Gamma = \int_S J \cdot \mathrm{d}S_\perp$
figure from arXiv:2108.04377.
Bubble radius is much thicker than the phase boundary thickness, so treat the surface as a thin wall.
Surface tension $\sigma$ at the nucleation temperature $T_N$ is the same as at the critical temperature $T_c$.
Expanding the Clapeyron relation $\frac{L}{T} = \frac{\mathrm{d} \Delta p}{\mathrm{d} T}$ for small $\Delta T = T_N -T_c$,
pressure difference is related to the latent heat (measured at the critical temperature) by
$$\Delta p = L (T_N - T_c)/T_c$$
Thin-wall with lattice data
Then the action of a bubble of radius $r$ is
$$ \begin{align} S_3(r) & = 4\pi r^2 \sigma - \frac{4\pi}{3} \Delta p r^3 \\
& = 4\pi r^2 \sigma - L\frac{(T - T_c)}{T_c} r^3 \end{align} $$
If we can generate 'bubble' configurations in a Monte Carlo simulation, we can use them as the starting point to measure the nucleation rate nonperturbatively.
Bubble nucleation rate on the lattice
General idea is that the nucleation rate is a product of:
Relative probability (density) of critical bubbles, $P_c$
Rate of change of the order parameter with time for a bubble being nucleated, the 'flux'
Number of times a critical bubble tunnels back and forth before settling back to one or the other phase, $\mathbf{d}$
Then $$\Gamma V = P_c \left< \frac{1}{2}\text{flux} \times \mathbf{d} \right> \approx P_c \frac{1}{2} \left< \text{flux} \right> \left< \mathbf{d} \right>.$$
Probability of critical bubbles
Probability density of a configuration lying in $\varepsilon$ relative to being in the metastable phase:
$$ P_c = \frac{P(|\theta - \theta_c| < \varepsilon/2)}{\varepsilon P(\theta < \theta_c)}.$$
Generating critical bubbles
Critical bubble configurations are heavily suppressed, and we need large volumes to see them:
Use multicanonical simulations so critical bubbles are not suppressed.
Save configurations that lie in the region $\varepsilon$.
Do those bubbles tunnel?
Given the candidate critical configurations, evolve them with the equations of motion in a heat bath backwards and forwards in time.
If they tunnel, then $\mathbf{d} = 1/N_\text{crossings}$, otherwise $\mathbf{d} = 0$.
The nucleation rate $\Gamma$ gives the probability
of nucleating a bubble per unit volume per unit
time.
More useful for cosmology is to consider the inverse
duration of the phase transition, defined as
$$ \beta \equiv - \left.\frac{d S(t)}{d t}
\right|_{t=t_*} \approx \frac{\dot \Gamma}{\Gamma}
$$
Phase transition completes when the probability
of nucleating one bubble per horizon volume is order
1
$$ S_3(T_*)/T_* \sim -4 \log \frac{T_*}{m_\mathrm{Pl}}
\approx 100
$$
Making further use of $\Gamma$
Using the adiabaticity of the expansion of the
universe the time-temperature relation is
$$ \frac{d T}{d t} = - T H $$
This gives, for the ratio of the inverse phase
transition duration relative to the Hubble rate,
$$ \frac{\beta}{H_*} = T_* \left. \frac{dS}{dT}
\right|_{T = T_*} = T_* \frac{d}{d T}
\left. \frac{S_3(T)}{T} \right|_{T=T_*}$$
If $\frac{\beta}{H_*} \lesssim 1$ then the phase
transition won't complete...
Nucleation - conclusion
Rate per unit volume per unit time $\Gamma$
computed from bounce actions $S(T) = \mathrm{min}\{S_3(T)/T ,S_4(T) \}$
Inverse duration relative to Hubble rate
$\frac{\beta}{H_*}$ computed from $\Gamma$, and controls
GW signal
To get $\beta$:
Find effective potential $V_\text{eff}(\phi,T)$
Compute $S_3(T)/T$ (or $S_4(T)$) for extremal
bubble by solving 'equation of motion'
Determine transition temperature $T_*$
Evaluate $\beta/H$ at $T_*$
Use $\beta/H_*$ as input
to compute GWs (c.f. $B$ earlier!)
Workshop section
Our lattice computation of the nucleation rate used a combination of multicanonical Monte Carlo and real time simulations.
At small enough supercooling, real time simulations are enough.
Interest in this as a method of studying phase transitions has grown in recent years, and computers are much faster than they were in the 90s.
Today you'll have a go at reproducing the nucleation rate from one early paper on the topic.
We have so far been using field theory
equations of motion.
Less tricky, but more abstract, are:
Boltzmann equations
Hydrodynamic equations
In particular, the hydrodynamic equations we get are
a valuable motivation for the rest of today's lectures
We will now look at how to arrive at these
higher-level approximations
Boltzmann equations: a reminder
What is a Boltzmann equation?
Phase space is positions $\mathbf{x}$ and momenta
$\mathbf{k}$.
Tells us how our distribution functions
$f_i(\mathbf{x},\mathbf{k})$ evolve.
Consists of four parts:
Time evolution $\partial_t
f_i(\mathbf{x},\mathbf{k}).$
Streaming terms in momentum and position space $\dot{\mathbf{x}} \cdot
\nabla_\mathbf{x} f +
\dot{\mathbf{p}} \cdot \nabla_\mathbf{p} f$
Collision $C[f]$
Boltzmann eqn. for distribution $f$
The Boltzmann equation is $$ \frac{\mathrm{d} f}{\mathrm{d} t} = \frac{\partial
f}{\partial t} + \dot{\mathbf{x}} \cdot \nabla_\mathbf{x} f +
\dot{\mathbf{p}} \cdot \nabla_\mathbf{p} f = -C[f].$$
This is a semiclassical approximation to the quantum
Liouville equations for all the fields
Only valid when the momenta of the fields is much
higher than the inverse wall thickness:
$$ p \gtrsim g T \gg \frac{1}{L_w}. $$
Very difficult to work with directly, so model the
distribution $f_i$ of each particle with a 'fluid'
ansatz.
Fluid approximation
As mentioned, fluid approximation sets the
scene for the rest of these lectures on the
electroweak phase transition
In short, we have $$ T_{\mu\nu}^\text{fluid} = \sum_i \int \frac{d^3
k}{(2\pi)^3 E_i} k_\mu k_\nu f_i(k) = w u_\mu u_\nu -
g_{\mu\nu} p $$
but we will try to justify this.
Deriving the fluid approximation
The flow ansatz is $$ f_i(k,x) = \frac{1}{e^X \pm 1} = \frac{1}{e^{\beta(x)
( u^\mu(x) k_\mu + \mu(x))} \pm 1} $$
with four-velocity $u^\mu(x)$, chemical potential
$\mu(x)$ and inverse temperature $\beta(x)$.
Substituting this ansatz into the Boltzmann
equations for the system yields (after much algebra!) a
(relativistic) Euler momentum equation
$$ u^\mu \partial_\mu u_\nu + \partial_\nu p = C.$$
Local thermal equilibrium
Plasma exerts a pressure on the bubble wall also in equilibrium (i.e. where the distribution functions are not perturbed)
This leads to the local thermal equilibrium (LTE) approximation
In LTE the wall velocity solution depends on only a few quantities: phase transition strength $\alpha$, ratio of enthalpies in each phase, and the sound speeds in each phase.
The field-fluid model
Energy conservation requires that
$$ \partial_\mu T^{\mu\nu} =
\partial_\mu (T^{\mu\nu}_\phi +
T^{\mu\nu}_\text{fluid}) = 0. $$
We are now ready to present the full model: $$ \begin{align} (\partial_\mu \partial^\mu \phi) \partial^\nu \phi - \frac{\partial
V_\text{eff}(\phi,T)}{\partial \phi} \partial^\nu \phi & = -\eta(\phi, v_\mathrm{w})
u^\mu \partial_\mu \phi \partial^\nu \phi \\
\partial_\mu (w u^\mu u^\nu) - \partial^\nu p + \frac{\partial V_\text{eff}(\phi,T)}{\partial \phi} \partial^\nu \phi & = + \eta(\phi, v_\mathrm{w})
u^\mu \partial_\mu \phi \partial^\nu \phi
\end{align} $$
Besides the (dimensionful) definition here, one
choice for $\eta$ that is well motivated is
$\tilde{\eta} \phi^2/T$.
This model is the basis of spherical and 3D
simulations. Can also obtain steady-state equations.
The field-fluid model: observations
Consider the fluid equation: $$ \partial_\mu (w u^\mu u^\nu) - \partial^\nu p + \frac{\partial
V_\text{eff}(\phi,T)}{\partial \phi} \partial^\nu \phi = \eta(\phi, v_\mathrm{w})
u^\mu \partial_\mu \phi \partial^\nu \phi $$
Away from the bubble wall, the right hand side goes
to zero. The left hand side has no length scale.
Therefore any fluid solution must be parametrised by
a dimensionless ratio, e.g. radius of the bubble to time
since nucleation - define $\xi = r/t$.
Fluid profiles will scale with the bubble radius:
they are large, extended objects!
Simulations with field-fluid model
Top: $\phi/\phi_0$
Bottom: fluid 3-velocity $V$
Spherical geometry; coordinate $\xi \equiv r/t$.
We'll return to this tomorrow
Runaway walls?
We have assumed that the wall reaches a terminal
velocity (less than $c$).
The energy density in gravitational waves (Isaacsson tensor) is
$$ \rho_\text{GW} \equiv t_{00} = \frac{c^2}{32 \pi G} \langle
\dot{h}_{ij}^\text{TT} \dot{h}_{ij}^\text{TT} \rangle.
$$
Averaging should be over space and time, but consider
$$ \rho_\text{GW} = \frac{c^2}{32\pi G V}\int \mathrm{d}^3 x \, \dot{h}_{ij}^\text{TT}
\dot{h}_{ij}^\text{TT},$$
and define a Fourier transform
$$
h_{ij}^\text{TT} (x) = \int \frac{\mathrm{d}^3 k}{(2\pi)^3}
\tilde{h}_{ij}^\text{TT}(\mathbf{k})e^{i\mathbf{k}\cdot \mathbf{x}}.
$$
GW power spectrum
Using the defined Fourier transform, we have
$$ \rho_\text{GW} = \frac{1}{32\pi G V} \int \frac{\mathrm{d}^3 k}{(2\pi)^3}
\dot{\tilde{h}}_{ij}^\text{TT}(\mathbf{k}) \dot{\tilde{h}}_{ij}^\text{TT}(-\mathbf{k}).$$
Can use this to
define the GW power per logarithmic frequency interval $$ \frac{\mathrm{d} \,
\rho_\text{GW}}{\mathrm{d} \, \log k } =
\frac{c^2}{32\pi G V} \frac{k^3}{(2\pi)^3} \int \mathrm{d}\Omega\,
\dot{\tilde{h}}_{ij}^\text{TT}(\mathbf{k})
\dot{\tilde{h}}_{ij}^\text{TT} (-\mathbf{k}) $$
This last quantity is what cosmologists often call the 'gravitational wave power spectrum'.
Computing GWs in a simulation
Evolve harmonic-gauge, position-space wave equation
$$ \nabla^2 h_{ij} (\mathbf{x},t) - \frac{\partial}{\partial t^2}
h_{ij}(\mathbf{x},t) = 8 \pi G T_{ij}^\text{source}(\mathbf{x},t)$$
using relevant $T^\text{source}_{ij}$ of 'source
system'.
Projection to TT-gauge requires expensive Fourier
transform, so
only project when measurement desired:
$$ h^{\text{TT}}_{ij}(\mathbf{k},t_\text{meas}) = \Lambda_{ij,lm}(\hat{\mathbf{k}})
h^{lm}(\mathbf{k},t) $$
Measure energy density or power spectrum $$ \rho_\text{GW}(t_\text{meas}) = \frac{1}{32
\pi G} \left< \dot{h}_{ij}^\text{TT} \dot{h}_{ij}^\text{TT} \right> $$
Redshift to present day.
Redshifting a cosmological background of GWs
Friedmann equations
Assume universe described (on cosmological scales) by Friedmann–Lemaître–Robertson–Walker
metric.
Universe contains a perfect fluid with energy density $\rho$ and pressure $p$.
Friedmann equations are $$ \begin{align}
\frac{\dot{a} + kc^2}{a^2} & = \frac{8 \pi G \rho + \Lambda c^2}{3} \\
\frac{\ddot{a}}{a} & = -\frac{4\pi G}{3}\left(\rho + \frac{3p}{c^2}\right) + \frac{\Lambda c^2}{3}
\end{align}
$$
where $a$ is the scale factor and $\dot{a} \equiv \frac{\partial a}{\partial t}$.
We will assume $\Lambda=0$ and $k=0$ (flat) (or nearly so).
Solving the Friedmann equations
If neither $k$ nor $\Lambda$ are significant, the solution is
$$a \propto t^{2/3(1+w)}$$
where $w = p/\rho c^2$ is the equation of state parameter.
After inflation, in the very early universe, everything was relativistic and behaved like
radiation, meaning $w=1/3$.
In other words, as time went by, the universe expanded ($a \propto \sqrt{t}$).
Redshifting a frequency
Assume we are interested in gravitational waves produced with frequency $f_*$ in the early
universe, at a time $t_*$, when the scale factor was $a(t_*)$.
Call our present day $t_0$, with scale factor $a(t_0)$. Then the present-day frequency $f_0$
is
$$ f_0 = \frac{a(t_*)}{a(t_0)} f_*.$$
Consider the entropy $s$. As the universe expands, it is conserved:
$$ s(t) a(t)^3 = \text{constant}. $$
Some statistical mechanics
We can relate entropy density to energy density, pressure and temperature:
$$ s = \frac{\rho + p}{T} $$
(with $p = \rho/3$ for a relativistic fluid).
Energy densities of relativistic bosons and fermions:
$$ \frac{\rho_\text{bosons}}{g} = \frac{\pi^2}{30} \frac{k_\mathrm{B}^4}{c^3 \hslash^3} T^4
\quad \frac{\rho_\text{fermions}}{g} = \frac{7}{8} \frac{\pi^2}{30}
\frac{k_\mathrm{B}^4}{c^3 \hslash^3} T^4 $$
per degree of freedom, respectively, for $k_\mathrm{B}T \gg mc^2$.
From now on work with units $k_\mathrm{B} = c = \hslash = 1$.
Entropy
So in Planck units
$$ \frac{\rho_\text{bosons}}{g} = \frac{\pi^2}{30} T^4 \quad \frac{\rho_\text{fermions}}{g}
= \frac{7}{8} \frac{\pi^2}{30} T^4 $$
Define $g_{*,s}(T)$, an effective number of degrees of freedom for entropy, and
write
$$ s = \frac{2\pi^2}{45} g_{*,s}(T) T^3.$$
Fermion dof's counted with factor ($7/8$, $T \gg m$).
Different 'kinds' of $g(T)$ for $n$, $p$ and $\rho$ (different factors; also more
important when $m \lesssim T$)
Redshifting
We can now write
$$ \frac{a(t_*)}{a(t_0)} = \left(\frac{s(t_0)}{s(t_*)}\right)^\frac{1}{3} =
\left(\frac{g_{*,s}(T_0)}{g_{*,s}(T_*)}\right)^\frac{1}{3} \frac{T_0}{T_*}.
$$
Number of relativisic degrees of freedom has not changed since recombination, so set $T_0 =
2.726 \, \mathrm{K} \equiv 234.4 \, \mu \mathrm{eV}$ (CMB temperature).
Number of effective degrees of freedom $g_{*,s}(T_r) = 3.91$.
Making it cosmologist-friendly
This is all well and good $$ f_0 = \left(\frac{3.91}{g_{*,s}(T_*)}\right)^\frac{1}{3} \frac{234.4 \,
\mu\mathrm{eV}}{T_*} f_* $$ but doesn't tell you how big the factor is typically.
So instead write (without loss of generality) $$ f_0 = \left(\frac{3.91}{100}\right)^\frac{1}{3}
\left(\frac{100}{g_{*,s}(T_*)}\right)^\frac{1}{3} \frac{234.4 \, \mu\mathrm{eV}}{100\, \mathrm{GeV}} \frac{100\, \mathrm{GeV}}{T_*} f_* $$
Assume some large-scale source produces waves some known (substantial) fraction of the
Hubble radius at the time:
$$ \lambda \lesssim H_*^{-1} $$
we can use the Friedmann equations to work out $H_*$.
The radiation energy density of the universe is
$$ \rho(T) = \frac{\pi^2}{30} g_*(T) T^4 $$
(this $g_*(T)$ is for energy density)
And the Hubble rate $H_* \equiv \dot{a}(t_*)/a(t_*)$.
The Friedmann equations again
Using our expressions $\rho(T)$ and $H_*$ in (one of) the Freidmann equations again:
$$ H_*^2 = \frac{8\pi G}{3} \rho_* = \frac{8\pi G}{3} \frac{\pi^2}{30}
g_*(T_*) T_*^4 $$
We then have
$$ H_* = \frac{1}{M_\mathrm{P}} \sqrt{\frac{\pi^2}{90}} \sqrt{g_*(T_*)} T_*^2,$$
where $M_\mathrm{P} = \sqrt{\hslash c /8 \pi G} = 2.44\times 10^{18}\, \mathrm{GeV}/c^2$
(reduced Planck mass).
Primordial Hubble rate
Casting $H_*$ in terms of some popular numbers:
$$
H_* = \frac{1}{M_\mathrm{P}} \sqrt{\frac{\pi^2}{90}} (100)^\frac{1}{2}
\left(\frac{ g_*(T_*)}{100}\right)^{\frac{1}{2}} (100\,
\mathrm{GeV})^2 \left(
\frac{T_*}{100 \, \mathrm{GeV}} \right)^2
$$
and momentarily going to SI units (convert energies and masses to Joules, divide by
$\hslash$),
$$
H_* = 1.90\times 10^{9} \left(\frac{ g_*(T_*)}{100}\right)^{\frac{1}{2}} \left(
\frac{T_*}{100 \, \mathrm{GeV}} \right)^2 \, \mathrm{Hz}.
$$
Redshifting a wavelength
We now know $H_*$ in SI units.
Let us take our primordial frequency to be some multiple $f_* = H_*/B$, with $B < 1$
We can then put our redshifting formula and Hubble rate formula together to get
$$ f_0 = 2.7\times 10^{-6} \frac{1}{B} \left(\frac{g_*(T_*)}{100}\right)^{\frac{1}{6}} \left( \frac{T_*}{100 \, \mathrm{GeV}} \right) \, \mathrm{Hz}. $$
Sometimes this is thought of as 'redshifting' the Hubble rate from the time of interest to
today.
Gravitational waves behave like radiation, so their fraction of the energy in the universe
does too:
$$ \rho_{gw, 0} = \rho_\text{gw} \left( \frac{a(t_*)}{a(t_0)} \right)^4
$$
... and we know how to compute $\frac{a(t_*)}{a(t_0)}$, so
$$
\begin{align}
\rho_{gw, 0} & = \rho_\text{gw}
\left( \frac{g_{*,s}(T_0)}{g_{*,s}(T_*)} \right)^{\frac{4}{3}}
\left(\frac{T_0}{T_*}\right)^4 \\
& = \rho_\text{gw} \left( \frac{g_{*,s}(T_0)}{g_{*,s}(T_*)} \right)^{\frac{4}{3}}
\frac{\rho_\text{rad,0}}{\rho_\text{rad}} \frac{g_*(T_*)}{g_*(T_0)}.
\end{align}
$$
using $\rho(T) = \frac{\pi^2}{30} g_*(T) T^4$ at the end.
Computing GW parameter $\Omega_\text{gw}$
We have
$$ \rho_{gw, 0} = \rho_\text{gw} \left( \frac{g_{*,s}(T_0)}{g_{*,s}(T_*)}
\right)^{\frac{4}{3}}
\frac{\rho_\text{rad,0}}{\rho_\text{rad}} \frac{g_*(T_*)}{g_*(T_0)}.$$
We now dress things up to look more cosmological:
Divide both sides by the present-day critical density $\rho_\text{c} \equiv 3
H_0^2/8\pi G$.
Write things in terms of density parameters: $\Omega_\text{gw} \equiv
\rho_\text{gw}/\rho_\text{c}$ for gravitational waves, and $\Omega_\text{rad} \equiv
\rho_\text{rad}/\rho_\text{c}$
for radiation.
Parametrising our ignorance
The critical density $\rho_\text{c} \equiv 3 H_0^2/8\pi G$ depends on the present day Hubble
constant $H_0$.
We would rather not depend on $H_0$, so multiply both sides by the reduced Hubble constant
$h$, where $H_0 = h \times 100 \, \mathrm{km} \, \mathrm{s}^{-1} \,\mathrm{Mpc}^{-1}$.
Planck and other big experiments/missions report cosmological parameters scaled by $h^2$
anyway.
The formula then becomes $$
h^2 \Omega_\text{gw,0} = h^2 \Omega_\text{rad,0} \left( \frac{g_{*,s}(T_0)}{g_{*,s}(T_*)}
\right)^{\frac{4}{3}}
\frac{g_*(T_*)}{g_*(T_0)} \frac{\rho_\text{gw}}{\rho_\text{rad}}.
$$
Filling in the numbers
We can plug numbers into $$
h^2 \Omega_\text{gw,0} = h^2 \Omega_\text{rad,0} \left( \frac{g_{*,s}(T_0)}{g_{*,s}(T_*)}
\right)^{\frac{4}{3}}
\frac{g_*(T_*)}{g_*(T_0)} \frac{\rho_\text{gw}}{\rho_\text{rad}}.
$$
Define $C = \rho_\text{gw}/\rho_\text{rad}$, fraction of (radiation-era) primordial
energy turned into GWs, $C \ll 1$.
With these quantities, we have
$$ h^2 \Omega_\text{gw,0} = 4.18 \times 10^{-5} \left( \frac{3.91}{g_{*,s}(T_*)}
\right)^{\frac{4}{3}}
\frac{g_*(T_*)}{3.36} C.$$
Making $\Omega_\text{gw,0}$ cosmologist-friendly
We cast
$$ h^2 \Omega_\text{gw,0} = 4.18 \times 10^{-5} \left(
\frac{3.91}{g_{*,s}(T_*)} \right)^{\frac{4}{3}}
\frac{g_*(T_*)}{3.36} C.$$
again in 'familiar' terms by setting $g_{*,s}(T_*) \approx g_*(T_*)$ (okay for high
temperatures).
This yields
$$ \begin{align}
h^2 \Omega_\text{GW,0} & \approx 4.18\times 10^{-5} \times
\frac{3.91^{\frac{4}{3}}}{3.36} \times 100^{-\frac{1}{3}} \left( \frac{100}{g_{*}(T_*)}
\right)^{\frac{1}{3}} C \\
& \approx 1.65 \times 10^{-5} \left( \frac{100}{g_{*}(T_*)} \right)^{\frac{1}{3}} C.
\end{align}$$
Power spectrum
Consider an isotropic signal with spectral shape
$$ \frac{1}{\rho_\text{rad}} \frac{\mathrm{d} \rho_\text{GW}(f)}{\mathrm{d} \ln f} = C
S(f,f_*),$$
normalised such that
$$
\frac{\rho_\text{GW}}{\rho_\text{rad}} = C \int \mathrm{d} f \frac{S(f,f_*)}{f}.
$$
Then we can directly apply the same formula to get
$$
h^2 \frac{\mathrm{d} \Omega_\text{GW,0}(f)}{\mathrm{d}\ln f} \approx 1.65 \times 10^{-5}
\left(
\frac{100}{g_{*}(T_*)} \right)^{\frac{1}{3}} C S(f,f_*).
$$
If we take a traditionally popular example for primordial GWs:
$$S(f,f_*) = N \frac{\left(f/f_*\right)^3}{1+\left(f/f_*\right)^4}.$$
then $N \approx 1.11$.
And if we suppose $C \sim 0.1$, then
$$
h^2 \frac{\mathrm{d} \Omega_\text{GW,0}(f)}{\mathrm{d}\ln f} \approx 1.49 \times 10^{-6}
\left(
\frac{100}{g_{*}(T_*)} \right)^{\frac{1}{3}}
\frac{\left(f/f_*\right)^3}{1+\left(f/f_*\right)^4}.
$$
(Very optimistic) examples
(Based on earlier table of numbers)
Day 4
Thermodynamics of phase transitions
How does LISA work?
Production of GWs from phase transitions
Thermodynamics
Motivation
In the previous section we described the various
layers of approximation up to the field-fluid model.
Now we will use that field-fluid model (and
steady-state results) to explore the macroscopic behaviour
of the wall.
This is important both for baryogenesis and also for
the GW power spectrum.
Further reading
Energy budget: Espinosa,
Konstandin, No and
Servant arXiv:1004.4187
Energy conservation across the wall, in the wall frame (in $z$-direction) implies $\partial_z T^{zz} = \partial_z T^{z0} = 0$.
Integrating these across the wall (from $+$ to $-$) yields (enthalpy $w$, 3-velocity $v$, relativistic $\gamma$ and pressure $p$)
$$
w_+ v_+^2 \gamma_+^2 + p_+ = w_- v_v^2 \gamma_-^2 + p_-, \qquad w_+ v_+ \gamma_+^2 = w_- v_- \gamma_-^2 $$
Add an equation of state, e.g. bag model equation of state (and the phase transition strength $\alpha$), and we can solve for $v_+$, which gives us a starting point on our solution curves.
Bubbles expand with finite velocity
($v_\mathrm{w}$)
Fluid shell forms around bubble
Latent heat $L$ ($+$ extra free energy difference) heats plasma...?
Object of this section is to quantify how much of
the latent heat ends up as kinetic energy.
One definition of $\alpha$
Define phase transition strength
$$ \alpha_L = \frac{L(T)}{g(T) \pi^2 T^4/30}
= \frac{\text{latent heat at $T$}}{\text{radiation energy at $T$}} $$
which tell us how much of the energy of the universe
was stored as latent heat in the phase transition.
NB: there are multiple definitions of $\alpha$ in the literature, differing by what
quantifies the 'available energy' (latent heat, trace anomaly, vacuum energy)
Larger $\alpha_T$ ⇒ stronger phase transition
Another, better definition of $\alpha$
Define the trace anomaly $\theta(T) = \frac{1}{4} (\epsilon(T) - 3p (T))$, where $\epsilon$ and $p$ are energy density and pressure of the plasma.
(trace anomaly can be calculated on the lattice using the same methods as we discussed yesterday)
Then $\alpha_\theta$ is defined as ('ms' = 'metastable') $$ \alpha_\theta = \frac{1}{\epsilon_Q(T)}\left(\theta_\text{ms}(T) - \theta_\text{stable}(T)\right); \quad \epsilon_Q(T) = \frac{3}{4}(\underbrace{\epsilon(T) + p (T)}_\text{enthalpy, $w$})$$
Agrees with $\alpha_L$ at small supercooling.
Computing the efficiency
But $\alpha_T$ does not say how much of $\mathcal{L}$
ends up as fluid kinetic energy
For that we define the efficiency $$ \kappa_\mathrm{f} = \frac{\overbrace{(\epsilon + p) u_i u_i}^{T^{00}_\text{fluid}}}{L}= \frac{\text{fluid
KE}}{\text{latent heat}}$$
Then $\kappa_\mathrm{f} \alpha_T$ is the fraction of the energy
density in the universe that ends up as fluid kinetic
energy.
Very roughly, $\kappa_\mathrm{f} \alpha_T \approx
\overline{U}_f^2$, the (weighted) mean square fluid velocity as the
transition completes.
Can be computed more accurately either from
spherical simulations or directly solving.
Thermal first-order transitions have a reaction
front
Reaction fronts can be deflagrations (generally
subsonic), detonations (supersonic) or hybrids (a
mixture).
The fluid reaches a scaling profile in $\xi = r/t$
based on the available latent heat and wall velocity.
From this, one can compute the efficiency
$\kappa_\mathrm{f}$ and hence how much of the energy in
the universe ends up in the fluid $\kappa_\mathrm{f}
\alpha_T$.
Spacecraft $j$ sends a laser with phase $\Phi_j$ to spacecraft $i$, which is received at
time $t$:
$$ \Phi_{i \leftarrow j} = \Phi_j(t - d_{ij}(t) - H_{ij}(t)) $$
where $d_{ij}$ is the light travel time, and $H_{ij}$ is an additional delay due to GWs
Taylor expanding about $t - d_{ij}$
$$ \Phi_{i \leftarrow j} = \Phi_j(t - d_{ij}) - \dot{\Phi}_j (t - d_{ij})H_{ij}(t),$$
Denote $\dot{\Phi}_i = \nu_i$, a frequency, and so the frequency of the received beam is
$$ \begin{align}\nu_{i \leftarrow j} \equiv \dot{\Phi}_{i \leftarrow j} & =
\left[1-\dot{d}_{ij}(t)- \dot{H}_{ij}(t)\right] \\
& \qquad \times \left[\nu_j(t-d_{ij}(t) -
\dot{\nu}_j(t-d_{ij}(t))H_{ij}(t))\right]\end{align}$$
On spacecraft $i$ this is compared with the local laser $\Phi_j(t)$,
$\Phi^\text{isc}_{ij}(t) \equiv \Phi_{i\leftarrow j}(t) - \Phi_i(t)$, to get
$$ \nu_{ij}^\text{isc} = (1-\dot{d}_{ij})\mathbf{D}_{ij}\nu_j - \nu_i -
(\mathbf{D}_{ij}\nu_j)\dot{H}_{ij}$$
Beatnote frequency
$\nu_{ij}^\text{isc}$ measures the frequency of the beats
Cannot neglect laser noise (fluctuations in $\nu_i$ and $\nu_j$): it will
always be bigger than the GW signal $(\mathbf{D}_{ij}\nu_j)\dot{H}_{ij}$.
Time-domain interferometry (TDI): take advantage of the same laser noise
appearing in different measurements to cancel it out.
Begin with the momentum-space Green's function
expression (assume source off at $t' < 0$) $$ h_{ij}^\text{TT}(\mathbf{k},t)=16\pi G \,
\Lambda_{ij,lm} \int_0^t \, \mathrm{d} t' \frac{\sin[k(t-t')]}{k} T_{lm}(\mathbf{k},t')
$$
If the source is slowly varying in space at low $\mathbf{k}$:
$$T_{lm}(\mathbf{k}) \to \text{const.}; \qquad k \ll
k_\text{max} $$
we get
$$ h_{ij}^\text{TT}(\mathbf{k},t) \approx 16\pi G \,
\Lambda_{ij,lm} \int_0^t \, \mathrm{d} t'
\frac{\sin[k(t-t')]}{k} T_{lm}(0,t') $$
Useful insight 1
If the source $T_{lm}(0,t)$ varies faster than the
$\sin[k(t-t')]$, the equation reduces to
$$ h_{ij}^\text{TT}(\mathbf{k},t) \approx 16\pi G \,
\Lambda_{ij,lm} \int_0^t \, \mathrm{d} t' T_{lm}(0,t')
$$
This gives
$$ \frac{\mathrm{d} \rho_\text{GW}(k)}{\mathrm{d} \,
\log \, k} \propto k^3 $$
In other words, at sufficiently long sub-horizon
scales, all that matters is how long the source is on for.
The power law is $k^3$.
This holds for any subhorizon physics.
Useful insight 2
If the source has an intermediate regime
where $T_{lm}(0,t)$ varies slower than
$\sin[k(t-t')]$, then the source stays in the integral,
and there is an additional $1/k$ factor
Therefore, in some cases we can expect a
$k^1$ power law at higher wavenumbers than the $k^3$ is
valid
Useful insight: graphical summary
Simulating first order phase transitions: simulations
Recap: what happens?
Bubbles nucleate (temperature $T_\mathrm{N}$, on timescale $\beta^{-1}$)
Bubble walls expand in a plasma (at velocity $v_\mathrm{w}$)
Reaction fronts form around walls (with strength $\alpha$)
Equation of motion is
$$ - \ddot{h}_{ij}(x,t) +\nabla^2 h_{ij}(x,t) = 16 \pi G
T^\text{source}_{ij}(x,t). $$
where the sources are
$$ T^{\text{source},\,\phi}_{ij} = \partial_i \phi
\partial_j \phi; \qquad T^{\text{source},\,\text{fluid}}
= w u_i u _j $$
Power law above peak is between $k^{-2}$ and
$k^{-1}$
(Spurious “ringing” due to simultaneous nucleation)
From $\phi$ and $u_\mu$ to $h_{ij}$ and $\Omega_\text{GW}$
As discussed, simply evolve:
$$ \square h_{ij}(x,t) = 16 \pi G
T_{ij}^\text{source}(x,t). $$
Note that when $T_{ij}^\text{source}(x,t) =
w(x)u_i(x)u_j(x)$ this is basically a convolution of the
fluid velocity power (assuming $w(x) \approx \bar w$)
Caprini, Durrer and
Servant
When we want to measure the energy in gravitational
waves, we do the projection to TT and measure:
$$ t_{\mu\nu}^\text{GW}= \frac{1}{32 \pi G}\langle
\partial_\mu h^\text{TT}_{ij} \partial_\nu
h^\text{TT}_{ij} \rangle; \quad
\rho_\text{GW} = \frac{1}{32 \pi G} \langle
\dot{h}^\text{TT}_{ij} \dot{h}^\text{TT}_{ij}
\rangle. $$
We can then redshift this to present day to get
$\Omega_\text{GW} h^2$.
Students: Anna Kormu,
Tiina Minkkinen, Mika Mäki, Riikka Seppä,
Satumaaria Sukuvaara
Postdocs and researchers:
Jani Dahl, Deanna C. Hooper,
Collaborators past and present:
José Correia, Daniel Cutting, Oliver Gould, Jonathan Kozaczuk, Mark Hindmarsh,
Stephan Huber, Lauri Niemi, Hiren Patel, Michael Ramsey-Musolf, Kari Rummukainen,
Tuomas Tenkanen, Essi Vilhonen
Understanding primordial gravitational waves, even with my focus on phase
transitions,
involves many areas of physics, not just GR or cosmology.
Early universe processes (such as phase transitions) can probe and constrain
BSM
physics
... but we need precise predictions of key parameters
$\Rightarrow$ lattice Monte Carlo simulations of phase transitions