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I was reading Sikivie's paper on Axion Cosmology: https://arxiv.org/abs/astro-ph/0610440 . It says the equation of motion for the axion field $a(x)$ in Friedmann–Lemaître–Robertson–Walker space-time is

$$D_\mu \partial^\mu a(x) + V_a′(a(x)) = (\partial_t^2 + 3\frac{\dot{R}}{R}\partial_t - \frac{1}{R^2}\nabla_x^2)a(x) + V_a′(a(x)) = 0$$ where $R(t)$ is the cosmological scale factor and $V$ is the potential.

Can someone tell me what $D_\mu$ is and how one might derive the right hand side of the above equation?

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2 Answers 2

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$D_\mu$ is just the covariant derivative acting on the vector $\partial^\mu a$. The middle expression is just expanding this derivative out in terms of the Christoffel symbols for the FLR metric. The actual eq of motion is the last equals sign.

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  • $\begingroup$ I tried expanding out $D_\mu$ using the -+++ sign convention for the FLRW metric, i.e., $g_{00} = -1$ and $g_{ij} = \delta_{ij} R^2(t)$. But I got $D_\mu\partial^\mu a = (-\partial_t^2 + 3\frac{\dot{R}}{R}\partial_t + \frac{1}{R^2}\nabla_x^2)a$. Am I using the wrong sign convention or did I do something totally wrong? $\endgroup$
    – Matrix23
    Oct 26, 2021 at 14:18
  • $\begingroup$ That I do not know, but it looks like a different metric convention since it's only signs that differ. $\endgroup$
    – mike stone
    Oct 26, 2021 at 17:31
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When applied to scalar functions, $Δ = D_μ ∂^μ$ is the Hodge-deRham operator associated with the metric $g$. Since $$\begin{align} Δa & = D_μ ∂^μ a = ∂_μ ∂^μ a + Γ_{νμ}^μ ∂^ν a = ∂_μ ∂^μ a + Γ_{μν}^ν ∂^μ a \\ Γ_{μν}^ν & = g^{νρ} \frac{1}{2} \left(∂_μg_{νρ} + ∂_νg_{ρμ} - ∂_ρg_{μν}\right) = \frac{1}{2} g^{νρ} ∂_μg_{νρ} = \frac{1}{2} \frac{∂_μg}{g} = \frac{∂_μ\sqrt{|g|}}{\sqrt{|g|}} \end{align}$$ then you may also write this operator as $$Δa = ∂_μ ∂^μ a + \frac{∂_μ\sqrt{|g|}}{\sqrt{|g|}} ∂^μ a = \frac{∂_μ\left(\sqrt{|g|}∂^μ a\right)}{\sqrt{|g|}} = \frac{1}{\sqrt{|g|}} ∂_μ\left(\sqrt{|g|}g^{μν}∂_νa\right).$$

If $$g_{00} = -1, \hspace 1em g_{i0} = 0 = g_{0j}, \hspace 1em g_{ij} = R^2 δ_{ij} \hspace 1em (i, j = 1, 2, 3),$$ where $R = R(t)$ is a function only of $t$, with $R(t) > 0$, and the coordinates are $\left(x^0, x^1, x^2, x^3\right) = \left(t, x, y, z\right)$, then $g = -R^6$, $\sqrt{|g|} = R^3$ and $$\sqrt{|g|}g^{00} = -R^3, \hspace 1em \sqrt{|g|}g^{i0} = 0 = \sqrt{|g|}g^{0j}, \hspace 1em \sqrt{|g|}g^{ij} = R δ^{ij} \hspace 1em (i, j = 1, 2, 3).$$ Therefore $$Δa = \frac{1}{R^3} \left(\frac{∂}{∂t}\left(-R^3\frac{∂a}{∂t}\right) + ∇·\left(R∇a\right)\right)$$ where $∇ = (∂/∂x, ∂/∂y, ∂/∂z)$. Thus, $$ Δa = -\frac{∂^2a}{∂t^2} - \frac{3}{R} \frac{∂R}{∂t}\frac{∂a}{∂t} + \frac{1}{R^2} ∇^2a. $$

If you rescale $g_{μν} → k g_{μν}$ by a constant $k$, then $\sqrt{|g|} → k^2 \sqrt{|g|}$ and $\sqrt{|g|} g^{μν} → k \sqrt{|g|} g^{μν}$. Therefore, $Δ → k^{-1} Δ$. The operator is scale-dependent ... which also means: sign-dependent. Choosing $k = -1$ resets the metric as $$g_{00} = 1, \hspace 1em g_{i0} = 0 = g_{0j}, \hspace 1em g_{ij} = -R^2 δ_{ij} \hspace 1em (i, j = 1, 2, 3),$$ and the Hodge-deRham operator as $$ Δa = \frac{∂^2a}{∂t^2} + \frac{3}{R} \frac{∂R}{∂t}\frac{∂a}{∂t} - \frac{1}{R^2} ∇^2a. $$ So, the authors are treating the metric as a proper-time metric, rather than a proper-length metric.

As a (propaganda) footnote, if you're writing the metric as a proper time metric then you could just as well treat the $R$ factor as being the actual $R$, not $R$ modulo any $c = 1$ assumption. Then, it is normalized as $R(\text{now}) = 1/c$, where $c = 299792458$ meters/second. That shows that this is secretly just an ordinary Minkowski metric ... but with a variable light speed $c(t) = 1/R(t)$ and line element $$ds^2 = dt^2 - \frac{dx^2 + dy^2 + dz^2}{c(t)^2}$$ where $s$ now denotes proper time; and that - with this understanding - the Hodge-deRham operator can just as well be written as: $$Δa = \frac{∂^2a}{∂t^2} - \frac{3}{c} \frac{∂c}{∂t}\frac{∂a}{∂t} - c^2 ∇^2a.$$ If you carefully apply this to the field $F_{μν}$, which is given in terms of the Maxwell vectors by $𝐄 = \left(F_{10}, F_{20}, F_{30}\right)$, and $𝐁 = \left(F_{23}, F_{31}, F_{12}\right)$, and to the response field, which is given by the tensor density $$𝔊^{μν} = ε_0 c \sqrt{|g|} g^{μρ} g^{νσ} F_{ρσ} + \frac{1}{2} θ ε^{μνρσ} F_{ρσ}$$ (where $ε_0$ denotes the permittivity of the vacuum) and in terms of the Maxwell letters by $𝐃 = \left(𝔊^{01}, 𝔊^{02}, 𝔊^{03}\right)$ and $𝐇 = \left(𝔊^{23}, 𝔊^{31}, 𝔊^{12}\right)$, then you'll get the following constitutive relations $$𝐃 = ε 𝐄 + θ 𝐁, \hspace 1em 𝐇 = \frac{𝐁}{μ} - θ 𝐄$$ with $$ε = ε_0 c R = ε_0 \frac{c}{c(t)}, \hspace 1em μ = μ_0 c R = μ_0 \frac{c}{c(t)}$$ where $μ_0 = 1/(ε_0 c^2)$ denotes the permeability of the vacuum. The axion $a$ is just $-θ$ up to an author-dependent positive constant factor and is just the chiral counterpart to permittivity, promoted to a field. The permittivity and permeability, themselves, are effectively fields too because of their metric-dependency. They embody the optics of warped space-time.

These fields satisfy the usual Maxwell Equations $∇·𝐃 = ρ$ for a charge density $ρ$ and $∇×𝐇 - ∂𝐃/∂t = 𝐉$ for a current density $𝐉$ and the corresponding electromagnetic wave speed is $$\sqrt{\frac{1}{εμ}} = \sqrt{\frac{c(t)^2}{c^2} \frac{1}{ε_0μ_0}} = c(t).$$ So $c(t)$ really, literally is the speed of light we're talking about there.

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