Pattern dark matter and galaxy scaling relations

Abstract

We argue that a natural explanation for a variety of robust galaxy scaling relations comes from the perspective of pattern formation and self-organization as a result of symmetry breaking. We propose a simple Lagrangian model that combines a conventional model for normal matter in a galaxy with a conventional model for stripe pattern formation in systems that break continuous translation invariance. We show that the energy stored in the pattern field acts as an effective dark matter. Our theory reproduces the gross features of elliptic galaxies as well as disk galaxies (HSB and LSB) including their detailed rotation curves, the radial acceleration relation, and the Freeman law. We investigate the stability of disk galaxies in the context of our model and obtain scaling relations for the central dispersion for elliptical galaxies. A natural interpretation of our results is that (1) ‘dark matter’ is potentially a collective, emergent phenomenon and not necessarily an as yet undiscovered particle, and (2) MOND is an effective theory for the description of a self-organized complex system rather than a fundamental description of nature that modifies Newton’s second law.

Appendices

Appendix A: The stress tensor for pattern dark matter

We now follows the discussion in [32] and compute the stress–energy–momentum $T^{\alpha \beta }$ corresponding to this solution using the Einstein–Hilbert prescription $T^{\alpha \beta } = \frac{2}{\sqrt{-g}} \frac{\delta {\tilde{E}}}{\delta g_{\alpha \beta }}$ where ${\tilde{E}}$ is the appropriate Lagrangian in curved space-time and $g = \mathrm {det}[g_{\alpha \beta }]$. The minimally coupled [34] pattern action ${\mathcal {S}}_\mathrm{{P}}$ is:

$$\begin{aligned} {\mathcal {S}}_\mathrm{{P}} = \frac{{\varSigma }^*_0 c^2}{k_0^3} \int \left\{ (\nabla ^\mu \psi \nabla _\mu \psi -k_0^2)^2 + (\nabla ^\mu \nabla _\mu \psi )^2\right\} \sqrt{-g} \ d^4x, \nonumber \\ \end{aligned}$$

(A.1)

where the metric $g_{\alpha \beta }$ has signature $(- \,+\, +\, +)$, $\nabla _\mu$ is the corresponding covariant derivative. To obtain the stress tensor in a (background) flat space time, it suffices to consider variations $g_{\alpha \beta } = \eta _{\alpha \beta } + t \rho _{\alpha \beta }$ and compute all the variations to first order in t. To this end, we record the following relations for the inverse metric $g^{\gamma \delta }$, the Christoffel symbols ${\varGamma }^\sigma _{\gamma \delta }$ and the quantities that appear in ${\mathcal {S}}_\mathrm{{P}}$:

$$\begin{aligned} g^{\gamma \delta }&= \eta ^{\gamma \delta } - t \, \eta ^{\gamma \alpha }\eta ^{\delta \beta } \rho _{\alpha \beta } + O(t^2) \\ {\varGamma }^\sigma _{\gamma \delta }&= \frac{t}{2}\, \eta ^{\sigma \xi } \left[ \frac{\partial \rho _{\gamma \xi }}{\partial x^\delta } + \frac{\partial \rho _{\delta \xi }}{\partial x^\gamma } - \frac{\partial \rho _{\gamma \delta }}{\partial x^\xi }\right] + O(t^2) \\ \sqrt{-g}&= 1 + \frac{t}{2} \eta ^{\alpha \beta } \rho _{\alpha \beta } + O(t^2) \\ \nabla ^\mu \psi \nabla _\mu \psi&= \eta ^{\gamma \delta } \partial _\gamma \psi \partial _\delta \psi - t \, \eta ^{\gamma \alpha }\eta ^{\delta \beta } \rho _{\alpha \beta } \partial _\gamma \psi \partial _\delta \psi + O(t^2) \\ g^{\gamma \delta } \nabla _\gamma \nabla _\delta \psi&= \Box \, \psi - t \, \eta ^{\gamma \alpha }\eta ^{\delta \beta } \rho _{\alpha \beta } \partial _{\gamma } \partial _{\delta } \psi \\&\quad - \frac{t}{2}\, \eta ^{\gamma \delta } \eta ^{\sigma \xi } \left[ \frac{\partial \rho _{\gamma \xi }}{\partial x^\delta } + \frac{\partial \rho _{\delta \xi }}{\partial x^\gamma } - \frac{\partial \rho _{\gamma \delta }}{\partial x^\xi }\right] \partial _\sigma \psi + O(t^2) \end{aligned}$$

The Einstein–Hilbert stress tensor is now given by

$$\begin{aligned} \left. \frac{\mathrm{{d}}}{\mathrm{{d}}t}{\mathcal {S}}_\mathrm{{P}}\right| _{t=0} = \frac{1}{2} \int T^{\alpha \beta } \rho _{\alpha \beta } \,d^4x + \hbox { boundary terms}. \end{aligned}$$

A straightforward but somewhat lengthy calculation now yields

$$\begin{aligned} T^{\alpha \beta }&= \frac{{\varSigma }^* c^2}{k_0^3}\left\{ -4(\eta ^{\mu \nu } \partial _\mu \psi \partial _\nu \psi -k_0^2) \eta ^{\sigma \alpha }\eta ^{\tau \beta } \partial _\sigma \psi \partial _\tau \psi \right. \nonumber \\&\quad \left. - 4\, \Box \psi \, \eta ^{\sigma \alpha }\eta ^{\tau \beta } \partial _{\sigma } \partial _{\tau } \psi \right. \nonumber \\&\quad + 2 (\eta ^{\alpha \tau } \eta ^{ \beta \sigma } + \eta ^{\beta \tau } \eta ^{\alpha \sigma } - \eta ^{\alpha \beta } \eta ^{\sigma \tau } ) \partial _\tau (\Box \psi \, \partial _\sigma \psi ) \nonumber \\&\quad + \left. \eta ^{\alpha \beta } \left[ (\eta ^{\mu \nu } \partial _\mu \psi \partial _\nu \psi -k_0^2)^2 + (\Box \psi )^2\right] \right\} \end{aligned}$$

(A.2)

We can now express the energy density $T^{\alpha \beta }$ for the (stationary) phase field $\psi (R)$ with respect to a normalized basis $\{{\mathbf {e}}_t, {\mathbf {e}}_R, {\mathbf {e}}_\theta , {\mathbf {e}}_\phi \}$ induced by (spatial) spherical polar coordinates $(r,\theta ,\phi )$.

We decompose the stress tensor into two pieces, $T_\mathrm{{s}}$ coming from the “stretching energy” with density $(|\nabla \psi |^2 - c^{-2} (\partial _t \psi )^2 - k_0^2))^{2}$ and $T_\mathrm{{b}}$ from the bending energy $(\Box \, \psi )^2$. We compute these quantities for a stationary, radial solution $\psi = \psi (r)$ to get

$$\begin{aligned}&T_\mathrm{{s}} \sim \frac{{\varSigma }^*c^2}{k_0^3} \begin{pmatrix} \tau _1 &{} 0 &{} 0 &{} 0 \\ 0 &{} \tau _2 &{} 0 &{} 0 \\ 0 &{} 0 &{}-\tau _1 &{} 0 \\ 0 &{} 0 &{} 0 &{} -\tau _1 \end{pmatrix} \nonumber \\&T_\mathrm{{b}} \sim \frac{{\varSigma }^*c^2}{k_0^3} \begin{pmatrix} \tau _3 &{} 0 &{} 0 &{} 0 \\ 0 &{} \tau _4 &{} 0 &{} 0 \\ 0 &{} 0 &{}-\tau _3 &{} 0 \\ 0 &{} 0 &{} 0 &{} -\tau _3 \end{pmatrix} \end{aligned}$$

(A.3)

where

$$\begin{aligned} \tau _1&= (\psi '(R)^2-k_0^2)^2 \nonumber \\ \tau _2&= (3 \psi '(R)^2 +k_0^2)(\psi '(R)^2- k_0^2) \nonumber \\ \tau _3&= -\psi ''(R)^2-\frac{2 \psi '(R) \left( R \psi '''(R)+4 \psi ''(R)\right) }{R} \nonumber \\ \tau _4&= \frac{8 \psi '(R)^2}{R^2}+\psi ''(R)^2-2 \psi '''(R) \psi '(R) \end{aligned}$$

(A.4)

We remark on the expected structure of $T_\mathrm{{s}}$ and $T_\mathrm{{b}}$, viz. the off-diagonal stresses should be zero from time-reversal and spherical symmetries of the solution $\psi$, and further $T^{00} = -T^{\theta \theta } = - T^{\phi \phi }$ since the metric has signature $(- \,+\, +\, +)$, and $\psi$ is independent of $t, \theta$ and $\phi$. Finally, the quantities $\tau _1, \tau _2, \tau _3$ and $\tau _4$ are constrained by the conservation of energy–momentum $\nabla _\mu T^{\mu \nu } = 0$. A calculation shows that, as expected, these 4 conditions reduce to just one constraint on $\psi (r)$, namely that $\psi$ should satisfy the Euler–Lagrange equation (4).

The stress tensor associated with the field $\psi$ will act as source for the curvature of space-time as we discuss in the body of this paper.

Appendix B: Bound states for Gaussian wells

Following the discussion in Sect. 5.1, we now estimate the ground state energies for spherical and oblate Gaussian wells, to determine the appropriate choices for $k_0$ for spherical (i.e. elliptic) and disk galaxies.

The 1d Gaussian well is given by a potential $V(x) = - V_0 e^{-\frac{x^2}{2 a^2}}$. Close to the minimum, this is approximated by the Harmonic potential $-V_0 + \frac{V_0 x^2}{2 a^2}$ suggesting that an estimate can be obtained by using the variational method with trial functions given by eigenfunctions for the Harmonic oscillator [80]. With the normalized wavefunction $\psi _0 = \left( \frac{\beta }{\pi }\right) ^{1/4} e^{-\beta x^2/2}$, we have, from the variational principle [81],

$$\begin{aligned} \int \left( \psi _0'^2 + V(x) \psi _0^2 \right) dx = \frac{\beta }{2} - V_0 \sqrt{\frac{2 a^2 \beta }{1+2a^2\beta }} \ge -k_0^2 \end{aligned}$$

where $-k_0^2$ is the ground state energy for the operator $-{\varDelta } + V$. For small $\beta > 0$, the bound is $\frac{\beta }{2} - V_0 a \sqrt{2 \beta }$ which can be made negative independent of how small $V_0$ is, i.e. we always have bound states. The optimal $\beta$ minimizes the bound, and is given by

$$\begin{aligned} \frac{1}{2}= & {} V_0 \sqrt{\frac{2 a^2 \beta }{1+2a^2\beta }} \left[ \frac{1}{\beta } - \frac{2a^2}{1+2a^2\beta } \right] \\= & {} a^2 V_0 \sqrt{\frac{2}{a^2\beta (1+2 a^2 \beta )^3}} \end{aligned}$$

It is clear that the optimal $\beta$ is in the “scaling form” $\beta = a^{-2} F(a^2 V_0)$. Considering the regimes $a^2 V_0 \gg 1$ and $a^2 V_0 \ll 1$ separately we get

$$\begin{aligned} a^2 \beta \approx {\left\{ \begin{array}{ll} 8(a^2V_0)^2 &{} a^2 V_0 \ll 1 \\ \sqrt{a^2V_0} &{} a^2 V_0 \gg 1 \end{array}\right. } \end{aligned}$$

In the regime $a^2 V_0 \gg 1$ we get the “deep potential” limit

$$\begin{aligned} k_0^2 > rsim V_0 - \frac{\sqrt{V_0}}{2a}, \end{aligned}$$

(B.1)

corresponding to the Harmonic oscillator $V(x) = -V_0 + \frac{V_0}{2a} x^2$. In the complementary regime $a^2 V_0 \ll 1$, we get cancellation at leading order and

$$\begin{aligned} k_0^2 > rsim 0 (aV_0)^2 + 32 a^6 V_0^4 + \cdots \end{aligned}$$

(B.2)

To investigate if this cancellation is specific to the form of the variational test function that was used, we also consider the normalized wavefunction $\psi _1 = \left( \beta ' \right) ^{1/4} e^{-\sqrt{\beta '}|x| }$, where the variational parameter $\beta '$ is chosen so that it is dimensionally consistent with the earlier choice in the definition of $\psi _0$. The variational principle gives

$$\begin{aligned}&\int \left( \psi _1'^2 + V(x) \psi _1^2 \right) dx\\&\quad =\beta ' - V_0 \,\sqrt{2 \pi a^2 \beta ' } \, \hbox {erfc}\left( \sqrt{2 a^2 \beta '} \right) e^{2 a^2 \beta ' } \ge -k_0^2 \end{aligned}$$

It is again clear that the optimal $\beta '$ is in the “scaling form” $\beta ' = a^{-2} F(a^2 V_0)$. Indeed this is immediate from dimensional considerations. Considering the regimes $a^2 V_0 \gg 1$ and $a^2 V_0 \ll 1$ separately we get

$$\begin{aligned} a^2 \beta ' \approx {\left\{ \begin{array}{ll} \frac{\pi }{2}(a^2V_0)^2 &{} a^2 V_0 \ll 1 \\ \frac{1}{2}\sqrt{a^2V_0} &{} a^2 V_0 \gg 1 \end{array}\right. } \end{aligned}$$

In the regime $a^2 V_0 \gg 1$ we get the variational bound

$$\begin{aligned} k_0^2 > rsim V_0 - \frac{\sqrt{V_0}}{a}, \end{aligned}$$

(B.3)

which is consistent in terms of the scaling of the first correction to the leading order behavior, but suboptimal, in comparison with the bound in (B.1). In the complementary regime $a^2 V_0 \ll 1$, we get

$$\begin{aligned} k_0^2 > rsim \frac{\pi }{2} (aV_0)^2 \end{aligned}$$

(B.4)

showing no cancellation at leading order and giving a result consistent with the limit $V(x) \rightarrow -\sqrt{2 \pi } aV_0 \delta (x)$.

The 3d isotropic Gaussian well is given by the potential $V(x_1,x_2,x_3) = -V_0 \exp \left( -\frac{x_1^2+x_2^2 +x_3^2}{2a^2}\right)$. Considering the product test function ${\varPsi }_0(x_1,x_2,x_3) = \psi _0(x_1) \psi _0(x_2) \psi _0(x_3)$ we get the variational bound

$$\begin{aligned} \int \left( |\nabla {\varPsi }_0|^2 + V(x) {\varPsi }_0^2 \right) d^3x = \frac{3\beta }{2} - V_0 \left( \frac{2 a^2 \beta }{1+2a^2\beta }\right) ^{\frac{3}{2}} \ge -k_0^2 \end{aligned}$$

For sufficiently small $\beta$, the bound is asymptotically given by $\frac{3 \beta }{2} - \sqrt{8} V_0 a^3 \beta ^{3/2}$, and is positive. Likewise, for large $\beta$, the bound is asymptotically equal to $\frac{3 \beta }{2} - V_0$ which is also positive. This reflects the well known fact that attractive potentials in 3d do not support bound states, unless the potential is sufficiently deep.

To estimate the critical value of $V_0$ that allows for a bound state, we exploit the well known connection between the 3d and 1d Schrödinger equations [81], that follows from the identity

$$\begin{aligned} \frac{\partial ^2 f}{\partial R^2} + \frac{2}{R} \frac{\partial f}{\partial R} = \frac{1}{R} \frac{\partial ^2}{\partial R^2} (R f), \end{aligned}$$

so that for every spherically symmetric eigenfunction ${\varPsi }(R)$ of the 3d operator $-{\varDelta } + V(R)$ with a spherically symmetric potential, the function $\chi (R) = R {\varPsi }(R)$ is an eigenfunction of the 1d operator $-\partial _{RR} + V(R)$ with the same energy.

The converse, however, is not true. In order for the kinetic energy $4 \pi \int |\nabla {\varPsi }|^2 R^2 \mathrm{{d}}R$ to be finite, we cannot have ${\varPsi }(R)$ diverging as 1/R near $R = 0$. We therefore need that the corresponding 1d wavefunction $R {\varPsi }(R)$ vanish at $R = 0$, i.e. we need a 1d bound state with a node at the origin to have a bound (ground) state for the 3d potential. The 1st excited state for the 1d gaussian well, which is the lowest energy state with a node at $R=0$, gives the energy of the ground state for the 3d gaussian well.

Fig. 12

The solid curve is energy level $\lambda _1$ of the first excited state of the 1d Gaussian potential $V(R) = \eta e^{-R^2/2}$. This state exists only if the well is sufficiently deep $\eta \ge \eta ^* \approx 1.3$. A quadratic fit (the dashed curve) shows that $\lambda _1 \approx 0.154(\eta -1.3)^2$

Figure 12 shows the numerically computed energy [35] of the first excited state for the potential $W(R) = \eta e^{-R^2/2}$. There is a critical value $\eta ^* \approx 1.3$ and an O(1) constant $c \approx 0.154$ such that the first excited energy level satisfies

$$\begin{aligned} \lambda _1 = a^2 k_0^2 \approx c (V_0 a^2 - \eta ^*)^2 . \end{aligned}$$

(B.5)

For applications to LSB disk galaxies, we also consider the anisotropic (oblate) potential $V(x_1,x_2,x_3) = -V_0 \exp \left( -\frac{x_1^2+x_2^2}{2a^2} - \frac{x_3^2}{2b^2}\right)$ with $b \ll a$. We now consider the product test function ${\varPsi } =\psi _0(x_1) \psi _0(x_2) \psi _1(x_3)$ to allow for the possibility that $V_0 a^2 \gg 1$ but $V_0 b^2 \ll 1$. We now get the variational bound

$$\begin{aligned}&\int \left( |\nabla {\varPsi }|^2 + V(x) {\varPsi }^2 \right) d^3x \\&\quad = \beta +\beta ' - \sqrt{2 \pi b^2 \beta ' } \, \hbox {erfc}\left( \sqrt{2 b^2 \beta '} \right) e^{2 b^2 \beta ' } \frac{2 a^2 \beta V_0}{1+2a^2\beta }\\&\quad \ge -k_0^2 \end{aligned}$$

Since $b^2 V_0 \ll 1$ it is suggestive that $\beta ' b^2 \ll 1$. We will assume this provisionally, and verify later that the assumption is valid. With this assumption, we have the variational bound

$$\begin{aligned} k_0^2 \ge \frac{2 a^2 b \beta \sqrt{2 \pi \beta '} V_0}{1+2a^2\beta } - \beta - \beta ' \end{aligned}$$

Optimizing over $\beta '$, we get that the optimal $\beta '$ is given by

$$\begin{aligned} b^2 \beta ' = \frac{\pi }{2} \left( \frac{2 a^2 b \beta }{1+2a^2\beta }\right) ^2 (b^2 V_0)^2 \end{aligned}$$

so that our assumption that $b^2 \beta ' \ll 1$ verifies. Also, we have the variational bound

$$\begin{aligned} k_0^2 \ge \beta \left( \pi \frac{2 a^2 \beta }{(1+2a^2\beta )^2} \cdot (a^2 V_0) \cdot (b^2 V_0) -1\right) \end{aligned}$$

(B.6)

From this bound, it follows that there exists a bound $\eta _\mathrm{{c}}'$ such that we have bound states for all $a b V_0 \ge \eta _\mathrm{{c}}'$. The optimal $\beta \sim a^{-2}$ and for $ab V_0 \gg 1$, we have $k_0 \sim b V_0$.