2. MHD results

The system considered is sketched in figure 1. The upstream field $\boldsymbol {B}_1$ and velocity $\boldsymbol {v}_1$ are normal to the front, but the downstream ones $\boldsymbol {B}_2$ and $\boldsymbol {v}_2$ are not. They make an angle $\theta _2$ and $\xi _2$ with the shock normal and by default, $\theta _2 \neq \xi _2$ (even though they will be found equal in the following).

Figure 1. System considered. The upstream has density $n_1$ and isotropic temperature $T_1$. Both the upstream field $\boldsymbol {B}_1$ and velocity $\boldsymbol {v}_1$ are normal to the front. The downstream has density $n_2$ and temperatures $T_{2\parallel }, T_{2\perp }$, parallel and perpendicular to the downstream field, respectively. The downstream field $\boldsymbol {B}_2$ and velocity $\boldsymbol {v}_2$ make an angle $\theta _2$ and $\xi _2$, respectively, with the front normal. The parallel and perpendicular directions are therefore defined with respect to the local magnetic field.

We here briefly remember the MHD theory for switch-on shocks. Due to the complexity of the forthcoming calculations, we treat only the case of a sonic strong shock, namely upstream temperature $T_1=0$, or equivalently, upstream sonic Mach number $\mathcal {M}_{s1}=\infty$.

For isotropic pressures in the upstream and the downstream, and $\theta _1=\xi _1=0$, the MHD conservation equations for strong shock and an adiabatic index of $\gamma =5/3$ read (see for example Kulsrud Reference Kulsrud2005, p. 141)

(2.1)\begin{gather} n_2 v_2 \cos \xi_2=n_1 v_1, \end{gather}

(2.2)\begin{gather}B_2 \cos \theta_2=B_1, \end{gather}

(2.3)\begin{gather}B_2 v_2 \sin \theta_2 \cos \xi_2 - B_2 v_2 \cos \theta_2 \sin \xi_2= 0, \end{gather}

(2.4)\begin{gather}\frac{B_2^2 \sin ^2\theta_2}{8 {\rm \pi}}+n_2 k_B T_2 + m n_2 v_2^2 \cos ^2\xi_2 = m n_1 v_1^2, \end{gather}

(2.5)\begin{gather}m n_2 v_2^2 \sin \xi_2 \cos \xi_2-\frac{B_2^2 \sin \theta_2 \cos \theta_2}{4 {\rm \pi}} = 0, \end{gather}

(2.6)\begin{gather}m n_2 v_2 \cos \xi_2 \left(\frac{5}{2}\frac{k_B T_2 }{m}+\frac{B_2^2 \sin ^2\theta_2}{4 {\rm \pi}m n_2}+\frac{v_2^2}{2}\right) -\frac{B_2^2}{4 {\rm \pi}}v_2 \sin \theta_2 \cos \theta_2 \sin \xi_2 = \frac{1}{2} m n_1 v_1^3, \end{gather}

where $m$ is the mass of the particles and $k_B$ the Boltzmann constant.

Equation (2.1) stands for the conservation of mass. Equation (2.2) for the conservation of the magnetic field normal component. Equation (2.3) for the vanishing of the $z$ component of the electric field. Equations (2.4) and (2.5) come from the conservation of the momentum flux (see Appendix A), and (2.6) from the conservation of energy.

By eliminating $v_2$ and $B_2$ thanks to (2.1) and (2.2), and then eliminating $T_2$ thanks to (2.4), the system is amenable to three equations,

(2.7)\begin{gather} \tan \theta_2-\tan \xi_2 = 0, \end{gather}

(2.8)\begin{gather}\mathcal{M}_{A1}^2 \tan \xi_2-r \tan \theta_2 = 0, \end{gather}

(2.9)\begin{gather}2 \mathcal{M}_{A1}^2 [(r-5) r+5-\sec ^2\xi_2]+r \tan \theta_2 \left(\tan \theta_2+4 \tan \xi_2\right) = 0, \end{gather}

in terms of the dimensionless density ratio $r$ and the Alfvén Mach number $\mathcal {M}_{A1}$,

(2.10)\begin{equation} \left. \begin{gathered} r = \frac{n_2}{n_1}, \\ \mathcal{M}_{A1}^2 = \frac{m n_1 v_1^2}{B_1^2/4{\rm \pi}}. \end{gathered} \right\} \end{equation}

The first equation imposes $\theta _2=\xi _2$. Replacing in the last two gives

(2.11)\begin{gather} \tan \theta_2 (\mathcal{M}_{A1}^2-r) = 0, \end{gather}

(2.12)\begin{gather}2 \mathcal{M}_{A1}^2 [(r-5) r+5-\sec ^2\theta_2]+5 r \tan ^2\theta_2 = 0 . \end{gather}

Equation (2.11) clearly defines two kinds of shocks.

1. (i) The first kind comes from $\tan \theta _2=0$, that is, $\theta _2=0$. Inserting it into (2.12) gives $r=1$ or $r=4$. The first option, $r=1$, is the continuity solution, where nothing changes between the upstream and the downstream. The second option is the parallel shock solution, with $r=4$ for a sonic strong shock and an adiabatic index $\gamma =5/3$.

2. (ii) Yet, (2.11) also allows for

   (2.13)\begin{equation} r = \mathcal{M}_{A1}^2, \end{equation}
   which is the MHD switch-on solution. Inserting it into (2.12) gives
   (2.14)\begin{equation} \cos^2\theta_2=\frac{3}{10 \mathcal{M}_{A1}^2-2 \mathcal{M}_{A1}^4-5}. \end{equation}
   The value of $\theta _2$ so defined is displayed in figure 2(a). Here, $\theta _2 \neq 0$ is only permitted within a finite range of Alfvén Mach numbers defined by $\cos ^2\theta _2<1$, that is, $1 < \mathcal {M}_{A1} < 2$. Note that instead of parametrizing $\theta _2$ by the Alfvén Mach number $\mathcal {M}_{A1}$, we choose the variable
   (2.15)\begin{equation} \sigma = \frac{B_1^2/4{\rm \pi}}{m n_1 v_1^2} = \frac{1}{\mathcal{M}_{A1}^2}. \end{equation}
   This $\sigma$ parameter allows for a straightforward comparison with PIC simulations where $\sigma$ is usually used instead of $\mathcal {M}_{A1}$ (see for example Sironi & Spitkovsky Reference Sironi and Spitkovsky2011; Bret Reference Bret2020). As a function of $\sigma$, $\theta _2 \neq 0$ is allowed for $\sigma \in [1/4,1]$.

Figure 2. (a) Value of $\theta _2$ from (2.14) in terms of $\sigma =\mathcal {M}_{A1}^{-2}$. Its maximum value is $\arccos \sqrt {2/5} \sim 0.56 ({{\rm \pi} }/{2})$. (b) MHD density jump $r$ in terms of $(\sigma,\theta _1)$ for $\theta _1 \in [0,{\rm \pi} /2]$. Only the blue line, which has $\theta _1=\theta _2=0$, was considered in Bret & Narayan (Reference Bret and Narayan2018). The red line is the switch-on solution (2.13), with $\theta _1=0$ but $\theta _2\neq 0$.

For a finite upstream temperature $T_1 > 0$, MHD switch-on solutions are also restricted to a range of upstream temperatures via $\beta _1=n_1k_BT_1/B_1^2 < 2/\gamma$, where $\gamma$ is the adiabatic index (Kennel, Blandford & Coppi Reference Kennel, Blandford and Coppi1989; de Sterck & Poedts Reference de Sterck and Poedts1999; Delmont & Keppens Reference Delmont and Keppens2011). (See in particular figure 3 in de Sterck & Poedts (Reference de Sterck and Poedts1999).) Since the present work is limited to $T_1=0$, it cannot explore this dimension of the switch-on solutions range.

While they have been produced in the laboratory (Craig & Paul Reference Craig and Paul1973), such shocks have been rarely detected in space due to the smallness of the parameter window that allows them. Feng et al. (Reference Feng, Lin, Chao, Wu, Lyu and Lee2009) reported the detection of a ‘possible’ interplanetary switch-on shock. Also, Farris et al. (Reference Farris, Russell, Fitzenreiter and Ogilvie1994); Russell & Farris (Reference Russell and Farris1995) reported the detection of one switch-on shock among the ISEE (International Sun-Earth Explorer; see Ogilvie, Rosenvinge & Durney (Reference Ogilvie, Rosenvinge and Durney1977)) data. The more recent review by Balogh & Treumann (Reference Balogh and Treumann2013) still refers to the results of Farris et al. (Reference Farris, Russell, Fitzenreiter and Ogilvie1994) as ‘the rare case of observation of a switch-on shock’ in its § 2.3.6.

Finally, it is interesting to compute the MHD density jump $r$ for any upstream angle $\theta _1$. This can be done solving the ‘shock adiabatic’ equation given in Fitzpatrick (Reference Fitzpatrick2014, § 7.21), and setting $T_1=0$. ($V_{S1}=0$ in the notation of Fitzpatrick (Reference Fitzpatrick2014).) The result is pictured in figure 2(b), in terms of $\theta _1 \in [0,{\rm \pi} /2]$ and $\sigma$. Only the blue line, which has $\theta _1=\theta _2=0$, was considered in Bret & Narayan (Reference Bret and Narayan2018). The red line is the switch-on solution (2.13), with $\theta _1=0$ but $\theta _2\neq 0$.

4. Properties of Stage 1

Due to the complexity of the calculations, we treat only the case of a sonic strong shock, namely $T_1=0$.

4.1. Conservation equations for anisotropic temperatures

The conservation equations for anisotropic temperatures in the downstream are established in Appendix A. Though with different notation, they can be found in Hudson (Reference Hudson1970); Erkaev et al. (Reference Erkaev, Vogl and Biernat2000). With $T_1=0$, they read,

(4.1)\begin{gather} n_2 v_2 \cos \xi_2=n_1 v_1, \end{gather}

(4.2)\begin{gather}B_2 \cos \theta_2=B_1, \end{gather}

(4.3)\begin{gather}B_2 v_2 \sin \theta_2 \cos \xi_2 - B_2 v_2 \cos \theta_2 \sin \xi_2= 0, \end{gather}

(4.4)\begin{gather}\cos ^2\theta_2 n_2 k_B T_{2 \parallel}+ \sin ^2\theta_2 n_2 k_B T_{2 \perp}+ m n_2 v_2^2 \cos ^2\xi_2 -\frac{B_2^2 \cos \left(2 \theta _2\right)}{8 {\rm \pi}} ={-}\frac{B_1^2 }{8 {\rm \pi}} +m n_1 v_1^2, \end{gather}

(4.5)\begin{gather}\sin \theta_2 \cos \theta_2 n_2 k_B(T_{2 \parallel}-T_{2 \perp})+m n_2 v_2^2 \sin \xi_2 \cos \xi_2-\frac{B_2^2 \sin \left(2 \theta _2\right)}{8 {\rm \pi}}= 0, \end{gather}

(4.6)\begin{gather}\left[v \left(\mathcal{A} \cos \xi+\mathcal{B} \cos \xi+\mathcal{C} \sin \xi\right)\right]_1^2= 0, \end{gather}

where

(4.7)\begin{equation} \left. \begin{aligned} \mathcal{A} & =\frac{1}{2} n k_B T_{{\parallel}}+n k_B T_{{\perp}}+\frac{B^2}{8 {\rm \pi}}+\frac{1}{2} m n v^2, \\ \mathcal{B} & ={-}\frac{B^2 \cos \left(2 \theta \right)}{8 {\rm \pi}}+ \cos^2\theta n k_B T_{{\parallel}}+ \sin^2\theta \,n k_B T_{{\perp}},\\ \mathcal{C} & = \sin \theta \cos \theta \, n k_B\left(T_{{\parallel}}-T_{{\perp}}\right)-\frac{B^2 \sin \left(2 \theta \right)}{8 {\rm \pi}}. \end{aligned} \right\} \end{equation}

In (4.6), the notation $[Q]_1^2$ stands for the difference of any quantity $Q$ between the upstream and the downstream.

With $T_1=0$, prescriptions (3.2) for the downstream temperatures in Stage 1 simply read

(4.8)\begin{equation} \left. \begin{gathered} T_{2\parallel} = T_e \cos^2\theta_2, \\ T_{2\perp} = \frac{1}{2} T_e \sin^2\theta_2. \end{gathered} \right\} \end{equation}

4.3. Stability of Stage 1

If unstable, Stage 1 is mirror or firehose unstable. The thresholds for these instabilities are given by (Gary Reference Gary1993; Gary & Karimabadi Reference Gary and Karimabadi2009)

(4.9)\begin{equation} \frac{T_{2\perp}}{T_{2\parallel}} \equiv A_2 = 1 \pm \frac{1}{\beta_{{\parallel} 2}}, \end{equation}

where

(4.10)\begin{equation} \beta_{{\parallel} 2} = \frac{n_2 k_B T_{2\parallel}}{B_2^2/(8{\rm \pi})}, \end{equation}

and the ‘+’ and ‘$-$’ signs stand for the thresholds of the mirror and firehose instabilities, respectively. From (4.8), we obtain the downstream anisotropy $A_2$,

(4.11)\begin{equation} A_2 = \tfrac{1}{2} \tan^2\theta_2. \end{equation}

For $\beta _{\parallel 2}$, we obtain, in Appendix C,

(4.12)\begin{equation} \beta_{{\parallel} 2} ={-} 2 \frac{r \sigma \tan^2\theta_2-2 r+2}{ r \sigma (\tan ^4\theta_2+2 ) }. \end{equation}

To assess the stability of Stage 1, we then proceed as follows.

1. (i) From (4.9), we plot the thresholds for the mirror and firehose instabilities in the $(\beta _{\parallel 2},A_2)$ plane.

2. (ii) Then, on the same graph, we plot the curves $(\beta _{\parallel 2},A_2)$ for the two non-trivial $\theta _2$-branches found in § 4.2.

The result is pictured in figure 4(a). In Bret & Narayan (Reference Bret and Narayan2018), Stage 1 had $A_2=0$ for the sonic strong shock case. Here, $A_2$ departs from 0 but remains small.

Figure 4. (a) Thresholds for the mirror (upper grey line) and firehose (lower grey line) instabilities. The plasma is unstable within the shaded areas. The loop shows the curves $(\beta _{\parallel 2}(\sigma ),A_2(\sigma ))$ for the two $\theta _2$-branches defined previously. The two branches start from the same point for $\sigma =0.432$, and both reach $A_2=0$ for $\sigma =1$. (b) Stability analysis of the blue branch for $\sigma \in [0.432, 1]$. It is found firehose unstable for $\sigma \in [0.82, 1]$.

It turns out that the orange branch pictured in figure 3(a), namely the one closest to the MHD solution, is stable for any $\sigma$. Yet, the blue one is slightly unstable in some $\sigma$ range. For this branch, the quantity $A_2 - (1-\beta _{\parallel 2}^{-1})$ is plotted in figure 4(b). It is negative for $\sigma \in [0.83, 1]$, indicating firehose instability. In this $\sigma$-range, the downstream will therefore migrate to Stage 2, on the firehose threshold.

As can be seen in figure 4(a), the blue branch Stage 1 is only slightly unstable. Consequently, the corresponding marginally stable Stage 2 is very close to the unstable states. This is confirmed below in § 5, where Stage 2 is analysed.

For now, to document the differences between our two branches, we further study Stage 1 by computing its entropy and its Alfvénic downstream velocity.

4.4. Entropy of Stage 1

The two branches for Stage 1 cannot be distinguished by their energy since they both fulfil the energy conservation equation (4.6), where the upstream energy is the same in both cases. Their energy densities are therefore identical. Yet, they can be distinguished on the basis of their entropy.

For a bi-Maxwellian of the form,

(4.13)\begin{equation} F= \frac{n}{{\rm \pi}^{3/2} \sqrt{a} b}\exp \left(-\frac{v_x^2}{a}\right) \exp \left(-\frac{v_y^2+v_z^2}{b}\right), \end{equation}

where $a=2k_BT_\parallel /m$ and $b=2k_BT_\perp /m$, the entropy reads

(4.14)\begin{equation} S ={-}k_B \int F \ln F d^3v = \frac{1}{2} k_B n [3 + \ln ({\rm \pi}^3 a b^2 )-2 \ln n], \end{equation}

where $n=\int F d^3v$. Using the subscript ‘b’ for the blue branch in figure 3, and subscript ‘$o$’ for the orange one, we get for the entropy difference per particle between the two branches,

(4.15)\begin{align} \frac{2}{k_B} \left(\frac{S_o}{n_o}-\frac{S_b}{n_b}\right) \equiv \Delta s & = \ln \left[ \frac{a_o b_o^2}{a_b b_b^2} \right] + 2\ln \frac{n_b}{n_o}, \nonumber\\ & = \ln \left[ \frac{T_{{\parallel} 2,o} \, T_{{\perp} 2,o}^2}{T_{{\parallel} 2,b}\, T_{{\perp} 2,b}^2} \right]+2\ln \frac{n_b}{n_o}, \nonumber\\ & = \ln \left[ \frac{T_{{\parallel} 2,o}^3\, A_{2,o}^2}{T_{{\parallel} 2,b}^3 \, A_{2,b}^2} \right]+2\ln \frac{n_b}{n_o}, \nonumber\\ & = \ln \left[ \left( \frac{T_{e,o} \cos^2\theta_{2,o}}{ T_{e,b} \cos^2\theta_{2,b} } \right)^3 \frac{A_{2,o}^2}{A_{2,b}^2} \right] + 2\ln \frac{n_b}{n_o}, \end{align}

where we have used $A_2=T_{\perp 2}/T_{\parallel 2}$ and then $T_{\parallel 2} = T_e \cos ^2\theta _2$ for both branches.

The numerical evaluation ($T_e$ is given by (B6)) of this quantity displayed in figure 5(a) shows that $\Delta s < 0$ for any $\sigma \in [0.432, 1]$. Therefore, $S_o/n_o < S_b/n_b$ for any $\sigma$. The orange branch in figure 3 has lower entropy than the blue one.

Figure 5. (a) Entropy difference $\Delta s$ as defined by (4.15), between the blue branch in figure 3 and the orange one. Here, $\Delta s < 0$ implies the orange branch has lower entropy than the blue one. (b) Value of $\mathcal {M}_{A2}$ in MHD (green) and for the two branches of our model.

4.5. Downstream Alfvénic Mach number of Stage 1

Another difference between the two branches lies in their respective Alfvénic Mach number, namely,

(4.16)\begin{equation} \mathcal{M}_{A2}^2 = \frac{m n_2 v_2^2}{B_2^2/4{\rm \pi}}. \end{equation}

From (4.2), we get $B_2=B_1/\cos \theta _2$. Then (4.1) gives $v_2 = n_1v_1/n_2\cos \xi _2 = n_1v_1/n_2\cos \theta _2$, since $\xi _2=\theta _2$ in our model as in MHD (see (2.7) for MHD and Appendix B for our model). We finally obtain, in terms of the dimensionless variables (2.10), with both our model and MHD,

(4.17)\begin{equation} \mathcal{M}_{A2} = \frac{\mathcal{M}_{A1}}{\sqrt{r}} = \frac{1}{\sqrt{\sigma r}}. \end{equation}

1. (i) In MHD, (2.13) has $r=\mathcal {M}_{A1}^2$, so that for the switch-on MHD shock, $\mathcal {M}_{A2} = 1$ (Goedbloed et al. Reference Goedbloed, Keppens and Poedts2010, p. 853).

2. (ii) In our model, the value of $\mathcal {M}_{A2}$ is pictured in figure 5(b) for the two branches represented in figure 3. Our two branches are found slightly sub-Alfvénic.

5. Properties of Stage 2

The firehose instability of the blue branch for $\sigma \in [0.83,1]$ requires studying the properties of Stage 2 when marginally firehose stable. The conservation equations are the same. However, instead of imposing prescriptions (3.2) for the temperatures, we now impose firehose marginal stability for the downstream, namely,

(5.1)\begin{equation} \frac{T_{2\perp}}{T_{2\parallel}} = 1 - \frac{1}{\beta_{{\parallel} 2}}. \end{equation}

The resolution of the system follows the same path as that described in Appendix B for Stage 1. It yields three equations for $\theta _2,\xi _2$ and $r$,

(5.2)\begin{gather} \tan \theta_2-\tan \xi_2= 0, \end{gather}

(5.3)\begin{gather}\frac{2 \tan \xi_2}{r}-\frac{\tan \theta_2}{\mathcal{M}_{A1}^2} = 0, \end{gather}

(5.4)\begin{gather}\mathcal{M}_{A1}^2 ({-}\sec ^2\xi_2+(r-5) r+5)+r \tan \theta_2 \tan \xi_2+r = 0. \end{gather}

Equation (5.2) imposes again $\theta _2 = \xi _2$. Replacing in (5.3) gives

(5.5)\begin{equation} \tan \theta_2 ( r - 2\mathcal{M}_{A1}^2 ) = 0, \end{equation}

which leaves two options only.

1. (i) $\theta _2=0$, which pertains to the parallel shock solution. Setting then $\theta _2=0$ in (5.4) then gives exactly the Stage 2 solution found in Bret & Narayan (Reference Bret and Narayan2018). (See (3.5) of Bret & Narayan (Reference Bret and Narayan2018) for $\chi _1=\infty$.)

2. (ii) The other option is,

   (5.6)\begin{equation} r=2\mathcal{M}_{A1}^2=2/\sigma. \end{equation}
   Inserting this result in (5.4) and solving for $\theta _2$ gives
   (5.7)\begin{equation} \cos^2 \theta _2 = \frac{1}{10 \mathcal{M}_{A1}^2 - 4 \mathcal{M}_{A1}^4 - 5}, \end{equation}
   reminiscent of (2.14) for the MHD case. Solutions can here be found for
   (5.8)\begin{equation} \left. \begin{aligned} \mathcal{M}_{A1} & \in \left[ \frac{\sqrt{5-\sqrt{5}}}{2}, \frac{\sqrt{5+\sqrt{5}}}{2} \right] \sim [0.83, 1.34], \\ \Leftrightarrow \sigma = \frac{1}{\mathcal{M}_{A1}^2} & \in \left[ \frac{4}{5+\sqrt{5}}, \frac{4}{5-\sqrt{5}} \right] \sim [0.55, 1.44]. \end{aligned} \right\} \end{equation}

The counterpart to switch-on shock is therefore recovered in our model for Stage 2 as well, still in a limited range of Alfvén Mach numbers.

Figure 6 is eventually the end result of the present work. Like figure 3(b), it features the density jump of the MHD switch-on solution, together with the two branches of our model. However here, the way Stages 1 and 2 fit together in the $\sigma$ unstable range is elucidated. Since the blue branch has been found firehose unstable for $\sigma \in [0.83, 1]$, it is replaced by Stage 2, namely (5.6), in this range. As expected, the corresponding density jump is very close to that of Stage 1 since the system is almost marginally stable in this range, while Stage 2 sits exactly on marginal stability.

Figure 6. Same as figure 3(b) but showing how Stages 1 and 2 fit together when accounting for the firehose instability of the blue branch for $\sigma \in [0.83, 1]$. Stage 2 density jump $r=2/\sigma$ is discontinued from $\sigma = 1$ since Stage 1 has no solution in this range.

In figure 6, the jump for Stage 2, namely $r=2/\sigma$, is shown in black and plotted within the full range (5.8) where it is defined. For $\sigma < 0.83$, the line is dashed because Stage 1 is stable, and hence defines the density jump. Then for $\sigma \in [0.83, 1]$, the blue branch is dashed since it pertains to the unstable Stage 1. There, the jump is now given by Stage 2 through $r=2/\sigma$. Beyond $\sigma = 1$, Stage 1 offers no solutions. Since in our scenario Stage 1 is the first state of the downstream after crossing the front, the shock cannot accommodate such values of $\sigma$ in the switch-on regime. For $\sigma > 1$, there is therefore no Stage 1 from where the system could jump to Stage 2, even though Stage 2 offers solutions. As a consequence, the black curve is dashed from $\sigma =1$ to $1.44$.