Deriving our starting equations

Starting with the momentum equation for a fluid element in ideal MHD (ignoring viscosity and gravity):

\rho \frac{d}{dt} \bold{u} + \rho (\bold{u} \cdot \bold{\nabla}) \bold{u} = - \bold{\nabla} P + \frac{1}{\mu}(\bold{j} \times \bold{B})

The 2nd term on the RHS is the Lorentz force (or more accurately force per unit volume). Since we don't follow charges or currents, we rewrite it as:

\bold{j}\times\bold{B}= (\bold{\nabla} \times \bold{B})\times \bold{B}

which using a vector identity is rewritten as:

(\bold{\nabla} \times \bold{B})\times \bold{B}=-\bold{\nabla}(\frac{B^2}{2 \mu}) + \frac{1}{\mu}(\bold{B}\cdot \bold{\nabla}) \bold{B}

the first term on the RHS is the same form as thermal pressure and so we call it the magnetic pressure. The second term is generally called the magnetic tension. It is the directional derivative of B along B, or in other words, disappears when B is straight and uncurved. Thus, it acts as a restoring force to straighten B out. It can be rewritten as:

\frac{1}{\mu}(\bold{B}\cdot \bold{\nabla}) \bold{B} = \hat{\bold{b}}\hat{\bold{b}} \cdot \bold{\nabla}(\frac{B^2}{2 \mu}) + \frac{B^2}{\mu} \frac{\hat{\bold{n}}}{R_c}

The first term on the RHS is a unit tensor dotted with the gradient of the magnetic pressure. The tensor is formed by the outer product of \hat{\bold{b}} with itself. \hat{\bold{b}} is the unit vector along the magnetic field line. Thus, if the field line was oriented along z,

\hat{\bold{b}}\hat{\bold{b}} =

0	0	0
0	0	0
0	0	1

If the gradient of magnetic pressure was along the field line, then that first term on the RHS would just be equal to the gradient of magnetic pressure. This would cancel with the other term for magnetic pressure in the Navier Stokes equation. Thus, we see that gradients in the magnetic pressure do not provide a force along field lines, i.e. do not provide an isotropic pressure force. Rather, they only act perpendicularly to the field lines, which is what we will consider next.

Say the gradient of magnetic pressure is perpendicular to the magnetic field lines, say for example the gradient is along y (field is pointing along z still). Then,

\hat{\bold{b}}\hat{\bold{b}} \cdot \bold{\nabla}(\frac{B^2}{2 \mu}) = \bold{0}

Under this situation the momentum equation is:

\rho \frac{d}{dt} \bold{u} + \rho (\bold{u} \cdot \bold{\nabla}) \bold{u} = - \bold{\nabla} P -\bold{\nabla}(\frac{B^2}{2 \mu}) + \frac{B^2}{\mu} \frac{\hat{\bold{n}}}{R_c}

leaving us with the magnetic pressure term discussed above, and a term that points perpendicularly to the field line toward the center of the curvature (remember the tension term as a whole only arises when the field line is bent). Here R_c is the radius of curvature. It represents the distance to which perpendicular lines drawn from the bent field lines would converge. This means if field lines aren't bent a lot, the radius of curvature would be large, thus coinciding with a smaller degree of magnetic tension. This would be expected for a small degree of bending in the field lines.

Now that we have the basics down, we outline the model for the ring.

Ring model

To model the ring, we imagine the following situation:

1. a uniform thermal pressure distribution in the ring (so ignore gradients in pressure)
2. steady state (so ignore time derivatives, implying the bulk fluid is at rest so throw out terms with u also)
3. the ring is puffed out from a spherical flow of gas expelled from the collision region
4. there is no field within the cylinder of the colliding flows themselves (which is a good approximation given how dynamically weak the field is there)
5. to track the ring we define 2 radii, the outer radius of the ring r_o, and the inner radius of the ring, r_i (both relative to the center of the collision region)
6. this ring contains the same amount of flux as it spreads out, by flux freezing. this means that as the ring grows the field strength is geometrically diluted.
7. as far as the ram pressure pushing this ring out, we assume the velocity of the ejecta is a constant with radius, and only the density is decreasing by geometrical dilution.
8. we take the ram pressure of the ejecta to be the incoming ram pressure. this is a reasonable approximation as the shocked sound speed should be of order the incoming velocity.
9. we consider the ring to be formed from a spherical distribution of field lines centered on the collision region. thus, the radius of curvature is just r, and the magnetic pressure gradient is along r, but perpendicular to the field directly above the collision region.
10. we ignore gravity out in the ambient since the freefall time for gas at those densities is much larger than the time it takes to form the ring — this is a good approximation.

This gives the following momentum equation for the ring (i.e. for r_i < r < r_o):

0 = -\bold{\nabla}(\frac{B^2}{2 \mu}) + \frac{B^2}{\mu} \frac{\hat{\bold{n}}}{r}

Now, projecting this equation onto the r-axis gives n-hat = -r-hat, and thus we have:

-\frac{d}{dr}(\frac{B^2}{2 \mu}) \hat{\bold{r}} = \frac{B^2}{\mu r} \hat{\bold{r}}

The solution to this differential equation is:

B^2 = B_0^2(\frac{r_0}{r})^2

where the field is the initial value in the unperturbed medium beyond the ring's outer radius r0.

More fully then, the total solution for B²(r) in this model is:

B^2 = B_0^2(\frac{r_0}{r})^2 ~for ~r_i<r<r_0
B^2 = B_0^2 ~for ~r>r_0
B^2 = 0 ~for ~r<r_i

Thus, we envision a ring of flux surrounding the colliding flows cylinder R that is carried outward away from the cylinder boundary. It comes to some steady state in the ambient by developing a magnetic pressure gradient that counterbalances magnetic tension. At the rings outer boundary, the magnetic pressure is balanced with the initial unperturbed magnetic pressure. At the rings inner boundary, the magnetic pressure is balanced with ram pressure from the colliding flows.

Now, the ram pressure as a function of r from the cylinder's center is given by:

P_{ram} = \rho v^2(\frac{R}{r})^2

(assuming spherical dilution of a constant radial velocity ejecta), where R is the radius of the CF cylinder.

Balancing ram pressure and magnetic pressure at the ring's inner edge gives:

\rho v^2(\frac{R}{r_i})^2 = \frac{B_0^2}{2\mu_0}(\frac{r_0}{r_i})^2

or

\boxed{r_0 = \sqrt{\beta_{ram}} R}

which for our sims gives

\boxed{r_0 \approx 6R}

Our ring has not yet reached steady state, so the outer boundary of the ring may come to rest at about 6R. What about for the inner radius of the ring? To get this we use flux-freezing, i.e.,

\theta_1 = \theta_2

where

\theta_1 = \int_{r_i}^{r_o} \frac{B_0 r_0}{r} 2 \pi r dr = 2 \pi B_0 r_o (r_o - r_i)

\theta_2 = \int_{R}^{r_o} B_0 2 \pi r dr = \pi B_0(r_o^2 - R^2)

this second integral is the flux as measured before the field has been distorted (i.e. when R=r_i).

Setting these 2 fluxes equal and solving for r_i gives

r_i = \frac{r_o^2 + R^2}{2r_o}

Plugging in for r_o found above gives

r_i = R \frac{(\beta_{ram} + 1)}{\sqrt{4 \beta_{ram}}}

which is approximately

r_i \approx R \sqrt{\frac{\beta_{ram}}{4 }}

or

\boxed{r_i \approx \frac{1}{2}r_o}

or

\boxed{r_i \approx 3R}