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Perpendicular MHD Shocks

The second special case is the so-called perpendicular MHD shock in which both the upstream and downstream plasma flows are perpendicular to the magnetic field, as well as the shock front. In other words,

$\displaystyle {\bf V}_1$ $\displaystyle = (V_1,\,0,\,0),$ $\displaystyle {\bf V}_2$ $\displaystyle = (V_2,\,0,\,0),$ (7.269)
$\displaystyle {\bf B}_1$ $\displaystyle = (0,\,B_1,\,0),$ $\displaystyle {\bf B}_2$ $\displaystyle = (0,\,B_2,\,0).$ (7.270)

Substitution into the general jump conditions (7.250)-(7.255) yields

$\displaystyle \frac{B_2}{B_1}$ $\displaystyle = r,$ $\displaystyle \frac{\rho_2}{\rho_1}$ $\displaystyle = r,$ (7.271)
$\displaystyle \frac{V_2}{V_1}$ $\displaystyle = r^{\,-1},$ $\displaystyle \frac{p_2}{p_1}$ $\displaystyle = R,$ (7.272)


$\displaystyle R = 1+ {\mit\Gamma}\,M_1^{\,2}\,(1-r^{\,-1}) + \beta_1^{\,-1}\,(1-r^{\,2}),$ (7.273)

and $ r$ is a real positive root of the quadratic

$\displaystyle F(r) = 2\,(2-{\mit\Gamma})\,r^{\,2}+ {\mit\Gamma}\left[2\,(1+\bet...
...ta_1\,M_1^{\,2}\right] r- {\mit\Gamma}\,({\mit\Gamma}+1)\,\beta_1\,M_1^{\,2}=0.$ (7.274)

Here, $ \beta_1= 2\,\mu_0\,p_1/B_1^{\,2}$ .

If $ r_1$ and $ r_2$ are the two roots of Equation (7.274) then

$\displaystyle r_1\,r_2= -\frac{{\mit\Gamma}\,({\mit\Gamma}+1)\,\beta_1\,M_1^{\,2}}{2\,(2-{\mit\Gamma})}.$ (7.275)

Assuming that $ {\mit\Gamma} < 2$ , we conclude that one of the roots is negative, and, hence, that Equation (7.274) only possesses one physical solution: that is, there is only one type of MHD shock that is consistent with Equations (7.269) and (7.270). Now, it is easily demonstrated that $ F(0)<0$ and $ F[({\mit\Gamma}+1)/({\mit\Gamma}-1)]>0$ . Hence, the physical root lies between $ r=0$ and $ r=({\mit\Gamma}+1)/({\mit\Gamma}-1)$ .

Using similar analysis to that employed in the previous section, it can be demonstrated that the second law of thermodynamics requires a perpendicular shock to be compressive: that is, $ r>1$ (Boyd and Sanderson 2003). It follows that a physical solution is only obtained when $ F(1)<0$ , which reduces to

$\displaystyle M_1^{\,2} > 1 + \frac{2}{{\mit\Gamma}\,\beta_1}.$ (7.276)

This condition can also be written

$\displaystyle V_1^{\,2} > V_{S\,1}^{\,2} + V_{A\,1}^{\,2},$ (7.277)

where $ V_{A\,1}=B_1/(\mu_0\,\rho_1)^{1/2}$ is the upstream Alfvén velocity. Now, $ V_{+\,1} = (V_{S\,1}^{\,2} + V_{A\,1}^{\,2})^{1/2}$ can be recognized as the velocity of a fast wave propagating perpendicular to the magnetic field. (See Section 7.4.) Thus, the condition for the existence of a perpendicular shock is that the relative upstream plasma velocity must be greater than the upstream fast wave velocity. Incidentally, it is easily demonstrated that if this is the case then the downstream plasma velocity is less than the downstream fast wave velocity. We can also deduce that, in a stationary plasma, a perpendicular shock propagates across the magnetic field with a velocity that exceeds the fast wave velocity.

In the strong shock limit, $ M_1\gg 1$ , Equations (7.273) and (7.274) become identical to Equations (7.260) and (7.261). Hence, a strong perpendicular shock is very similar to a strong hydrodynamic shock (except that the former shock propagates perpendicular, whereas the latter shock propagates parallel, to the magnetic field). In particular, just like a hydrodynamic shock, a perpendicular shock cannot compress the density by more than a factor $ ({\mit\Gamma}+1)/({\mit\Gamma}-1)$ . However, according to Equation (7.271), a perpendicular shock compresses the magnetic field by the same factor that it compresses the plasma density. It follows that there is also an upper limit to the factor by which a perpendicular shock can compress the magnetic field.

next up previous
Next: Oblique MHD Shocks Up: Magnetohydrodynamic Fluids Previous: Parallel MHD Shocks
Richard Fitzpatrick 2016-01-23