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Perpendicular Shocks

The second special case is the so-called perpendicular 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,
$\displaystyle {\bf V}_1 = (V_1,\,0,\,0),$ $\textstyle \mbox{\hspace{1cm}}$ $\displaystyle {\bf V}_2 = (V_2,\,0,\,0),$ (954)
$\displaystyle {\bf B}_1 = (0,\,B_1,\,0),$ $\textstyle \mbox{\hspace{1cm}}$ $\displaystyle {\bf B}_2 = (0,\,B_2,\,0).$ (955)

Substitution into the general jump conditions (933)-(938) yields
$\displaystyle \frac{B_2}{B_1}$ $\textstyle =$ $\displaystyle r,$ (956)
$\displaystyle \frac{\rho_2}{\rho_1}$ $\textstyle =$ $\displaystyle r,$ (957)
$\displaystyle \frac{V_2}{V_1}$ $\textstyle =$ $\displaystyle r^{-1},$ (958)
$\displaystyle \frac{p_2}{p_1}$ $\textstyle =$ $\displaystyle R,$ (959)

where
\begin{displaymath}
R = 1+ \Gamma\,M_1^{\,2}\,(1-r^{-1}) + \beta_1^{-1}\,(1-r^2),
\end{displaymath} (960)

and $r$ is a real positive root of the quadratic
\begin{displaymath}
F(r) = 2\,(2-\Gamma)\,r^2+ \Gamma\,[2\,(1+\beta_1)+ (\Gamma-...
...a_1\,M_1^{\,2}] \,r- \Gamma\,(\Gamma+1)\,\beta_1\,M_1^{\,2}=0.
\end{displaymath} (961)

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

Now, if $r_1$ and $r_2$ are the two roots of Eq. (961) then

\begin{displaymath}
r_1\,r_2= -\frac{\Gamma\,(\Gamma+1)\,\beta_1\,M_1^{\,2}}{2\,(2-\Gamma)}.
\end{displaymath} (962)

Assuming that $\Gamma < 2$, we conclude that one of the roots is negative, and, hence, that Eq. (961) only possesses one physical solution: i.e., there is only one type of MHD shock which is consistent with Eqs. (954) and (955). Now, it is easily demonstrated that $F(0)<0$ and $F(\Gamma+1/\Gamma-1)>0$. Hence, the physical root lies between $r=0$ and $r=(\Gamma+1)/(\Gamma-1)$.

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

\begin{displaymath}
M_1^{\,2} > 1 + \frac{2}{\Gamma\,\beta_1}.
\end{displaymath} (963)

This condition can also be written
\begin{displaymath}
V_1^{\,2} > V_{S\,1}^{\,2} + V_{A\,1}^{\,2},
\end{displaymath} (964)

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 Sect. 5.4. 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 which exceeds the fast wave velocity.

In the strong shock limit, $M_1\gg 1$, Eqs. (960) and (961) become identical to Eqs. (945) and (946). 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 $(\Gamma+1)/(\Gamma-1)$. However, according to Eq. (956), 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 Shocks Up: Magnetohydrodynamic Fluids Previous: Parallel Shocks
Richard Fitzpatrick 2011-03-31