14. EQUILIBRIUM: GENERAL CONSIDERATIONS, THE VIRIAL
THEOREM
In MHD we are often interested in situations of equilibrium, or force balance. This
is because, in magnetic fusion we seek to confine a hot plasma for a very long time.
Clearly, a state of equilibrium is a minimum condition for fusion to occur.
We distinguish between the cases stationary and non-stationary equilibrium. In
stationary equilibrium, the flow velocity vanishes, and the condition for force balance is
dV / dt  V / t  F /   0 . In non-stationary equilibrium there is a finite flow velocity,
so the condition is V / t  0  F /   V  V . From now on, when we talk about
equilibrium we mean stationary equilibrium (unless otherwise noted).
In hydrodynamics, stationary equilibrium is relatively simple. Consider the case of a
fluid in a gravitational field. The condition for stationary equilibrium is p  g  0 .
In one spatial dimension, with g  gê x ,
dp
  g ,
dx
(14.1)
or
x
p(x)  p0  g  ( x )dx .
(14.2)
x0
We can find the pressure if the density profile is specified, or if there is a relationship
p  p( ) . In the latter case dp / dx  Cs2 d  / dx , where Cs2  dp / d  is the square of the
sound speed. If p and  are related linearly, then Cs2  constant , and the solution is
(x)  0egx /C
2
s
.
(14.3)
For the case of a non-stationary equilibrium in hydrodynamics, we have
V  V  p , so that finite flow can lead to non-uniform pressure. In one dimension,
this is just
d 
1 2
 p  Vx   0 ,
dx
2
(14.4)
or p  Vx2 / 2  constant . This is a special case of Bernoulli’s theorem.
In ideal MHD the pressure gradient can be balanced by the Lorentz force. This is the
basis for “magnetic confinement”. This situation is of great importance and must be
studied in more detail. The condition for stationary equilibrium in ideal MHD is
p  J  B .
(14.5)
We note immediately that B  p  0 and J  p  0 , so that the pressure gradient must
be perpendicular to both the magnetic field B and the current density J . The first
condition means that the pressure must be constant along the direction of the magnetic
field. This means that magnetic field lines lie everywhere within regions of constant
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pressure, and implies the possibility that these regions could be two-dimensional
surfaces. The second condition means that the existence of a current that is not parallel to
the magnetic field requires a pressure gradient, and vice versa.
We will immediately be more general. In MHD the momentum evolves according to

V    T ,
t
(14.6)
so that the condition for equilibrium can be expressed as
T  0 ,
(14.7)
Tki
0 ,
xk
(14.8)
or,
where, with V  0 ,

B2 
1
Tik  Tki   p 
 ik  Bi Bk

2 0 
0

(14.9)
is the total stress tensor. It is convenient to rewrite Tik as

B2 
B2 Bi Bk
Tik   p 


ik
2 0 
0 B 2

.
(14.20)
Then we can define the effective perpendicular and parallel pressures as
p  p 
B2
,
2 0
(14.21)
pP  p 
B2
,
2 0
(14.22)
and
so that pP  p  B2 / 0 . The stress tensor can then be written as

Tik  p ik  pP  p
BBB
i
k
2
.
(14.23)
Therefore, in MHD the stress tensor acts as though the pressure were anisotropic. For
the special case B  Bê z , the components of the stress tensor are
Txx  p
Tyx  0
Tzx  0
Txy  0
Tyy  p
Tzy  0
Txz  0
Tyz  0
Tzz  pP
Since p  pP, the fluid feels a greater pressure perpendicular to the magnetic field than
it does parallel to the field. This behaviour is completely different from our usual
experience with an unmagnetized fluid.
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We now consider the expression

x
T
xiTik   Tik i  xi ik ,
xk
xk
xk
 Tik ik ,
 Tii ,
(14.24)
where we have used the equilibrium condition, Equation (14.8). (Note that Tii is the
trace of the tensor T , i.e., the sum of the diagonal elements.)
Now introduce an isolated fluid permeated by a magnetic field. In equilibrium,
Equation (14.24) must hold. Integrating over a volume and applying Gauss’ theorem, we
have
 T dV  —
 x T dS
ii
V
i ik
k
,
(14.25)
S
V is the volume if integration and S is the surface area of that volume (not necessarily
the same as the volume and surface of the fluid). Evaluating the terms in Equation
(14.25), we find


B2 
B 2  BB 
3
p

dV

x

p

  2 0 
—
S  2 0  I  0   dS .
V 


(14.26)
The left hand side of this equation is positive definite. We now seek to estimate the sign
and magnitude the right hand side.
We consider the case where all the current density J and the pressure p are
completely within the fluid, and the surface S is completely outside the fluid. The
magnetic field arises from currents within the plasma only. This is shown in the figure.
We now take S   , so that only terms involving B contribute to the right hand side of
Equation (14.26). The vector potential due to currents within the fluid is
3
A(r) 
0
4
J(r  )
 r  r  dV 
.
(14.27)
V
If r is far from the source r  , we can expand 1 / r  r as

1
r  r
 
 ..... .
r  r
r r  r 3
(14.28)
Inserting this expansion into Equation (14.27), and using the fact that, since   J  0 ,
 J(r)dV   0
,
(14.29)
V
we find that A ~ 1 / r 2 , B ~ 1 / r 3 (since B    A ), and B2 ~ 1 / r 6 . The other term is
x  dS , which scales like r 3 , so the right hand side of Equation (14.26) scales like 1 / r 3 ,
which vanishes as r   .
Then, as S   , the left hand side of Equation (14.26) remains positive, while the
right hand side goes to zero. This contradiction means that, under the stated conditions,
Equation (14.8) cannot be satisfied, and equilibrium is impossible! This important and
general result is called the Virial Theorem. It says that a magnetized fluid cannot be in
MHD equilibrium under forces generated by its own internal currents. It implies that
any MHD equilibrium must be supported by external currents.
Of course, the Virial Theorem is satisfied in all laboratory experiments that contain
external coils to produce magnetic fields. It may not be satisfied under astrophysical
conditions. However, we already know that the universe is a dynamic place!
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